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Physics for Chemists Ruslan P. Ozerov Department of Physics Faculty of General Technological Sciences D.I. Mendeleev University of Chemical Technology Moscow, Russia The School of Biomedical Biomolecular and Chemical Science University of Western Australia Western Australia Crawly, Australia
and Anatoli A. Vorobyev Department of Physics Faculty of General Technological Sciences D.I. Mendeleev University of Chemical Technology Moscow, Russia
Amsterdam ● Boston ● Heidelberg ● London ● New York ● Oxford Paris ● San Diego ● San Francisco ● Singapore ● Sydney ● Tokyo
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Elsevier Radarweg 29, PO Box 211, 1000 AE Amsterdam, The Netherlands The Boulevard, Langford Lane, Kidlington, Oxford OX5 1GB, UK First edition 2007 Copyright © 2007 Elsevier B.V. All rights reserved No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means electronic, mechanical, photocopying, recording or otherwise without the prior written permission of the publisher Permissions may be sought directly from Elsevier’s Science & Technology Rights Department in Oxford, UK: phone (⫹44) (0) 1865 843830; fax (⫹44) (0) 1865 853333; email:
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Contents Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiii Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xv Recommendations to the Solution of the Physical Problems . . . . . . . . . . . . . . . . . . . . . . . . . . xvii
1 Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2.1 Kinematics of a material point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2.2 Kinematics of translational movement of a rigid body . . . . . . . . . . . . . . . . . . 12 1.2.3 Kinematics of the rotational motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 1.3 Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 1.3.1 Newton’s first law of motion: inertial reference systems. . . . . . . . . . . . . . . . . 16 1.3.2 Galileo’s relativity principle: Galileo transformations . . . . . . . . . . . . . . . . . . . 18 1.3.3 Newton’s second law of motion: Momentum . . . . . . . . . . . . . . . . . . . . . . . . . 20 1.3.4 The third Newtonian law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 1.3.5 Forces classification in physics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 1.3.6 Noninertial reference systems. An inertia force: D’Alembert principle . . . . . . 33 1.3.7 A system of material points: internal and external forces . . . . . . . . . . . . . . . . 35 1.3.8 Specification of a material points system . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 1.3.9 The dynamics of rotational motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 1.4 Work, Energy and Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 1.4.1 Elementary work of a force and a torque . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 1.4.2 Power. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 1.4.3 Kinetic energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 1.4.4 A force field. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 1.4.5 Potential energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 1.5 Conservation Laws in Mechanics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 1.5.1 Conservation law of mechanical energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 1.5.2 Momentum conservation law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 1.5.3 Angular momentum conservation law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 1.5.4 Potential curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 1.5.5 Particle collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 1.6 Einstein’s Special Relativistic Theory (STR) (Short Review) . . . . . . . . . . . . . . . . . . . 90 Problems/ Tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 2 Oscillations and Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Definitions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Kinematics of Harmonic Oscillations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Summation of Oscillations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v
105 105 106 113
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2.3.1 Summation of codirectional oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.2 Summing up two codirectional oscillations with slightly different frequencies: beatings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Dynamics of the Harmonic Oscillation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 Differential equations of harmonic oscillations. . . . . . . . . . . . . . . . . . . . . . . . 2.4.2 Spring pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.3 The mathematical pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.4 A physical pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.5 Diatomic molecule as a linear harmonic oscillator . . . . . . . . . . . . . . . . . . . . . 2.5 Energy of Harmonic Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Damped Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 Forced Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8 Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8.1 Introductory remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8.2 An equation of a plane traveling wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8.3 Wave energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8.4 Acoustic Doppler effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9 Summation of Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9.1 Superposition of waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9.2 Standing waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9.3 String oscillations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9.4 Group velocity of waves: wave package . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems/ Tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3 Molecular Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Kinetic Theory of Ideal Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.1 Introductory remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.2 Distribution function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.3 An ideal gas model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.4 General equation of an ideal gas. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.5 Absolute temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Distribution of Molecules of an Ideal Gas in a Force Field (Boltzmann Distribution) . . . 3.2.1 An ideal gas in a force field: Boltzmann distribution . . . . . . . . . . . . . . . . . . . 3.2.2 Barometric height formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.3 Centrifugation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.4 Boltzmann factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Distribution of the Kinetic Parameters of an Ideal Gas’ Particles (Maxwell Distribution) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.1 The Maxwellian distribution of the absolute values of molecule velocities . . . . 3.3.2 The kinetic energies Maxwellian distribution of molecules. . . . . . . . . . . . . . . 3.4 First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.1 Equipartition of energy over degrees of freedom . . . . . . . . . . . . . . . . . . . . . . 3.4.2 First laws of thermodynamics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.3 Heat capacity of an ideal gas: the work of a gas in isoprocesses . . . . . . . . . . . 3.4.4 Heat capacity: theory versus experiment. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 The Second Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.1 Heat engines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
113 117 118 118 118 119 121 129 131 133 138 145 145 147 151 154 156 156 157 160 163 165 166
169 169 169 172 174 175 177 178 178 180 183 185 186 186 193 194 194 195 197 204 205 206
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3.5.2 The Carnot cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.3 Refrigerators and heat pumps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.4 Reduced amount of heat: entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.5 Clausius inequality and the change of entropy for nonequilibrium processes . . . 3.5.6 Statistical explanation of the second law of thermodynamics . . . . . . . . . . . . . 3.5.7 Entropy and disorder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 A Real Gas Approximation: van der Waals Equation . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.1 An equation of state of a van der Waals gas . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.2 Internal energy of the van der Waals gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.3 A Joule–Thomson effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 Elements of Physical Kinetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7.2 Transport processes: relaxation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7.3 Transport phenomena in ideal gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7.4 A macroscopic representation of a transport coefficient . . . . . . . . . . . . . . . . . 3.7.5 Diffusion in gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7.6 Heat transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7.7 Viscosity or internal friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7.8 A transport phenomena in a vacuum condition . . . . . . . . . . . . . . . . . . . . . . . . Problems/ Tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
207 210 211 214 219 220 221 221 226 227 230 230 230 231 233 235 237 238 243 245 247
4 Dielectric Properties of Substances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Electrostatic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.1 General laws of electrostatics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.2 Strength of an electrostatic field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.3 The Gauss law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.4 Work of an electrostatic field force and potential of an electrostatic field . . . . 4.1.5 Electrical field of an electric dipole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Dielectric Properties of Substances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.1 Conductors and dielectrics: a general view . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.2 Macroscopic (phenomenological) properties of dielectrics . . . . . . . . . . . . . . . 4.2.3 Microscopic properties of dielectrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.4 Three types of polarization mechanisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.5 Dependence of the polarization on an alternative electric field frequency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.6 A local electric field in dielectrics. Lorentz field . . . . . . . . . . . . . . . . . . . . . . 4.2.7 Clausius–Mossoti formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.8 An experimental determination of the polarization and molecular electric dipole moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems/ Tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
251 251 251 252 259 273 276 280 280 282 284 286
5 Magnetics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 General Characteristics of the Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.1 A permanent (direct) electric current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.2 A magnetic field induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
293 294 296 298 301 303
305 305 305 309
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5.1.3 The law of a total current (ampere law) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.4 Action of the magnetic field on the current, on the moving charge . . . . . . . . . 5.1.5 A magnetic dipole moment in a magnetic field. . . . . . . . . . . . . . . . . . . . . . . . 5.1.6 Electromagnetic induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Magnetic Properties of Chemical Substances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.1 Atomic magnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.2 Macroscopic properties of magnetics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.3 An internal magnetic field in magnetics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.4 Microscopic mechanism of magnetization . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Magnetically Ordered State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.1 Ferromagnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.2 Domains: magnetization of ferromagnetics. . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.3 Antiferro- and ferrimagnetics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Displacement Current: Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems/ Tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
318 320 327 328 331 332 333 334 336 344 344 347 349 350 358 360
6 Wave Optics and Quantum–Optical Phenomena . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Physics of Electromagnetic Optical Waves. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 An Interference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.1 Superposition of two colinear light waves of the same frequencies . . . . . . . . . 6.2.2 Interference in thin films . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Diffraction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.1 Huygens–Fresnel principle: Fresnel zones . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.2 Diffraction on one rectangular slit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.3 Diffraction grating . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.4 Diffraction grating as a spectral instrument . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.5 X-ray diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Polarization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.1 Polarized light: definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.2 Malus law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.3 Polarization at reflection: Brewster’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.4 Rotation of the polarization plane. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.5 Birefringence: a Nichol prism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 Dispersion of Light . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6 The Quantum-Optical Phenomena . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6.1 Experimental laws of an ideal black body radiation . . . . . . . . . . . . . . . . . . . . 6.6.2 Theory of radiation of an ideal black body from the point of view of wave theory: Rayleigh–Jeans formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6.3 Planck’s formula: a hypothesis of quanta—intensity of light from wave and quantum points of view . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6.4 Another quantum-optical phenomena . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.7 The Bohr Model of a Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems/ Tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
361 361 369 369 370 377 378 379 381 383 385 386 386 387 388 389 391 395 398 398 402 404 407 416 421 422
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7 Elements of Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Particle-Wave Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.1 De Broglie hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.2 Electron and neutron diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Heisenberg’s Uncertainty Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Wavefunction and the Schrödinger Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3.1 A wavefunction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3.2 The Schrödinger equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3.3 Standard requirements that the wavefunction should obey . . . . . . . . . . . . . . . 7.4 Most General Problems of a Single-Particle Quantum Mechanics . . . . . . . . . . . . . . . 7.4.1 A free particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.2 A particle in a potential box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.3 A potential step . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.4 A potential barrier: a tunnel effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.5 Tunnel effect in chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 The Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.1 The Shrödinger equation for the hydrogen atom . . . . . . . . . . . . . . . . . . . . . . . 7.5.2 The eigenvalues of the electron angular moment projection Lz . . . . . . . . . . . . 7.5.3 Angular momentum and magnetic moment of a one-electron atom. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.4 A Schrödinger equation for the radial part of the wave function; electron energy quantization. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.5 Spin of an electron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.6 Atomic orbits: hydrogen atom quantum numbers . . . . . . . . . . . . . . . . . . . . . . 7.5.7 Atomic orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.8 A spin–orbit interaction (fine interaction) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6 A Many-Electron Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6.1 Types of electron’s coupling in many-electron atoms . . . . . . . . . . . . . . . . . . . 7.6.2 Magnetic moments and a vector model of a many-electron atom. The Lande factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6.3 The atomic terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6.4 Characteristic X-rays: Moseley’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.7 An Atom in the Magnetic Field: The Zeeman Effect . . . . . . . . . . . . . . . . . . . . . . . . . 7.8 A Quantum Oscillator and a Quantum Rotator. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.8.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.8.2 Quantum oscillators: harmonic and anharmonic . . . . . . . . . . . . . . . . . . . . . . . 7.8.3 A rigid quantum rotator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.8.4 Principles of molecular spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems/ Tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
423 423 423 424 428 432 432 433 434 435 435 436 441 442 445 447 448 451
8 Physical Principles of Resonance Methods in Chemistry. . . . . . . . . . . . . . . . . . . . . . . . 8.1 Selected Atomic Nuclei Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.1 A nucleon model of nuclei . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.2 Nuclear energy levels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
497 497 497 499
452 457 460 461 462 467 468 469 470 473 473 477 480 480 481 486 491 494 495
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8.1.3 Nuclear charge and mass distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.4 Nuclear quadrupole electrical moment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Intraatomic Electron–Nuclear Interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.1 General considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.2 Coulomb Interaction of an electron shell with dimensionless nucleus . . . . . . . 8.2.3 Coulomb Interaction of an electron shell with a nucleus of finite size: the chemical shift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.4 The nuclear quadrupole moment and the electric field gradient interaction . . . 8.2.5 Interaction of a nuclear magnetic moment with an electron shell . . . . . . . . . . 8.2.6 Atomic level energy and the scale of electromagnetic waves . . . . . . . . . . . . . 8.3 -Resonance (Mössbauer) Spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.1 Principles of resonance absorption . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.2 Resonance absorption of -rays: Mössbauer effect . . . . . . . . . . . . . . . . . . . . . 8.3.3 -Resonance (Mössbauer) spectroscopy in chemistry . . . . . . . . . . . . . . . . . . . 8.3.4 Superfine interactions of a magnetic nature . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Nuclear Magnetic Resonance (NMR) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4.2 Use of nuclear magnetic resonance in chemistry . . . . . . . . . . . . . . . . . . . . . . 8.5 Abilities of Nuclear Quadrupole Resonance. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.6 Electron Paramagnetic Resonance (EPR) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems/ Tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
500 501 502 502 504 504 506 507 507 508 508 510 513 515 516 516 517 525 526 528 529
9 Solid State Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Crystal Structure, Crystal Lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Electrons in Crystals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.1 Energy band formation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.2 Elements of quantum statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.3 Band theory of solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Lattice Dynamics and Heat Capacity of Crystals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.1 The Born–Karman model and dispersion curves. . . . . . . . . . . . . . . . . . . . . . . 9.3.2 The heat capacity of crystals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4 Crystal Defects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4.1 Point defects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4.2 Dislocations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5 Transport Phenomena in Liquids and Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.6 Some Technically Important Electric Properties of Substances. . . . . . . . . . . . . . . . . . Problems/ Tasks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Answers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
531 531 537 537 540 544 545 545 550 561 561 563 567 571 577 578
Appendix 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 581 Appendix 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 589 Appendix 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 591
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Appendix 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 595 Appendix 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 599 Glossary of Symbols and Abbreviations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 605 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 609
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All miracles of nature, no matter how extraordinary they are, have always found their explanation in the laws of physics Jules Verne Journey to the Center of the Earth
Preface In this new century, the development of science, technology, industry, etc., will require new materials and devices, which will in many respects differ from those of the past. Even now, there are many such examples. Certainly, the main foundation of these achievements is science, primarily physics, which enables the solid building of chemical, biological and atomic technologies, etc. In general, this book is a textbook on physics, but takes the above circumstances into account. It is aimed at students and scientists in the field of technology (chemical, biological and other branches of sciences), who will be working in the times ahead. The book differs substantially from standard physics textbooks in its choice of subjects, the manner of its presentation, selection of examples and illustrations as well as problems to be solved by the reader. The book contains problems important for chemists such as the language of potential curves and the essence of the theory of molecular collisions, and a large part is devoted to molecular physics with classical Boltzmann and Maxwell statistics, transport phenomena, etc. In a special part, the dielectric and magnetic properties of molecules are considered from the point of view of their structure. Optics is also covered in order to give the reader some idea of how its laws can be used for molecular structure analysis. Quantum mechanics is presented in an adapted form, aimed at a description of atomic and molecular spectroscopy. A special chapter describes tunneling both as a general phenomenon and as a mechanism of chemical reactions. Special attention is paid, also in an adapted form, to inter-atomic fine and superfine interactions, which are the basis of many modern and productive physical methods in the field of atomic and molecular structural investigations. Solid-state theory is presented on the basis of quantum statistics in order to form a bridge to their properties. A new technological field—nano-scale technology—is touched on here. In our opinion, no other textbook covers the sophisticated modern subjects mentioned above in such an acceptable form. Physics always operates with certain models—simplified representations of real systems. The ideal gas model is one such example. Despite its variety of real gas properties, this simple model assists in understanding the behavior of more complex systems using more complex factors permitted within the model, and it provides numerical results. For example, the introduction of additional interactions leads to van der Waals’s gas and allows further inclusion of virial factors, which in turn make the model more universally applicable to all gases. When using the model, the level of required accuracy has to be defined and on that basis, an appropriate model can be selected. The authors have aimed to make the subject matter easy to master; therefore many theoretical approaches in the text have been presented in an adapted manner, while the more strict proofs are given separately throughout each chapter in the Examples, in the Problems/Tasks at the end of each chapter, and particularly in the Appendices. A set of important constants is given in Appendix A1 to facilitate the solution of the problems. Units of measurement of physical values are also listed there. xiii
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The text is further enhanced by the illustrative material, including selected drawings, graphs and tables, which are an inseparable part of the book and greatly simplify understanding of the text. The book will be particularly useful for students, not only in the narrow area of their future profession but also in allowing them a broader glimpse of the surrounding world. In our opinion, this is necessary to encourage in the young people of the twenty-first century a firm perception of the world as an objective reality. The book pays a good deal of attention to the laws they need to learn in order to acquire new knowledge and to use if to expand human possibilities both in industrial and spiritual spheres of activity. The book’s nine chapters provide a description of the main laws of mechanics, statistical physics, thermodynamics, physics of dielectrics and magnetics, wave optics, quantum mechanics and physics of electronic shell of atoms, solid-state physics, physics of electromagnetic waves and physical methods of investigation. It contains a large amount of comprehensive information, useful for everybody in all stages of tutorial, practical and scientific activity. For successfully understanding the book, the reader should have a knowledge of the mathematical laws of, and some experience regarding, operation with vectors, differentiation and integration of elementary functions and others. Mathematics is the language of physics: the faults in mathematics must be considered as the faults in physics. The book is the fruit of our long experience at the Mendeleev University of Chemical Technology (MUChT) in Moscow. The results we have achieved have had a great influence on the content of the book and the problems chosen. One of us (RO) participated in publishing a textbook on physics under the auspices of the “State Program of Education and Science Integration” (“Physics in Chemical Technology” in collaboration with Professor E.F. Makarov from the Institute of Chemical Physics of Russian Academy of Sciences). Although it appeared in a very limited edition, which meant that the book wasn’t available for purchase, it has nevertheless greatly influenced the publication of this new work. One of the authors (RO) has spent a relatively long time at The University of Western Australia (UWA) in Perth, and this too has had a significant influence on the content and style of the book. R.P. Ozerov A.A. Vorobjev
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Acknowledgments It happens that both co-authors of this book, learning together in the same institute (Moscow Institute of Physical Engineering) and working thereafter for a very long period of time in the same department of the Moscow University of Chemical Technology have now been separated by long distance, corresponding only by telephone and e-mail. Therefore this part of the acknowledgment is expressed mostly by one of us, namely RPO. Firstly I would like to give my thanks the late Professor Edvard N. (Ted) Maslen, my colleague on the commission on Electron Density of the International Union of Crystallography; he strongly supported an acceptance of my previous student Dr. Victor A. Streltsov to a postdoctoral position at the Crystallography Centre of the University of Western Australia, which has resulted in an profitable collaboration with the D.I. Mendeleev Institute of Chemical Technology in Moscow. Ted and Professor Sid Hall made provision for a very favorable and productive investigation into X-ray crystallography. I greatly appreciate the initiative that Professor Sid Hall showed when he undertook on the position to assist me, especially considering the unusual circumstances. Dr. Alexandre N. Sobolev jointed the project a little latter; though the skill he has acquired in UWA made him a high-class specialist in X-ray crystallography. His skill and determination are extremely valued and his endless commitment enabled me to complete this book. I have never met a more kind and generous specialist like Professor Alan White. He was always forthcoming with valuable advice and support. I would also like to acknowledge Professor Brian N. Figgis who helped me immensely with my previous book “Electron density and chemical bonding in crystals.” Dr. Lindsay Byrne—a rare find as an NMR guru—is a devoted man with much enthusiasm and was always willing to help. I appreciate a lot Professor Ian McArthur for his attention and professional help. AV and me would like to express our gratitude to Professor Vitalii I. Khromov for his help and valuable advice. I would dearly like to thank the head of the School of Biomedical, Biomolecular and Chemical Sciences, The University of Western Australia, Professor Geoff Stewart and Mrs Leigh Swan for the pleasant and prosperous working environment they provided. Without this comfort and assistance I would not have been able to finish this project. I express my gratitude to my friend Mrs Suzanne Collins—the extra class specialist in English teaching—for her valuable help and advice. I am greatly obliged to all member of my family who helped me at all stages of my work.
Professor Ruslan P. Ozerov Professor Anatoli Vorobyev
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Recommendations to the Solution of the Physical Problems Readers are recommended to begin solving the problems by writing down the given data in the normal way. On the left-hand side, all the data should be written in a column as accurately as possible. At this point, it is useful to translate all the given date to the SI system of units in order to avoid confusion at the end. In the majority of cases, it is also very useful to make a competent analysis of the task, choose a reference system, and make the drawing indicating all the details correctly; with the proper indication of all details; it can be said that a reliable drawing is 50% of the solution. The solution should be carried out, as a rule, in a general form, i.e., without intermediate numerical calculations down to the final answer; this means that the symbol for the physical value sought should remain on the left-hand side of the answer, while the symbols for the physical values of the given conditions and the necessary physical constants should be on the right-hand side. The values of physical constants are listed in Appendix 1. It is also useful to accompany the decision with a brief explanation, both physical and mathematical, to state the physical ideas behind the solution. Moreover, the mathematical treatment should also be explained: if the definite integral is treated, the limits should be explained. In square root calculations it is desirable to explain whether both roots should proceed or one root should be rejected and why? The final answer in a general form should be marked by any way. We especially want to emphasize that a general solution is of greater significance and value: it means that not just the particular problem has been solved, but a real task. The results can then always be used to solve similar problems without starting the treatment again from the very beginning. The analysis of the dimension of the result is one of the important stages of the solution; it permits one to be confident of the result. It is necessary to carry out calculations keeping the desired accuracy in mind; more often three significant figures will be sufficient. The numerical answer should be written down as a number increased to the proper power of 10 or using multiple prefixes, e.g., A = 2.56⫻107 J and/or 25.6 MJ. It is important that one should be certain that a reasonable result has been achieved, i.e., the speed of a body does not exceed the speed of light, or its size does not exceed the size of the universe, etc. We wish our readers all success!
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–1– Mechanics
1.1
INTRODUCTION
An enormous number of physical events and phenomena are taking place around us all the time: the movement of all types of transport (bicycles, cars, trains, airplanes, etc.), building activity, athletes in competition, rain falling, wind blowing, water flowing, earthquakes and a wide range of other phenomena. All of these are performed at speeds much smaller than the velocity of light (c ⬇ 3 108 m/sec) and at scales much greater than atomic scales (⬃1010 m). All are described by classical mechanics, based mainly on Newton’s laws. This does not, of course, exclude the existence of other phenomena described by other physical branches. Quantum mechanics deals with the world of atoms and molecules, their transformations and accompanying changes in property. The overwhelming majority of them are invisible to the naked eye, but experience shows the following to be true: all materials, though differing in their characteristics, consist of a limited number of various particles— atoms and molecules. This is the world of so-called quantum mechanics. We can indirectly observe these phenomena manifest themselves, but for their investigation and understanding, a special knowledge is needed. To continue this analysis, we can mention one more branch of phenomena that manifest themselves at velocities close to the velocity of light; this is the more exotic area of classical and quantum relativistic physics.
1.2
KINEMATICS
Kinematics is the branch of mechanics that explores the motion of material bodies from the standpoint of their space–time relationships, disregarding their masses and the forces acting on them. 1.2.1
Kinematics of a material point
For a description of a point’s motion in space and time, a reference system should be chosen. The reference system is a collection of instruments: the time-measuring device (e.g., a watch) and the bodies conditionally considered as being fixed in space with respect 1
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to which the motion is considered. Time, a continuously changing scalar value, is measured by a watch, and cannot be negative. In problems of kinematics time is usually taken as an independent variable (or argument), the rest of the parameters being considered as functions of time. For different problems the reference system can be chosen either in the form of Cartesian coordinates, or as a cylindrical or spherical coordinates system. A moving point describes a certain continuous line in space that is referred to as a trajectory. In a number of problems the path itself will define the motion (for instance, its rails will dictate the motion of a railway carriage). At a certain instant, corresponding to a certain body motion, tangent unit vectors—principle normal and binormal vectors—are taken as natural axes. In the following we will consider only plane motion, so there is no need for a binormal vector. The principle normal is perpendicular to the tangent and is directed to the center of curvature. The direction of the tangent and normal unit vectors will be denoted as and n. Let us recall some information about the line curvature (trajectory). The tangent lines assigned by vectors 1 and 2 at two adjoining points A and B of the plane form an angle (Figure 1.1) to be drawn, which is referred to as the angle of contingence. If we then make the distance AB shorter, an arc AB l aspires to zero. At the limit /艎, it gives the trajectory curvature K in a given point: lim
K ᐉ ᐉ
at
AB 0.
The reciprocal value 1/K is the curvature radius in point A. In fact, a circle’s curvature is equal to its radius; the curvature radius of a straight line is infinity. The simplest object in mechanics is called a material point (MP); this implies a body whose size in the framework of a given problem can be considered to be negligibly small. Another definition of an MP is that it is a point that possesses a mass. Different objects in different problems can be considered differently: the molecules acting on a vessel’s wall can be imagined as an MP, the earth moving around the sun may, in some instances, also be treated as an MP. However, the same objects in different problems cannot be considered
1 B ∆
A
2
n
Figure 1.1. The trajectory curvature.
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as MPs: e.g., molecules in molecular spectroscopy rotating around their center of mass (CM), the earth rotating around its geographical axis, etc. An important task in kinematics is to assign an equation of motion, i.e., to construct the necessary mathematical equations that are sufficient to determine the MP’s position in space at any instant of time. In the Cartesian coordinate system such an equation is the time dependence of the radius vector r(t); three scalar equations x(t), y(t) and z(t) correspond to one vector equation. If a point in a time interval t moves from point A to B along an arc l (Figure 1.2), the vector r r2 r1 is referred to as the displacement, whereas the length of the arc AcB is the distance travelled. If one takes one’s car in the morning, travels some distance during the day and then returns the car to the garage, the overall day displacement is equal to zero, whereas the distance travelled is the non-zero speedometer indication. The distance travelled and the displacement can coincide in two cases: when the movement occurs along a straight line or at t → 0. The equation 冓冔
r . t
(1.2.1)
allows us to calculate the average speed at a time interval t. The instant velocity is given by the equation lim
t0
r dr r(t ). t dt
(1.2.2)
V
c A
B
∆r
r1
r2
O
Figure 1.2. A displacement vector ∆r and distance travelled AcB.
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The velocity at a given point is a physical value, numerically equal to the time derivative from the radius vector of the MP in the reference system under consideration. Remember that for brevity of writing, the time derivative function is denoted by a point above the letter, expressing a given function. Where the direction of the vector is concerned, in the limit of the movement of point B to point A the secant will coincide with the tangent to the trajectory in point A. Consequently, an instant velocity vector is directed along the tangent to the trajectory, and the modulus is the time derivative from the function, expressing the law of point movement. As usual, the point radius vector r(t) can be decomposed upon the orts r(t ) x(t )i y(t ) j z(t )k,
(1.2.3)
(t ) r(t ) ix (t ) jy (t ) kz(t ).
(1.2.4)
ix jy kz ,
(1.2.5)
and therefore,
The velocity vector is
where x, y and z are its projections onto the coordinate axes: x x (t ), y y (t ), z z(t )
(1.2.6)
The modulus of the velocity vector is the square root sum of their projections’ squares: x2 y2 z2
(1.2.7)
Acceleration is the change of the velocity vector in time. If, in the time interval t, an MP displaced along the trajectory and a change in velocity and its direction had taken place then 2 1. The mean acceleration in the t interval is then 冓 a冔
. t
(1.2.8)
The acceleration at a given time instant (instantaneous acceleration) is the limit of the ratio (/t) at t → 0.
a lim
d (t ). t dt
(1.2.9)
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or a ix jy kz ,
(1.2.10)
(because orts are in this case independent of time). In another form a ia x ia y kaz ,
(1.2.11)
where ax, ay and az are projections of the a vector onto the coordinate axes. Comparison of eqs. (1.2.6) and (1.2.11) gives a x x (t ) x(t ); a y y (t ) y(t ); az z (t ) z (t )
(1.2.12)
and correspondingly a a x2 a y2 az2 .
(I.2.13)
At curvilinear movement the velocity vector is the product , where is a tangent ort. Because of the fact that the point is moving along a curvilinear trajectory and “draws” the unit vector behind, its position is also dependent on time. In this case:
a
d d d . dt dt dt
(1.2.14)
Expression (1.2.15) shows that acceleration is the sum of the two vectors: the first is directed along the tangent and is equal to the first derivative of velocity and the second term depends on the change of in time. To determine the magnitude and direction of the second term, we need to find the meaning of the derivative d/dt. Let the direction of the velocity vector at two adjacent positions separated by time interval t be specified by orts 1 and 2 (Figure 1.3). Then the change of the vector in the time interval t can be expressed by vector 2 1. We shall consider the derivative d/dt as a limit of a ratio / for t ; 0. We find the value of vector magnitude from the triangle (/2) ACD: sin , then sin , at t → 0 numerically t → , since 2 2 2 the unit-vector magnitude is unity and sin(/2) ⬇ /2 at ^ 1. Then d lim lim t 0 t 0 dt t t
(1.2.15)
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/t by the arc Multiplying both the numerator and denominator of the function length l we obtain d ᐉ ᐉ lim lim lim dt t0 t ᐉ t0 ᐉ t0 t Let us consider both these limits. Since an angle is the angle of contiguity, the lim(/l) K is equal to the curvature of a curve at a given point, i.e., to curvature radius . The second limit is the velocity magnitude lim
to
ᐉ d ᐉ . t dt
Thus, d K . dt
(1.2.16)
To determine the direction of the vector (d/dt) we shall draw a straight line from point A parallel to and examine the value of an angle ⬔CAE at the limit t → 0. As can be seen from Figure 1.3, an angle ⬔CAE ⬔CAF /2 (/2) /2). At t → 0 the → 0 whereas ⬔CAE → ( /2). Therefore, the vector (d/dt) contiguity angle lim(/t) will be directed along the normal to the center of curvature at point A; it can be presented as d n. dt
(1.2.17)
C 1 ∆ A
2
∆ D
F Ε
n ∆/2
derivative. Figure 1.3. Calculation of the d dt
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Returning to expression (1.2.14), we can write a
d 2 n dt
(1.2.18)
Therefore, the total acceleration a in curvilinear movement can be separated into two parts: the first is the tangent acceleration a
d dt
(1.2.19)
an
2 .
(1.2.20)
and the second is
The tangent part influences the absolute velocity magnitude whereas the normal part changes the direction of the velocity vector. The square of the total acceleration can be written as 2
⎛ 2 ⎞ ⎛ d ⎞ ⎜ ⎟ ⎜ ⎟ . ⎝ dt ⎠ ⎝ ⎠ 2
a
2
a2 an2
(1.2.21)
The expressions derived are valid for the general movement of the MP along the curve with an arbitrary regime of velocity change. Let us consider some particular cases: Rectilinear movement: , an 0, a a Uniform movement along a circle: const., a 0, a n(2/R) Nonuniform movement along a circle: const., a 苷 0, an 苷 0. In the case of uniformly alternating movement a const. 苷 0, an 0; it is possible to derive the general expressions. Integration of an expression (d/dt) a over time gives (t ) ∫ a(t )dt C at 0 ,
(1.2.22)
where 0 is the initial speed (at t 0). The distance travelled can be derived from (1.2.22) by repeated integration as
x(t ) x0 0 t where x0 is the initial coordinate (at t 0).
at 2 , 2
(1.2.23)
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EXAMPLE E1.1 A car moving uniformly covers a distance of 100 km at a speed 60 km/h, although on the way back it travels at a speed of 40 km/h. Determine the average speed of the car. Solution: At first glance the answer is simple: 50 km/h. However, this is incorrect. The average speed is the total distance travelled (200 km) divided by the total time spent (100/60) (100/40) 4.16 h. Therefore, the average speed is (200/4.16) ⬇ 44 km/h.
EXAMPLE E1.2 The movement of a MP along an x-axis is described by the equation x A Bt CT 3, where A 4 m, B 2 m/sec, C –0.5 m/sec3. For the instance of time t1 2 sec determine: (1) the MP coordinate x1, (2) an instant velocity 1, and (3) an instant acceleration a1. Solution: (1) To find the point coordinate one should substitute time t for the instant time t1x1 A Bt1 Ct13. Inserting the given values we obtain: x1 4 2.2 0.5 23 4 m. (2) To find an instant speed at any time we should differentiate a coordinate on time: (dx/dt) B 3Ct12. Introducing B, C and t1 we obtain: 1 4 m/sec. The sign shows that at that very moment the point moves in a negative direction on the x-axis. (3) To find acceleration as a function of time we should take the second time derivative from coordinate: a (d2x/dt2) (dx/dt) 6Ct. To find the instant acceleration at t1 we should introduce the given data and obtain the result. a1 – 6.0 5.2 –6 m/sec2; the sign shows that the movement is decelerative.
EXAMPLE E1.3 The movement of an MP along an x-axis is described by equation x A Bt Ct 2, where A 5 m, B 4m/sec, C –1.0 m/sec3. Draw a graph of x(t) and the distance travelled S(t). Solution: For the drawing of the graph of the point coordinate time dependence x(t), we find characteristic values of movement: initial and maximum coordinates The initial coordinate corresponds to the moment t 0, its value equals x(0) A 5 m. The point reaches maximum height corresponding to the moment when the point starts to move back (speed changes sign). We can find this moment having equated to zero the time first derivative from coordinate: (dx/dt) B 2Ct 0, wherefrom t –(B/2C) 2 sec. The maximum coordinate xmax x(2) 9 m. The time instant t when x 0 can be found from equation x ABtCt2 0. Solving the quadratic equation we obtain t (2 3) sec. The negative value does not satisfy the problem. Therefore t 5 sec. Using the data obtained we can draw the
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graph of coordinates’ dependence on time. The distance travelled and the coordinate coincide until the point stops; from this time the point goes in opposite direction and its coordinate diminishes; however the distance travelled continues to grow (Figure E1.3).
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2
4
6
t
–5
EXAMPLE E1.4 A mortar is installed on a hill at a height of H 60.0 m above ground level. It fires a missile at an initial angle of 60° to the horizon. The missile’s initial velocity is 0 80 m/sec. Derive: (1) the kinematical equations of the missile’s flight x(t) and y(t); (2) the equation of trajectory y(x); (3) the expression for projections x(t) and y(t) on the coordinate axes x and y and the time velocity dependence (t); (4) the velocity dependence on time (t) on absolute value and direction; (5) the absolute values of tangential 冟a冟 and normal 冟an冟 acceleration dependence on time (derive a corresponding formula and execute calculations): (6) the maximum height ymax of flight; (7) the time of the missile’s flight ; (8) the range L of missile; (9) the missile’s velocity (on modulus and direction) at the moment of falling on the ground; (10) the curvature radii trajectory 1 and 2 at the moment of falling and at the highest point of flight, respectively. Solution: To solve this problem we have to begin with the choice of reference frame. The motion of the missile is subject to a constant acceleration g directed downward. Therefore, the flight trajectory is a plane (two-dimensional). Choose a Cartesian system xOy in such a way that the x-axis is horizontal and the y-axis is vertical; the flight will occur in this plane. The origin is superposed with the earth
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surface, axis x directed horizontally and y vertically upward. Accept the missile as an MP. The movement in this case can be separated into two independent components: along axes Ox and Oy. A movement along axis Ox is uniform with a speed x 0 cos ; however along axis Oy it uniformly accelerates with initial coordinate y0 H and initial speed oy 0 sin and acceleration ay g (see inset in Figure E1.4).
y
x(τ)=L an H
a g
x()=L x
O an
a g
Thus, (1) Kinematical equations for mine movement projected on axes x and y can be written:
x(t ) 0 t cos and
y(t ) H 0 (sin )t
gt 2 . 2
(2) An equation of the trajectory can be obtained by excluding time from the kinematical equation for the missile’s movement: sin x , then y( x ) H 0 x 0 cos 0 cos g g 2 x 2 H (tan ) x 2 x2 . 2 20 cos 20 cos2
since t
(3) The velocity (t) projection on coordinate axis can be found by time differentiation of x(t) and y(t) as x (t )
dx dy Ox 0 cos const., y (t ) 0 sin gt. dt dt
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(4) The dependence of velocity (t) on time, in vector form, can be obtained in the form (t) ix(t) jy(t). Then the velocity modulus is 冷 (t ) 冨 x2 y2 02 2 gt0 sin g 2 t 2 . The velocity (t) direction can be determined by an angle between this vector and the axis Ox. It can be seen from a Figure E1.4 that tan (t )
y (t ) x (t )
⎛ sin gt ⎞ 0 sin gt , therefore (t ) arctan ⎜ 0 . 0 cos ⎝ 0 cos ⎟⎠
(5) Since the total acceleration is constant (i.e., g) the moduli of tangential 冟a冟 and normal 冟an冟 components (as can be seen from Figure E1.3) will be equal to: a g sin and an g cos , where sin (y /x) and cos (x /y ) or a (g(oy gt) / (2o 2gtoy g2t2)1/2 ) and an n(goy /(2o 2gtoy g2t2)). (6) The highest point of the flight ymax can be found from the kinematical equations y(t) and y(t). From the first dependence one can find the time of ascent tasc, from the second—the maximal ascent ymax y(tmax). In the upper trajectory point y 0, therefore Oy gtasc 0, then tasc (Oy /g). Therefore: ymax H
2 2 2 2 Oy Oy gOy gtasc H 0 sin 2 H 245m 2g 2 g 2g
(7) The missile’s total flight time can be found from the fact that in the moment of its drop y(t) y() 0, i.e., H 0 sin α (g2/2) 0 and 2 (20 sin /g) (2/g) H 0. Solving the quadratic equation regarding τ we can arrive at 1,2 12
20 sin 2 sin 2 2 gH 0 2 2 , g g g
1g 冢0 sin 02 sin 2 2 gH 冣
(Since the time cannot be negative we should accept sign “”). Executing calculations we obtain 14.9 sec. (8) The missile’s range L can be found by inserting τ into the x(t): L x() Ox 596m. (9) The modulus of velocity at the moment it hits the ground can be found using the equation given in point (4) substituting a running time on as 2 () 兹 苶苶 苶苶0苶g苶si 苶n苶苶 苶 苶g2苶苶2 87 m/sec 0 2
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or, converting this expression to a form 兹苶 02苶 苶苶2g苶H 苶, one can obtain the same result. Since tan (y /x), then () arctan 62.5°; a minus sign shows that the velocity vector makes with the x-axis an angle , counts off in a negative direction, i.e., clockwise (Figure E 1.4, insertion). (10) To find curvature radius one can use expressions n (2/) wherefrom ( 2/an), and an g cos , and is the speed at the moment of hitting the ground: (τ). Therefore,
1
02 20 g g 2 2 02 2 gH 1.67 m, g cos g cos
however, at the highest point 2 2Ox/g 163 m. Note that the curvature at the maximum height is approximately 100 times less than at point of hitting the ground. By solving a problem in this way, we can then use all these particular equations in future.
1.2.2
Kinematics of translational movement of a rigid body
A body in which the distance between two arbitrary points remains at constant temperature, unchanged by any motions or interactions, is referred to as an ideally rigid body (IRB). During the translational motion of the IRB, any segment inside of it remains parallel to itself at any time. With such motion the displacement, velocity and acceleration of any point of the IRB are the same at any time. Therefore, many characteristics of the IRB’s translational movement can be described by the motion of a single body’s point with a mass equal to the mass of the whole body moving with velocity (acceleration) in any point of the body. The best point to choose is the centre of mass (CM) (see below). 1.2.3
Kinematics of the rotational motion
Rotational movement is widespread in nature, no less (but can be even more) than translation motion. Indeed, the motion of electrons around the nucleus (within the Bohr atomic model) and the earth around the sun, the rotation of a gyroscope, the rotation of numerous details and assemblies in technology and industry, the rotation of a wheel (this genius invention of mankind)—all of these are examples of rotational motion. The rotational motion of the IRB around a motionless axis Oz in which all points of the body are moving in parallel planes, making circles with their centers lying on a single straight line coinciding with the z-axis, is referred to as the rotational motion of the IRB. When rotating, all points of the IRB have linear velocities differing in size and direction, depending on the point distance from the axis of rotation. So, for a description of rotational motion we should introduce angular kinematic features unique to the whole body: angular displacement, angular velocity and angular acceleration. Let us restrict ourselves to the case of IRB rotation around an axis whose space position does not change in time.
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Angular displacement Consider a body revolving around axis Oz. Select in the body a point, not lying on the axis of rotation (point A in Figure 1.4; the body itself is not shown in the figure). In accordance with the definition of rotational motion, this point while moving describes a circle with a radius R, the center of which (O) is lying on an axis Oz. While rotating, vertical planes drawn through the axis of rotation and any body point turn on the same angle. Let the plane (and the body) turn on an angle d. This angle is referred to as the angular displacement. The angular displacement is a vector, coinciding with the axis of rotation, whose direction is defined by the right-handed system. Remember that this rule concludes that if the right screw reconciles with the axis of rotation and turns it in a direction complying with the rotating body, the translational direction of the screw movement along the axis of rotation . complies with the direction of the vector d Vectors whose directions are aligned with the rotation direction are called axial vectors. is an axial vector, the modulus of which is equal to the ratio of Angular displacement d arc dS and radius R and the direction of which coincides with the rotation axis in accordance with the right screw rule. d
dS k, R
(1.2.24)
k being the ort of rotation axis Oz. The value to which the limt → 0 tends is called the angular velocity :
d (t ). dt
(1.2.25)
z
d
k 0 R
d A
dl
. Figure 1.4. Elementary angular displacement vector d
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∆ > 0
∆ < 0
(a)
(b)
Figure 1.5. Relationship between angular velocity and angular acceleration.
Angular velocity is a first time derivative from the vector of angular displacement. It shows the speed of angular displacement changing with time. Angular velocity is also an axial vector, which coincides in direction with the angular displacement vector d. The value to which the limit (/t) tends is called angular acceleration: d k k, t0 t dt
lim
(1.2.26)
Angular acceleration is also an axial vector. The different mutual orientations of angular velocity and angular acceleration are presented in Figure 1.5: when the angular velocity is rising (d 0), then the direction of the angular acceleration vector coincides with the former (both are directed along the axis of rotation); if the angular velocity decreases (d 0), then the direction of the angular acceleration vector is opposite to the angular velocity. If the axis of a body rotation changes its orientation in the course of time, some interesting effects appear which are unfortunately beyond the scope of our consideration here. There exists a linear relationship between angular and translation features. It can be seen in Figure 1.4 that dl Rd, i.e. d (dl/). Time derivation (d/dt) (1/R)(dl/dt) (1/R), i.e., R.
(1.2.27)
The relationship between angular acceleration and linear tangent acceleration a can also be obtained. The modulus of angular acceleration is (d/dt), where (/R); then (d/Rdt). Since (d/dt) is a point linear (tangential) acceleration a then (a/R) or a R.
(1.2.28)
The last formula connect the linear and angular characteristics (Figure 1.6). They can be given in vector form: [r ],
(1.2.29)
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z
ω
k
O r
υ
Figure 1.6. Relationship between vectors of angular and linear velocity.
and [a r ].
(1.2.30)
Using eqs. (1.2.27) and (1.2.28) we can obtain by integration the dependencies of (t) and (t) like (1.2.22) and (1.2.23) valid for uniformly accelerated rotation 0 t ,
(1.2.31)
and
0 0 t
t2 . 2
(1.2.32)
where 0 and 0 are angular characteristics at the initial instant of time. The structure of these expressions is equivalent to those obtained for linear motion (eqs. (1.2.22) and (1.2.23)). However, rotational motion is distinct from linear because of the fact that it is periodical. In rotation, the values of are repeated in a certain time interval. If these intervals are constant (uniform rotation, const.), the period T, a duration of one full turn (on 360°), and accordingly rotating frequencies, i.e., numbers of full repetition in the unit time (v n 1/T ), can be used. Bearing in mind that one turn corresponds to an angular displacement equal to 2 radian, we can introduce an angular velocity 2v 2/T. The difference between angular vector features of revolution and the corresponding features of linear motion lies in the fact that angular vectors are directed not along the linear motion of each point of the IRB, but along the axis of rotation (perpendicular to their planes of motion). Many remarkable characteristics of rotational motion are bound up with this circumstance (refer to Section 1.3.9, Figure 1.17 and Appendix 2).
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In general, the arbitrary motion of the IRB can be presented as a combination of translation motion of an MP with a mass equal to the mass of the whole IRB and located in the center of inertia (refer to Section 1.3.7, see below), and rotation of the body’s points around the center of inertia. EXAMPLE E1.5 A disc of a radius R 10 cm starts to rotate with angular acceleration 0.1 rad/sec2 around a motionless axis, perpendicular to the disc’s plane passing its geometrical center. Determine at the time instant 12 sec after the beginning of the disc rotation: (1) an angle of the disk turns (an angular displacement); (2) the number of complete revolution Ntot; (3) the net turn angle ; (4) the distance traveled by any point A of the disk crown S along an arc; (5) the angular speed value and (6) a frequency of rotation n at this moment. Solution: (1) For a uniformly accelerated rotation a kinematical equation is (1.2.32), where (t) is a turning angle for time instance t, 0 and 0 are the initial angle and angular speed; in our case 0 0 and 0 0. Therefore (t) ( t2/2). Introducing the numerical values and execute calculations for the time instant we obtain () (0.1122)/2 rad 7.20 rad (413°). (2) We can find the number of revolutions N by dividing the previous result by 2, i.e. N (/2) 7.20/(2 3.14) 1.15 revolutions. Since the number of revolutions is an integer then Ntot 1. (3) The net turn angle can be found as a difference between the final turn angle minus the 2 integer value: (720 2 1) rad 0.917 rad (52.6°). (4) The total distance S traveled by point A along an arc can be found multiplying the turning angle by the radius R:S R 7.20 0.1 m 0.72 m. (5) To determine the disk angular speed at the time instance one first should take the time derivative of an angular displacement
d d ⎛ t 2 ⎞ ⎜ ⎟ t . dt dt ⎝ 2 ⎠
For t we obtain () . Execute the calculations 0.1 12 1.20 rad/sec. (6) The instant frequency of rotation n() can be obtained as n (/2) (1.20/2) 0.19 sec1. 1.3 DYNAMICS Dynamics deals with the study of a body’s motion with definite mass under the action of applied forces. 1.3.1
Newton’s first law of motion: inertial reference systems
Generally speaking, the same physical events can be described differently in different reference systems. Undoubtedly, we would wish to find a reference system in which the laws
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of different physical phenomena have the simplest expression. On the other hand, it would be interesting to find in the surrounding world a system that would be at absolute rest so that any motion could be considered with respect to this system. Is it possible to find such a system? To answer this question we shall analyze the simplest form of motion—the motion of a free body. A body is called free if it is at such a distance from all other bodies that their effect on it is negligible. (Such a body and such a motion is actually a physical abstraction since it cannot be fully realized. Nevertheless this model has played a very important role in the development of physics, from Aristotle to Galileo and Newton). So, experiencing no external effect, the free body must move rectilinearly and uniformly. Such a motion cannot be achieved in any reference system but only in the so-called inertial one. A reference system is referred to as inertial if the free body moves in relation to it with constant velocity—in magnitude and direction. Besides, if one of the reference systems moves relative to another, additional effects can appear. All these questions are the subjects of different theories of relativity, realizing the relationships between physical laws in reference systems moving relative each other (refer to Section 1.6). The classical theory of relativity is based on Galileo’s and Newton’s hypotheses. Their main feature is separation and independence the space and time and their independence. The laws of the classical theory of relativity appear from mankind’s everyday experience of isotropic and uniform space (all directions are equivalent and space metric is constant everywhere) and independence of time intervals from the reference system (an interval in Moscow is the same as in London). These laws prove to be perfectly justified in the case of motion of a material body with velocities 0
(a)
b 2
2 A 0
∆K > 0
U+K=const
∆U < 0
∆K > 0
therefore ∆T > 0
therefore ∆T< 0 (a)
r2
(b)
Figure 3.26 (a) and (b) An explanation of the Joule–Thomson effect.
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For any gas, the sign of the Joule–Thomson effect depends on temperature and pressure. The positive effect for each gas is observed only in the limited interval of temperatures and pressures. For each gas there are values of temperature and pressure at which the Joule–Thomson effect is equal to zero (no temperature changes occur at gas expansion in vacuum). These points (Ti, pi) are called points of inversion. At these points, the influence of forces of attraction is completely compensated for by the influence of repulsion forces; consequently the gas temperature does not change. The set of inversion points forms an inversion curve in a p–T diagram. Figure 3.27 presents the inversion curve for nitrogen. It can be seen that, to a given value p, two points of inversion can occur. The curve of inversion outlines two points of inversion for which a positive Joule–Thomson effect is observed. Values for the upper and lower inversion points for some gases at various pressures are given in Table 3.2. For the majority of gases, the upper point of inversion lies above room temperature. Hydrogen and helium are an exception. The Joule–Thomson effect finds important practical applications in the techniques of gas fluidization, when the gas is throttling over a wide range of pressures from 2×107 to 105 Pa. Successive repetition results in reduction of the gas temperature down to its boiling point. If this procedure is applied to air, first oxygen gas is condensed (90 K) and then nitrogen (77 K). The boiling point of hydrogen is 20 K and helium 4.2 K. All these gases are widely used in various technical equipments, scientific explorations and in various crystallographic technologies.
p(10 MPa)
3
2
1
0
73
573 T,K
Figure 3.27 An inversion curve of the Joule–Thomson effect. Table 3.2 Upper and lower temperature inversion points for some gases (in K) Gas
p (in 105 Pa)
Upper
Lower
He H2 Air CO2
1 113 150 18–100
23.6 192.7 583 2050
140 249
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3.7 3.7.1
ELEMENTS OF PHYSICAL KINETIC
Introduction
In the previous sections we have considered ideal gases in a state of thermodynamic equilibrium. Such systems are stationary, i.e., their parameters do not change in time. In this section we shall consider macroscopic systems removed from an equilibrium state and aspiring to return to it. The area of physics that deals with this process is referred to as physical kinetics. To describe quantitatively, such a process is possible only within the framework of the model of an ideal gas. The application of the results of this research to real gases and even to liquids can be provided on a qualitative or semiqualitative level. Although such analysis is carried out with such approximation, it can be usefully applied to nonideal systems. In physical kinetics, there are two approaches to the study of physical phenomena: empirical phenomenological and microscopic at the molecular level. In the first of these, problems are investigated from a macroscopic point of view without considering the detailed atomic mechanism. In the second method the behavior of systems is investigated from microscopic standpoint on the basis of molecular representations. Both methods should yield the same results as the description of the same phenomena. A system can be removed from a condition of thermodynamic equilibrium by external influence. For example, one can inject another gas into a certain point of the predominant gas and, due to thermal (chaotic) molecular movement the concentration of the second component will tend to spread over the whole volume. Sooner or later it will be equal. One can heat gas locally in one area of the volume and the gas temperature will also start to equalize over the whole volume due to molecular chaotic movement. From the resulting examples it is clear that the thermal movement of molecules plays an active role in reaching equilibrium. We shall examine below the phenomena of alignment caused exclusively by this factor—chaotic movement of molecules, i.e., at a molecular level. This does not mean that there are no other mechanisms of alignment; however, we will avoid them at the moment. 3.7.2
Transport processes: relaxation
Transport phenomena appear in different substances, mostly gases, because of chaotic molecular movement within them and permanent collisions with an interchange of their kinetic properties. During chaotic molecular movement and collisions, a transfer of energy and momentum and masses diffusion can take place and there is hence a gradual alignment. In this way a system can reach a state of equilibrium. The process of a system returning back to equilibrium is called relaxation. Let us evaluate the speed of the relaxation process from a mathematical point of view. We shall suppose that the displacement rate (d/dt) of the return to equilibrium is proportional to the displacement itself and is opposite to it in sign: (d/dt)k, where k is a coefficient of proportionality. Integration of this equation gives ln (t ) kt ln C
(3.7.1)
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or, (t ) C exp(kt ).
(3.7.2)
Constant C can be found from the initial displacement. If at t 0 the maximum displacement 0 is taking place C 0 and eq. (3.7.2) takes the final form: (t ) 0 exp(kt ).
(3.7.3)
The time for which the displacement will decrease in e time is called the time of relaxation. This concept can be applied to a number of phenomena: radioactive fusion, damped oscillation, etc. Let us find the physical sense of factor k. If we write eq. (3.7.3) for the time , the displacement becomes: () 0 exp(k). Therefore, (()/(0)) exp(k) e1. Therefore, k 1 or k (1/). Hence, ⎛ ⎞ (t ) 0 exp ⎜ ⎟ . ⎝ t⎠
(3.7.4)
This equation is valid for many relaxation phenomena, not only for transport properties in gases. We must emphasize that the transport phenomena can take place either in a closed system (where the system is removed from an equilibrium state and recovered coming back to it due to chaotic molecular movement), or in open stationary system where nonequilibrium can be induced permanently due to an external influence. In both cases, the relaxation consists of the redistribution of the uniformity over the whole system. 3.7.3
Transport phenomena in ideal gases
We define an ideal gas (see Section 3.1.3) as a gas whose molecules move from collision to collision without interaction. As a first rough approximation, we shall represent molecules as solid spheres with a certain diameter d. Enter an important characteristic of the gas, namely the free path length, i.e., the average distance , which a molecule moves freely from one collision to another. We can calculate the average time between molecule collisions as t /, where is the average speed of the thermal movement of the molecules (to keep things simple in this section, we shall not write the sign on averages). The average number of collisions of a molecule for 1 sec will be written as: 1 y
. t
(3.7.5)
The shortest distance between the centers of the molecules on impact is referred to as the effective diameter of molecules d (see Section 1.5.4, Figure 1.34). During its
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movement, a molecule (which we consider as a solid spheres) collides with those molecules whose centers lie at distances smaller than d from the line of its movement (Figure 3.28 inset). The value d2 is the target area in which the center of the molecule should lie in order to guarantee the collision. This area is called a collision cross section. On collision, each molecule performs a zigzag path. If one rectifies the molecule’s path one can assert that in the unit of time, the molecule describes the cylinder volume with a height equal to the speed of the molecule and with a cross-section d2. The collision will take place with those molecules whose centers lie inside this cylinder. We shall consider that all molecules are at rest except those, which we follow up. Then the number of collisions in a unit of time will make
d 2 yn,
(3.7.6)
where n is the molecule concentration. Taking into account the movement of other molecules leads to an additional term 兹2苶. Then expression (3.6.6) will have the form:
2 d 2 yn,
2 d 2 yn.
(3.7.7)
From comparison of eqs. (3.7.5) and (3.7.7) it can be deduced that
1 2n
1 2 d 2 n
.
(3.7.8)
Consider the relation of the free path length on the parameters of an ideal gas state. In an isothermal process p nT then ⬃ 1/p. So for nitrogen under normal conditions 107 m, at pressure 1 Pa 102 m, at pressure 104 Pa a molecule on average runs a long distance without collision (about 103 m). If n const., (isochoric process) should
def = πd 2
d
d
Figure 3.28 A path traveled by a molecule.
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not depend on temperature. However, practice shows that increases a little if the temperature rises; this dependence is described by empirical law:
T T C
(3.7.9)
where is the free path length at T . 3.7.4
A macroscopic representation of a transport coefficient
Molecules in a closed volume can differ in their characteristics (mass, momentum, energy, etc.); assume that the spatial distribution of property values initially is nonuniform. Due to the thermal movement of the molecules and their collision, a process of restoration of uniformity in the volume can take place. We began a general analysis of these phenomena, then at particular processes: diffusion, heat conductivity and internal viscosity. Denote G a transferable property and let it be distributed along an x-axis nonuniformly (Figure 3.29). We shall fix a point x0 on the x-axis and an area S normal to axis x. In time
G(x)
G2 −G1≈− dG 2 dX
x0−
x0+ x0
υ0∆t
x
υ0∆t
Figure 3.29 Macroscopic consideration of transport phenomena.
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t, due to thermal motion, a number of molecules will pass through this area from right to left and from left to right; both numbers being identical. However, each molecule will transfer that “quantity” of a property G, which corresponds to the last collision of the molecule. Due to nonuniformity of distribution G(x), it will cause resulting transport G through the given area S which we shall designate q. We shall consider q 0 if the property G is transferred to the positive direction of the x-axis and vice versa. At the same time it is easy to understand that the sign q should be opposite to the sign on the gradient of the G-function, i.e., to the sign of derivative dG/dx. Thus, from the most general macroscopic consideration, it follows that q ⬃
dG( x ) S t . dx
Let us introduce a macroscopic (phenomenological) coefficient , which we denote as a transport coefficient. Then q
dG( x ) St . dx
(3.7.10)
The expression (q/t)(1/S) is a flow of the G property. Let us determine the microscopic meaning of coefficient . We will take two layers of the gas, parallel to the area S at distances from both sides of point x0. Let G1 and G2 be the values of G in these layers. All molecules allocated in parallelepipeds (see Figure 3.29) and moving in the definite direction (to area S ) at a speed and in time t will cross the area S in both directions. As the molecules move chaotically, the number of molecules moving in each side is 1/6 of their total number. Their number is equal, but they carry different amounts of property G. All molecules moving in a positive x-direction carry property G1whereas in a negative direction they carry property G2. The total transport of the G-value through the area S is q 1 1 ny (G1 G2 ). t S 6
(3.7.11)
Assuming that the gradient of the G-value is small, we can write (G1G2)(dG/dx)2. Hence, q 1 1 dG ny. t S 3 dx
(3.7.12)
Comparing eqs. (3.7.10) and (3.7.12) we arrive at 13 yn.
(3.7.13)
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The significance of this coefficient consists of the fact that it connects the macroscopic (technical) value with microscopic one ( and ). Notice that this result is obtained from the comparison of macroscopic (phenomenological) and microscopic approaches.
3.7.5
Diffusion in gases
Imagine a gas in which a certain constant difference of concentration exists in the various fixed points (a nonequilibrium, but stationary system), whereas the temperature at any point in this system is the same and remains constant. Thus, there is a resulting flux of gas molecules referred to as stationary diffusion. We would like to emphasize that this flux is caused only by the chaotic movement of molecules and not for any other reasons. Another situation arises when the difference of concentration is not constant; the system aspires to equalize concentration; this will be nonstationary diffusion. In any case, the driving force of molecular diffusion is the concentration gradient. We shall restrict ourselves to self-diffusion, i.e., diffusion of similar molecules distinguished by any insignificant (for diffusion) property (e.g., labeled by radioactive mark) in the environment of the same nonradioactive molecules. In fact, the diffusion of carbon oxide in nitrogen can be considered as self-diffusion because the molecules of these gases differ insignificantly with regard to their size and mass. Our task is to find a relationship between the macroscopic and microscopic characteristics of diffusion. We shall consider a simple one-dimensional case: imagine a vessel in the form of a long thin cylinder (an x-axis is directed along one axis of the cylinder) containing a mixture of two gases; the concentration gradient of one of the gases is artificially maintained constant. Assume that the masses and the sizes of the molecules of both gases are identical: (m1 m2 m). Therefore, the molecules can acquire identical average speed u; free path length can be taken as equal to 1/(兹2苶n), where n n1 n2 is the total average concentration of molecules in the volume. A comparison of what we obtained from a macroscopic consideration and that of microscopic one looks like
j
q G D , St x
(3.7.14)
where j is the mass diffusion flux and ⬅ D is the diffusion coefficient. In this form, this equation is referred to as Fick’s law. The transport value G(x) in this case is the relative concentration of the chosen molecules n1, that is G(x) (n1(x))/n. Comparison with the eq. (3.7.13) shows that
j
n ( x) q ⎛ 1⎞ ⎜ ⎟ ny 1 . ⎝ 3⎠ St n
(3.7.15)
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By comparing eqs. (3.7.13) and (3.7.14), we can further derive the microscopic expression for the macroscopic diffusion coefficient D: D 冢 13 冣 y.
(3.7.16)
It follows from this expression that the physical sense of the diffusion coefficient D consists of the fact that it shows the number of molecules that diffuse through a unit area in a unit time and at a unit gradient of relative concentration. Data on key parameters of diffusion as well as other transport phenomena, i.e., of heat conductivity and viscosity, are shown in Table 3.3. If molecules differ considerably in their masses and the dimensions mentioned above, the calculations demand specification. More detailed examination shows that the process of diffusion is determined by the speed of the fastest (smallest) molecules, whereas for effective cross section determination, it is the larger molecules. For the nonstationary diffusion, it is possible to estimate the time t for which there is an alignment of concentration (reduced in e times) from a dimension consideration. In fact, is defined only by the character of distribution of molecular masses in the initial instant of time and gas property. The initial state is defined by the dimension of heterogeneity area L. There is only one combination from D and L that has a dimension of time, namely Table 3.3 Transport phenomena characteristics Phenomenon
Main laws Transferred value
Driving force
Diffusion
q dc( x ) D St dx Fick’s law Mass
dc( x ) dn1 ( x ) dx dx Concentration gradient
Transfer coefficients
1 D y 3
Heat conductivity
q dT ( x ) St dx Fourier’s law Energy
dT ( x ) dx Temperature gradient
1 y(C )sp 3
Internal friction
q du( x ) St dx Newton’s law Momentum
du( x ) dx Velocity gradient
1 y 3
Coefficient’s value Units Gases Liquids Resin, glass Crystals
m2/sec
W/(m)
104106 1091010
104105
10161018
100 102
Non-SI units: *Poise (P) 0.1(N sec/m2); **Stokes (St) 104(m2/sec).
1 y 3
N sec/m2* m2/sec** 103105 ⬃102 ⬃106 ⬃1015
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⬃L2/D. The time in gases can be estimated under the order of value: at D⬃104 m2/sec and L ⬃ 0.1 m, the order of value is 103 sec, i.e., diffusion is a slow process even in gases and in liquids it is slower by some orders. 3.7.6
Heat transfer
If, at different points of any gas in a closed volume for a short time, a different temperature is created and the gas is then left to itself, the temperature at all gas points is equalized owing to the process of heat transfer. From a macroscopic point of view, the phenomenon of heat transfer in gases consists of temperature transport from hotter to colder places. Within the framework of the molecular kinetic theory, the process of heat transfer consists of the fact that molecules from a heated site of the gas, where they have large kinetic energy, transfer the energy to cooler areas via collisions; a flow of heat is thus created. In reality, in gases and liquids this phenomenon is usually accompanied by heat transport by the steam of a gas or a liquid initiated by their density difference, i.e., so-called convection. However, we will now consider heat transfer exclusively from the point of view of molecular kinetics. By the way, it is very difficult to subdivide these two processes. The transfer of heat caused by the thermal (chaotic) movement of microparticles is referred to as a heat transfer phenomenon. Thus, in the general transport formula (3.7.10), q is the flux of heat. The transferable value in this case is the amount of heat Q or G(x) CVT(x) where CV is the heat capacity of a substance. This heat is produced by the total kinetic energy of the molecules. Hence the heat transport description can be given as: dq 1 dT ( x ) , dt S dx
(3.7.17)
where is the heat transfer coefficient. In this form, this equation is referred to as Fourier’s law. The minus sign shows that the heat transfer is directed against the temperature increase (against the temperature gradient). From a microscopic point of view, the driving force is the molecular averaged kinetic energy gradient: d具典/dx. Taking into account that 具典(i/2)T, we obtain: d 冓冔 i dT i dT N A iRm dT dT (CV ) m . sp dx dx 2 dx 2 dx N A 2 N A m dx
(3.7.18)
where sp index specify the specific heat capacity. From the other side a value of the particle’s flux is equal to (dq/dt)(1/S)(1/3)n (d具典) / (dx). Substituting here the value d具典/(dx) from eq. (3.7.18), we obtain: (dq/dt) (1/S) (1/3)(CV)spmn(d具T典) / (dx). From this equation the heat transfer coefficient is derived 13 y(Cv )sp ,
(3.7.19)
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where mn is the density of the substance. The physical meaning of the factor of heat transfer consists of the fact that it defines the amount of heat transferred in a unit of time through a unit of perpendicular area S⬜ and at a unit of temperature gradient. Analyzing expression (3.7.19), we can see that the heat transfer coefficient depends on the structure of the gas’s molecules (as depends on the molecules’ number of degrees of freedom). In isothermal processes, does not depend on gas concentration and pressure ( ⬃ n and ⬃ 1/n). In the isochoric process depends on temperature as 兹T 苶. The independence of on pressure seems strange at first sight: in fact it is well known that for manufacturing high quality Dewar vessels (thermoses), the presence of air in the space between the walls reduces the quality of the vessel (outflow of heat): the less pressure between the walls of the vessel, the higher the Dewar quality. However, the result obtained seems to contradict this statement. In fact, there are no contradictions in the result, there is incorrectness in its interpretation: at low pressure when becomes commensurable with the space between the walls, dependence of on (1/n) disappears and becomes independent on n (and on p) as will be shown in Section 3.7.8. 3.7.7
Viscosity or internal friction
Viscosity is a property of gases, liquids and solids to resist a flow under the action of external forces. We shall consider in more detail the viscosity of gases. In Figure 3.30, the laminar flow of gas or a liquid is schematically shown; the division of the overall flow into layers is also shown. Due to viscosity, the speed of movement of various gas layers is different: because of its interaction with the unmovable wall’s border, the edge layer is zero and due to the external pressure increases to a central line. The microscopic description of internal viscosity of gases is based on the fact that all molecules in the limit of its layer participate in the macroscopic motion with the momentum, where u is the macroscopic velocity of a molecule with the layer. However, owing to their chaotic movement, molecules jumping over one layer to another transfer the macroscopic momentum m u. Such jumping can influence the macroscopic speed of layers: molecules from faster layers accelerate slower layers and vice-versa. It looks like one layer renders a friction on its neighbor. Thus, viscosity in gases is the phenomenon of the chaotic momentum transfer of the macroscopic movement from layer to layer in a flowing gas. Consider now the law to which the phenomenon of viscosity submits. For this purpose, we shall consider the behavior of a gas between two flat parallel plates (Figure 3.30b). Let one of the layer be resting and the other moves with a constant speed u parallel to the plane of the plates. If there is a viscous medium between the plates in order to move this plate with constant speed u, it is necessary to apply some constant force (directed along the speed) since the medium will exhibit resistance to such movement. Accordingly there appear tangent forces between the separate layers in the medium. Experience shows that this force F is proportional to the speed of plate u and to the area of plates S and is inversely proportional to the distance between the plates x. In a limit at x 0 ⎛ du ⎞ F ⎜ ⎟ S, ⎝ dx ⎠
(3.7.20)
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X F dx S
(a) X
(b)
u(x)
Figure 3.30 The viscous flow: (a) streamline motion of gases and liquids at a laminar flow, (b) friction between adjacent layers: the Newton’s law of internal friction.
i.e., force F, which needs to be applied to move the two layers of gas over each other, is proportional to the area of contact of layers S and to the gradient of relative speed (du/dx) perpendicular to the moving layers. This is Newton’s law of internal friction, is the coefficient of dynamic viscosity of the media in which the movement takes place. To clarify its physical meaning, we shall increase the left- and right-hand parts of eq. (3.7.20) by t. In this case, we have Ft (du/dx)St. On the left-hand side we substitute Ft, which is equal to p (eq. 1.3.12), i.e., ⎛ du ⎞ p ⎜ ⎟ S t , ⎝ dx ⎠
(3.7.21)
where p is the change in momentum of a flux element due to the change in the speed of movement. The coefficient of dynamic viscosity is numerically equal to the momentum of macroscopic movement, which is transferred in a time unit between adjacent layers of flux at a speed gradient along the x-direction equal to unity. In the phenomena of viscosity,
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the transferred quantity is the momentum of the macroscopic movement of the molecules’ layer G(x) mu(x). Substituting this value in eq. (3.7.10) we obtain: q 1 du( x ) , nym St 3 dx
(3.7.22)
冢 13 冣 y.
(3.7.23)
therefore,
For dynamic viscosity in SI units, the coefficient of viscosity of that media is that at a speed gradient equal to unity through area S in 1 m2, the total molecule momentum 1 N sec / m2 kg m / sec is transferred. Thus the unit of viscosity in SI units is N sec/m2 kg/(m sec). Another widely used unit of viscosity is g/(cm.sec) (poiseuille or poise (P)) (in honor of J.L.M. Poiseuille). In tables, viscosity is usually expressed in centipoise (cP). The ratio between units is 1 kg/(m sec)10 P. Besides the coefficient of the dynamic viscosity , the coefficient of the kinematical viscosity is used in technology; there is a relation between the two coefficients: the kinematical coefficient is the ratio of the dynamic viscosity to the medium density /. The coefficient of the kinematical viscosity is measured in stokes (St): 1 St 1 cm2/sec. In SI the unit of kinematical viscosity is m2/sec (1 m2/sec 104 CT). Some viscosity factors are given in Table 3.3.
EXAMPLE E3.15 A free path flight of CO2 gas at normal conditions is 40 nm. Determine the molecule’s average speed 具典 and the number of impacts z a molecule undergoes in 1 sec. Solution: The average molecule speed can be found according to expression (3.3.7´). Substituting the known and given values we obtain 具典 362 m/sec. The average number of impacts can be found according to equation z / (the averaging sign is omitted here). Substituting the given values in this equation, we arrive at z 9.05 109 sec1.
EXAMPLE E3.16 Two horizontal disks with identical radii R 20 cm settle one over another so that their z-axes coincide. The distance between the disks is d 0.5 cm. The lower disk rotates around the common axis (refer to Figures E3.16a and b) with frequency n 10 cm1, but the upper disk is kept motionless. Find the torque M working on the upper disk. Both disks are in air, the coefficient of the dynamical viscosity of air being 17.2 Pa sec.
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z Driven disk
R d
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(a)
R
dr r
(b)
Solution: Due to the viscosity of the air that is between the disks, the rotational moment of forces is transferred from the bottom to the upper disk. Together with rotation of the bottom disk, the adjoining air layer comes in movement. In order to find the moment of friction forces working on the upper disk, we shall allocate an elementary thin layer of air with thickness dr. The bottom air layer rotates together with the bottom disk with angular velocity . Accordingly, the linear speed of air particles is u r. We shall take advantage of Newton’s law of external friction (see Section 3.6.4, eq. (3.7.20) and Figure 3.30) and make some simplifications (proceeding from a condition of the problem). We shall take the velocity gradient to be equal to (du/dz) ⬇ (u/d) (u/d) (since u2 is equal to 0). We shall express the force of viscous friction by an elementary tangent force dF and the small area
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of air layer touching as dS 2 rdr. Accordingly, Newton’s law for our particular problem will be u dF 2 rdr. d We can find the elementary moment of friction forces relative to the z-axis working on the part of the allocated air layer having increased a tangent force dF on a shoulder r, i.e., ⎛ n⎞ dM dF r 4 2 ⎜ ⎟ r 3 dr. ⎝ d⎠ To find the total force moment, we have to integrate dM over the whole disk area, i.e., find the overall contributions from all elementary layers dr, that is to integrate the elementary force moment upon the radius in the limits from 0 to R. We obtain R
n R4 R4 ⎛ n⎞ M 4 2 ⎜ ⎟ ∫ r 3 dr 4 2 n . ⎝ d⎠ 0 d 4 d Checking the dimension of the result, we obtain mN m, which corresponds to the torque dimension. Let us express all values in SI units and execute calculations: MF3.1421.72105 (0.2)4/510—3 mN m 0.543 mN m (milli-Newton m). EXAMPLE E3.17 Define how many times the coefficient of diffusion of gaseous hydrogen differs from that of gaseous oxygen; both gases are under identical conditions. The effective diameters of the hydrogen and oxygen molecules are 0.27 and 0.36 nm, respectively; the molar masses M of oxygen and hydrogen molecules are 16 and 1, respectively. Solution: The coefficient of diffusion D is expressed through physical characteristics of molecules by the formula (3.7.16): D (1/3)具典, where 具典 is the average molecules speed (3.3.7 ) and the mean free path. Therefore, we can write the diffusion coefficients for oxygen and hydrogen (to an order of 1/3): D1
1 8RT 3 M1
1 2 d12 n
and D2
1 8RT 3 M 2
1 2 d22 n
.
In the equations, we took into account that the conditions (n and T ) of both molecules are identical. Therefore the ratio of coefficients is as follows: 2
D1 M 2 ⎛ d2 ⎞ . D2 M1 ⎜⎝ d1 ⎟⎠
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Substituting the given values we arrive at: 2
D1 32 ⎛ 0.36 ⎞ ⎜ ⎟ 7.11 D2 2 ⎝ 0.27 ⎠ 3.7.8
A transport phenomena in a vacuum condition
If the free path length of the gas molecules becomes commensurate (or more) with the linear size of the vessel, the condition of the gas is referred to as a vacuum. This means that the process of intermolecular collisions, which usually plays a predominant role in establishing a balance is no longer of any importance; the interaction between the molecules and the vessel’s walls is more important in a vacuum. During its last collision with the surface of a hot wall, a molecule acquires additional kinetic energy and carries it, without sharing it with other molecules, to the surface of another wall to heat it. This kind of heat transfer is characterized by an absence of temperature gradient inside the gas (as molecules move the whole distance at the same speed); all occasions we have discussed so far have vanished. In order to estimate heat transfer in a vacuum, we shall consider a simple experiment. Imagine a chamber divided in two by a partition wall with a hole of area S in this dividing wall (Figure 3.31). The temperature of the left wall is T1 and of the right is T2. Let T1 T2. On impact with wall 1 a molecule obtains energy iT1/2. Moving toward wall 2 along an axis x and reaching it the molecule gives the acquired energy to wall 2. The amount of energy will make: ⎛ 1⎞ ⎛ i ⎞ q ⎜ ⎟ ny ⎜ ⎟ (T1 T2 )S. ⎝ 6⎠ ⎝ 2⎠
(3.7.24)
As the factor similar to the heat transfer coefficient in the given expression appears as (1/6)n(i/2); it is proportional to concentration (and hence to pressure, contrary to what we obtained earlier in Section 3.7.6). This can be expected as this fact is well-known
T1
S
T2 X
Figure 3.31 A scheme of vacuum experiments. Effusion.
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from human experience. In fact, the quality of Dewar vessels (thermoses) depends significantly on the gas concentration between walls (residual molecule concentration): the less molecule concentration, the better the thermal protection. Notice that this phenomenon starts to operate only when becomes commensurate (or more) with the distance between the walls. At a distance of 1 cm, the pressure should be no higher than 0.10 Pa. A further reduction in pressure reduces the heat transfer between the walls under the linear law. This distinction from the usual transport phenomena is not the only one for rarefied gases. The flow of molecules at outflow differs significantly too (it is referred to as Knudsen’s flow.) Viscosity or internal friction in high vacuum is absent as there is no collision of molecules. In spite of the fact that pressure does not influence the individual molecular movement, it does influence the current. Let us consider a gas outflow through an aperture under these conditions; i.e., molecular effusion (effusion is the slow outflow of gases through small apertures.) We shall restrict ourselves to the case of isothermal effusion. Refer to the scheme presented in Figure 3.31: two vessels where gas is at different pressures p1 and p2, The effusion flow will take place through the aperture. If p1 p2, the resulting flow will be directed from left to right. In the time unit the number of molecules crossing from left to right can be estimated as N1 n1S/6. The opposite stream is less as concentration is less: N2 n2S/6. Hence, the resulting stream is N N1 N 2
yS (n1 n2 ) ⎛ 1 ⎞ 8 T 2 ⎛ 1⎞ ⎜ S p⎜ ⎟ S p. ⎟ ⎝ 3 ⎠ Tm ⎝ 6T ⎠ m 6
(3.7.25)
One can see from this equation that effusion depends on molecular mass. This circumstance is used for separation of gas isotope mixes. At transmission of a gas mix stream through a porous partition with fine pores, there is an enrichment of the mixture by molecules to lighter isotope. Repeating the process results in a mixture enriched mainly by a definite isotope. With different temperatures in both chambers, the ratio of concentrations n1/n2 will be other than at normal pressure. In fact, in normal conditions, it follows from the formula p nT that n1/n2 T2/T1. Another ratio is observed in vacuum. Equating the number of particles moving in the time unit in two sides, we can obtain: N1N2n11Sn22S and finally arrive at the ratio (n1/n2)(2/1)(兹T 苶2苶 / 兹T 苶1苶), which differs from normal conditions. Notice that the laws marked in this section are relevant only to ideal gases. For gases that cannot be considered as ideal, or for liquids, the laws obtained should be used carefully. The method of measuring small pressures is based on the effusion phenomenon (104–105 Hg mm or 0.1–0.01 Pa), if the diameter of the aperture is small in comparison with the free path length. Otherwise, the gas outflow in the unit time occurs under the laws of hydrodynamics; the gas volume that is penetrated in the time unit is ⬃(1/兹 苶), where is the gas density. This allows one to determine the density of gases measuring time of outflow through small (0.1–0.01 mm) apertures.
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PROBLEMS/TASKS 3.1. How many atoms contain in 1 g weight each of gases (1) helium, (2) carbon, (3) fluorine, (4) polonium? 3.2. The cylinder contains gas at temperature t1 100°. To what temperature t2, is it necessary to heat gas so that its pressure p has increased twice? 3.3. Assume that dry air consists only of oxygen and nitrogen. In this approximation air mass fractions of both gases are w10.232 and w20.768, respectively. Define the relative molar weight Mr of air M(air)rM/k (M is air molar weight; k103 kg/mol). 3.4. A centrifuge rotor with a radius a 0.2 m is filled with atomic chlorine at T 3000 K. The chlorine consists of two isotopes 37C and 35C. The share of isotope 37 C atoms make is w1 0.25. Determine shares w1 and w2 of atoms of both isotopes near the centrifuge’s rotor if it is rotated at an angular speed of 104 rad/sec. 3.5. Knowing the Maxwell distribution of a molecule’s speed (3.3.6) derive function f(u) where u is the molecule’s relative speed u(/prob), i.e., expressed in dimensionless units. Apply this distribution to find that the particular molecule has the probability w of having a speed that differs from half the most probable speed prob/2 by not more than 1%. 3.6. Determine the relative number N/N of ideal gas molecules that have a speed in the limits from zero to 0.1 of the most probable speed prob 3.7. Find the relative number of ideal gas molecules whose kinetic energy differs from the average energy 具典 by not more than 1%. 3.8. How many times will the maximum of the f() function change if the gas temperature T increases two times? Illustrate the result graphically. 3.9. Find a number of all collisions N per t 1 sec between all hydrogen atoms in a volume of V 1 mm3 at normal conditions. 3.10. Find the average time 具典 of the free path flight of oxygen molecules at T 250 K and pressure p 100 Pa. 3.11. Calculate the diffusion coefficient of an ideal gas at (1) normal conditions, (2) p 100 Pa and temperature T 300 K. 3.12. Calculate the dynamic viscosity coefficient of oxygen at normal conditions. 3.13. Oxygen of mass m 160 g was heated for T 12 K, and expended an amount of heat Q 1.76 kJ. By which process did the heating take place: at constant V or at constant p? 3.14. Water vapor is expanded at constant pressure p. Find the work of expansion A, if the amount of heat consumed is Q 4 kJ. 3.15. Nitrogen is heated at constant pressure by an amount of heat Q 21 kJ. Determine the gas work A and the change of its internal energy U. 3.16. An ideal gas performs a Carnot cycle. The cooler temperature T2 is 290 K. By how much is the cycle efficiency e increased if the heater temperature increases from T 1 400 K to T 2 600 K? 3.17. An ideal gas performs a Carnot cycle. The work of isothermal expansion of gas is A1 5 J. Define the work A2 of isothermal compression if thermal efficiency e is 0.2.
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3.18. As a result of isochoric heating of hydrogen in weight m 1 d, the gas pressure increased by two times. Determine the change of gas entropy S. 3.19. A piece of ice m 200 g in weight taken at t1 10°C is heated to t2 0°C and fused; the water formed is heated to t 10°C. Define the entropy change at these processes. 3.20. Water in weight m 36 g is at boiling temperature at normal atmospheric pressure. Determine change U of the internal energy at its full evaporation. Specific heat of evaporation is L 2.26 J/kg. Ignore the volume of liquid water. 3.21. A gas mixture contains 1 2 mole of helium and 2 3 mole hydrogen. Determine the specific heat capacity Cp,sp of the mixture. M14103 kg/mol and M2 2103 kg/mol. 3.22. Determine the adiabatic index of partly dissociated nitrogen gas if its dissociation ratio is 0.2. 3.23. A mixture of gases consists of argon of mass m18 g and nitrogen of mass m24 g. Determine the specific heat capacity CV,sp of the mixture. (M140103 kg/mole, M228103 kg/mol). 3.24. Determine the dissociation rate of oxygen gas if its specific heat capacity is CV,sp 727 J/(kg K). (Moxg 32103 kg/mol). 3.25. A gas mixture consists of equal masses of neon and oxygen. Determine the specific heat capacity of this mixture. (M120103 kg/mol—neon, M232103 kg/mol—oxygen). 3.26. A gas mixture consists of an equal number of moles of argon and oxygen. Determine the adiabatic index . 3.27. A gas mixture of helium and hydrogen are at equal conditions. Find the adiabatic index of the mixture consisting of V1 4 L of helium and V2 3 L of hydrogen. 3.28. Equal volumes of gas neon and oxygen are at equal conditions. Determine the specific heat capacity CP,sp of the mixture. (M120103 kg/mol, M232103 kg/mol). 3.29. Determine the dissociation rate of the partly dissociated chlorine gas if the adiabatic index is (CP /CV) 1.48. 3.30. Identical particles of dust of mass m41019 g are in air as a suspension. The temperature of the air is T 284 K. Define the relative change of concentration (兩n兩/no) of particles of dust at levels with a height difference of h 1 m. 3.31. Air is in a uniform field of gravity. Assume that the temperature of the air is uniform and equals T 290 K. Define the relative change in air pressure (兩p兩/po) at levels with a height difference of h 8.5 km. Assume the molar mass of air to be 29 103 kg/mol. 3.32. Oxygen gas is in a uniform field of gravity. Assume the temperature of oxygen is identical in all layers and is T 300 K. Define the relative change in gas pressure (兩p兩/po) at levels with a height difference of h 80 m. Molar mass of the oxygen is 32103 kg/mol. 3.33. The rotor of a centrifuge is filled with air and finely dispersed particles of weight m 1019 g. The temperature of the air is T 286 K. Find the ratio of the concentration of particles (n/no) at the walls of the rotor and in its center if the rotor revolves with a frequency 8 sec1 and its radius is r 25 cm. 3.34. A concentration no of nitrogen molecules near to the ground surface is equal to 2 1019 cm3. What is the change in concentration 兩n兩 of the molecules at height
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Answers
3.35.
3.36.
3.37.
3.38.
3.39.
3.40.
3.41. 3.42.
3.43. 3.44.
3.45.
247
h 10 m? The temperature of the air is T 280 K. The molar weight of nitrogen is 28 103 kg/mole. Identical particles of dust are suspended in air. The temperature of the air is T 300 K. Under the action of a gravitational force, the relative change of the concentration of suspension particles (兩n兩/no) in levels with a height difference h 3 cm reaches a value of 0.01. Determine the particles’ mass m. Identical particles of dust of mass m 1 1018 g are suspended in air. The temperature of the air is T 300 K. At what distance h are the layers in which the ratio (no/n) of the concentration of the particles is equal to 4? Carbonic acid is in a uniform field of gravity. The temperature of the gas is T 300 K. At what height h does the relative change of the molecule concentration (兩n兩/no) of gas reach 1%? The molar weight of the carbonic gas is 44 103 kg/mol. The rotor of a centrifuge is filled with nitrogen at temperature T 300 K. The rotor radius r is 20 cm. At what frequency of the rotor rotation n (sec1) will the gas pressure on its wall exceed the gas pressure in the center of the rotor by 2%? The molar weight of nitrogen is 28 103 kg/mol. The rotor of an ultracentrifuge is filled with radon at temperature T 305 K. The radius of the rotor is r 20 cm. At what frequency of the rotor rotation n (sec1) will the pressure on its walls be times higher than the pressure in the center of the rotor? The molar weight of radon is equal 0.222 kg/mol. Determine the molecular thermal conduction coefficient of saturated water vapor at temperature T 373 K (100 ºC). The effective molecular diameter d is 0.30 nm, molar mass M 18 103 kg/mol. Find the dependence of the diffusion coefficient on molar mass M at p constant and T constant. Find the average time between oxygen molecule collisions at p 1 Pa and T 300 K. The effective diameter of oxygen molecule d is 0.36 nm, mole mass M 32 103 kg/mol. Determine the ratio (1/2) of the viscosity coefficients of oxygen and hydrogen at similar conditions if the ratio of effective diameters of their molecules is (d1/d2) (4/3). Helium of mass m 0.5 g is in a high-pressure vessel with volume V 20 L. Find the average path length of the helium atoms. The effective diameter d of a helium atom is 0.22 nm, molar mass is M 4.0 103 kg/mole. Oxygen is at a temperature of T 300 K. At what pressure p will the average molecule free path distance be N 103 times bigger that its effective diameter d 0.36 nm? ANSWERS
3.1. (1) 1.50 1023; (2) 5.02 1022; (3) 3.17 1022; (4) 2.87 1021. 3.2. t2
p2 (t1 To ) To 473 C(To 273 C). p1
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3.3. Mair
1 ⎛ w1 w2 ⎞ ⎜⎝ M M ⎟⎠ 1 2
28.9 kg mol.
3.4. 28% and 72%. 3.5. f (u) du 3.6. 3.7. 3.8. 3.9. 3.10. 3.11. 3.12. 3.13. 3.14. 3.15. 3.16. 3.17. 3.18. 3.19.
4
exp(u2 )u2 du; w 4.39 103.
N/N 7.52 107. w N/N 9.3 103. Decreases three times. N 1.57 1021. 具典 147 ns. (1) 90 105 m2/sec, (2) 0.061 m2/sec. 18 Pa sec. At p constant. A 1 kJ. A 6 kJ. U 15 kJ. e 1.88 e. A2 4 J. S 7.2 J/K. S 291 J/K.
3.20. U
m ( ML RTk ) 75.2 J. M
3.21. C p,sp 3.22.
5 1 7 2 R 9.20 kJ (kg K).
1 M1 2 M 2 2
7 3 1.46. 5
m1 m 5 2 M1 M2 R 3.23. CV,sp 455 J (kg K). m1 m2 2 3
3.24.
2M CV,sp 5 0.6. R
⎛ 3 5 ⎞R 636 J (kg K). 3.25. CV,sp ⎜ ⎝ M1 M 2 ⎟⎠ 4 3.26.
(i1 2) (i2 2) (12 8) 1.5. i1 i2
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3.27.
249
5V1 7V2 1.52. 3V1 5V2
3.28. CP,sp 3.29.
i1 i2 4 R 6 R 959 J (kg K). M1 M 2 2 M1 M 2
5 7 0.263. 3
3.30.
冷 n 冨 1 exp(gh T ) 1 e1 0.632. no
3.31.
冷 p 冨 1 exp(gh RT ) 1 e1 0.632. po
3.32.
冷 p 冨 (gh RT) 0.01. po
3.33.
n 2 m 2 r 2 exp e 2 7.39. no T
3.34. 冷 n 冨
gh no 2.36 1016 cm3 . RT
3.35. m (T gh )
冨 n 冨 1.411022 kg. no
3.36. h(T/mg) 1n (no/n)0.588 m. 3.37. h ( RT g )
冷 n 冨 57.8 m. no
3.38. n
1 r
冷 p 冨 RT 47.5s1 . 2 po
3.39. n
1 r
RT ln 100 s1 . 2
3.40.
i 3 d 2
3.41. D( M ) ⬃
RT 22.9 mW mK. M 1 M
.
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3.42.
3.43.
Page 250
MRT 4 d 2 pN A
1 d22 2 d12
3.44.
3.45. p
16.1s.
M1 2.25. M2 MV
2 d 2 mN A T 2 d 3 N
1.24 m.
20 kPa.
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–4– Dielectric Properties of Substances
Electric and magnetic interactions play an enormous role in chemistry and chemical technology; they govern the processes taking place in atoms and molecules, crystals, electrolysis, surfaces of solids electrolyzing of dielectric polymer materials and others. Because the electric field in molecular systems has a very complex structure, for the convenience of the reader, we will give the nomenclature of electric fields at the beginning of the chapter. Let us start by briefly presenting the main information on electric charges and the characteristics of the fields they create.
4.1 4.1.1
ELECTROSTATIC FIELD
General laws of electrostatics
The main electric charge carriers are charged particles—electrons and protons. They carry a charge |e| 1.6 1019 C, electrons being negative and protons positive. In neutral atoms and molecules, negative electron charge is compensated for by a positive nuclear charge. By removing one or more electrons from an atom, one can make a positive (monoor multicharged) ion and, conversely, by adding electrons, it is possible to create a negative ion from an originally neutral particle. In such a way, positive and negative carriers of charge are created (in an electrolyte, for instance). When an excess or lack of electrons is created in a body, the body becomes charged, carrying a charge Q. The value Q is always proportional to |e|. However, at large Q (in comparison with |e|), this discontinuity is not exhibited, so it is possible to consider the charge changing continuously. The distribution of charge in a body can be described by the function (r) such that (r)dV dq, where dq is the charge, comprised in the volume dV. Function (r) is called the volume charge density. Charge distribution on a surface is described by surface charge density (r)dS dq. Charge distribution along a line gives a linear charge density (l) so that dq (l)dl. (In the first two cases, r is a radius vector of the elements dV and dS, and l is a coordinate of a point measured along the charged line.) Electric charge is invariant with respect to Galileo transformations. If an electric system is closed, its total algebraic charged is preserved. 251
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The force of interaction of two point charges q and Q in a vacuum is defined by the Coulomb’s law: F
1 Qq r , 4 0 r 2 r
(4.1.1)
where r is the radius vector, drawn from one charge to another (|r| is the distance between charges) and 0 is the electric constant. Factor 1/40 is included in the Coulomb’s law in order to correspond to the International System: then q is measured in coulomb, r in meter and F in newton; Maxwell equations (Chapter 5.4) also appear very rational in this case. Coulomb’s law has much in common with Newton’s law of gravity: in both laws, the same functional dependence on the distance (1/r2) is present. There is a profound physical meaning in this fact which we will consider below. There exists however an essential difference: attraction between dissimilar charges and repulsion between similar charges are automatically taken into account in the Coulomb’s law (i.e., Q and q are taken with their signs); however, in Newton’s law, the negative sign of force always corresponds to the masses’ attraction. Comparing these two laws, we notice that Coulomb interaction is billions of times greater than Newton’s law. If we calculate, for instance, the gravitational attraction of electrons and protons with their electric interaction, we would arrive at the enormous value of an order 1040 times. Therefore, in atomic systems gravitational interaction is not taken into consideration. Note that eq. (4.1.1) is applicable only to the interaction of point charges; an exception is the interaction of two spherically symmetric balls uniformly charged over their volumes and/or surfaces (see below). Eq. (4.1.1) cannot be applied directly to the calculation of interaction of nonregularly charged bodies. In this case the extensive charges have to be subdivided into elementary volumes dV; knowing (r), find dq, then find the Coulomb interaction dF between elementary charges and then integrate these elementary forces over the volumes of both charges. The exception is the case when charges are spread over symmetrical bodies; then it is useful to apply mathematical methods to this problem, e.g., the Gauss theorem (refer to Section 4.1.3). 4.1.2
Strength of an electrostatic field
An electrostatic field is created by motionless (in the given system of coordinates) charges. If a charge q is located at any point near another charge Q, the Coulomb force (eq. (4.1.1)) is acting, and it can be accepted that charge Q creates an electrostatic field (see eq. (1.4.4)), in which a charge q is situated. If Q is a spherically symmetric body, the field also possesses spherical symmetry, i.e., is central (Figure 1.23). It is convenient to describe a field by the force characteristic —strength of an electrostatic field E. At a point A, the vector magnitude E is numerically equal to force, acting on the positive dimensionless (point) unit charge.
E
F 1 Qr ; q 40 r 2 r
(4.1.2)
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E A
Figure 4.1 Force lines of a nonuniform electrostatic field and orientation of an electric field strength vector at a point A.
The force F(r) can be substituted by the strength E(r): F(r ) qE(r ),
(4.1.3)
with which the field acts on the point charge q. The electrostatic field can be described by the so-called force lines (Figure 4.1): at a given point A the force is directed along the tangent to a line and the vector magnitude 冟F冟 being numerically equal to force, acting on the unit positive point charge placed in this point of space. If q is equal to unity, E is equal to F. This permits us to investigate a field configuration by placing a probe point unit charge q at different points of the field and, measuring vector force F, to find the distribution of E. Consequently, the electrostatic field can be depicted by the force lines. Taking into account the general law, these lines are drawn in such a way that the tangent to them at any point gives the direction of vector E at that point, and the density of lines per cross-sectional area gives the magnitude of E. Looking at a picture of force lines, it is easy to judge the configuration of a particular field, the direction and the magnitude of vector E at every point of space, gradient E, etc. In Figures 4.2a and b, the central fields of positive and negative charges are depicted, and in Figure 4.2c, a uniform field (the plain capacitor with plates of infinite extent) is represented. Electrostatic fields obey the general physical principle of superposition (first given in Section 2.9.1): if the electrostatic field is created by several charges, the field of every charge is created irrespective of the presence of other charges. The mathematical expression for this case is reduced to the total field E as the geometric sum of strength of fields created by every point charge N
E ∑ Ei , i1
where N is the total number of charges.
(4.1.4)
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+
+
–
(a)
(b)
–
(c)
Figure 4.2 A central field of the point charges (a and b) and a uniform field of a plane condenser (c).
The field created by an extended body can be obtained by integration over the whole volume of a body V: E ∫ dE. V
The integral from vector quantity can be calculated, having expanded the vector onto components along the unit orts vectors E i∫ dE x j∫ dE y k ∫ dEz .
(4.1.5)
It is very important that the vector E (as a polar vector) has its own axis of symmetry directed along it (vector E can be rotated around itself at any angle, i.e., it possesses an axis of symmetry of an “infinite” order). This means that, if the charged body creating an electrostatic field has an axis of symmetry, the vector of the field strength E is directed necessarily along this axis. Hence, when dealing with the problem of calculating the strength of electrostatic field E it is strongly recommended first to discover the charged body symmetry: if the charged body has a symmetry axis, the overall strength E must by all means be directed along this axis! There is then no need to derive Ey and Ez: they automatically become equal to zero. From the formula (4.1.5), only one item of sum remains
E i∫ dE x .
(4.1.6)
There is no problem calculating the integral from the scalar function. EXAMPLE E4.1 The electric field is created by two point charges Q1 and Q2. The distance between the charges is d. Determine the strength of electrostatic field E at a point which is removed at a distance r1 from the first charge and r2 from the second. The numerical values are Q1 30 nC, Q2 10 nC, r1 15 cm and r2 10 cm. Solution: According to the superposition principle of electric fields, each charge creates a field, irrespective of the presence of other charges. Therefore, the intensity
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of a field at a specified point can be found as the vector sum of the fields, created by two charges separately, E E1 E2. The strength of the electrostatic field can be determined by eq. (4.1.2). The vector relations can be seen in Figure E4.1. The E1
A
E
−
r1
E2 r2 −Q2
+Q1
partial field strength created by each charge can be determined according to the Coulomb’s law; they are E1
| Q1 | 4 0 r12
and E2
| Q2 | 4 0 r22
.
The sum can be found according to the cosine theorem E E12 E22 2 E1 E2 cos, whereas cos can be found from triangle (see figure)
cos
E
1 4 0
d 2 r12 r22 0.25. 2r1r2
Q12 r14
Q22 r24
2
| Q1 | | Q2 | r12 r22
cos .
Substituting all data and executing calculations, we arrive at E 1.67 104 16.7 kV/m.
EXAMPLE E4.2 Three identical positive charges Q1 Q2 Q3 1 nCb are located at the vertex of a flat equilateral triangle. What negative charge should be placed in the center of the triangle that would counterbalance the forces of mutual repulsion of the positive charges in the vertexes (Figure E4.2)?
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+ Q2
− Q4
r1 F4
F3
+
F
Q1
+ r
Q3
F2
Solution: All three charges located in the vertexes of a triangle are in identical states. Therefore, for the solution of the problem, it is enough to find out what charge is necessary to place in the center of the triangle so that one of the charges, for example, Q1, would be in equilibrium. According to the principle of superposition each charge is exhibited irrespective of the presence of the others. Therefore, the charges will be in equilibrium if the vector sum of forces acting on them is equal to zero, F2 F3 F4 0 F F4 where F2, F3 and F4 being the forces with which charges Q2, Q3 and Q4 act on charge Q1; F is the resultant of forces F2 and F3. Since forces F and F4 are codirectional they can be substituted by scalar sum F F4 0 or F4 F. The force F can be presented as a sum F2 F3, and as F2 F3 we can write F4 F2兹苶2苶 (1苶 苶 co苶 s 苶苶 )*. Applying Coulomb’s law and taking into account that Q2 Q3 Q4, one can find Q1Q4/40r 12 Q12/40r 2兹苶2苶 (1苶 苶 co苶 s 苶苶), wherefrom Q4
Q1r12 r2
2(1 cos )ⴱ .
From geometrical considerations, we can obtain r1 r/(2 cos 30°) r/ 兹3苶. Since cos cos 60° 0.5, the formula * takes the form Q4 Q1/ 兹3苶. Taking into account that Q1 is 1.0 nCb, the charge Q4 is 0.58 nCb. Note that such electrostatic equilibrium is unstable.
EXAMPLE E4.3 The field is created by a uniformly charged thin rod of length l with a linear charge density . Define the strength of an electrostatic field at a point A lying on the line of the charged rod at a distance d from its end. Numerical values are: 1.2 C/m, l 10 cm and d 5 cm. Solution: Let us draw an x-axis along the rod, and let the origin be placed at the beginning of the rod (Figure E4.3). Because the problem has an element of symmetry (an axis of infinite order collinear to the direction of the rod), the vector of the field strength must certainly lie on this axis.
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y dx
A
E
X
X
a
l
Since the electrostatic field is created by a continuous body (rod), it is not possible to apply Coulomb’s law directly to the solution. Therefore, we have to allocate an elementary charged segment on the rod, to calculate intensity created by it and to integrate this with the length of the whole rod. The element of the rod dx carries charge dx. The elementary field strength is calculated using eq. (4.1.2) dE
dx 1 . 4 0 (ᐉ a x )2
Integration gives ᐉ
ᐉ
E
dx 1 ⎛ ᐉ ⎞ . 4 0 ∫0 (ᐉ a x )2 4 0 (ᐉ a x ) 0 4 0 ⎜⎝ a(a ᐉ) ⎟⎠
It is easy to prove that the dimension corresponds to the field strength dimension. There are other ways to choose the origin, though all choices will give the same result.
EXAMPLE E4.4 One-third of the circle’s circumference of radius R 20 cm made of a dielectric material is charged uniformly with linear charge density 1 106 C/m. Determine the strength of electrostatic field E created by this piece of charged arc in its center—point O (Figure E4.4).
y dE
dEy j
d dl
i dEx x
O
R
/3
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Solution: It is impossible to apply Coulomb’s law directly to the whole part of the ring because it is extensive (not a point). We must divide an arc body into pieces and apply the law mentioned to each elementary piece. Notice that the whole problem has an axis of symmetry of the second order passing through the middle of the arc and the arc center O. This means that the resulting electric strength vector is directed along this axis (why?). Therefore, we direct an axis y along an axis of symmetry. Axis x is of no significance; let it be perpendicular to the y-axis. We allocate an elementary piece dl, bearing charge dq, dq dl. This charge creates an elementary field strength dE at the point O. This vector should be projected onto the y-axis in order to be able to make further integration over the whole charged arc. The projection dE on y-axis is . dE y dᐉ cos 4 0 R 2 d cos 4 0 R The component dEx can be ignored, because we definitely know that after integration the x-component should give zero (“why?” check up by integration). We have to express two variables l and through one, let it be : d dl/R. Hence 3
E
∫
3
3
dE y 2 ∫ dE y 0
2 4 0 R
3
∫ cos d 20 R 0
(pay attention to the integration limits, to coefficient 2 and to the way of measuring the value). To substitute the given and the physical constants, we arrive at the value E 兹苶3/20R; executing the calculations we obtain E 2.18 kV/m.
EXAMPLE E4.5 Along a ring of radius R 1 m, a charge Q 1 C is uniformly distributed. Find an electric field strength E(h) in points lying on the axis that pass through the center of the circle perpendicular to the plane of the ring. Solution: As the field strength E is a polar vector, therefore in this problem it must certainly coincide with the coordinate axis z. The linear charge density is Q/2R. Allocate an elementary vector dl at the ring. This segment carries the charge dq dl Qdl/2R; at point A it creates a field 冟dE冟 dq/40r2, where r is the distance from dl to the point A. Vector dE is directed along the line joining the element dl with the point A. Since we a priori know the general direction of the electric field strength E (along the axis z), we have to project the dE vector on the z-axis and integrate projections (since an integral over perpendicular components must be equal to zero). It can be seen from Figure E4.5 that dEO NdEN cos , cos being equal to h/ 兹苶h2苶 苶 R2苶. Therefore dE
h d ᐉ. 4 0 (h R 2 )3 2 2
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Z
dE dEz A
h
r
O R dl
The integration dE over the whole ring gives:
E
Q h . 2 4 0 (h R 2 )3 2
This general expression permits us to calculate the electric strength in any point of x-axis. In particular, it is very useful to use this example to investigate the extreme conditions. In fact, we know that the field in the center of the ring is zero. This corresponds to the limit h → 0: indeed at h 0 the relation obtained gives E 0. From the other side, at h R, or at R → 0, E ⬃ Q/h2 which corresponds to Coulomb’s law.
EXAMPLE E4.6 A thin half-ring carries a charge uniformly distributed along the ring with a linear charge density 10 nC/m. Determine the electric potential in the center of ring O. Solution: A ring element dl carries a charge dl. This charge produces an electric potential equal to d dq/40R dl/40R in the point O. After integration of the scalar values over the whole half-ring (/2 /2), we arrive at /40 282 V. Note that potential calculation is less troublesome than the field strength!
4.1.3
The Gauss law
A vector flux through an area and vector circulation along a closed contour is the basic characteristic of a vector field in vector algebra. The application of these concepts to an electrostatic field appears extremely fruitful.
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We apply this concept, which we first met in Section 2.8.3, to a flux and density of flux. Accordingly, the elementary flux d of the electrostatic field strength E through an elementary area dS is d E dS
(4.1.7)
(see eq. (2.8.18)). Having written down an expression of scalar product and then attributing cos first to E and then to dS, we obtain: d E dS cos En dS E dSn ,
(4.1.8)
where En is the projection of the vector E onto a normal n to the elementary area dS, and dSn is the projection of dS on a plane, perpendicular to E (refer to Figure 2.24). It can be seen that the flux d is subject to change not only due to E magnitude, but also due to the mutual position of vectors E and dS (due to the change of cos ) from E dS up to E dS, i.e., positive and negative. If the field is produced by a point charge, the flux d of the field strength is proportional to a solid angle d . The part of space confined by a conical surface (Figure 4.3) is referred to as a solid angle. A measure of a flat angle d is the ratio of the length of an arc dl of a circle, drawn by an arbitrary radius r about a point O, to the value of the radius mentioned, i.e., d
dᐉ ; r
(4.1.9)
accordingly a measure of a solid angle d is the ratio of an elementary area of a spherical surface dS, drawn by any radius r around a point O, to the square of the radius r2 mentioned (Figure 4.3):
d =
dS dSn = 2 . r2 r
(4.1.10)
dl d 0 dS
r
r
0 dΩ
Figure 4.3 A measure of a solid angle.
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(Replacement of dS by dSn is permissible because the area dS is always perpendicular to the radius.) When dSn is numerically equal to r2, the solid angle is equal to 1 sr (steradian). 2 Similar to the way in which the total flat angle is equal to dl/r 2r/r 2 rad, the 0 total solid angle is equal to
冕
养
dSn 4r 2 2 4 sr. r2 r
(4.1.11)
Gauss law asserts that: the flux of an electrostatic field strength vector through any closed surface S is equal to the algebraic sum of the charges confined by surface S divided by 0. We shall call this virtual (imaginary) surface the Gauss surface. Note that we ourselves choose the form and size of a Gauss surface! Let us prove the theorem taking a single point charge Q as an example. We place it in a point O and cover this charge by a closed spherical Gauss surface S with the center in the charge Q position (Figure 4.3). Let us find the value of the elementary flux d of the vector E through the elementary surface dS: d E dSn. The magnitude E is defined by expression (4.1.2) and therefore
d
1 Q 1 Q 2 dS r d 40 r 2 n 40 r 2
(4.1.12)
(we used here the expression (4.1.10)). After integrating over the total solid angle, we arrive at
养 En dS 养 S
S
Q 1 Q d 4 0 4 0
养 d S
1 Q, 0
or, keeping in mind expressions (4.1.7) and (4.1.8), 养 EdS s
1 Q. 0
(4.1.13)
Let us comment on the expression derived. In this expression the point charge Q is comprised of a Gauss surface S. It was taken in spherical form, but it can also be a irregular shaped surface. On the right-hand side the charge Q can be seen which we encompassed by the Gauss surface (Figure 4.4). If the charge is outside, the flux through the Gauss surface is equal to zero (Figure 4.5) since part of the flux is negative and the other part is positive (because of various orientations of a normal n to the surface relative to vector E). This is the case for all those charges which lie outside the Gauss surface; they can be excluded from consideration.
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S
E dS n dΩ
r
Q >0
Figure 4.4 The Gauss theorem: Q is a charge, S is a closed Gauss surface, dS is an elementary area and n is a unit vector to dS. n E
O Q>0
E
n
Figure 4.5 The zero total flux of the electric field strength for a particular case when an electric charge is outside the Gauss surface.
Values Q can be determined if distribution of the charge in a given problem is known. If the field is produced by a sum of N individual charges, then according to the superposition principle 养 E dS s
1 0
N
∑ Qi .
(4.1.14)
i1
If the field is set up by a volume charged body, the expression can be modified as follows: 养 E dS s
1 (r ) dV , 0 V∫
(4.1.15)
where (r) is a charge density distribution function. If the field is set up by a charged surface with the surface charge density (r), the Gauss law looks like: 养 E dS s
1 (r ) dS. 0 S∫ⴕ
(4.1.16)
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If the source of field is a charged line, the Gauss law can be given as 养 E dS s
1 0
∫ (l ) dl,
(4.1.17)
L
where (l) is the linear charge density distribution function and l is the point coordinate along the charged line. The body volume V, charged surface S and the length L of the charged line correspond to those parts of the charged physical bodies that are encompassed by the Gauss surface. Repeat once more that in Gauss law the total charge laying inside of the Gauss surface is present on the right side which is equal being divided by o to the electrostatic field strength flux on the left side. The Gauss law is valid for all forms of charged bodies and any Gauss surface. However, its most fruitful application is in calculating the field when the problem possesses some symmetry. In this case the skilful choice of the Gauss surface form permits one to achieve essential simplification: to provide that E En const., to take E out of an integral and to integrate only upon a surface, or, in general, to arrange the field strength being En 0 onto a part of a surface. Examples of application of the Gauss law given below will show calculations of fields of the charged physical body possessing some symmetry. These problems could be solved using only the principle of electric fields’ superposition. However, this way is troublesome; it demands integration upon a volume. Application of the Gauss law allows many problems to be solved “in a single line”. EXAMPLE E4.7 An electrostatic field is created by two parallel, infinitely large, charged plates with surface charge density 1 and 2. Define the electrostatic field strength created by these two plates between and behind them. Numerical values are 1 0.4 C/m2 and 2 0.1 C/m2. Solution: According to the principle of superposition of electric fields, each charge creates a field irrespective of the presence of other charges. In this case it concerns the charged planes. Therefore, the intensity of a field in specified areas E can be found as the vector sum of the fields, created by two planes separately: E E1 E2. The absolute values are in this case: E1 1/20 and E2 2/20, E1 being higher than E2. The sign of each term depends on two peculiarities: the sign of the plate charge (objective characteristic) and the choice of the axis direction (subjective characteristic). Let us divide the problem into three parts (I, II and III) and choose an x-axis direction, we recommend that it is usually accepted from left to right (Figure E4.7). The superposition principle offers us the choice of the direction and absolute values of particular fields. Therefore, we recommend beginning the solution of the problem with the determination of E(I) and E(III) E (I) E (III)
1 (I) (III) . 0 2
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E
1
2
x
Between the plates the field is directed oppositely from both plates and therefore E(II)
1 | 1 2 | . 0 2
The situation presented in this example can rarely be met in practice. Much more popular is the plane condenser, which differs from that given here by the fact that 1 2; the electric field inside the condenser is E /0, there is no field outside the condenser to say nothing on the edge effects (see Figure 4.11).
EXAMPLE E4.8 Two concentric spheres of radius R1 6 cm and R2 10 cm, respectively, carry charges Q1 1 nC and Q2 0.5 nC. Find the electric field strength at points r1 5 cm, r2 9 cm and r3 =15 cm (Figure E4.8). Draw a graph of E(r). | E | = En
E
E
n n
Q2
R2
E,V/cm
I
2500
Q1 R1 III
II
I
II
O 900 III
450 (a)
(b)
0
R1
R2
r
Solution: The three points mentioned are correspondingly disposed just inside the inner sphere (domain I, r1 < R1), between the two spheres (domain II, R1 < r2 < R2) and outside the spheres (domain III). The two spheres really exist carrying a definite charge. Apply the Gauss theorem to solve the problem. First, we have to solve the
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situation inside the smallest (internal) sphere. One should choose an imaginary sphere with r1 < R1 and write down the Gauss theorem for this domain: wEn dS 0. Because En on the sphere is constant, En 冟E冟 E, then Ew dS E4r 2 0; since r 2 is not zero it means that E1 ⬅ 0*. In domain II, R1 r2 R2, the Gauss surface should be chosen having R1 R R2. In this case the whole first real sphere occurred inside the Gauss surface. Therefore, wEndS Q1/0, E4r2 Q1/0 and E (Q1/40)(1/r2)**. (Sometimes the sphere charge is given through surface charge density and the whole charge of the sphere is 4R2. Note that R is a number, but r is a variable.) In order to find the E(r) function in the domain III, the Gauss surface radius should be larger than R2; it incorporates both spheres. Therefore, EwdS (Q1Q1)/4r 32 ***. The values of E’s presented in the graph can be obtained from the marked functions E2 and E3. Three marked equations permit us to draw the graph. Try to find yourself the values presented in Figures E4.8a and b.
EXAMPLE E4.9 An electron with zero initial speed has passed an electrical potential difference between a cathode and an anode U0 10 kV and entered a space between horizontal plates of the flat condenser on line AB parallel to plates (Figure E4.9); the condenser is charged up to potential difference U1 100 V. A distance d between plates is equal 2 cm. The length L1 of the condenser plates in direction AB is equal 20 cm. Determine distance BC (BD DC) between fluorescent screen spots at the distance from the condenser’s end to a screen is L2 1m. L1 Cathode A
–
L2
0
B 0
M + + + + + + + 1
Anode
l1 D
l2
C Screen
Solution: An electron movement inside the condenser consists of the two components: (1) by inertia, along the line AB with the constant speed 0 acquired earlier by action of accelerating potential difference between anode and cathode U0 which an * m02 electron has passed up to the condenser eU0 and (2) a uniform acceleration 2 movement inside the condenser in a vertical direction due to an action of constant electric field. After leaving the condenser at point M, an electron moves uniformly and rectilinearly.
冢
冣
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One can see from the figure that the sought distance consists of two values BC l1 l2, one of them l1 being a distance on which the electron is displaced in a vertical direction during movement thought the condenser, the other one l2 between a point on the screen DC (see figure). Let’s estimate separately l1 and l2. The value l1 can be found using the formula of travel distance in the uniform acceleration movement l1 at2 /2 **, where a is the acceleration acquired by the electron under the action of the condenser’s constant electric field and t is the time of its action. According to the Newtonian second law, acceleration is a F / m (the force F which acts on the electron in condenser, m being its mass). In its turn, F eE eU1 /d. The time of the electron flight inside the condenser l1 we shall find by the formula of a uniform movement t L1 / 0. Consequently, substituting values from other expressions into the formula * we can obtain l1 U1L12 / (4dU0). We can find the length of a segment l2 from corresponding triangles MDC and then build on vectors 0, 1 and assuming that the deviation from AB-direction is small, l2 (1L2 / 0)**, where 1 is the electron speed in a vertical direction at point M; and L 2 is a distance from the condenser end up to the screen. We can find the speed 1 using the formula 1 at which in view of expressions for a, F, and t will become
1
eU1 L 1 . dm0
Having submitted the expression derived into the formula **, we obtain l2 eU1L1L2 / (d m02) or, having replaced 0 from equation **, we found l2 U1L1L2 / 2 dU0. For the required distance BC l1 l2, we shall finally arrive at
BC
U1 L 12 U1 L1L 2 U1 L1 ⎛ L1 ⎞ L 2⎟ . ⎠ 4 dU 0 2 dU 0 2 dU 0 ⎜⎝ 2
Substituting all given values into the last expression and having made calculations, we arrive at BC 5.5 cm. Let us start with calculation of a field set up by a single charge Q. The field of the point charge depends only on the distance from a charge to a point of observation and does not depend on the direction, i.e., the field is spherically symmetric. Choose an arbitrary point A (arbitrary means that it is typical and not distinguished from any other). We shall carry out a spherical Gauss surface through this point at the radius r with the center in charge Q (Figure 4.6). We shall write down for this sphere S the Gauss law in the form sEn dS Q/0. In this expression, En is the projection of vector E on the normal vector n to the sphere surface. Any line passing through point O is the axis of symmetry because any turn around it superposes a system with itself.
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E, r A n
∝1/r
E =En
Q
E ∝1/r 2 r
Figure 4.6 The electric field of a point charge. Point A is a point of observation, the dashed line represent a Gauss surface.
Therefore, the vector E coincides with the symmetry axis. It means that |E| En. Besides, as a consequence of the symmetry of the problem, E const. at any point of the Gauss sphere. Therefore, wsEdS EwsdS 40r 2 . The right-hand side of the equation is equal to Q/0. So, E4r2 Q/0 or E Q/40r 2. This coincides with eq. (4.1.2). This coincidence is quite natural because while deriving the Gauss law, we use for simplicity the point electric charge. It is worth mentioning here that the Gauss law is valid for all fields whose strength falls as 1/r2. Obviously, only in this case the r2 terms in the nominators and denominators of the expression (4.1.12) are cancelled. We know two such fields: electrostatic and gravitation; in these two cases the Gauss theorem is valid. Let us now apply the Gauss law to consideration of a field set up by both uniformly charged spherically symmetric bodies (sphere) and a ball uniformly charged upon the surface, the radius of spheres being R in both cases. This means that there is no charge inside the body. The charge surface density can be denoted as Q/S, where S is the surface area. In both cases we have to signify two domains: domain I is the region inside the bodies (rA < R, where rA is the distance from the center of spheres to an arbitrary chosen point A) and domain II corresponds to outer space (Figure 4.7). As was valid above, any line passing through the center of the spheres is the symmetry axis and |E| En const. In the first domain 0 r R, the Gauss integral is sEn dS 0. As En |E|, then EwsdS E4r2 0 and since r 2 苷 0, then E ⬅ 0. In domain II, r > R, the arbitrary point A has to be chosen outside the spherical bodies. The total charge Q will here be fully encompassed by the Gauss surface. Therefore
E ∫S dS E 4r 2
Q 0
and
E=
Q 40r
2
⎛ 1 ⎞ ⎛ 4R 2 ⎞ ⎝ 4 0 ⎟⎠ ⎜⎝ r 2 ⎟⎠
=⎜
=
R2 . 0r 2
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r O R A
E(r)
(r)
E(r)
R
r
Figure 4.7 The electric field strength E(r) and potential (r) of the surface-charged sphere.
The result is depicted in Figure 4.7. It is apparent that the strength values on the internal and external sides of the surface are different, i.e., the field strength on the charged surface experienced a break. In a similar way it is possible to obtain the values of the field strength of several spherical surfaces with the same center (e.g., for a spherical condenser). If the sphere is charged homogeneously ( Q/V const.), the field inside is not equal to zero. As before, the problem has a spherical symmetry; therefore, the Gauss sphere should also be chosen in spherical form. With such a choice, |E| En const. at Gauss sphere. Accordingly, the arbitrary point A is chosen in two domains: inside the sphere (r < R) and outside it (r > R). To find the electrostatic field strength dependence at the distance r from the charged sphere center inside it, we shall take advantage of eq. (4.1.15). We shall find the charge density by dividing overall charge by the volume Q/(4/3)R3 3Q/4R3. Then 养 En dS E 养 dS E 4r 2 s
s
3Q 4 3 r , 40 R3 3
wherefrom
E (r )
Q 40 R3
r.
Pay attention that E ⬃ r! This graph is given in Figure 4.8.
(4.1.18)
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r1 R
O
A
E
E~r E~1/r 2 E~r/
r
R
Figure 4.8 Field of a uniformly charged ball.
Calculation of the field outside the sphere is easier: the Gauss surface encompasses the whole sphere. Then
养 En dS E 4r 2 s
Q 0
and
E
Q 40r 2
.
As previously seen E is proportional to 1/r2. If there is a material with definite (see below), the dependence E(r) should contain term 1/ (Figure 4.8). Let us generalize the result. In all three cases of spherical symmetry, a function E(r) at r R is described by dependence E ~ 1/r2. It means that an observer who is outside the sphere at any point r, knowing the fragment of the measured dependence E(r) cannot judge whether the field is created by the point charge, a sphere
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uniformly charged over its surface or the uniformly charged solid ball. In all three cases dependence E(r) is identical! Moreover, in the case of spherical symmetry nothing prevents “swiping” the sphere or ball to a point and to consider a field as if it is created by point charge! Sometimes it essentially facilitates a problem. Remember that we have already done this in Chapter 1: in studying the force and potential energy of a gravitational field (a field of the earth), we dealt with formula F(r) GMm/r2 and U(r) GMm/r, where r is the distance from the center of the earth. We swiped the whole mass of earth to a point (see Section 1.4.5). The possibility of the generality of the results is in identical dependence on Coulomb and Newton forces from distance r! Calculate now a field created by a body with cylindrical symmetry, for example, infinite and uniformly charged cylinder (for simplicity—over a surface) with linear charge density dQ/dl const. (Note that there are no such cylinders or infinite planes in nature. This physical problem is equivalent to the condition where the cylinder has finite length L, however, we consider a field near to the charged surface, i.e., at r L . Then it is possible to neglect the edge effects and solve the problem for an infinite cylinder.) Let us analyze a problem that has cylindrical symmetry, i.e., the axis of symmetry coincides with the axis of the cylinder; this means that in calculating the field strength, we deal with E (not with E), it depends only on the distance from a cylinder axis to a point of observation r (but not from r!). Furthermore, any straight line perpendicular to the cylinder axis and crossing it is an axis of symmetry of the second order (the turning of the infinite cylinder around the axis on superposes it with itself). This means that vector E should be directed along such a straight line. Choose accordingly a closed Gauss surface in the form of a coaxial cylinder of length l (cyl) and end surfaces (end) perpendicular to the cylinder axes (Figure 4.9). Then the integral over the Gauss surface will be separated into three parts:
养 En dS s
∫ En dS 2 ∫
cyl
En dS.
end
The last item is equal to zero because En 0 at end surfaces (E ⊥ n). Over all cylindrical surface, E ⱍⱍ n; therefore
∫ En dS ∫ En dS 2r ᐉE. S
cyl
At r < R (not depicted in the scheme), E 0 (according to the same consideration why the field inside an empty sphere is absent). At r R, E2rl Q/0 l/0. Thus, E(r) /20r. A graph of this dependence is presented in Figure 4.9.
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R n E r
l
E
n A
E
~1/r r
R
~-ln r r
Figure 4.9 Field and potential of a system with cylindrical symmetry.
Let us now calculate a field created by an infinite plane (i.e., with linear sizes much larger than the distance from the plane to an observation point), the plane being uniformly charged ( = dQ/dS = const.). Any straight line perpendicular to the infinite plane is the axis of symmetry (because any turn around it will impose a plane on itself). Therefore, E in any point should be directed along this axis, i.e., perpendicular to the plane. In this case, it is expedient to choose the Gauss surface to be a cylinder with the generatrix perpendicular to the plane, as is shown in Figure 4.10. Therefore, the electric strength flux through a side of the cylinder surface is zero, whereas the flux through two ends is
养 En dS 2 s
∫
En dS 2 ESend .
Sbot
According to the Gauss theorem, this expression is equal to the charge inside the Gauss surface, i.e., Send. So 2ESend = Send/0, i.e., E /20 const (see Figure 4.10)
E
. 2 0
(4.1.19)
It can be seen that the field strength near to plane does not depend on distance (!). Negative value of E to the left of the plane means that vector E is directed opposite to the axis x.
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n
+
E n
n
E x S
S
S E +/ 2 0
–/2ε 0 x
Figure 4.10 Field of a uniformly charged plate.
+
–
+
–
d
d
Figure 4.11 A Uniform condenser field.
The system of two similar planes carrying an equal charge with an equal charge density represents a plane condenser, provided they carry charges of opposite signs. According to the superposition principle, the field in the condenser is an algebraic
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dm=υtρdS υt
S dS
Figure 4.12 A Gauss theorem and a liquid’s flow.
sum of the fields created by each plane separately (Figure 4.11). So, the field inside the plane condenser is
E 2 E
; 0
(4.1.20)
however, the field outside it is zero. The plane condenser is a source of a uniform electric field. In summary, we make two more remarks on the Gauss theorem. The first concerns the physical nature of the equation. Because in eq. (4.1.13) an electric charge is presented on the right-hand side, the theorem asserts that the source of an electrostatic field is electric charge. Another concerns the general meaning of the theorem, wider than only electrostatics. Imagine, for instance, a flow of a liquid in a pipe (Figure 4.12). Each particle of the liquid moves with speed u. The mass of the liquid will cross a surface dS in time t, making dm ut dS. The mass of the liquid m s t dS t s dS will pass through surface S in time t, and if S and dS are directed parallel to each other, m t dS. On one hand, these equations allow us to calculate the mass of the liquid at its flow in pipes (at known distribution of speed, at turbulent or laminar current), and, on the other hand, to formulate a criterion for liquid incompressibility, 养sdS constant. If there is a source of liquid inside the Gauss surface, 养sdS j, where j m/t is a source power. 4.1.4 field
Work of an electrostatic field force and potential of an electrostatic
Coulomb forces are central and, consequently, conservative; the field of these forces is potential (refer to Section 1.4.4, Figure 1.29 and eq. (1.4.24)). Indeed, by definition, an elementary work dA of a force F on a displacement dl is determined as: dA F dl qE dl qE dl cos, where E Q/40r 2 and dlcos dr. Then dAqQdr/40r2 and A12(qQ/40) r r21 dr/r 2. Therefore,
A12
qQ ⎛ 1 1 ⎞ , 40 ⎜⎝ r1 r2 ⎟⎠
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i.e., the work in fact does not depend on the distance traveled but on r1 and r2, the radius vectors of the initial and final points. If a charge starting from the initial point returns to it, the work of electrostatic forces is equal to zero: Aq LEl dl 0, where El E cos is a projection of vector E on the displacement vector dl. Then 养LEl dl 0: circulation of the electrostatic field strength along the closed trajectory is equal to zero. This is the definition of the potential character of a force field. From equation 养LEl dl 0 it follows that the force lines of an electrostatic field cannot be closed (otherwise with a positive detour the integral would be essentially positive). They start from the positive charges and end at negative charges (or pass into infinity). The work of electrostatic field forces can be presented as a potential energy decrease, A U1 U2 U. From a comparison of this equation with that given above, the potential energy of a charge q in a field of another charge Q can be written as U(qQ/40r)C. If one accepts that at r ∞ the potential energy U(∞) is equal to zero, then C 0 and the potential energy is finally
U
qQ . 40r
(4.1.21)
The electrostatic potential is a value numerically equal to the potential energy of a unit positive point charge placed into the given field point. Then U = q .
(4.1.22)
The point charge potential is therefore
Q . 4 0 r
(4.1.23)
Inasmuch U q , then
A12 U1 U2 q( 1 2 ) q .
(4.1.24)
If one assumes (∞) 0, then A1∞ q 1; therefore, 1 A1∞ /q. This means that the potential is numerically equal to the force work done on the displacement of the unit positive point charge from the given point to infinity. The field potential is the scalar power characteristic of an electrostatic field. The potential of a field having been created by a system of motionless charges obeys the superposition principle, i.e., each charge creates a field irrespective of the presence of other charges in the space. As the potential is a scalar value, the potential in any point of the field, created by the system of charges, is equal to the algebraic sum of potentials of fields created by each charge individually: N
(r ) ∑ i (r ), i1
where N is the total number of charges.
(4.1.25)
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From the point of view of the potential distribution, the electrostatic field can be graphically characterized by surfaces of equal potential, i.e., by equipotential surfaces with constant (Figure 4.13). The force lines are always normal to them. Because electric field strength E(r) and potential (r) are functions of a point radius vector r, there should be an interrelation between them. Let us find it. The elementary work of a field force can be determined as either dA F dl qE dl or dA dU q d . Comparing them, we can arrive at El d /dl. If the field is created by a spherically symmetric charge, E and are both r-dependent; therefore
Er (r )
d (r ) . dr
(4.1.26)
Ex ( x)
d ( x) . dx
(4.1.27)
In the case of linear dependence
In the general case, r(x, y, z), E(r) and (r) are interconnected by ⎛ d
E(r) ⎜
⎝ dx
i
d d ⎞ j k⎟ grad (r), dy dz ⎠
(4.1.28)
i.e., the electrostatic field strength is equal to the electric field potential gradient taken with opposite sign. One can see that the field strengths’ vector is always normal to equipotential surfaces (Figure 4.13). It follows from eqs. (4.1.26)–(4.1.28) that the function (r) cannot be a discontinuous function. In fact, if a function undergoes a break (for instance, in the point x0), then, according to eq. (4.1.27),
E ( x0 ) lim , x0 x since in this point x → 0, but tends to a finite difference; this means that E(x0) is equal to infinity. This cannot be the case. = const
Figure 4.13 Force lines and equipotential surfaces.
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The relation established between the electrostatic field strength and potential allows us to find dependence (r) knowing the function E(r) for all examples considered above. In fact, it follows from eqs. (4.1.26) and (4.1.27) that (r ) ∫ E (r )dr C and ( x ) ∫ E ( x )dx C. Substituting here the already known functions E(r) and E(x), we can calculate functions (r) and (x). The constant C reflects the fact that the function , as well as potential energy, is known to be within a constant; it can be found from boundary conditions. The choice of C can provide a continuity of function . Thus, the potential for the problems considered can be obtained. Corresponding graphs are depicted in Figures 4.6, 4.7, 4.9–4.11. In order to calculate the potential difference , integration should be carried out in certain limits. So, for instance, the potential difference in a plane condenser is d
d
0
0
∫ E (x )dx E ∫ dx Ed , where d is the distance between the plates. The reverse problem (knowing , calculate E) can be obtained as well. This method will be used, for example, when we calculate an electric dipole field (because potential calculation is easier than calculating the field strength since the former is a scalar value, whereas the latter is of vector quality). 4.1.5
Electrical field of an electric dipole
A system consisting of two charges, equal in absolute value and opposite in sign, is referred to as an electric dipole. The vector drawn from a negative charge to a positive one l is called a dipole arm. The vector p equal to a product of the charge and arm is referred to as electric dipole moment (Figure 4.14): p ql.
(4.1.29)
It can be seen that the unit of dipole moment measurement is C m. This is a very large value and therefore a significantly smaller one is used in chemistry, namely 1 Debye (D): 1 D is equal to 3.33 1030 C m. When a dipole field is studied at a distance much larger than the dipole arm, it is called the point dipole. In spite of the fact that a dipole seemed to be an electrically neutral particle, it produces an electric field; this field has different properties compared to a field of
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point charges. In order to describe a dipole field, we use the superposition principle: , i.e., ( q 4 0 r ) ( q 4 0 r ) q (r r ) 4 0 rr , We consider the point dipole (l r) (Figure 4.14); therefore, rr ⬇ r2, r r ⬇ l cos which means that
(r , q )
qᐉ 4 0 r
2
cos
p cos . 4 0 r 2
(4.1.30)
In order to determine the strength of the dipole field at point A, the relationship between strength and potential can be used. We use a polar coordinate r and with a polar axis coinciding with the dipole arm direction (Figure 4.15). Component Er (projection E on the radius vector r) is Er
2p d cos . dr 4 0 r 3
A r
−
l + +q
p
–q
Figure 4.14 An electric dipole model.
Er
Eθ
A r+
r– r
l cos −
p
+
l
Figure 4.15 Dipole electric field calculation.
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The E component perpendicular to r is equal to E
d p sin , r d 4 0 r 3
where r d is the length of an arc with radius r. Then, the total electric field strength at point A is E Er2 E2
p 13 cos2 . 4 0 r 3
(4.1.31)
The potential and strength of the dipole field are defined not by a value of charge q but by the dipole electric moment p |q|l. With distance, the dipole potential and strength decrease faster ( ~ 1/r2, E ~ 1/r3) than those of a field of a point charge (1/r and 1/r2, respectively). A configuration of the dipole electric field is shown schematically in Figure 4.16. In a uniform electric field a torque MF aspires to turn the dipole moment p to be oriented along the field E. A force couple MF is equal to the product of force F |q|E and the force arm l sin (Figure 4.17), i.e., M |q|El sin pE sin or, in vector form, M F [ pE].
(4.1.32)
The vector M is directed perpendicular to the plane of drawing from a spectator. The resulting force in the uniform electric field is zero. In order to calculate the potential energy of a dipole in the uniform electric field we use the recipe given in Section 1.4.5. To calculate the potential energy we should first z E
p
Figure 4.16 A dipole electric field configuration.
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find the elementary work of the force couple (dA dU, where dA MF d is the force couple work and dU is the increment of the potential energy). Then
U ∫ M F d ∫ pE sin d, after integration we can obtain U() pE cos C. If we accepts U(/2) 0, then C 0 and U pE cos or in vector form U (pE).
(4.1.33)
A graph of dependence U() is given in Figure 4.18. When the vector p coincides with vector E, the dipole potential energy acquires the lowest value ( 0, Umin pE): the dipole is in a stable equilibrium; the force couple is also equal to zero. When the vector p is oriented perpendicular to the vector E, the dipole energy is equal to zero ( /2, U 0). When the vector p is directed opposite to the vector E, the dipole energy is maximum ( , Umax pE), the dipole is in an unstable equilibrium: any deviation from this state leads to turn-off of the dipole. In a nonuniform electric field, forces acting on dipole charges are not equal in absolute value and direction. In Figure 4.19, a graph of a nonuniform field (dE/dx < 0) with a dipole p in this field is depicted. Both the force couple MF and resulting force F will act on the dipole.
F+ P
l.sin
F–
Figure 4.17 Electric dipole in a uniform electric field.
U +pE
0
/2
–pE
Figure 4.18 A dipole potential energy in a uniform electric field.
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F+ +
x
p F–
–
Figure 4.19 An electric dipole in a nonuniform electric field.
To calculate a force projection on x-axis, we can use eq. (1.4.33) where the relation of the force projection with the field nonuniformity is given (Fx ∂U/∂x); in our instance, U pE cos and then Fx p
E cos . x
(4.1.34)
If a dipole in the electric field is in a stable state (cos 1), then the force will draw the dipole into an area of stronger field (for Figure 4.19, Fx 0). The ratio (eq. (4.1.34)) shows that an interaction force can exist even between neutral molecules, though having the dipole moments. In fact, if an electrically neutral polar molecule creates the nonuniform field described by expression (4.1.31) and another molecule with a dipole moment p is in this field, a dipole–dipole interaction appears between them.
4.2 4.2.1
DIELECTRIC PROPERTIES OF SUBSTANCES
Conductors and dielectrics: a general view
From the point of view of electric properties, all substances can be divided into two main classes—conducting and nonconducting an electric current. Metals, their alloys and a small number of chemical compounds with metal character of interatomic interactions relate to the class of conductors. The second class includes other substances and represents the overwhelming majority. Conductivity is defined by the presence of free charge carriers in a substance; their absence determines dielectric properties. So, dielectrics are substances in which there are no free charges capable of covering long distances in the substance (in comparison with molecular sizes). Depending on their molecular structure, all dielectrics can, in turn, be divided into two large groups—polar and nonpolar. In polar dielectrics, molecules themselves represent the electric dipoles with the electric moment p; it appears due to displacements of electric charges from positions of their equilibrium in free atoms as a result of chemical bonding (e.g., H2O, HCl and NH3). The molecular dipoles of polar dielectrics participate in thermal motion; this can be translational motion (in gases and liquids), oscillation
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about equilibrium positions (solids and liquids) and rotation around the center of mass. As a result, the dipole electric moments p are chaotically distributed along directions N (Figure 4.20a). For the whole dielectric, the vector sum i1pi is equal to zero. Therefore, in spite of the fact that each molecule possesses a dipole moment, the whole sample does not. Molecules of nonpolar dielectrics do not possess a dielectric moment. This means that the positions of the positive and negative charge centers in the molecules coincide. Examples of nonpolar molecules are H2, CCl4, C6H6, CH4, etc. In this case, the macroscopic dielectric sample does not possess a dipole macroscopic moment either. However, when placed in an external electric field E0, all dielectrics regardless of their molecular properties are polarized, i.e., pi becomes nonzero, the dielectric acquires a macroscopic dipole moment (Figure 4.20b). For the quantitative description of dielectrics, the notion of the polarization (degree) or the polarization vector is introduced; numerically, it is equal to the electric dipole moment of a unit volume. Accordingly
1 V
N
∑ pi .
(4.2.1)
i1
In this expression V is a so-called physically infinitesimal volume, i.e., a volume containing enough dipoles that the vector sum in eq. (4.2.1) adequately reflects the macroscopic dielectric state of this volume, but, simultaneously, small enough that within the limits of this volume the polarization degree can be considered as uniform. It is important to notice once more that the sum in eq. (4.2.1) is of a vector character. However, in the case when all N (identical) moments are directed along one and the same axis (we shall choose an x-axis), expression (4.2.1) can be written as x
∑ px Npx np V
V
x
np,
(4.2.2)
where p px (because all vectors p are focused along axis x).
E0 = 0
(a)
E0 > 0
(b)
Figure 4.20 Mutual orientation of a dielectric’s molecules in the absence (a) (random orientation) and in a presence of an external electric field (b).
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There are two approaches to the description of dielectric properties: phenomenological which includes the description of macroscopic dielectric characteristics (subject to direct measurement) and microscopic which includes the analysis of dielectric behavior taking place at an atomic and molecular level. We shall begin by considering dielectric behavior in an external electric field from the macroscopic point of view. 4.2.2
Macroscopic (phenomenological) properties of dielectrics
Consider now what occurs with a dielectric if it is placed in an external field E0 (Figure 4.21), created, for example, by a plane condenser. As was mentioned above, the dielectric polarizes. All charges inside a volume remain mutually compensated; however, there is no compensation near surfaces. Bound charges with density are created at the edges of a body. These charges cannot be taken from the dielectric surfaces; therefore, they are referred to as bounded charges (opposite to those free charges which are on capacitor plates and form a field E0 (E0 free /0) (4.1.19)). The surface-bounded charges produce another charged condenser-like “plate” which creates the electric field E1 directed opposite to the external field E0 (Figure 4.21). This field is referred to as a depolarized field or an electric field of bounded charges. The electric field E inside dielectrics can be treated as a superposition of two fields: the external field E0 and the depolarized field E1, i.e., E E0 E1, or in scalar form E E0 E1 .
(4.2.3)
Thus, the resulting average field E in the dielectric body is always less than external field E0. This field is referred to as an average macroscopic field E in the dielectric; the average field represents the result of superposition of an external field and electric fields of bounded charges. The value equal to a ratio of the strength E0 of an external electric field to the strength E of an average field in the dielectric is referred to as dielectric permeability of the dielectric medium.
– '
− − −+ − − − − − − − − − −
−+
E0 . E
−+
E
+ + −+ −+ + + + + + + + + E1 + +
(4.2.4)
'
E0
Figure 4.21 Dielectric in an external electric field E0: formation of surface–bounded charges with surface density is shown.
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Namely this value has to be introduced into the Coulomb’s law (4.1.1) in order to determine a force acting between two point charges in a medium with dielectric permeability . F
1 Qq r . 4 0 r 2 r
(4.2.5)
There is a certain correlation between the polarization and the surface-bounded charges’ density. To establish this, we must imagine that we have cut in a flat dielectric plate an elementary volume (e.g., in the form of a cylindrical rod) along the field force lines perpendicular to the dielectric’s surfaces (Figure 4.22). Assume the face area to be S, the cylinder length l and the bounded charges density . Therefore qS and the induced electric dipole moment p ql Sl. Then the polarization of the allocated dielectric rod shall be found relating the electric moment p of the rod to its volume V, i.e., p/V. As an elementary volume is V Sl (see Figure 4.22) and p Sl, then by definition (see (4.2.1))
ⴕ Sᐉ ⴕ. Sᐉ
So, the surface density of the bounded charges is numerically equal to the dielectric polarization. ⴕ.
(4.2.6)
As experiments show, the isotropic dielectric polarization is proportional to the electric field strength E and coincides with it directionally ( ~ E). In the SI, the relation between and E is written as 0 E,
(4.2.7)
where is dielectric susceptibility. Note that E in this expression is the strength of the average electric field (4.2.3).
E0 +'
–'
+∆q
–∆q
∆S
l
Figure 4.22 Relationship of the bounded charges’ field and a body polarization.
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Dielectric susceptibility characterizes the dielectric from the point of view of its ability to be polarized in the electric field. The dielectric susceptibility in the SI is numerically equal to dielectric polarization divided by 0 of the dielectric polarization in the field of the unit electrostatic field strength. Thus, the ability of a dielectric to be polarized is characterized by two values: dielectric permeability and dielectric susceptibility . There must be an interconnection between them. Indeed, the average field in a dielectric, according to (4.2.3), is E E0 E1. From the other side, E0 E and E1 /0. Then E E /0. Taking into account (4.2.6), we can write E E /0 E 0E/0, and therefore 1 .
(4.2.8)
The numerical values of dielectric permeability of various substances change over a wide range and depend on the frequency of the external electric field. We shall discuss this question in more detail below. For the majority of nonpolar liquids, dielectric permeability is 2–2.5; for polar liquids, it is significantly higher at 10–80. For the majority of solids, 1.5–2.5; however, for ferroelectrics (see Section 9.6), achieves a value of 104. For gases, differs a bit from unity. In order to characterize an electric field inward the dielectric, it is convenient to use one more value similar to electric field strength, namely a vector of an electric induction (or a vector of electric displacement) D. In isotropic dielectrics, D 0E. If does not depend on E, values D and E are proportional to each other. From the above-mentioned relationships, it follows: D 0 E . If E depends on whether dielectrics are present in the field and what their dielectric permeability is, D does not depend on these circumstances and does not change its value at the transition from vacuum to dielectrics (remember that this concerns isotropic dielectrics). If depends on E, as it sometimes takes place (e.g., in ferroelectrics, see Section 9.6), then D nonlinearly depends on E, remaining independent of the presence in space of other dielectrics. 4.2.3
Microscopic properties of dielectrics
Among the microscopic properties of a dielectric, the basic place is occupied by the molecular dipole moment p; this depends, in turn, on atomic structure and chemical bonding. In Figure 4.23 examples of molecules are given and both directions and values of the dipole moments (in D) are specified. Note that the concept of molecular dipole moments is one of the most important in chemistry. Some main principles connecting a molecule structure with its dipole moment can be noted. Firstly, the polarity of multiatomic molecules depends on the polarity of a given chemical bond and their mutual arrangement. On a simplified level, the electric dipole moment of a multiatomic molecule can be considered as a result of geometrical summation of the individual moments of each bond. Such an approach is based on the additive property of the total dipole moment: each chemical bond being considered as a dipole
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– p =1.84 D
H2O +
+ –
HCl
p=1.03 D + –
NH3
+
+
p=1. 48 D
+
p= 0
+ CO2
+
–
Figure 4.23 Molecular models and dielectric properties of some molecules.
moment, the total moment is obtained as a vector sum of the moments of a particular bond. The transferability of chemical bonds polarity is used here as well: it is accepted that the dipole moment is the specific property irrespective of in which particular molecule these bonds participate. At the same time, if a molecule possesses the center of symmetry (e.g., C6H6, CO2, etc.), its dipole moment a priori is equal to zero. This statement follows from the fact that the polar dipole moment vector is incompatible with the center of symmetry because any vector itself does not possess that center. Secondly, the molecular dipole moment strongly depends on the charge transfer from one atom to another. So, the diatomic molecules consisting of identical atoms, due to the symmetry of the electron pair arrangement, do not possess polarity at all; their electric dipole moment is equal to zero. The diatomic molecules consisting of different atoms are, in most cases, polar. The polarity of the bond is determined by the electron affinity of the constituent atoms. The greater the difference in the electron density, the more polar is the bond. The polarity reaches the highest value at a purely ionic bond. Thus, the electric dipole moment characterizes the degree of ionicity of a chemical bond. For example, the dipole moment of halogen–hydrogen bonds increases from HI to HF according to increase of electronegativity of the halogen atoms (see Table 4.1). Molecular dipole moments can be calculated by modern methods of quantum chemistry from the first principles (for small molecules). Experimentally dipole moments can be derived from X-ray diffraction experiments using a function of electron density distribution (r). Every dielectric at microlevel represents a discrete structure in which molecules or atoms are distributed in ordered (in crystals) or chaotic (in gases and liquids) manner. Therefore, the electric field strength at any point of the dielectric is the superposition of an external field and the fields of all neighboring molecules. The structure of an electric field in a dielectric is highly nonuniform; the value of an average field obtained above represents a highly averaged picture. It means that the electric field strength really acting on a given molecule is not equal to the averaged field: we should consider some effective field, referred to as a local field. The strength of the local field Eloc is defined as the geometrical sum of the strengths of an external field E0 and the total field of all the dielectrics’ dipoles except for the dipole being considered.
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Table 4.1 Electric dipole moments of some of halogen–hydrogen molecules Molecule
HF HCl HBr HI
Dipole moments 10−30 C m
D
6.37 4.33 2.63 1.30
1.91 1.03 0.79 0.39
If an isotropic nonpolar dielectric is placed in an external electric field, the local field will act on each molecule. Displacement of charges of different signs will take place and a dipole will be created. Such a dipole is called an induced dipole. Its value is proportional to the strength of the local field. In SI units, this dependence is p 0 Eloc .
(4.2.9)
The value characterizes the ability of a molecule to be polarized in an electric field, and is referred to as polarizability. Formula (4.2.9) permits us to rewrite the expression for dielectric polarization. Starting from (4.2.2), we arrive at np n 0 E loc .
(4.2.10)
We can compare the expression for polarization with that derived earlier (see eq. (4.2.7)) on the basis of macropresentations. Analysis shows that they are identical when E ⬇ Eloc, i.e., when the depolarizing field E1 is rather small. This corresponds in particular to a low molecule concentration. Then n.
(4.2.11)
Of course, this is valid for low dielectric polarization. The connection between and n at significant polarization is given in Section 4.2.7. 4.2.4
Three types of polarization mechanisms
There are several mechanisms of dielectric polarization, each being directly dependent on the molecules’ structure. Let us distinguish a deformation and an orientation polarization. In the first case the molecule deformation takes place when imposing it in the external field. Deformation can touch both the single atom and the whole molecule. We shall consider each of these mechanisms separately.
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Electronic polarization Relates to the deformation type of polarization and appears as a result of the displacement of an atomic electron shell relative to the nucleus which, accordingly, leads to an induced dipole moment. Certainly, such a polarization mechanism takes place in all systems containing atoms, i.e., in all dielectrics; however, in a pure state it can be seen in the case of nonpolar homo-atomic molecules and atoms. Let us estimate the value of electronic polarization. As an example, take a simplified model of an atom. Assume that the electron charge Ze is uniformly distributed around the nucleus in a sphere of radius R, i.e., with constant electron density Q/((4/3)R3)ZNeN/((4/3)R3). In the absence of an external electric field, the nucleus is in the center of a spherically symmetric electronic cloud, the centers of positive and negative charges coincide; the atom does not possess a dipole moment. We can impose an origin with negative charge center. Let us place an atom in an external electric field. Under the action of this field, the centers of charges of both signs will shift away from each other and an induced dipole moment will appear (Figure 4.24) p Z | e | l,
(4.2.12)
where Z|e| is a nuclear charge and that of the electron cloud as well and l is the displacement value of charges; the direction of dipole moment p coincides with the direction of external field Eloc. A nucleus displaced from its center will be under the action of two competitive forces: the action of a local electric field, Fⴕ Z e Eloc ,
(4.2.13)
and an internal atomic electric field, Fⴖ Z e
Z |e | l, rZ e 3 0 4 0 R 3
(4.2.14)
where l is the dipole arm; the latter force F tries to remove the nucleus in the initial central position. Because the forces are balanced, F F and ElocZ|e| Z |e|(Z|e|/40 R 3)l, wherefrom, after cancellation on Z |e|, it follows that Z⏐e⏐l 40 R3Eloc. The expression for
R
F" F' O l
+
E
Figure 4.24 A model of the electron polarization.
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the dipole moment is written on the left-hand side. Therefore, p 40 R3E loc. Comparing this expression with eq. (4.2.9), we can conclude that the electron polarization coefficient is el 4R 3 .
(4.2.15)
It can be seen that the atomic electron’s ability to be polarized is proportional to its volume. Although the result has been obtained at a certain level of simplification, it is qualitatively correct in describing a real picture of electron polarization. It can be seen from Table 4.2 that a cube of atomic radius increase and an electronic polarizability change in parallel. Let us estimate to an order of magnitude the value of electronic polarizability. We estimate 3 1010 m for atomic radius, then ⬃ 4R3 ⬃ 10 1030 m3 ⬃ 1029 m3 which coincides with the values of electronic polarizability of middle-sized atoms. The electron polarization is determined by the atomic electron shell. Therefore, it does not depend on temperature. Note that it consists of a shift of a light part of the atom regarding heavy nuclei and therefore possesses low persistence. Atomic (ionic) polarization Is observed in heteroatomic molecules in which atoms, due to different electronegativity, endure a redistribution of electron density. Therefore, each atom appears carrying an excessive (positive and/or negative) charge (Table 4.3). This charge is referred to as effective atomic charge and is defined in terms of an electron charge, i.e., q/|e|. Certainly, the sum of all over the whole molecule is equal to zero. Depending on the effective charges and interatomic distance, the hetero-atomic molecules exhibit a permanent dipole moment even in the absence of an external electric field Table 4.2 Electron polarizability of some atoms Atom
Polarizability coefficient, el (10−30 m3)
He Ne Ar Kr Xe
2.3 4.7 16 25 41
Cube of atomic radius, R3 (10−30 m3) 0.78 1.4 3.6 4.8 6.9
Table 4.3 The effective charges of atoms in selected molecules Substance
Atom
Substance
Atom
Na2O MgO Al2O3 P2O5 SO3 Cl2O7
O O O O O O
−0.81 −0.35 −0.31 −0.13 −0.06 −0.01
NaF NaBr NaCl MgO MgBr2 MgCl2
Na Na Na Mg Mg Mg
0.58 0.83 0.92 1.01 1.38 1.5
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+ +
− – l
–
−
+ F=qE +
F=− ∆l
E ∆l
Figure 4.25 Upper (without the external field) and lower (a molecule in the external field).
p |q|l. However, under the action of an electric field, there is additional displacement of the atoms relative to each other. Therefore, the interatomic distance is enlarged and the dipole moment increased. An additional (induced) electric dipole moment appears which causes the additional polarization. In order to estimate the atomic polarization, we can use a model: consider a molecule consisting of two atoms with effective charges and (Figure 4.25) bounded by a quasi-elastic force with rigidity coefficient . In the absence of the external electric field, its dipole moment is p |e|l (Figure 4.25, upper). Under the external field action, the interatomic distance l undergoes an enlargement l (Figure 4.25, lower). Two acting forces (|q|E and l) are balanced in absolute value: l |qef|Eloc. We can find the interatomic distance enlargement l (|qef|/)E. Due to the action of the local electric field, the molecules acquire an additional dipole moment p q l
qef2 Eloc .
(4.2.16)
Denote the polarization coefficient by at. Then the additional dipole moment can be rewritten as p at0E, wherefrom at p/0Eloc. Substituting ∆p according to (4.2.16), we arrive at
at
qef2 2 e 2 . 0 0
(4.2.17)
Evaluation of at, for a HCl molecule, for instance, gives: at 0.2|e|, 10 N/m, the coefficient of atomic polarization is at((0.2 1.6)2 1038)/(1011 102)⬇1029 m3. This is a bit lower than el for the same molecule. At moderately high temperatures when electronic density and, accordingly, interatomic forces can be considered as constants, the atomic polarization does not depend on temperature. The nuclear subsystem participates in atomic polarization and, therefore, atomic polarization possesses greater persistence than the electronic one. Orientation polarization Consider a polar dielectric, each molecule of which possesses an inherent, permanent dipole moment p. In the absence of an external field, dipoles are oriented chaotically due to the
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molecules’ thermal motion; this means that the dielectric’s polarization (4.2.1) is zero (Figure 4.21). In an external field, due to the action of force pairs (refer to Section 4.1.5 and formulas (4.1.32) and (4.1.33)), each dipole will acquire a tendency to be oriented by the field, but the thermal motion will prevent it. The general problem of the behavior of an ensemble of permanent dipoles (magnetic and electric) in an external permanent field was solved by the French scientist P. Langevin. Here, we shall take advantage of his unsophisticated dielectric model which arrived, nevertheless, at the correct result. Consider a molecular system consisting of the permanent moments p with concentration n. Accept a model in which dipoles are focused with equal probability along three coordinate axes, so 0. In the absence of an external field in any direction (positive or negative) of each coordinate axis, n/6 of all molecules will be directed. Assume now that operating on each molecule is a field whose direction coincides with the positive direction of the axis y (Figure 4.26). This means that each molecule will obtain a potential energy U pE pE cos , where is the angle between vectors p and E. For molecules whose dipole moments are directed along the field ( 0, cos 1), U pE. For molecules whose dipole moments are directed oppositely ( , cos 1), U pE. Assume also that dipoles directed along axes x and z do not participate in polarization. The competition between ordering tendency of an external field (in our case, this action is described by the value of potential energy U) and disordering tendency caused by thermal chaotic molecule movements (with energy T ) is described by the Boltzmann factor (see Section 3.2.4). This competition results in the fact that the concentration of molecules with dipoles oriented along the field (n) and oppositely (n) will be different and equal n
n ⎛ pE ⎞ exp ⎜ loc ⎟ ⎝ T ⎠ 6
and n
n ⎛ pE ⎞ exp ⎜ loc ⎟ . ⎝ T ⎠ 6
Since the concentration of molecules whose dipoles are oriented in a positive and negative y-axis direction will be different, the resulting dipole moment appears in a positive direction. According to (4.2.2), it is equal, y (n n)p. We will restrict our consideration to weak fields for which pEloc T (the potential energy of dipole interaction with an external field U is significantly less than the energy
z
E
n–p x
n+p
y
∆np
Figure 4.26 To the model of a orientational polarization.
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of the molecules’ thermal motion T). In this case, an exponent function can be decomposed in a series: we will limit ourselves to only the first two terms: n⬇ (n/6)(1(pEloc/T)) and n⬇ (n/6)(1(pEloc/T)). Then an excess of dipole concentration directed along the electric field will be n nn (n/3)(pEloc/T)) and the polarization will be
n pEloc np2 np2 p Eloc 0 Eloc . 3 0T 3 T 3T
(4.2.18)
The averaged effective dipole moment counting on a single molecule can be obtained by dividing by n: ef
p2 Eloc . n 3T
The orientation polarization can be characterized by the coefficient of orientation polarization op. Then p op0Eloc, wherefrom
op
p2 . 3 0T
(4.2.19)
Evaluation of or for an HCl molecule gives p 1 D 3.33 1030 C m and T 300 K
or
((1 3) 1029 )2 1028 m 3 , (3 1011 ) 1023 300
which is higher than el. The value of or is inversely proportional to absolute temperature and is proportional to the square of the molecular dipole moments. In extreme conditions (in very high fields or at very low temperatures when pE T— something that is technically very difficult to achieve), all dipoles are built along an external field; further increase does not essentially change polarization, the polarization reaches saturation. The polarization value in this condition depends only on dipole moment value p and concentration n. In Figure 4.27 a graph of polarization versus external electric field strength is given. At pET this function is linear (eq. (4.2.18)), the higher the molecular dipole moment and the lower the temperature, the steeper is the linear function, i.e., tan d/dE. In the other limiting case (pE
T), polarization is constant ( constant, saturation state). A smooth curve connects these two extremes. It is obvious that in polar dielectrics all three mechanisms of polarization (two of deformation and one of orientation) are exhibiting simultaneously. The total polarization of isotropic polar dielectrics el, atomic at and orientation or is a sum: el at or. The total polarization can also be given as the sum (el at or )n 0 E loc ,
(4.2.20)
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4. Dielectric Properties of Substances np
tg =
p2 30κT E
Figure 4.27 A relationship of a dielectric polarization and the external electric field strength E at orientational polarization.
a molecule can be characterized by a total averaged polarization coefficient el at or .
(4.2.21)
In different molecules each term can exhibit differently depending on structure and electron density distribution as well as syntheses, temperature, experimental methods, etc.
EXAMPLE E4.10 An electric dipole with a moment p 2 nC m is in a uniform electric field E 30 kV/m; vector p is oriented at 60° to E. Determine the work A of external forces to rotate the dipole at an angle 30°. Solution: From the initial position the dipole moment can be turned on /6 30° in two directions: (1) clockwise at an angle 1 0 /3 /6 /6 or (2) anticlockwise at 2 0 /3 /6 /2. In the first case, the rotation occurs under the action of inner forces, therefore the work is negative; in the second case, only external forces can do the turn and correspondingly the work is positive. The work can be determined by two methods: (1) direct integration and (2) using the relation between work and potential energy change. The first method comes to the integration dA M d pE sin d in the limits from 0 to
0
0
A ∫ pE sin d pE ∫ sin d; executing the calculation we obtain A pE(cos 0 cos ). Accordingly, the clockwise rotation gives A pE(cos 0 cos 1) 21.9 J and in anticlockwise rotation, A pE(cos 0 cos 2) 30 J. We think that the second method is preferable. In fact, A U2 U1 (see Section 1.4.5) and A pE(cos 0 cos ). This coincides with the previous expression.
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EXAMPLE E4.11 In an iodine atom at a distance r =1 nm from an -particle, an induction dipole moment p 1.5 1032 C m appears. Find the polarization coefficient of the iodine atom. Solution: The polarization coefficient can be found according to eq. (4.2.9), p/0Eloc, where Eloc is the field in which a given atom occurs. In this case, the field created by an -particle is a local field. Therefore, it is equal to Eloc E 2冟e冟/40r2. Combining the last two expressions, we arrive at 2r2p/ 冟e冟. Executing the calculation, we obtain 5.9 1030 m3.
EXAMPLE E4.12 Krypton is under a pressure p 10 MPa at a temperature T 200 K. Find (1) the dielectric permittivity of krypton and (2) its polarization in an external field E0 1 MV/m. Krypton polarizability is 4.5 1029 m3. Solution: (1) Expression (4.2.26) is suitable for solving this problem (1)/ (2) n/3, where n is krypton concentration. Find the dielectric polarizability from this equation: (1(2/3)n)/(1(1/3)n). The concentration is equal to n p/T; therefore, (3T2p)/(3Tp). Substituting all data in the expression, we arrive at 1.17. (2) In the uniform electric field krypton polarizability can be given by eqs. (4.2.2) and (4.2.10) np and p 0nEloc. To rewrite the local field via the external field for nonpolar substances (Eloc((2)/3)E0) and np/T 3.61027 m3, we obtain p 0 ((2)/3)nE0. Using all data obtained we arrive at p 1.30 106 C/m2. 4.2.5 Dependence of the polarization on an alternative electric field frequency Polarization is a measure of a dielectric’s “response” to the action of an external electric field. In a static electric field all the molecular dipoles align along the external field (we will not take the thermal motion into account at the moment). To measure total polarization in the static field is easy: having placed a dielectric sample in a condenser, its capacitance depends on the dielectric permeability of the material between the plates (C C0, C0 being the capacitance of an “empty” condenser, remember the ratio between susceptibility and permittivity ). In an alternating electric field the molecular dipoles should have time to turn as a whole when the electric field changes its direction. While field frequency is not so high, the dipoles have time to reorient, following the change of field direction. In a field of high frequencies when ⬃ 106–107 Hz due to the molecules’ inertia, they begin to delay, being unable to follow the electric field reorientation. The higher the frequency, the greater is the delay. At very high frequencies the molecules will not be able to reorient and the polarization comes to naught.
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We can conclude that in static electric field, all molecules regardless of their origin participate in overall polarization; therefore, such measurements give the sum of three polarization mechanisms (refer to (4.2.21) and Figure 4.28 in which a range of frequencies is given). When the electric field frequency reaches a value of 1011 Hz, the molecules are unable to turn over because of their great inertia; therefore, the orientation polarization is switched out. Atomic polarization does not need the molecules’ reorientation; in a high frequency field, the induced oscillations of atoms along the field direction take place. This will be preserved up to a frequency of approximately 1012 Hz. At this value, atomic oscillations also vanish and only electron oscillations (the lightest part of atoms) remain. At ⬃ 1013–1015 Hz, only electrons can accomplish their oscillations. This frequency corresponds to oscillations of light vectors, magnetic and, mainly, electric. At such frequencies, only the electron part remains; at higher frequencies, even the electron part disappears. The frequency dependence explains why the dielectric permeability of water measured in a static electric field is 81, but at optic frequencies is only 1.77: in the first case all polarization mechanisms are participating in the polarization but in the latter case only electron polarization takes place. Otherwise the polarization is exhibited in solids. The ability of a molecule to change orientation or oscillate depends essentially on its geometrical form and interaction forces with its neighborhood. If the form of the molecule is close to spherical and its electric moment is not high, it can rather easily change its orientation (e.g., molecular group CH4). Molecules HCl and H2O are less symmetric; in solids they have some steady orientations and rather slowly pass from one to another. The average time of such a transition is referred to as relaxation time. The value reciprocal to the relaxation time is referred to as relaxation frequency. When the frequency of an external alternative electric field is higher than the relaxation frequency, the system will no longer react to the action of such a field. The relaxation time depends on temperature and the aggregate substance state. So for water (t 200 °C), 3 1011c; however, for ice (t 20 °C), 103c. 4.2.6
A local electric field in dielectrics. Lorentz field
A local or microscopic electric field is a field acting on the given dipole in a dielectric. The strength of a local field is the geometrical sum of external field strength E0 and the or at el 109
1010
1011
1012
1013
1014
1015
1016
Figure 4.28 Dielectric polarization in the alternating field of different frequencies.
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complex field ∑Ei created by all dipoles, except that on which this field operates (Eloc E0 ∑Ei). This field strongly changes within the limits of intermolecular distances and, owing to thermal molecular motion, changes as well in time. In fact, it is impossible to calculate Eloc according to the formula presented because of the enormous number of dipoles. However, if the dielectric is polarized homogeneously, i.e., the polarization at any dielectric point is the same in magnitude and direction, it is possible, at a certain approximation, to calculate its value. Without carrying out a detailed calculation (because of its complexity), we shall, nevertheless, note the general remarks that allowed H.A. Lorentz to solve this problem. His method is as follows. Remember that the further the dipole from the point of consideration, the less is its contribution to the local field (the dipole field falls down as E ~ 1/r3, eq. (4.1.31)) but the larger their number. This allows him to replace summation upon individual dipoles onto integration, i.e., calculate a local field macroscopically. Lorentz suggested allocating in a homogeneously polarized dielectric a sphere of rather small radius, inside which there exist a large but, nevertheless, limited number of dipoles. The center of the Lorentz sphere should coincide with the point of observation (Figure 4.29). In an external field, polarization will take place, including the allocated sphere. We shall mentally remove the originally allocated sphere from the dielectric body. There appears a spherical cavity with charges distributed on its internal surface. The strength of the local field can now be given by four terms Eloc E0 E1 E2 E3, where E0 is the external field, E1 is the depolarizing field created by bounded charges distributed on an external dielectric surfaces, E2 is the field created by the bounded charges on the internal surface of the spherical cavity (Lorentz field) and E3 represents the field of the nearest neighbors; in isotropic dielectrics (in gases, liquids and isotropic crystals), this field is equal to zero. The values E0 and E1 have already been considered above, and calculation of the strength of the field E2 is the subject of our consideration. Integration over the internal cavity surface gives a field E2 which is referred to as a Lorentz field:
E2
1 . 3 0
(4.2.22)
Correspondingly, the local field strength can be written as: E loc E
–
. 3 0
(4.2.23)
+
0
E0
Figure 4.29 A dielectric’s body polarization.
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In scalar form, it can be given as Eloc E
1 . 3 0
(4.2.24)
This field should be presented in all expressions in which a local field appears. 4.2.7
Clausius–Mossoti formula
For rarefied dielectrics, eq. (4.2.11) connecting the macroscopic characteristic with the microscopic one , which in turn provides access to the analysis of the molecule properties, was given. In more complex cases of the more condensed matter at n noticeably larger than unity, the given simple ratio is not fair. In order to find the proper ratio in more dense substances we should substitute in the expression (4.2.10) the local field Eloc by eq. (4.2.24): ⎛ 1 ⎞ n n 0 Eo n 0 ⎜ E n 0 E . ⎟ 3 0 ⎠ 3 ⎝ Solving this equation relative to , we obtain
n 1 ( n 3)
0 E.
If we accept n 1, it goes to (4.2.10), i.e., n. However, if this is not the case, then we have to compare it with eq. (4.2.7). The comparison gives
n 3n , 1−(n 3) 3n
or n . 3 3
(4.2.25)
Substituting here (4.2.8) accordingly, we arrive at 1 1 n . 3 2
(4.2.26)
This is one of the forms of the Clausius–Mossoti law. It connects the macroscopic value of the susceptibility with polarizability of molecules.
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One can express a molecule concentration n in eq. (4.2.26) via the Avogadro constants NA and a molar volume M/: n NA/(M/) (/M)NA. Then the Clausius–Mossoti formula can be written as M 1 1 NA . 2 3
(4.2.27)
Clausius and Mossoti independently obtained eq. (4.2.27) for nonpolar dielectrics in the middle of the 19th century. In 1912, using Langevin theory for magnetization of paramagnetic substances, Debye obtained for polar dielectrics a connection between polarization and electric dipole moments. He suggested a concept of orientation polarizability of polar molecules and has generalized the Clausius–Mossoti equation for the case of polar dielectrics. This Debye–Langevin formula is usually written as M 1 1 ⎛ p2 ⎞ NA, ⎜ def 2 3 ⎝ 3 0T ⎟⎠
(4.2.28)
where def is the deformational polarization (see Section 4.2.4) or the polarization of elastic displacement, which is the sum of electronic el and atomic at polarizations. Taking into account all three types of polarization mechanisms, the Debye–Langevin formula can be written as: p2 ⎞ M 1 1 ⎛ ⎜ el at NA . 2 3 ⎝ 3 0 KT ⎟⎠
(4.2.29)
The Debye–Langevin formula is applicable to polar dielectrics at definite restrictions. It achieves good fulfillment for gases and vapors at low pressure, and for highly dissolved solutions of polar liquids in nonpolar solvents. This formula is of great importance in the interpretation of molecular structures. Being written as M 1 1 N , 2 3 ( el at ) A it successfully describes nonpolar gases at low and average pressures (500 kPa and lower), can be applied approximately for nonpolar gases at elevated pressures (above 500 kPa) and nonpolar liquids, and is good enough for crystals with face-centered lattice if atoms possess only electronic polarization and approximate for ionic crystals with a cubic lattice (see Section 9.6). As was already mentioned in Section 4.2.5, the polarizability of molecules depends on the frequency of the alternative electric field, especially at high frequencies. In the Maxwell electromagnetic theory, the ratio between a refraction index n and the dielectric permeability of substances is given. For low-magnetic substances, n 兹苶 . If in eq. (4.2.27) we substitute by n2 and take into account that at optical frequencies
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( ⬃ 10151016 Hz) all polarizability mechanisms, except the electronic one, are suppressed, it is possible to write M n 2 1 1 el N A . n2 2 3
(4.2.30)
This is the H.A. Lorentz–L. Lorentz formula. The right-hand side of this equation is a molar refraction R. It is fair for gases, nonpolar liquids and isotropic crystals (with cubic lattices). It is approximately applicable to nonpolar liquids at relatively high frequencies when the orientation polarization is not exhibiting.
EXAMPLE E4.13 The density of liquid benzene is 899 kg/m3 and its refraction index n 1.50. Determine the benzene electron polarizability el. Solution: The Lorentz–Lorentz formula (4.2.30) can serve us in solving this problem. Using it, we can solve it relative to el
el
3 M (n2 1) . N A (n2 2)
In this equation all the entries are known except for mole mass M. Since the atomic composition of benzene is C6H6, the benzene relative mole mass is 78. Therefore, the molar mass is M 78 10−3 kg/mol. Substituting all the values, we arrive at el 1.27 10−28 m3. 4.2.8 An experimental determination of the polarization and molecular electric dipole moments Experimental determination of microscopic characteristics is based on the Debye–Langevin (eq. (4.2.28)) and Lorentz–Lorentz (eq. (4.2.30)) equations. They connect macroscopic molecular characteristics, measured directly in physical experiment, with microscopic ones. Measuring dielectric permeability in an electrostatic field, the molar polarization can be found: 1 (el at or )N A , 3 from which the total polarizability of a molecule can be obtained 3
NA
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or 3
M 1 . N A 2
(4.2.31)
Measurements of a refraction index n in an optical range of frequencies allows us to find the molar refraction R, and consequently, the molecule electron polarizability using eq. (4.2.30) (R elNA/3) or
el 3
M n 2 1 . N A n2 2
(4.2.32)
The difference between and el gives the sum: el or el 3
R . NA
(4.2.33)
There are two ways of separating atomic and orientation polarizabilities: either taking advantage of the frequency dependence of the polarizability or using the temperature dependence of the molar polarization (or ~ 1/T ). In the latter case the temperature dependence of molar polarization is removed and a graph (1/T ) (Figure 4.30) is drawn. The segment under the horizontal dashed line gives that part of the molar polarization which does not depend on temperature 1 el at (el at ) N A . 3
(4.2.34)
The value el can be determined by refractometric experiments (4.2.33) and at can be found according to formula (4.2.34) at 3
el at el . NA
(4.2.35)
∆=or def =def +
p2 30
.
1 T 1/T
1/T
Figure 4.30 A relationship of a molar dielectric polarization versus reciprocal temperature.
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The orientation polarization op, corresponding to a given temperature, can also be found from the graph: the angle coefficient tan is determined by op. Indeed, tan
1 or or N A T , (1T ) 1T 3
from which the orientation polarizability can be found or
3 tan . N AT
(4.2.36)
The tan value is determined by expression / (1/T) m3 C/mol. Therefore, the slope has the same dimension. Note that op decreases with decrease in temperature. Knowing or, we can determine the molecular dipole moment. According to eq. (4.2.19), p230Tor; substituting further op according to eq. (4.2.36), we can obtain p2(9/NA)0 tan , from which ⎛ ⎞ p 3 ⎜ 0 tan ⎟ ⎝ NA ⎠
1 2
(4.2.37)
.
We would like to underline once more that dielcometry is a relatively simple though rather powerful, method of chemical structure investigation.
EXAMPLE E4.14. Thin semi-infinite rod is charged uniformly with linear density 1 C/m. At a distance r0 20 cm from the end of the rod perpendicular to it a point O is located (Figure E4.15). Calculate the electric field strength created by the charged rod in point O. y K
dE
dEy
N dl M
O dEx
d r0 K N dl M
L
x
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Solution: Allocate a piece of the rod with a charge dQ dl in an arbitrary point (1dl / 40) of the rod. This charge create a field in the point OdE , r being the r2 distance MO from the elementary charge to the point mentioned. Denoting angles MOL and, consequently MON, as we have r r0 / cos and dl r d / cos 2 . Substitute these equations into formula for E, we can obtain 冷dE冨 d . 40r0 Decompose the vector 冷dE冨 into two components: dEy dE cos and dEx dE sin . Proceeding further, we can find dEy (cos / 40r0 )d and dEx (sin / 40r0 )d . Integration over the limits 1 0 and 2 /2 (both ends of the semi-infinite rod) gives 2
Ey
∫ 0
cos and E x d 4 0 r0 4 0 r0
2
∫ 0
sin . d 4 0 r0 4 0 r0
兹苶2 After summing these two vector components, we arrive at the final value E . 40r0 1069109兹苶 2 Substituting the numerical values we obtain E 7.05 kV/m 20102 1 (keeping in mind that numerically 9109 ). 4πε0
PROBLEMS/TASKS 4.1. There are two similar small balls of mass 1g each. Find the electric charge q which should be given to the balls in order to compensate for the force of their mutual Newtonian attraction. 4.2. In the semiclassical theory of the hydrogen atom, an electron is supposed to travel around a proton along a circular orbit. Determine its speed if the orbit radius is r 53 pm and the frequency of the electron’s revolution. 4.3. In vertexes of an ideal hexagon with a side length a 10 cm, charges Q, 2Q, 3Q, 4Q, 5Q and 6Q (Q 0.1C) are situated. Find the force F, which acts on a point charge Q placed in the center of the hexagon. 4.4. A thin rod of l =10 cm in length is uniformly charged with =103 nC/m. On an extension of it at a 20cm from its end, a point charge Q 100 nC is placed. Determine the interaction force between the rod and the charge. 4.5. A thin ring of R 10 cm radius carries a uniformly distributed charge Q 102 nC. On a plane perpendicular to the ring, just over its center, is a point charge Q1 =10 nC at heights of (1) l1 20 m and (2) l2 2m. Determine the interaction force of the ring and the point charge.
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4.6. There are two coaxial cylinders made of thin metal foil of radius R and 2R. They carry uniformly distributed charges with density and − ( 60 nC/m2). (1) Using the Gauss theorem find the electric field distribution inside both cylinders and between them (areas I and II) and outside them (area III). (2) Calculate the field strength at the point E(3R). Draw a graph E(r) in all areas and calculate the characteristic (border) values. 4.7. There are two concentric sphere made of thin metal foil of radius R and 2R. They carry uniformly distributed charges −2 and ( 0.1C/m2). Using the Gauss theorem find the electric field distribution inside both cylinders (areas I and II) and outside them (area III). Calculate the field strength at the point E(3R). Draw a graph E(r) in all areas and calculate the characteristic values. 4.8. In an area extending between two half-circled rings of radii R and 2R (R 10 cm), a charge Q 20nC is uniformly distributed. Find in the central point O (the center of the rings) the field E(O). 4.9. An infinite thin wire is charged with a linear density 0.2 µC/m. The wire is bent at right angle. Determine point A is denoted on a continuation of one of the wire side at a distance ro15 cm from the corner. Determine at this point the field E. 4.10. An electric dipole p 0.4 C m is in a uniform electric field of strength E 25 kV/m at an angle 1 /6. Find the work of external forces at dipole reorientation (2 7/6). 4.11. An electric dipole p 200 nC m is in a uniform electric field of strength E 50 kV/m at an angle /3. Find the change of its potential energy U at its rotation anticlockwise at the angle 2π/3. 4.12. An electric dipole p 0.2 C m is in a nonuniform electric field with dE/dx −10 kV/m2 oriented against the electric field E and electric field gradient. Find the force direction and calculate its Fx value. 4.13. Two HCl molecules with the same orientation of their electric dipole moments p 1.91 D in value are at a distance r 5 nm from each other. Considering the molecules as point ones, determine the potential U energy of their interaction. 4.14. Two polar molecules SO2 (p 1.60 D) are at a distance r 8 nm from each other. Considering them as a point dipole, determine the force of their interaction. 4.15. Argon is under normal conditions in an electric filed E 30 kV/m. Determine the shoulder l of the induced dipole moment of an Ar atom if the dielectric permeability at the same state is 1.000554. 4.16. The dielectric susceptibility of the gas Ar under normal conditions is 5.54 10−4. Find the dielectric permeabilities 1 and 2 of liquid (1 1.40 g/cm3) and solid (2 1.65 g/cm3) argon. 4.17. What minimum velocity min should a proton possess in order to reach the surface of a metallic sphere charged up to 400 V? 4.18. From point 1 on the surface of an infinite negatively charged cylinder ( 20 nC/m) of radius R, an electron starts with zero velocity. Find the electron kinetic energy K at point 2 which is at a distance of 9R from the cylinder surface. 4.19. Knowing the electric dipole moment p1 of a chlorobenzene (phenyl chloride) molecule (C6H5Cl) (p1 1.59 D), find the dipole moment of ortho-dichlorobenzene (orthodi-phenyl chloride) p2.
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4.20. A xenon atom (the polarizability of which is 5.2 10−29 m3) is at a distance r 1 nm from a proton. Determine the xenon atom’s induced electric moment p. ANSWERS 4.1. Q 2m兹 苶苶 0G 苶 86.7 1015C (G is a gravitational constant). 4.2.
e 4 0 mr
2.19 106 m/sec; n (2r) 6.58 1015sec1 .
4.3. F
6Q 2 54 mN. 4 0 a 2
4.4. F
Qᐉ 1.5 mN. 4 0 (ᐉ a )a
4.5. (1) F1
QQ1ᐉ 1 4 0 ( R 2 ᐉ21 )3 2
15.7 kN; (2) F2
EA
2.26 KVm. 3 0
4.7. (1) E I 0(r R); EII (r )
(2) EA
2.25 N.
R R ( R r 2 R); EIII (r ) (r 2 R). 0r 0r
4.6. (1) E I 0 (r R); EII (r )
(2)
QQ1 4 0 ᐉ22
2R 2 2R 2 ( R r 2 R ); E ( r ) (r 2 R). III 0r 2 0r 2
2 2.51 kVm. 9 0
4.8. E
Q ln2 5.29 kVm. 3 2 0 R 2
4.9. E
5 26.8 kVm. 4 0 r0
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4.10. A12 2pE cos 17.3 mJ. 4.11. U pE[cos − cos( )] 15mJ. 4.12. Fx p(dE/dx)cos 2 mN. 4.13. U 4.14. F
p2 5.82 1024 J ( 36.4 eV). 2 0 r 3
3 p2 3.74 1016 N. 2 0 r 4
4.15. ᐉ
( 1) 0 E 1.9 1018 m. nnorm e Z
4.16.
3 M 2Vom ; 1 1.51; 2 1.61. 3 M Vom
4.17. min 4.18. K
3e
e 2 o
2m
0.24 106 msec.
ln10 828 eV.
4.19. p2 2 p1 cos . 6 4.20. p 6.6 10−31 C/m.
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–5– Magnetics
Before describing the magnetic properties of a chemical substance we will briefly discuss some of the properties of a magnetic field itself. Whereas the sources of an electrostatic field are motionless electric charges, the sources of a magnetic field are moving charges, i.e., an electric current. Let us consider some characteristics of a permanent electric current and the conditions of its maintenance.
5.1 5.1.1
GENERAL CHARACTERISTICS OF THE MAGNETIC FIELD
A permanent (direct) electric current
An electric current can run in a substance in which free charges (current carriers) are present. They can be either electrons (in metals, for example) or ions (in liquids or solid electrolytes). Such substances are referred to as the conductors of an electric current. However, if a conductor is brought into an external electrostatic field there will be only an instant displacement of charges according to electrostatic laws; this will lead to the creation of an internal electrostatic field in a conductor directed opposite to the external electrostatic field and numerically equal to it; the current instantly stops. Therefore, inside a conductor the electric field is always zero. This means that additional conditions to support the current flow are necessary. We shall consider these a little later, but first we shall introduce some ideas about the electric current. The ordered motion of electric charges is referred to as an electric current (a current of conductivity). The scalar value I determined by a total charge dQ having run through a cross-section of the conductor in a unit time interval dt, i.e., I
dQ , dt
(5.1.1)
is referred to as a current strength or simply current. The motion of positive charges is accepted as the current direction, the current flow generally being opposite to the electron movement. If the current magnitude does not change with time it is referred to as a permanent (constant, direct) current. If the current strength changes with time it is referred to as an alternating current. 305
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The current I is a macroscopic characteristic of a particular conductor. For the distribution of an electric current throughout the conductor section a vector of current density j is introduced. The vector of current density j is directed along the carrier’s motion and is numerically equal to the electric charge current, which in the unit time crosses a unit conductor area perpendicular to the carrier velocity (see Figures 2.19 and 2.20). Thus, the modulus of a j-vector is equal to the ratio of the current strength dI through an elementary area dS located in a given conductor point perpendicular to the direction of the ordered carrier motion: j dI/dS⊥. To attach to the current density a vector character, we can multiply j to the unit vector of a direction of the carrier motion /; therefore, .
(5.1.2)
I ∫ j dS,
(5.1.3)
j
dI dS⬜
The integral current strength is therefore
S
where S is the conductor cross-section. Let us establish correlation between the microscopic current characteristics, the density of current j, the concentration of current carriers n and the average velocity of carrier motion (current speed of drifts). Let charge Q be transferred in time t through the cross-section S, perpendicular to the ordered carrier movement (Figure 5.1). By definition, the density of a current j is numerically equal to the ratio I/S. The charge Q is equal to the product of a single carrier charge and their full number in the volume V S⬜ l. Therefore j q nlS⬜/tS⬜ q n. Because current density vector j and the velocity of positive curriers are codirectional, hence j qn冬 冭.
(5.1.4)
Taking into account the possible movement of both positive and negative charges (for instance in an electrolyte) the total current density is j qn 冬 冭 qn 冬 冭.
q +
S⊥
j ∆l = t
Figure 5.1 Relationship of a current density j, carrier concentration n and drift velocity .
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Now derive Ohm’s law in differential form. At a fragment of a uniform conductor in an electric circuit, the current I is proportional to voltage loss U and inversely proportional to its resistance R: I U/R. Consider an elementary cylindrical volume in the vicinity of an arbitrary point inside the conductor (Figure 5.2) with the cylinder generatrix parallel to current density vector j. A current jdS flows through the cross-section of the cylinder. The voltage loss U on the cylinder ends is equal to Edl, where E is the strength of the electric field in the vicinity of the given point. This means that dI jdS
Edl Edl dS, R dl
where is the specific resistance and R(dl/dS). Since the charge carriers at every point move parallel to vector E, the Ohm’s law in differential form acquires the form j E/ E, where 1/ is the specific electroconductivity. Therefore, the current density at any current point j(r) is equal to the product of specific electroconductivity and the electric field strength E(r): j(r ) E(r ).
(5.1.5)
This is Ohm’s law in differential form. Another characteristic of a current is the mobility of the current carrier b, which is the average speed acquired by the current carrier in a field of unit electric field strength. If charges have average speed in a field E then, by definition, their mobility is b /E. The mobility b can be expressed through the specific electroconductivity and carrier concentration n. As the current density is j nq and j E, therefore, E nq. Having divided both parts of equality by E we shall obtain: nqb or b
. nq
(5.1.6)
Let us now consider the conditions that can maintain the electric current in a closed electric circuit. In fact, the current in a circuit can exist only if external forces maintain at the conductor ends a constant voltage difference (to say nothing about superconductivity). Therefore, in the closed circuit, along with the parts in which positive charges move along the decreasing potential , there should be parts where the positive charges move against
j dS E dl
Figure 5.2 Derivation of the Ohm law in differential form.
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the potential loss, i.e., against the forces of an electrostatic field. Figure 5.3 shows a closed circuit with a part where the electrostatic field is acting (1-a-2) and a part (2-b-1), where the so-called extraneous (outside) forces operate. The extraneous forces are of a different, nonelectrostatic nature; they can be of chemical (galvanic cells), induction (electrogenerators), or of thermal origin, etc. These extraneous forces are capable of maintaining the ordered movement of carriers against Coulomb forces. There exists a field of extraneous forces, characterized by the extraneous force Fext having field strength Eext. By analogy with eq. (4.1.2) we can write E ext
Fext . Q
(5.1.7)
The extraneous forces produce the work on charges moving along a circuit. The physical value equal to the work of the extraneous forces produced in moving a positive unit charge Q from point 1 to point 2 is referred to as electromotive force (EMF) Thus, 2
12
Aext ∫ E ext dl. Q 1
Thus, for the closed circuit A Q ∫ E,ext dl. Therefore, the work of the EMF along the closed circuit is Q养El dl. Both extraneous and Coulomb forces act on the charge moving along the closed circuit. The work produced is AQ养E dl, where E symbolizes the sum of the extraneous Eext and the Coulomb Ecol field strengths, i.e., AQ养(EextEcou)dl. Since 养Ecoudl0, hence AQ养Eext dlAext. Therefore A/Q, i.e., the work on the unit positive charge along the closed circuit, i.e., the EMF force is ∫ E dl,
dA = Eextra⋅dl 2
b
1 a
dA = Ecol⋅dl
Figure 5.3 A closed circuit with extraneous forces.
(5.1.8)
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where E is the combined strength of both fields. If we consider the charge motion in a limit 2 2 from point 1 to point 2, then A12Q1 EextdlQ1Ecoudl, and consequently A12 Q12 Q(1 2 ).
(5.1.9)
The value U12 is numerically equal to the work of both fields, electrostatic and extraneous, at the displacement of a unit positive charge from point 1 to point 2, referred to as voltage loss (voltage) at a given section of circuit U12 IR12 12 (1 2).
(5.1.10)
If the circuit is open ended (I 0, U12 0) then 12 (2 1), i.e., the EMF is equal to the potential difference on the clumps of a current source. 5.1.2 A magnetic field induction The electric current (moving charges) creates a magnetic field in the surrounding space. This field affects the charges (currents) moving in it. Thus, the interaction of two currents has an electromagnetic character. Ampere’s experiments showed that two parallel infinite currents I1 and I2 running in one direction attract each other, whereas currents directed oppositely repelled (Figure 5.4). The interaction force falling at a unit conductor length, f, is inversely proportional to distance b between them, i.e., f⬃(I1I2/b). The given statement is the essence of Ampere’s law. In SI this law takes the form
f
0 2 I1 I 2 , 4 b
(5.1.11)
where 0 is the permeability constant (refer to Appendix 1).
F
F
I1
F I2
(a)
F I1
I2
(b)
Figure 5.4 Ampère’s law: interactions of currents, (a) currents are codirectional and (b) antidirectional.
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It can be imagined that one conductor creates the magnetic field and the other in this field acquires its action. Consider first how an electric current creates a magnetic field and then how the magnetic field acts on the conductor with a current. When we analyzed an electrostatic field it was convenient to use a unit point positive charge (probe charge), which we imagined to be placed in each point of the field and measured the force acting on it; this was the electric field strength E. It is impossible to apply such a procedure in the case of a magnetic field because there are no magnetic charges (monopoles) in nature. It is necessary to use a multipole of the next order, i.e., a dipole, placing it at different points in the magnetic field and measuring the torque acting on it (refer to Appendix 3). In order to carry out such a magnetic field analysis, we need to use a model. Such a model is a magnetic dipole moment, which is a small (ideally, a point) flat contour with a current (a probe contour). Orientation of the probe contour in space is determined by a vector n normal to the contour (Figure 5.5). Placing the probe contour into every point one can see that in a given point it accepts a definite position: the normal vector n is oriented in a strictly determined direction. Experience also shows that the action of the magnetic field on the contour is associated with the value IS (where I is the current flowing around an area S). This contour M IS is called the magnetic moment of a contour. The magnetic moment of a contour is the vector value directionally coinciding with the normal n to the contour and determined by the contour current by a right-hand system rule, so MMn ISn.
(5.1.12)
Accept the direction of the probe contour normal n freely oriented in a given point of the magnetic field as a direction of the force line. The force characteristic of the magnetic field is a vector B in the direction of the normal vector n; its absolute value is determined by the maximum torque acting on a contour in this point. Then,
B
M max . M
(5.1.13)
Vector B Bn is referred to as a magnetic field induction vector.
n M
I
Figure 5.5 An elementary magnetic moment.
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The magnetic field is schematically presented in full analogy to the electric field: the B vectors at any point are parallel to the tangent to force lines and their density is proportional to the 兩B兩 value. For the description of the magnetic field we require one more characteristic—field strength H. In vacuum, the two characteristics differ only by a constant, namely, B 0H; however inside of a body magnetic characteristics mentioned differ noticeably (refer to Section 5.4). Remember that the two characteristics appear also in the description of the electric field properties (see Section 4.2.2). The magnetic field obeys the general principle of superposition (see Section 2.9.1). With reference to the case given it can be formulated as follows: the magnetic field created at a given point of space by any current does not depend on whether there are other sources of a field (other currents) in this space or not. Owing to the vector character of the magnetic field, the total induction of a system of currents is equal to the vector sum of particular field inductions, which are created by each current separately: N
B∑B , i
(5.1.14)
i1
and when current is distributed continuously B ∫ d B,
(5.1.15)
L
where dB is the induction created by the elementary current Idl. An empirical method of calculating the induction of a magnetic field at some point in space if the distribution of currents is known was suggested by Biot and Savart; a corresponding law relates a current element Idl (I is a scalar current running in the conductor element dl) to an induction dB at a point A, the latter being assigned by the radius vector r drawn from the element dl to point A (Figure 5.6): dB
0 I [dl r ] . 4 r 3
A multiplier 0/4 entered into the law formula in the SI.
I
dl r
A + dB
Figure 5.6 Biot–Savart law.
(5.1.16)
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We can illustrate the application of the Biot–Savart law with two important examples. Calculate a field that creates a circular current (Figure 5.7) in its center point O. The magnetic induction dB at the point O is created by an element of current Idl accordingly. The vector product in (5.1.16) assign elementary vector dB: it is directed perpendicular to a plane of a drawing and is related to the current element by the right-handed system rule. The vector dB is directed along the same axis (a symmetry axis of a ring) independently of where on the circuit the current element is chosen; therefore, integration (5.1.15) can be executed in a scalar form. Proceeding from the general formula (5.1.16), keeping in mind that the angle between vectors dl and r remains /2, we arrive at dB( 0/4 )(Idl/R2), where r R is the ring radius. The circular current integration gives B 0 ∫ dB L
0 I 4 R
2 R
∫
dl
0
0 I . 2R
(5.1.17)
Next, calculate a field created by the current I running along a rectilinear section of the conductor. In Figure 5.8 conductor MN and an observation point A at a distance b from it are presented alongside the current element dl and a corresponding vector r. Wherever the element dl is chosen, the vector dB is directed along the same direction (perpendicular to the plane of the drawing). Therefore, integration (5.1.15) can be executed in a scalar form. Applying (5.1.16), we obtain
dB
0 Idl sin . 4
r2
There are several variables in this expression; they should be expressed by a single variable. We shall choose for the integration variable an angle (see Figure 5.8). Expressing r and dl through the distance b and an angle : r
b rd bd , dl 2 . sin sin sin
Having substituted these expressions in the formula for dB, we obtain dB(rd /4 )(I/b) sin d , and consequently
B
0 I 2 I sin d 0 (cos 2 cos 1 ). ∫ 4 b 4 b
(5.1.18)
1
Angles 1 and 2 are defined by the extreme points M and N.
dB 0 r
dl
Figure 5.7 Application of Biot–Savart law to the calculation of the magnetic field of a circular current.
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M
313
A + dB
b α
d r
r d
dl
N
Figure 5.8 Application of Biot–Savart law to the calculation of the magnetic field of a direct current.
Figure 5.9 Magnetic force lines of a direct current in planes perpendicular to the current.
For the infinite direct conductor wire (where 1 0 and 2 and the difference cos0 cos is 2): I I (5.1.19) B 0 2 0 . 4 b 2 b The magnetic field of a direct current in a plane, perpendicular to a conductor, is depicted in Figure 5.9 by the force lines. They appear as concentric circles; their density decreases as the distance increases. EXAMPLE E5.1 Along two parallel indefinitely long wires identical currents I 60 A flow in the same direction. The wires are located at distance d 10 cm from each other. Define the magnetic induction B at a point A (Figure E5.1) at distance of r1 5 cm from one conductor and r2 12 cm from another. B B2 A
r1 D + l
B1 r2 d
+ C l
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Solution: In order to find the magnetic induction at the specified point A we should define the directions of induction vectors B1 and B2 created by each conductor separately and then combine two vectors B B1 B2. We can find the module of total induction according to the cosine theorem B B12 B22 2 B1 B2 cos ⴱ . Induction values B1 and B2 are expressed accordingly by the current strength I and distances r1 and r2: B1( 0I/2 r1) and B2( 0I/2 r2). Substituting these values in * we obtain B
0 I 2
1 1 2 cos . r12 r22 r1r2
Find cos using the cosine theorem
cos
r12 r22 d 2 0.576. 2r1r2
Substituting all constants and given values we arrive at 286 T.
EXAMPLE E5.2 Determine the magnetic field induction B produced by a section of infinitely long wire at point A at a distance r0 20 cm from the center of the wire segment (Figure E5.2). The flowing current I 30 A, the segment length l 60 cm.
dl
r d
1 d
r
l
r0 I
2
A
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Solution: The magnetic field induction B can be calculated according to eq. (5.1.18). The scalar form of this equation is dB( 0I sin d/4 r2). There are three variables in this equation: , r and l. It is more convenient to integrate over an angle provided all the variables are expressed through this angle. So drd /sin . Substituting this ratio into the first equation we obtain 0 sin rd 0Id dB ; 4 r2 sin 4 r besides r is also a variable and should be expressed through r (r0/sin ) and therefore dB 0I sin d /4 r0. This expression should be integrated over variable . dB
0 I 4 r0
2
0 I
∫ sin d 4 r0 (cos 1 cos 2 ) . ⴱ
1
Note that in the case of the symmetrical position of wire cos 1 cos 2, therefore the formula * is ( 0I/2 r0) cos 1. We need to define angle . It can be seen from Figure E5.2 that cos
2
r02 冢 2 冣2
4r02 2
.
Therefore, B
0 I I cos 1 0 2 r0 2 r0
4r02 2
.
Substituting all data into the formula obtained we arrive at B 2.49 105 NA m 24.9 T.
EXAMPLE E5.3 A long wire with a current I 50 A is bent at an angle 2 /3. Determine induction B at a point A (Figure E5.3). The distance d is equal to d 5 cm. I 1 B + A
I
0 r0 π−
2
d
2
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Solution: A bent wire can be considered as consisting of two semiinfinite pieces. According to the principle of magnetic fields superposition, the magnetic induction B at a point A is equal to the geometrical sum of the magnetic inductions B1 and B2, i.e., the fields created by two wire pieces 1 and 2, B B1 B2. The field magnetic induction B2 is zero: it follows from the Biot–Savart law, according to which in points lying on an axis of a conductor, dB 0 ([dlr] 0). Therefore, we need to find only B1. In Section 5.1.2 this problem was considered in detail and eq. (5.1.18), i.e., B( 0 I/4 r0) (cos 1cos 2) was derived, where r0 is the shortest distance from the wire to point A (the length of the perpendicular descended from point A on the wire). In the case considered 1 → 0 (the wire is infinite, cos 1 1), 2 2 /3 (cos 2 2 /3 1/2). The distance is r0 d sin( ) d sin ( /3) d(兹苶3/2). Correspondingly, 2 0 I ⎛ 3 0 I 1⎞ . ⎜⎝ 1 ⎟⎠ 2 4
d 4 d 3
B1
Executing calculations we obtain B 34.6 T. Vector B is directed perpendicular to the drawing in downward direction.
EXAMPLE E5.4 A wire in the form of a thin half ring of radius R 10 cm is in a uniform magnetic field (B 50 mT) perpendicular to magnetic force lines. A current I 10 A flows along the wire. Find the force acting on the half ring. y
B X
jdFy
I
dF
dl
id Fx
R j
d
0
x
i
Solution: (See first the Section 5.1.4). (Let us arrange the wire in a plane of the drawing and direct the coordinate axes as is represented in Figure E5.4. On the wire allocate an elementary section with a current Idl. On this area an Ampere force dF I[dl B] (5.1.24) operates. Let us divide the elementary force into two components dF i dFx j dFy. The force acting on the whole conductor can be found by the integration F i ∫ dFx j∫ dFy , L
L
where the symbol L indicates that the integration is taken over the whole half ring length.
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Using symmetry of the problem we can a priory write L dFx0. Therefore the whole force depends only on y-component dFy dF cos , where dF is the modulus of the force dF, dF IBdl. Expressing dl through R and (dl Rd ) we can write dFy IBR cos d . Integration over a quarter of ring (multiplied by 2) gives
/2 Fj IRB2 0 cos d and |F | 2IBR. It can be seen that the force is directed along the y-axis. Executing calculations gives F 0.1 N. EXAMPLE E5.5 A current I 80 A is flowing along a thin conducting ring of radius R 10 cm. Find the magnetic induction B at the z-axis perpendicular to the circle crossing the ring center at an arbitrary point of the axis. Then find the magnetic induction at a point r 20 cm (Figure E5.5). z
dBz dB dB⊥
r
dl
Solution: Since the z-axis is perpendicular to the ring plane and passes the center of the ring, it is a symmetry axis L . This means that the induction vector certainly must be codirectional to the z-axis and only the Bz component gives contribution to the field induction B. First, derive the general expression for B(z). Use the Biot–Savart law dB( 0/4 )(I[dlr]/r3) to determine the dBz component: dBz 冟dB冟 sin dB (R / 兹苶 R2苶 苶 Z2苶). The vector product [dl.r] is perpendicular to the plane fixed by vectors dl and r. Then dBz
0 I dl R , 2 2 2 4 R z ( R z 2 )1 2
where sin R/(R2z2)1/2. The only variable is dl. The integration over the whole circle gives
Bz
0 I 2 R R 0 IR 2 . 2 2 32 2 4 ( z R ) 2( z R 2 )3 2
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This is the general result. To make sure of the result we can use the border conditions B(0) and B( ). We know that B(0) is equal to ( 0I/2R) (see eq. (5.1.17)); by assuming z 0 we arrive at the correct result. The magnetic field diminishes at z → , which is also in agreement with our result. In order to find the field at an equidistant point r we can use the result obtained, accept r2 x2 R2 and substitute it into the general result; therefore B( 0IR2/2r3). Executing calculations, we arrive at 62.8 105 T or 628 T.
5.1.3
The law of a total current (Ampere law)
The sign of the potential character of a force field is the equality to zero of the circulation of the field intensity vector along any closed contour. Let us see whether the magnetic field is potential, i.e., whether the integral 养LBdl is equal to zero or not. Consider the simplest case when a magnetic field is created by a linear conductor with current I. The magnetic force lines in this case are the concentric circles lying in parallel planes, perpendicular to the conductor, with their centers on the linear conductor (Figure 5.9). Choose for simplicity contour L coinciding with one circular force line of any radius R (Figure 5.10). Then the circulation of the vector B along the contour L will be equal to
∫ B dl L
0 I I Rd 0 ∫ 2 R L 2 R
2 R
∫
dl.
0
Therefore, by using eq. (5.1.19):
∫ B dl 0 I .
(5.1.20)
L
Since circulation of the magnetic field induction is not zero the magnetic field is not potential. (Notice that in the above integral dl is an element of a contour L but not current.) To obtain this ratio for a noncircular contour of any form is not a difficult task. Expression (5.1.20) is the essence of Ampere’s law: circulation of the induction vector along a closed contour L is equal to the current multiplied by 0 comprised by this contour.
L I B dϕ
dl
Figure 5.10 An Ampere law consideration.
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(This means that if the current passes outside the contour chosen, this particular current does not contribute to the total current.) A field satisfying the condition (5.1.20) is referred to as a nonpotential field. Expression (5.1.20) is also referred to as Ampere’s law to emphasize the unity of the phenomena of interaction of currents with each other and with the magnetic fields. If contour L comprised N currents then, according to the principle of superposition, the circulation of a vector B is equal to their algebraic sum N
Ii , ∫ B dl 0 ∑ i1
(5.1.21)
L
the current is considered positive if it corresponds to the clockwise rule, otherwise it is considered negative. If the current is distributed nonuniformly across the conductor this law can be rewritten as
∫ B dl 0 ∫ j(r)dS, L
(5.1.22)
S
where a surface S is resting on contour L (Figure 5.11).
S
L Figure 5.11 A surface rested on a contour loop.
Let us apply Ampere’s law to calculate the induction of a magnetic field created by a solenoid. Remember that a coil that has been reeled up by thin wires without misses on the cylinder (Figure 5.12) is referred to as a solenoid. We shall choose a rectangular contour 1–2–3–4, depicted in the figure. Then circulation along the whole contour can be divided into four integrals:
∫ B dl ∫ Bl dl ∫ L
L
12
Bl dl
∫
Bl dl
23
∫
Bl dl
34
∫
Bl dl.
41
It can be seen that the integrals along segments 2 3 and 41 are zero since the angle between B and dl is /2. The integral on the segment 34 is also zero because this segment can be chosen far enough from the solenoid where B 0. Therefore,
∫ Bl dl ∫ L
12
Bl dl 0 ∑ I i 0 nlI , 12
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1
2
4
3
Figure 5.12 Calculation of a solenoid magnetic field strength.
where l is the length of segment 1–2, n is a number of turns over a unit solenoid length, I is the current in the solenoid. Therefore, B 0 nI ,
(5.1.23)
i.e., the induction inside an infinitely long solenoid is proportional to the overall current running onto the length unit. The magnetic field inside the solenoid is uniform. In this respect the solenoid plays the same role as a plate condenser plays in electrostatics. 5.1.4
Action of the magnetic field on the current, on the moving charge
Comparing the expression of the Ampere law (5.1.11) with the expression for a field created by an infinite conductor (5.1.19), one can imagine that current I1 (Figure 5.4) is creating the magnetic field B1 which in turn acts on a neighboring conductor with current I2; then f12 B1I2. Accordingly, f21 B2I1. If the conductor is not rectilinear, we should consider a conductor element dl and then f(dF/dl)IB or, finally, in the vector form d F I [ dl B ].
(5.1.24)
Figure 5.13 shows the arrangement of the vectors describing the field B, the conductor element dl and force acting on the conductor dF. Since the differential expression (5.1.24) is obtained starting from the Ampere law (5.1.11) the force dF is accordingly referred to as Ampere force. Another manifestation of the Ampere force is the action of a magnetic field on a moving charge. Moreover, from the point of view of electromagnetic dynamics all macroscopic (ponderomotive) forces can be reduced finally to the forces applied to the electric charges included in this body. We shall reduce force FA, which operates on the whole conductor, to a force that
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I dl B dF
Figure 5.13 Vector disposition according an Ampere law in the differential form.
operates on each single charge moving inside the conductor. For this purpose allocate the elementary cylinder of a volume dV dlS inside a conductor so that its generatrix is parallel to the direction of the carriers’ motion (Figure 5.14). According to eq. (5.1.24) force dFA, acting on this cylinder, is equal to dFA I [dl.B] jS [dl.B]. On the other hand, the current density j can be submitted according to eq. (5.1.4). Taking into account that j is codirected with u and, accordingly, with dl, the force dFA can be rewritten as dFAqnSdl[ B]qdN[ B], where dN is the number of electric carriers in the allocated cylinder. If the overall force is divided by the number of carriers (dFA/dN), force acting on a single charge is obtained Flor
d FA dN
q[ B].
(5.1.25)
The force acting on a single charge moving with a speed in a magnetic field B is referred to as a Lorentz force. The sign (direction) of the force depends on the sign of the moving charge, i.e., from sign q in eq. (5.1.25). From the expression defined for the Lorentz force Flor it can be seen that it is always perpendicular to the particle velocity. Therefore, a Lorentz force does not produce work. It follows that it is impossible to accelerate the particles by means of a Lorentz force, i.e., an electric field is required to do so. (Nevertheless, the Lorentz force is used in accelerators to make a motion cyclic.) The Lorentz force defines the movement of charged particles in a magnetic field. If the particles enter the magnetic field in a plane perpendicular to the induction B, the Lorentz force will act perpendicular to both vectors and B. In the absence of any other force, the Lorentz force is centripetal, and a circular movement will occur (Figure 5.15). Write the equation of Newton’s second law for this case qBmanm(2/R). The radius of a circle R can be derived from this expression R
1 , B ( q m )
(5.1.26)
where q/m is the specific particle charge. The period T of the circular particle motion turns out to be
T
2 R 1 2
. B(q m)
The period does not depend either on the particle radius R or on the speed.
(5.1.27)
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dl
dF
Figure 5.14 To a Lorentz force derivation.
+
+ Fπop
B +
+
Figure 5.15 Movement of a charge in a plane perpendicular to a magnetic field induction.
If the particle velocity is directed at an angle to a vector B (Figure 5.16, where B is directed along the z-axis), vector has to be projected in two directions: perpendicular ⊥ and parallel
to the induction vector B. Accordingly, the component ⊥ defines the circular motion of the particle and another component
determines its uniform motion along axis z since the Lorentz force for this component is zero. This results in the particle’s spiral movement. The basic spiral increment h is defined as h ||T 2
cos , B(q m)
(5.1.28)
whereas the circus radius R depending upon the perpendicular constituent ⬜ sin is R
sin . B(q m)
(5.1.29)
Therefore, the particle entering the magnetic field moves with a winding-up movement on the magnetic force lines. The charged particle movement described above permitted the development of a powerful instrument – the mass spectrometer: a device for “sorting” ions on their specific charge q/m. Such an opportunity is extremely tempting for modern chemistry for the analysis and synthesis of new substances and for many other problems. The basic scheme of a mass spectrometer is shown in Figure 5.17. The gas to be analyzed enters a vacuum chamber at a point S and is ionized (by any method, for example, by an electron beam impact). Between points S and A the potential difference is applied and ions are accelerated by an electric field. Passing an aperture (at point A) all ions possess identical energy, but not speed. To select ions with identical speed from a beam a speed filter is used in which both forces, Coulomb’ (Fcou qE) and Lorentz’ (Flor qB1), operate perpendicular to
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z
h
B
Fxop
r
Figure 5.16 Movement of a charge at an arbitrary orientation of a field induction B and a charge velocity vector. C
S
υ
A
+ B1 × + E −
∅
∅
qυB1 + + × υ × × υ ×
−
× −
×
qE
×
×
× R2 A1 R1 × × K
× × B2
× υ ×
×
×
×
×
Figure 5.17 Scheme of a mass spectrometer.
each other; the filter allows the singly ionized particles with speed E/B1 to pass through. This filtered beam goes to another point A1 in the chamber where the ions start to move in a circular trajectory with a radius depending on the q/m ratio. In a collector (or a film, or special detector device) located on line A1C ions with different specific charges q/m fall in different points. In Figure 5.18 the mass spectrum of the air is shown, where along abscissa axis values A/q (A is a mass number, q is a particle charge) are plotted, whereas along the ordinate a relative number of the given molecules in the object under investigation are shown. The construction of the modern mass spectrometer differs significantly from the one described above, although it is based on the same principle.
EXAMPLE E5.6 An electron is projected into a uniform field of induction B (B 30 mT) with its velocity vector ( 2 106 m/sec) making an angle of 30o with B. It begins
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x3
324
A+
(N216)+ x3
x100
(N14N15)+
x30
(arbitrary units)
x300
N (OTH. eΠ)
x10
O2+
44
40
x3
OH+
H2O+
34 32
29 28
20
x3
A2+
x3
CO2
x3
x3
(O16O18)+
+
N+ (N22+)
O+ (O22+)
18 17 16
14
A/q
Figure 5.18 Air mass spectrum (every peak relates to a definite ion; numbers over intense peaks show how many times its height was curtailed in order to present all peaks in one spectrogram).
to move along the helix (refer to Figure 5.16). Find the radius R of the helix and its pitch h. Solution: A Lorentz force is acting on the electron changing the direction of travel of the electron. Let us divide electron’s velocity into two components || and ⬜ relative to vector B: ⱍⱍ sin and ⬜ cos . The helix radius can be found from the second Newton law: F man, therefore F 冟e冟 ⬜B and an ⬜2/R. Then 冟e冟⬜B m(⬜2/R) and further R(m⬜/冟e冟B)(m sin /冏e冏B). Substituting the given values we obtain R 0.19 mm. The helix pitch is h ⱍⱍ T, where T 2 R/⬜. Substituting this value into an expression for h we obtain ⎛ 2 R|| ⎞ ⎛ 2 R cos ⎞ h⎜ or h ⎜ 2 R tan . ⎟ ⎝ sin ⎟⎠ ⎝ ⬜ ⎠ Calculation shows that the pitch is h 2.06 mm.
EXAMPLE E5.7 An electron enters a uniform magnetic field of induction B 0.03 T and begins to move along a circle of radius R 10 cm. Determine the electron’s speed . Solution: The second Newton law can be applied to the movement of the electron along a circle (m2/r)冏e冏B; its momentum can be found from the expression p m 冏e冏Br. However, relativistic laws should be used in this case (as we will
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see in the end). In this case p(m0c /兹1 苶 苶 苶2苶) (the electron’s velocity is included in the value ). Solve the last expression relative to :
p
m0 c
2
⎛ p ⎞ 1 ⎜ ⎝ m0 c ⎟⎠
2
冷 e 冨 Br m0 c ⎛ 冷 eBr 冨 ⎞ 1 ⎜ ⎝ m c ⎟⎠
2
.
0
Our solution can be simplified; calculating separately the value appears equal to (冏e冏Br/m0c) 1.76. Therefore, the value can be calculated ( 0.871) and then the velocity found c 2.61 108 m/sec. The electron moving with such a velocity is the relativistic one. EXAMPLE E5.8 An alpha particle runs through the accelerating potential difference 104 V and projects into two fields crossing at right angles – magnetic B 0.1 T and electrostatic E 10 kV/m. Find the charge/mass ratio of this particle if, when traveling in these fields perpendicular to both the fields, it does not diverge from a rectangular motion. Solution: In order to find the ratio it is useful to apply the relationship between the electrostatic forces work qU and the change of its kinetic energy m2/2. From this equality it follows that (q/m)(2/2)*. The special arrangement of the Coulomb and Lorentz fields provides the equality of their action and straight-line motion (Figure E5.8). Therefore, qE qB. This equation permits us to find particle speed as E/B. Inserting this equality into the * fraction, we arrive at (q/m)(E2/2B2). Executing calculations, we obtain (q/m)4.81107 C/kg. Let us check the dimension of the result: (E2/B2)((1B/m)2/1BT2) (1B A2/1B N2)1J C/(1N sec)21C m/(1N sec)21C/kg. Dimension is just the specific charge. z B
Flor
E O
υ x
Fc
y
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EXAMPLE E5.9 In one plane with an infinite direct wire is a frame with sizes as shown in Figure E5.9. Along the wire flows a current I 50 A. Find the magnetic induction flux d through the frame in square S al . x
dx
I
l + B a
a
Solution: In full analogy with the electrostatic strength flux the induction of the magnetic field is d B dS. In our problem induction of the magnetic field depends on the distance from the wire x, therefore the flux is
冕
2a
d B(x)dx. Since B(x)( 0 I/2x) the flux is equal to ( 0lI/2)ln 2. Executing a
the calculation we arrive at 4.5 Wb.
EXAMPLE E5.10 An infinite wire is bent as is shown in Figure E5.10. The circle radius is R 10 cm and the current flowing along it is I 80 A. Determine the magnetic induction B at a point O. 3
O
1
2
π
x B = B1 + B2 + B3
2
l
1
Solution: Divide the current wire into three pieces: 1, 2 and 3. The magnetic field induction is the vector sum B B1 B2 B3. The first segment does not produce a magnetic field at point O since dB 0 ([dl r] 0), according to Biot–Savart law along the whole segment piece. Two components remain both giving codirectional
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induction perpendicular to the drawing plane; they can be summed as scalar values. The half-ring induction B2 can be found according to eq. (5.1.17) taking only half of it: B2( 0I/4R). Induction B3 can be found according to eq. (5.1.18) taking into account that 1 /2 and 2 ; therefore B3 ( 0 I / 4 R). Then B B2 B3
5.1.5
0 I 0 I 0 I ( 1). Therefore B 3.31104 T 331 T. 4 R 4 R 4 R
A magnetic dipole moment in a magnetic field
Just as an electrostatic field acts on an electric dipole moment (refer to Section 4.1.5), a magnetic field acts on a magnetic dipole moment. At a significant difference of working forces the results are very similar. Let us first consider a homogeneous magnetic field; we accept a magnetic dipole moment as a rectangular hoop (Figure 5.19). If the contour is oriented so that vector B is parallel to its plane, the sides having length b will not fill any action of the Ampere force because the vector product in eq. (5.1.24) is zero. The forces acting on the side a are F IaB (refer to eq. (5.1.24)), the force couple renders the torque rotating moment of the contour M F Fb IaBb ISB M B or in vector form: M F [M B ].
(5.1.30)
This expression corresponds to eq. (4.1.32). The torque MF aspires to turn the contour so that the magnetic moment M as a vector turns along the field. On two sides having length b the Ampere forces act oppositely and will stretch (or compress) the contour but not rotate it. It is also possible to show that formula (5.1.30) is valid for a contour of any form and, hence, can be used regardless of the form of the magnetic moment M. The potential energy of a magnetic moment in a magnetic field can be calculated according to the recipe given in Section 1.4.5. Taking into account that in this case MF is the moment of external forces, we obtain dU M F d M B sin d
MF a
B
F n
F b
M
Figure 5.19 A frame with a current in a uniform magnetic field.
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5. Magnetics
and after integration U ∫ M F d M B cos C. Accept C to be zero at 0. Thus, finally in vector form U (M B ).
(5.1.31)
The graph is just the same as for an electric analog (Figure 4.18). In a nonuniform field besides the torque, a force F will act on the magnetic dipole moment (providing the moment is turned into a stable position with 0 and cos 1). To calculate the force we can take the advantage of eq. (1.4.32) describing it as dependent on potential energy. Then (for a one-dimensional case, B B(x)) Fx
U dB M cos . x dx
(5.1.32)
It follows from this equation that the magnetic moment M can be dragged in or pushed out of the magnetic field depending on the angle (cos 1) and the sign of magnetic field gradient. 5.1.6.
Electromagnetic induction
The concept of a vector flux d through a surface dS has been given in Sections (2.8.3) and (4.1.3). Being a particular case of a more general theory of a vector field, the same concepts can be applied to a magnetic field as well. An elementary flux d of a magnetic field induction vector B through the surface dS is equal to a scalar product of B and dS d (B dS).
(5.1.33)
Depending on the angle between a normal n to the surface dS (Figure 2.20) and the induction vector B, a flux d can vary in limits BdS. In general, flux through surface S is defined by the integration ∫ (B dS ) ∫ Bn dS. S
(5.1.34)
S
In 1812 the English physicist Michael Faraday made a discovery, which has significantly influenced the development of all mankind. Having made a real conducting contour, confined to a surface S, he established that by changing the flux (5.1.34) an electric current appeared in the circuit. By numerous experiments, Faraday established that current value does not depend on the way the flux changes but on the speed of this change. The mathematical form of Faraday’s law is extraordinary simple: i
d . dt
(5.1.35)
The minus sign in this expression corresponds to the general physical law of inertia: the induction current in a contour is always directed in a way that opposes the reason of its appearance. This statement is referred to as the Lenz rule.
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To exemplify Faraday’s law, let us imagine a simple experiment. We shall create a closed electric contour with motionless rails and a metal axis with wheels moving on them (Figure 5.20). On this “construction,” impose a magnetic field perpendicular to the plane of the image. Move the axis at a uniform speed . At the same time and with the same speed begin to move electric carriers creating a current in the metal axis and in the circuit in general. The force F directed along the axis and equal to q[ . B] will operate on the carriers. The action of this force is equivalent to the action of the electric field E [ . B]. This field is not electrostatic because it has been created in a different way – the movement of charges in a magnetic field. The circulation of the electric field strength E along the contour will produce EMF in the contour (see eq. (5.1.8)): i ∫ E dl ∫ [ B ]dl. L
(5.1.36)
L
Only the movable part of the contour creates EMF, therefore 2
i ∫ E dl ∫ [ B ]dl. L
1
Supposing that both the axis movement and the magnetics field are uniform, we can obtain 2
i B∫ d B. 1
Multiplying and dividing this intermediate expression by dt we derive
i
B dt . dt
1 i
E' B +
υ
l
dl 2
Figure 5.20 A modeling of a Faraday magnetic induction law.
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Taking into account that ldt dS and BdS d , we obtain i(d /dt), which coincides with eq. (5.1.35). The sign appeared in this expression after analysis of the vector disposition and accounting for the negative sign of electron. The integral 养LE dl is not zero since E is not an electrostatic field. It represents quite different, solenoidal (or vertex or curl) electric field. The acting force F is also nonCoulomb in origin; in contrast, it can be related to the extraneous one. Therefore, we can proceed by adding to the nonzero integral 养LE dl the zero’s addition 养LEdl and write the expression ⴱ i ∫ E d.
(5.1.37)
L
Here, however, E* is the strength of both the conservative and (potential and nonpotential) electric fields. Furthermore, if the left-hand part of eq. (5.1.35) is written in the form (5.1.37) and using eq. (5.1.34), we can arrive at an important expression which is referred to as a Maxwellian equation ⎛ B ⎞
d
∫ E d dt ∫ Bn dS ∫ ⎜⎝ t ⎟⎠ n dS. ⴱ
L
S
(5.1.38)
S
At the left-hand side the integral is taken on any contour L, whereas the surface S is fixed by the already chosen contour L: the surface S rests on the contour L (see Figure 5.11). From this Maxwellian equation it follows that any change of a magnetic field (the righthand side of the equation) generates an electric field (the left-hand side of the equation). If a conducting wire is drawn along the contour L an electric current would occur in it. If contour L is in vacuum, along it the electric field would be excited. Once again we would like to emphasize that an induction electric field is not electrostatic. In fact, the source of an electrostatic field is motionless electric charge whereas in producing a nonpotential electric field the source of the field is an alternating magnetic field. This field, induction by origin, is solenoidal (i.e., is not potential) and certainly possesses other than electrostatic field properties. Faraday’s law is one of the general laws of electrodynamics. In an alternating magnetic field, induction leads to the excitement of EMF. It defines the mutual induction of one conductor onto another. However, even if there is only one conductor with an alternative current a force appears that renders back the current value state. A magnetic flux that penetrates its own contour with current I and generates current variation, is referred to as its own, intrinsic magnetic flux and is designated as S. This flux is not influenced by a change of counter orientation; it is firmly connected with a contour. Since, according to Biot–Savart law, B is proportional to I, therefore it is also proportional to c LI .
(5.1.39)
The coefficient L is called the self-inductance of the counter. It describes the relation between an alternating current and the intrinsic magnetic field produced by it.
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If the counter consisted of N windings and the magnetic flux penetrated all of them without omission, then the total magnetic flux linkage is equal to N c or LIN L I.
(5.1.40)
Excitation of an induction in a closed counter when a current change is taking place is referred to as self-induction. It is equal to the speed of flux linkage change taken with the sign minus: S
d dt
or for solenoids without a ferromagnetic core: S L
dI . dt
(5.1.41)
In order to calculate the inductance of a long solenoid, we can use an expression for the total magnetic flux (Ampere law) LI and N c nlBS (where n is the number of solenoid windings on a unit length and l is the total solenoid length; the stroke at L is omitted). Since the solenoid magnetic induction is B 0nI (see eq. (5.1.22)), hence 0 n2 lSI . Therefore, L 0 n 2V .
(5.1.42)
It can be seen that the solenoid inductance depends quadratically upon the number of windings on a unit length and is proportional to the solenoid volume. In the absence of a ferromagnetic core the inductance is constant and is not dependent on the current strength.
5.2
MAGNETIC PROPERTIES OF CHEMICAL SUBSTANCES
From the point of view of their reaction to an external magnetic field, all substances are referred to as magnetic. Its chemical structure defines the magnetic properties of a substance. All magnetic materials can be divided mainly into three main classes: diamagnetic, paramagnetic and magnetically ordered substances. Diamagnetics are pushed out of an external nonuniform magnetic field; they consist of those molecules that do not possess their own (i.e., “intrinsic”) magnetic dipole moments. Paramagnetics are drawn into the external nonuniform magnetic field; they consist of molecules possessing inherent magnetic dipole moments in the absence of an internal magnetic field. Among the magnetically ordered substances are ferromagnetics and ferrimagnetics, highly reacting on an external magnetic field. There is also a class of antiferromagnetic substances, weakly reacting on an external field. Further we shall describe in more detail the nature of all these substances and consider those characteristics that describe macro- and microproperties of magnetics.
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Moreover, we shall follow approximately the same logic we used for the description of dielectrics. In particular, we should establish where the magnetic properties originated from. It is logical to connect magnetic properties with magnetic dipole moments of atoms and molecules. 5.2.1
Atomic magnetism
In semiclassical Bohr theory an atom is represented as consisting of a nucleus and electrons traveling along circular stationary orbits. This motion can be characterized by an orbital angular momentum: the orbital angular (mechanical) moment Ll (see Section 1.3.9). An intrinsic electron angular momentum is also considered; sometimes it is described by electron rotation around its own axis (see Chapter 6.7 and Section 7.5.5); it is referred to as electron spin with its own angular momentum Ls. Orbital and spin states are sometimes imagined as circular electric currents. These currents create orbital and spin magnetic dipole moments. More advanced notions have been developed in quantum mechanics, but these semiclassical representations are very distinct and useful at this point. There are three sources of the magnetic properties of substances: (1) electron spin; (2) orbital electron motion; (3) change of the electron orbital angular momentum at the imposition of an external magnetic field. The first two can explain paramagnetism, and the third can be used in considering diamagnetism. A nuclear magnetic moment is very weak in comparison with orbital and spin electron magnetic moments; thus it can be temporarily neglected here. However, nuclear magnetism will be closely considered in Chapter 8 because the nuclei participate strongly in resonant methods of investigations in chemistry. We shall take advantage of the semiclassical Bohr theory as it provides an elementary model for understanding the physical essence of the phenomenon. We will begin by calculating a inasmuch as the Bohr theory permits us to do it very easily. The gyromagnetic ratio is the ratio of the magnetic and mechanical moments is referred to as giromagnetic ratio (Figure 5.21). Remember that the angular momentum L of an MP (an electron, in our case) relative to an origin (nucleus) is defined as a vector product L [rp], where r is the electron radius vector relative to the nucleus, and p the linear electron momentum (p m). On traveling along a circular orbit the linear velocity is perpendicular to r ( ⬜ r), then Ll ⱍLlⱍ mr. The direction of the vector Ll is defined by the rule of vector product (the right-hand screw rule).
Ll
I
e
− +
r M
Figure 5.21 A gyromagnetic ratio for orbital electron movement in an atom.
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The circular current caused by the electron traveling along the orbit produces an orbital magnetic moment Ml, the direction of which is also defined by the right-hand screw rule (Figure 5.21). The module of the orbital magnetic moment is Ml IS (ⱍeⱍ r2/T ), where T is the period of revolution. Thus, the gyromagnetic ratio is Ml 冷 e 冨 r 2 Ll T m yr
Canceling r and taking into account that (2 r/T ) we arrive at Ml 1 冷 e 冨 . 2 Ll m
(5.2.1)
Ml 冷e冨 , g 12 Ll m
(5.2.2)
Eq. (5.2.1) can be rewritten as
where g is the gyromagnetic ratio in the unit 12 (ⱍeⱍ/m). In this unit the orbital gyromagnetic ratio is equal to 1 (gorb 1). In their outstanding experiments, Einstein and de Haas showed that for spin the gyromagnetic ratio is equal to 2 (gsp 2) and consequently (Ms /Ls)2 12 (ⱍeⱍ/m). This gyromagnetic ratio anomaly is a source of some of the most interesting and important phenomena and is used, in particular, in many physical methods of chemical substances’ research. 5.2.2
Macroscopic properties of magnetics
The magnetization of a substance is quantitatively characterized by magnetization (, which is numerically equal to the magnetic moment of a volume unit (
∑ Mi . V
(5.2.3)
Alongside with magnetization of a unit volume, a specific and mole magnetization are considered as well. The specific magnetization (magnetization of a mass unit) is equal to: ( sp
1 ∑ Mi , m
(5.2.4)
where m is a mass of the physically infinitesimal volume V (refer to Section 4.2.1). Having replaced m on V where is a substance density we shall obtain ( sp 1 (.
(5.2.5)
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Mole magnetization (magnetization of one mole) is (M
1 N ∑ Mi , v i1
(5.2.6)
where is the number of moles in a physically infinitesimal volume m/M, M is a molar mass. Alongside with a magnetic induction B one more value characterizes the magnetic field H, a strength of the magnetic field, is used; in isotropic magnetic the magnetization ( is proportional to magnetic field strength H, that is ( H
(5.2.7)
where is the scalar value referred to as a magnetic susceptibility. The magnetic susceptibility characterizes ability of substance to be magnetized in a magnetic field. As ( and H have identical dimension is dimensionless value. For diamagnetic materials 0, its value is ⬃105107 , for paramagnetic materials 0, its value is ⬃103106; for ferromagnetic ⬃ 103105. One can see that diamagnetic weakly and oppositely magnetized in an inner magnetic field and therefore pushed out from it. Paramagnetic weakly magnetized too but positively. Ferromagnetic magnetized very strong and intensively drawn in magnetic field. Alongside with magnetic susceptibility of the unit volume a specific magnetic susceptibility sp is often used in practice sp((sp/H ) the relation being exist sp(1/). The same is for molar magnetic susceptibility M which is equal to M
(M H
M
M .
(5.2.8)
and
5.2.3
An internal magnetic field in magnetics
One more magnetic characteristic, a magnetic permeability , is usually introduced; it shows how much the magnetic induction in a magnetic B is larger then that of the external magnetic field B0: B B0 .
(5.2.9)
This means that this value should be substituted into the Bio-Savare law (5.1.16) dB =
0 I [dl r ] 4 r 3
(5.2.10)
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B0 l (
Figure 5.22 A magnetic in an external magnetic field; a circular micro-molecular current and surface currents can be seen.
Consider now a sample of a cylinder form placed in an external magnetic field with induction B0. Let the cylinder by a cross section S and length l is oriented along the external field force lines (Figure 5.22). Under an action of a field all of molecular currents (vectors of the magnetic dipole moments) will be ordered in the field (along or opposite, it does not matter at the moment). At averaging, inside of the magnetic cylinder molecular currents will be mutually canceled. Not compensated micro-currents will be left on the cylinder’s surface. The picture remained a solenoid and can be considered as a certain total macroscopic current, flowing the cylinder over (with a magnetization current Jm). We can introduce a value of current linear density ℘ being equal to ℘Jm / l. The situation is really just like the solenoid with the magnetic field induction B1 inside. The field inside the solenoid is defined by eqn (5.1.23). Considering the magnetized cylinder as the solenoid we can calculated the induction B1 superimposed on the external field B1 0 nI 0℘
(5.2.11)
Find the relation of the magnetic magnetization ( and the surface density of the current ℘. According to eqn (5.2.3), the magnetization is the magnetic moment of the unit volume. Therefore for this case ( = ∑ Mi V IS (Sl ) ℘,
(5.2.12)
i.e. the magnetization of a piece of the magnetic is numerically equal to the linear density of a surface current. The expressions obtained allow one to find a ratio outside and inside of the magnetic and to establish the ratio between a magnetic susceptibility and magnetic permeability. The macroscopic field in a substance is characterized by a magnetic induction which is the geometrical sum of magnetic inductions of the external B0 and internal B1 fields, i.e. B B0 B1 .
(5.2.13)
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Having replaced in this expression B0 and B1 according to (5.2.9) and (5.2.11), and taking also into account a collinear arrangement of all three vectors, we obtain B = 0 H 0℘ 0 H 0 (.
(5.2.14)
It shows that the strength of the magnetic field inside of the magnetic differs from that of an external field on the value of magnetization. Writing further the magnetization according to (5.2.7) one can obtain B = 0 H 0 H 0 (1 ) H .
(5.2.15)
On the other hand, according to (5.2.9), B = B0. Therefore 1
(5.2.16)
and the induction inside of isotropic magnetic is B 0 H.
(5.2.17)
This expression corresponds to the definition of the magnetic susceptibility (5.2.9). It follows that H
B ( 0
Let’s note here a distinction of the induction and the strength notions. The induction of a magnetic field according to (5.2.9) depend on substance property, however the strength of the field outside and that of inside of the magnetic is the same. Moreover, the strength doesn’t depend at all on the sample magnetic properties (i.e. on ) (H B/ 0 0 H/ 0 H). At the same time the induction varies at transmitting from one magnetic to another. Therefore at calculations of magnetic circuits to use the strength H is more convenient.
5.2.4
Microscopic mechanism of magnetization
It has already been mentioned that from the point of view of their magnetic properties, we can distinguish three main classes of substance: diamagnetic, paramagnetic and magnetically ordered substances. We shall now consider the same question from a microscopic point of view, i.e., which processes cause magnetic properties and how these properties are related to their chemical structure. We shall start with a diamagnetic.
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Diamagnetics magnetize opposite to an external magnetic field and are pushed out of it. The magnetic susceptibility of a diamagnetic is thus negative and depends neither on temperature, nor on the strength of the magnetic field. Diamagnetic properties are defined by the electron atomic orbit. It is easier to begin with one-electron atom. In a magnetic field the electron orbit precesses in the same way as a spinning top in a gravitational field (refer to Appendix 2). This precession arises because an atom possesses both magnetic and angular (mechanic) momentums (refer to 1.3.57 and Figure 1.19). Find the angular velocity of orbit precession. Let the atom possess an angular momentum L and magnetic moment , directed opposite to each other (see Figure 5.23). In an external magnetic field B, excited along an axis z, a torque MF will operate MF [B], directed perpendicular to vectors and B (Figure 5.24). Under the action of this torque, vector L in time dt will acquire an increment dL MF dt and, accordingly, L (t dt) L(t) dL. The vector dL is perpendicular to vector L and therefore ⱍL ⱍⱍ ⱍLⱍⱍ. Thus, the action of torque MF changes the direction of vector L, but not its length. Thereof, the plane in which the axis z and vector L lie, will turn by an angle d
d
dL Mdt . L sin L sin
Since M |MF| MB sin , hence
d
M B sin M dt Bdt. L sin L
z B
d L sin
L dL
L'
Figure 5.23 An electron orbit precession.
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B
M
−e
r′ d dL
−e
r′
I′ s=πr′2
Figure 5.24 Origin of a diamagnetic moment.
Dividing both sides by dt, we obtain the angular velocity of the electron orbit precession L
d M B dt L
Using the gyromagnetic ratio the orbit angular velocity can be found:
L g
1 冷e冨 B 2 m
(5.2.18)
This expression can be applied both to orbit and spin precession taking the g-factor into account. The angular velocity is named after J. Larmor and the magnetic precession is also referred to as Larmor precession. In particular, for orbit precession, the angular velocity is 1冷e冨 L B. (5.2.19) 2 m Additional electron movement caused by orbit precession leads to the excitation of an equivalent circular current I (Figure 5.24). This current induces the magnetic moment M, which is the diamagnetic moment. Irrespective of the direction of the
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torque vector MF or of the direction of induction of an external magnetic field B, the induced diamagnetic moment M is always directed against B. This is the origin of the diamagnetic effect. Therefore, the minus sign is present in the expressions for the diamagnetic moment. The additional electron movement occurs along a circle smaller than r (the electron radius designated in Figure 5.24 as r ). A circular electric current corresponds to this movement
I
e L e2 B. 2
4 m
Strictly speaking, the radius (r ) is not the radius of an electron orbit; besides, it depends on the inclination angle . As an electron orbit can be inclined to the induction direction (to an external field), averaging over all these parameters allows us to obtain 冬(r )2 冭 2 冬(r )2 冭, 3 where 具(r)2典 is the average value of the square of the electron distance from the z-axis. Having substituted this value in the expression by M, we shall obtain for one-electron atom
M
e2 2 冬r 冭 B. 6m
(5.2.20)
Summing up the expression obtained for all electrons in a multielectron atom, we shall find its induced moment as M
e2 kZ 2 B ∑ 冬rk 冭, 6 m k1
(5.2.21)
where Z is the number of electrons in an atom. If we now increase this value by Avogadro number NA we shall obtain the value of the mole magnetizations
( MN A
N A e2 kZ 2 B ∑ 冬rk 冭. 6m k1
Multiplying numerator and denominator by 0 and comparing the expression obtained with eq. (5.2.7), we shall find the molar magnetic susceptibility of multielectron atoms
M 0
N A e2 kZ 2 B ∑ 冬 rk 冭. 6m k1
(5.2.22)
It can be seen that the more electrons in an atom and the larger the radius of electron orbits, the greater is the diamagnetic susceptibility. Substituting here values of fundamental physical values and accepting the radius of atoms ⬃1010 m, we obtain M ⬃ 107108 m3/mol which corresponds well to experiment for molar susceptibilities of diamagnetics.
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The molar susceptibility of a number of compounds is presented in Table 5.1. Notice that diamagnetism is inherent in all substances without exception. Paramagnetics magnetize in the direction of an external magnetic field and are drawn into it. The magnetic susceptibility of a paramagnetic is positive, depending on the temperature and value of the magnetic field strength. Paramagnetism is inherent in substances whose molecules possess permanent magnetic moments regardless of magnetic field. We shall connect all these experimental facts with the microscopic properties of paramagnetics. In the absence of an external magnetic field, the magnetic moments of paramagnetic molecules are disordered in space by the action of chaotic thermal movement (Figure 5.25a). This means that the vector sum for magnetization (5.2.3) is zero; hence, the magnetization ( is also zero. When a paramagnetic substance is brought into the magnetic field, each magnetic moment aspires to be guided in the field’s direction; however, the molecule’s thermal movement prevents it from doing so. A balance is established (Figure 5.25b); as a result the vector sum in eq. (5.2.3) becomes distinct from zero, the substance is magnetized. A factor which should be taken into account when describing the competition of the ordering action of a magnetic field (whose energy is U( )) and the disordering tendency due to chaotic movement (with averaged energy T ) is the Boltzmann factor: exp(U( )/T). It is necessary to take all these circumstances into account when considering the magnetization process.
Table 5.1 Values of molar susceptibilities of some diamagnetic compounds Substance
M, 1011 m3/mol
Substance
Helium (He) Neon (Ne) Argon (Ar) Krypton Xenon (Xe)
2.4 8.2 25 40 84
Silver (Ag) Bismuth (Bi) Glass (SiO2) Methane (CH4) Naphthalene (C10H8)
(a)
M, 1011m3/mol 27 350 50 76 240 (perpendicular to the molecule plane)
(b)
Figure 5.25 A paramagnetic (a) outside a magnetic field (Mi 0) and (b) inside a magnetic field (Mi 0).
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An analysis of all these processes was carried out by the French physicist P. Langevin. He considered the behavior of a system of magnetic dipole moments in an external magnetic field. (An identical problem exists in dielectric physics when describing the orientation polarization, see Section 4.2.4.) The essence of both phenomena consists of the competition between the two processes: the aspiration of a field to direct the dipole moments along the field direction and the action of chaotic thermal movement interfering with it. At each temperature a compromise is achieved. Finding this compromise is the essence of the Langevin theorem. (The Langevin theorem is presented in full in Appendix 4.) The result of this theorem is an expression for mole magnetic susceptibility of a paramagnetic substance. At ( B/T) 1 the mole paramagnetic susceptibility has been found to be
M
0 N A 2 3 T
(5.2.23)
(see eq. (A4.10) in Appendix 4). It can be seen from this formula that M is inversely proportional to temperature: M
C T
(5.2.24)
This dependence is referred to as Curie’s law. Comparing eqs. (5.2.23) and (5.2.24), we can find a constant C, which is also named after Curie: C=
0 NA M2 . 3
(5.2.25)
Substituting into formula (5.2.23) the values of fundamental physical constants and the value of the spin magnetic moment, we can obtain M ⬇ 108 m3/mol, which is compatible with most experimental data (see Table 5.2 below). In very highly magnetic fields and/or at very low temperatures ( B T ) the field can orient all the magnetic moments of all molecules in parallel; this results in saturation, i.e., further increase of the field intensity cannot appreciably change the magnetization of the sample since all moments are already parallel along the field ( M N A M.
(5.2.26)
Table 5.2 The molar magnetic susceptibility of some paramagnetics Substance
M, 1011 m3/mol
Substance
M, 1011 m3/mol
Sodium (11Na) Aluminum (13Al) Lithium (3Li) Vanadium (23V) Oxygen (O2)
2.0 2.1 3.1 37 430
MnSO4 Fe2O3 NiSO4 FeCl2 Dysprosium (66Dy)
1.7 4.8 5.0 16 150
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In an intermediate area where B⬇T one dependence (m(H) smoothly passes to another (Figure 5.26). The curve repeats for orientation polarization (Figure 4.28). Consider now a number of examples that will allow us to see how the structure of chemical substances influences their magnetic properties. In the creation of diamagnetic properties the main role is played by the electron orbit. This means that diamagnetism is inherent to all substances without any exception. All atoms possess diamagnetic properties but they cannot always be measured because, as a rule, diamagnetism is masked by a larger paramagnetic effect. Accordingly, diamagnetism becomes apparent in substances consisting of atoms with completely compensated magnetic moments. This takes place in noble gases (He, Ne, Ar, Xe and Rn). As an example consider a neon atom: it has 10 electrons in its electron shell (1s22s22p6). Figure 5.27 shows the electron distribution among quantum cells. Notice that for s-electrons the orbital magnetic moment is zero (l 0). The total magnetic moment of s-shells is also zero because all quantum cell are occupied by pairs; the p-shell is also filled completely. So the total magnetic moment of the neon atom is zero. In this case diamagnetic properties can be exhibited and measured. Ions Na and Cl whose electron configurations coincide with Ne and Ar will also be diamagnetic. On the other hand, the neutral atoms Na and Cl possess magnetic moments as there is one noncoupled 3s electron in the Na atom and a 3p electron in the Cl electron shell. While forming a chemical compound, the Na atom’s 3s electron passes to the Cl atom and a NaCl molecule with ionic bond is formed. Since Na and Cl magnetic moments do not possess noncoupled spins, the molecule NaCl is diamagnetic. Note that, in general, the majority of chemical compounds are diamagnetic. In particular, this is true for the ionic compounds of the type considered and covalence compounds with nonsaturated bonds. By way of example, we can consider a CCl4 molecule. Upon formation of this molecule, the carbon atom, having two noncoupled 2p electrons, is excited (Figure 5.28); 2s electrons are dicoupled and, as a result, sp3-hybridizations are generated; four equivalent sp3-hybrid orbits arise, with two electrons on each bond. These orbits form chemical bonds with the Cl atom. Thus, in the molecule CCl4 there are no free noncoupled electrons and consequently it is diamagnetic. mB0>>T
( (sat mB0 n2
Solution: Let us allocate a narrow light beam from the light falling on the film. The path of this beam in the case when the angle of incidence is is shown in Figure E6.2. At point A, the beam is in part reflected and refracted. The reflected beams AS1 and ABCS2 fall on a convex lens S1S2 and interfere at point F. As the parameter of refraction of air n1 1 is less than the refraction index of the film, which in turn is less than the glass refraction index, in both rays reflection occurs in total without phase change. Since the light is maximally weakening the optical path length l2n2 – l1n1 (AB BCn2 – ADn1 should be equal to an odd number of halfwavelengths (AB BCn2 – ADn1(2k 1) (/2). If the incidence angle tends to zero, AD → 0 and AB BC → 2d and we obtain 2dn2 (2k 1) (/2). Then the film thickness is d [(2k1)]/4n. Taking k 0, 1, 2, 3, we arrive at a number of possible values for d: d0
3 0.11 m, d1 3, d0 0.33 m,etc. 4 n2 4 n2 EXAMPLE E6.3
A monochromatic light of wavelength falls onto a nearly parallel glass wedge (two plate) with a very small wedge angle , normal to its sides. An interference picture appears. It consists of a sequence of light and dark strings (see Figure 6.9). On a length of wedge l 1 cm, 10 strips are observed. Define the refracting angle of the wedge (Figure E6.3).
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dk+10 −dk
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l
dk
k + 10
k
k+9
k+1
2
l
Solution: A parallel beam of light falling normally onto the wedge reflects from both sides of the wedge, upper and lower. Both beams are coherent and practically parallel; though an optical path length difference is created, therefore an interference picture is observed. Dark strips are visible on those sites of the wedge for which is multiple to an odd number of half-wavelengths: (2k 1)(/2) with k 0, 1, 2…. The geometrical difference of paths length is 2dn cos; losses of /2 appeared at the reflection from the upper side of the wedge and should be added. Therefore, for dark strips we have (2k 1)(/2) 2dkn cos (/2), where n (n 1.5) is refractive index of glass, dk is the thickness of the glass wedge in a point of dark strip. The incident angle is assumed to be zero. Therefore after simplification, we obtain 2dkn k*. Let the thickness for any dark strip be dk and the thickness of glass in the point k 10 is dk 10, the distance between them being l. Then ( expressed in radians)
dk10 dk . ᐉ
Calculating values of d’s from expression * and substituting them into the last expression we arrive at 5/(nl). The angle sought is then 2 104 rad. This angle in degrees is 2 104 2.06 105 41.2 .
6.2 6.2.1
AN INTERFERENCE
Superposition of two colinear light waves of the same frequencies
In Section 2.3.1 a summation of two oscillations of the same frequencies and propagating in the same direction was considered analytically and using a vector diagram. It was also shown in Section 2.8.3 that the oscillation intensity is proportional to the square of amplitude. Apply now the conclusions mentioned to the calculation of the light intensity at waves imposed in any fixed point of space; let it be x0. At the fixed coordinate the equation of a running wave
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E (x0, t) E0 cos ( t–kx0) transforms into the equation of oscillation E(t) E0 cos ( t ), where product kx0 is included as a certain number into the phase . A splintered wavetrain traveled in two parts, 1 and 2; then they superpose (Figure 6.6) enhancement or weakening of the light intensity. The expression for the square of total amplitude (2.3.1) is 2 2 E02 E0,1 E0,2 2 E0,1 E0,2 cos( 1 2 ).
(6.2.1)
Consider a number of cases of the addition of two light waves with equal amplitude (i.e., at E0,1 E0,2) at various phase differences ( 1 2) . If light waves from two different independent sources are summed, the average value of cos is equal to 0. This 2 means that if in expression (6.2.1), E 20 2E1,2 , i.e., the intensity of light will increase twice, there will be an enhancement of intensities. If remains constant the two waves are coherent (refer to Section 2.9.1). At (2m 1) (m is an integer), cos 1 and the resulting light amplitude and intensity in the given point are equal to zero. At 2m, cos 1 and the resulting light amplitude is twice as high as the amplitudes of each of the initial waves, i.e., an enhancement of amplitudes takes place; intensity thus grows four times. The phenomenon of redistribution of intensity of light in space on imposing two or several coherent waves is referred to as the interference. Interference phenomena (including diffraction phenomena) are the direct consequence, and proof of, the wave nature of light. 6.2.2
Interference in thin films
The practical realization of two coherent light sources is very difficult (it can be achieved, for instance, with the use of lasers). However, there is a relatively simple way to carry out an interference. It consists of splitting a single light beam into two components by reflection from a pair of mirrors and then superposing them in a single point; they will interfere, thus P1
n1 Ο
M
n2
P2
Figure 6.6 Reflection and refraction ray paths of two split parts of a single wavetrain in two media: 1 and 2; P1 and P2 are mirrors.
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a splintered wave “interferes with itself.” The basic scheme of such an experiment is submitted in Figure 6.6. At point O, on the border of the two media with the refraction indexes n1 and n2, a wave is splintered in two parts. With two mirrors P1 and P2 both parts go to point M at which they interfere. The speed of propagation of the two beams, due to different media properties, is c/n1 and c/n2. At point M, the two parts of the wave will superpose with each other with constant shift in time equal to S1/1 S2/2, where S1 and S2 are geometrical lengths of the path traveled by the two parts of the wave. The oscillations of electric field strength at point M will be E0,1 cos (t – S1/ 1) and E0,2 cos (t– S2/2). The square of the resulting oscillation amplitude at point M is ⎡⎛ S ⎞ ⎛ S ⎞ ⎤ 2 2 E0 E0,1 E0,2 2 E0,1 E0,2 cos ⎢⎜ 1 ⎟ ⎜ 2 ⎟ ⎥ . ⎣⎝ y1 ⎠ ⎝ y 2 ⎠ ⎦
(6.2.2)
Since 2/T and c/n, the expression in square brackets is equal to (2/cT )(S2n2–S1n1) (2/o)(S2n2–S1n1). The product of the path traveled S and refraction index n is referred to as optical path length denoted by . Keeping in mind that cT 0 (0 being the wavelength in vacuum),
2
. 0
(6.2.3)
This expression joins the phase differences and the optical length traveled in the splintered wave. defines the interference effects. Indeed, cos 1 corresponds to the maximum intensity since (2/0) 2m. From this, the condition of the intensity maximum can be derived:
m
(6.2.4)
The largest diminishing of the light intensity corresponds to cos 1, i.e., (2m 1). Then (2m 1) (2 /0) or
(2m 1)
0
2 .
(6.2.5)
It is easy to see that the summation of waves described above with fourfold enhancement of intensity corresponds to the displacement of the two “parts” of the splintered wave from each other by the difference in lengths equal to the integer wavelengths (or, accordingly, to the phase difference 2m). The complete extinction of the wave’s intensity is observed at the displacement of the two wave parts on the wavelength half (on an odd number of the wavelength half, i.e., (2m 1)). Consider as an example the interference of light at the reflection from thin films (or from a thin plane-parallel plate; Figure 6.7). The direction of a beam falling on the film is shown in the figure by an arrow. Splitting of the wavetrains occurs in this case at partial reflection
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of each part of it on the upper (point A) and the lower surfaces (point B) of the film. We shall consider that the light beam goes from air and leaves after a point B into air (with air refraction index equal to unity) whereas the parameter of this film material is equal to n. Every wave of the beam falling at an angle at point A is split into two parts: one of them is reflected (beam AD) and the other refracts (beam AB). At point B every wave of the refracted beam is split again: part is reflected from the lower film surface and part refracts leaving the film. At point C the wave is again split, but we will follow that wave part which leaves the film at the same angle as beam AD. The two reflected beams are gathered by a lens (not shown in the picture) at one point. Being parts of the same primary wave the beams are coherent and can participate in the interference, the intensity being dependent on the difference of their optical traveled lengths (or differences in phases). The phase difference in waves 1 and 2 is accumulated in traveling along path lengths AD and ABC. The optical path length is (AB BC)n – AD, where AB BC 2d/cos and AD 2d sin sin /cos . Remembering that sin n sin , then (2dn/cos ) (1 – sin2) or 2dncos . Since angle is usually given in problems but not , it is more convenient to present the value in the form
2 d n2 sin 2 .
(6.2.6)
When defining the conditions of light intensity (maximum and/or minimum), it was necessary to equate the value to the integer or half integer to the number of wavelengths (eqs. (6.2.4) and (6.2.5)). However, as well as estimating the optical path length difference
, it is also necessary to analyze the opportunity of the loss of half a wavelength during reflection. This depends on a specific condition, namely, whether the media from which the reflection occurs is more or less dense. So, if the film with n 1 is surrounded with air with n 1, the loss of half-wavelength occurs at point A (Figure 6.7). If the film is on a surface of a medium whose reflection index is higher than for the film material, the loss
D
A
C
d B
Figure 6.7
Ray paths in a thin film.
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of half a wavelength occurs at two points—A and B; as in this case the whole wavelength is accumulated, this effect should not be taken into account at all. It follows, that specific tasks demand individual consideration. The main principle consists in finding the whole difference of optical lengths , to consider the possible loss of half a wavelength at reflection, if necessary, to add (or subtract, it does not matter) it to (or from) and to bring it into correlation with the conditions presented. In the case of a film in air represented in Figure 6.7, the condition of maximum interference looks like
2 d (n2 sin 2 )1 2
0 m 0 . 2
(6.2.7)
Since the refraction index depends on the wavelength (see Section 6.5), the interference conditions are qualitatively different. Therefore, the film will decompose falling light in a spectrum, i.e., in falling white light the thin film always looks as if it has been painted. We all have met examples of this: observing multicolored soap bubbles or an oil stain on the surface of water. Consider now the example of a thin air wedge. This wedge is opposite to the thin film picture (Figure 6.8). A plate with well-polished surfaces lies on another, similarly perfect plate. At a definite place between the two plates a thin subject (e.g., a thin wire) is introduced, so an air wedge is formed. Consider a beam of light falling normally onto the upper plate. We shall accept that there is no divergence at surface points on reflection and refraction, keeping in mind that the wedge angle is very small. Admit that A is a point on the lower plate where the optical path length between plates is equal to integer m of wavelengths plus /2 (due to reflection from the optically more dense lower plate); two reflected waves are nearly parallel to each other. Suppose that there is the condition of maximum interference intensity at this point. An equation describes this condition (factor 2 appears because the beam runs the distance twice): 1⎞ ⎛ 2 m ⎜ m ⎟ . ⎝ 2 2⎠
(6.2.8)
If we look at this picture from above (for this purpose a simple optical system is required), it is possible to see geometrical strips in which, at certain m, light (or dark)
A
B
Figure 6.8 An air wedge.
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Figure 6.9 Lines of constant thickness.
strips are formed. Along this strip condition (6.2.8) is fulfilled, i.e., along it the backlash of air has the same thickness. Such strips referred to as strips of equal thickness. Provided that the plates are made carefully, the strips of equal thickness are represented by parallel straight lines. If, however, there are defects in the plates, the appearance of the strips changes appreciably, and the position and form of the defects develop clearly. Fringes of equal thickness are shown in Figure 6.9: in an air wedge a narrow stream of warm air is produced, the density of which and, accordingly, the refraction index, differ from the values for cold air. The curvature of the lines of constant thickness is visible. If a convex lens touches a perfect flat plate, at a favorable ratio of the lens curvature radius, light wavelength and the presence of an optical magnifying system, so-called Newton’s rings can be observed. They represent the fringes of equal thickness in the form of concentric circles.
EXAMPLE E6.4 A vivid example of strips of equal thickness is Newton rings. They appear when a lens of a radius of curvature R lying on a carefully processed glass plate is irradiated with monochromatic light in the wavelength . Determine the radius of the mth ring.
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incident beam R reflected rays 1
2
R−dm
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lens
dm glass (a)
(b)
Solution: A geometrical diagram and the interference picture are presented in Figure E6.4a. The condition of interference maxima is 1⎞ ⎛ dm ⎜ m ⎟ . ⎝ 2⎠ 2
The air refraction index is assumed to be unity. In Figure 4a, it can be seen that
dm R R 2 rm2 R R 1
rm2 R2
,
where r is the radius of the mth ring. Taking into account the fact that the radius of curvature R is much larger than the size of the interference picture, we can expand the last equation into binominal series limiting ourselves by two terms ⎡ 1 r2 ⎤ r2 d R R ⎢1 . ⎥ 2 ⎣ 2R ⎦ 2R
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The ring’s radii r can be deduced as 1⎞ ⎛ r ⎜ m ⎟ R ⴱ. ⎝ 2⎠ The character of rings—light or dark—depends on the loss of /2 at the reflection from the more dense matter: in each particular case this needs to be derived separately. In our problem, we use a light falling from above, therefore, there is only one reflection from a denser media (from a glass plate). Therefore, for the light ring 1⎞ ⎛ rm ⎜ m ⎟ R . ⎝ 2⎠ 2 An overall picture of Newton’s rings is depicted in Figure E6.4b. If the light is directed from below (and the results are observed from above), there will be no loss of a half-wavelength because the reflection from the denser media takes place twice: from the lens and from the glass plate).
EXAMPLE E6.5 Find the radii of the second r2 dark, and fifth r5 light Newton rings in a monochromatic light 0 0.56 m provided the lens radius is R 1.2 m. Solution: Using the star equations from the previous example, keeping in mind that the air refraction index is 1 and taking into account that we are first searching for the radius of the dark ring, we can write (2m1)(/2) (rm2 /R)(0 /2), there7 m 1.16 mm. fore, rm m R. Therefore, at m 2, r2 2 1 .2 5 .6 10 For the light ring, m0 (rm2 /R) (0/2) and rm (2 m 1)R ( /2 ). Executing 7 .5 .2 (5 .6 /2 )1 0 1.74 mm. calculations, we obtain r5 (2 1)1 The phenomena of interference find wide application in chemistry and the chemical industry. In particular, they are used in interferometry in defining the refraction indexes of substances in their three states: solid, liquid and gaseous. There is a large number of various interferometers which differ by their assignment. Let us illustrate the determination of the refraction index of substances by a simple interferometer intended for the measurement of the refraction indexes of liquid and gaseous substances (Figure 6.10). Two completely identical thick plane-parallel glass plates A and B are fixed in parallel to each other. The light from source S falls onto the surface of plate A at an angle close to 45°. As a result of its reflection from both sides
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S
l 1 A
K1
K2
B 2 L
P
Figure 6.10 The diagram of a simple interferometer.
of plate A, two parallel beams 1 and 2 are produced. Running through two identical glass cells K1 and K2, these beams fall onto plate B and are again reflected from both its sides and are gathered at a point P by a lens L. At this point, they interfere, and the interference strips are examined with an ocular, which is not shown in the figure. If one of the cells (e.g., K1) is filled by gas with a known absolute refraction index n1 and the second with a substance with a measured refraction index n2, the optical path length difference between plates will be equal to (n1n2) l, where l is the cell length. With the help of a special device, the displacement of the interference strips concerning their position with empty cells can be observed. Displacement is proportional to the difference (n1n2), which allows one to determine one parameter knowing another. We note that while there are rather low requirements as to the accuracy of the measurement of the strips’ position, the relative accuracy in defining the refraction indexes can achieve values of 106–107. This accuracy enables the study of small impurities in gases and liquids, measurement of the different dependences of the refraction indexes on temperature, pressure, humidity, etc. There are still many other designs of interferometer construction, intended for various physical measurements. In particular, using a specially designed interferometer, Michelson and Morley in 1881 established the independence of light speed from the speed of its source (refer to Section 1.6). Einstein took this fact as a principle of his Special Theory of Relativity.
6.3
DIFFRACTION
Diffraction is a set of phenomena arising from the propagation of light in a media with pronounced heterogeneity and consisting of light deviations from the laws of geometrical
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optics. Diffraction leads to light deviating from rectilinear distribution, bending around opaque obstacles and penetrating into an area of geometrical shadow. 6.3.1
Huygens–Fresnel principle: Fresnel zones
Taking into account experiments in which light exhibited its wave nature C. Huygens assumed that each point of a primary light wavefront serves in its turn as a source of secondary spherical wavelets. The new position of the wavefront will be the enveloping surface of these secondary waves (Figure 6.11); in turn, each point of the secondary wavefront is again the source of the next generation of waves and so on. In Figure 6.12, this principle is illustrated with an example where a light wave is passed through an aperture; it can be seen that, due to secondary waves, light can penetrate into the area of geometrical shadow. These phenomena are only exhibited in an appreciable measure
envelop of secondary waves at time instance t+dt
wavefront at instant of time t
r =c∆t wavefront at instant of time t+dt
Figure 6.11 Huygens principle (the wavefront at time instants t and t t is shown, each point is the source of the second waves).
diffracted beams
areas of geometric shadow
Figure 6.12 Diagram of penetration of the diffraction radiation into the area of geometrical shadow.
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when the diameter of the aperture is commensurable with the light wavelength: only in this case do the angles of diffraction appear to be measurable (see below). A.J. Fresnel enlarged this principle by assuming the same laws of interference which had already been developed for primary waves were also applicable to secondary waves. It is rather difficult to calculate the distribution of intensity in a diffraction picture. However, a method allowing an essentially simplified calculation of the diffraction effects, at least at a qualitative level, has been offered: Fresnel has suggested mentally breaking a wave surface into zones, the distance from respective points of which up to the sighting point differs from the previous one by /2. In this case light waves from the adjacent zones are in an antiphase (because of the shift by /2); this leads to the mutual cancellation of such waves; in other words, the adjacent zones extinguish each other. The method has been successfully used to solve different problems of wave optics, in particular, in the explanation of rectilinear distribution of light. We shall take advantage of this principle by considering the diffraction on a slit. Two kinds of diffraction can be distinguished: diffraction in parallel light rays from a plane front wave (referred to as Fraunhofer diffraction), and diffraction in converging beams (Fresnel diffraction). Here we will consider only the Fraunhofer diffractions. The scheme of this diffraction is presented in Figure 6.13: point S marks the light source, a condenser lens K provides a parallel light beam, and lens L with a focal length f concentrates the result of the diffraction at an angle at a screen point P. A central ray O and axis sin along which the figure is expanded are shown. 6.3.2
Diffraction on one rectangular slit
We use the Fresnel zone principle for qualitative consideration of the Fraunhofer diffraction on a single rectangular slit. By definition, each following zone extinguishes the previous one. This means that if, at the slit width d and wavelength at an angle , an even number
S
K
L f
screen P
0
sin
Figure 6.13 Diffraction in parallel rays (Fraunhofer diffraction).
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of Fresnel zones opens, all zones extinguish each other; in this direction the intensity of diffraction becomes equal to zero. On the contrary, if at another angle , an odd number of zones open, there should be a maximum in the spectrum. It can be seen in Figure 6.14 that if the condition is even, the zone number corresponds to the length of a segment MN dsin on which the integer of wavelengths is stacked (the even half-wavelengths). This is an indication of the minimum intensity. Mathematically, it looks like d sin ,
(6.3.1)
where k is an integer which shows the diffraction order. The maximum intensity appears when an odd number of half-wavelengths stack up in the segment MN d sin (2 1) . 2
(6.3.2)
In Figure 6.15, an experimental diffraction spectrum on a single slit is schematically depicted: at 0 in a direct beam the maximum is seen because only one zone is opened
d
N
M
Figure 6.14 Fraunhofer diffraction from a single slit (the angle corresponds to eight Fresnel zones half of them faintly marked in the figure).
I
− 2 d
− d
0
d
2 d
sin
Figure 6.15 Light intensity distribution after single slit diffraction.
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(the whole slit width). On changing in both directions the maximums and minimums alternate in an orderly way. The quantitative solution of this plot shows that ⎡ ⎤ A0 sin ⎢ d sin ⎥ ⎣ ⎦, A ⎡ ⎤ ⎢⎣ d sin ⎥⎦ the square of which gives the so-called interference function I (sin ) A 2 .
(6.3.3)
This function describes the intensity distribution at the diffraction on a single slit (Figure 6.15). At the diffraction on a single slit the intensity of diffraction rapidly decreases with angle. 6.3.3
Diffraction grating
A diffraction grating is composed of a large number of identical, regularly distributed alternating transparent strips on an opaque flat carrier. A constant of the diffraction grating b is the distance between corresponding points of two adjacent strips (Figure 6.16).
N N
M
Figure 6.16 Diffraction from a diffraction grating: e is a grating constant, N is the number of slits, eN is the total width of the grating, is the diffraction angle; the faintly marked zones in Figure 6.14 are now completely nontransparent.
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For simplicity we shall consider that the grating consists of transparent and opaque strips of identical width. We shall designate N as the number of transparent and opaque pairs. Then the general width of the grating will be bN. Compare the condition of minimum intensity for one slit (eq. 6.3.1) with that for diffraction grating. Imagine that we could close every second Fresnel zone at the diffraction on a single slit. Therefore all open zones, having no “antagonists,” make a full contribution to the diffraction spectrum. Figure 6.16 presents a scheme to illustrate this idea. What was the slit width d, is now bN, the minimum condition (6.3.1) transforms into the maximum condition eN sin or esin k/N , where k and N are integers. There can be two cases, the most important is when is divisible by N, i.e., when (/N) m, where m is a simple integer. In this case, a so-called main maximum of order m is obtained; it corresponds to diffraction maximum when all transparent slits are “in a phase.” The main maximum condition can be written as: b sin m.
(6.3.4)
The second case is when in expression (/N) both numbers are integers, but are not divisible by each other. This gives the so-called subsidiary maxima of small intensity, which are obtained due to diffraction only on a single-grating slit. As a result, the spectrum consists of a rear strong main and many weak subsidiary maxima, as shown in Figure 6.17a. The intensity of the diffraction maxima (eq. (6.3.2) and (6.3.3)) increases N2 times in comparison with one slit, and the maxima width decreases by 1/N. The condition of the main maximum (6.3.4) is of primary importance. It shows that for a given diffraction grating (at fixed b), a different wavelength gives maxima at different points of the spectrum. This is the basis of the use of diffraction gratings in optical spectroscopy.
m=−3
m=−2
m=−1
m=0
m=1
m=2
m=3
a
m=4
m=−4 m=−3 m=−2 m=−1
b
m=2 m=1
m=3
m=0 −60
−40
−20
0
20
40
60
Figure 6.17 Spectrum of white light as viewed in a grating instrument. The different orders of spectra identified by the order number m are shown separated vertically. The central line in each order corresponds to 0.55 m.
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If white light falls on a diffraction grating each of the main maxima is broadened. The width of a white spectrum is defined by boundary wavelengths 0.4–0.7 m of visible light. Spectrum of adjacent orders can sometimes overlap. This overlapping is shown in Figure 6.17b where, for clarity, spectra of different orders are given on a different vertical level. It can be seen that spectra of zero-, first- and second-orders exist separately, whereas spectra of the third- and fourth-orders are partially overlapped. Notice that a diffraction grating can also be used in a reflecting position. 6.3.4
Diffraction grating as a spectral instrument
Spectroscopy is the method of studying the composition and structure of a substance or the control of technological processes (refer to Chapter 7, Section 7.8). The main stage in spectroscopy is the decomposition of electromagnetic radiation in a spectrum on the wavelength or frequency. Optical spectroscopy deals with the optical range of electromagnetic radiation, including UV and IR. The basic units of optical spectrometers are either a prism or a diffraction grating. The most important characteristics of the quality of a spectral device are dispersion D and resolution R. Distinguish an angular and linear dispersion. The value numerically equal to the ratio of the angular distance between spectral lines to the difference of wavelengths of these spectral lines is referred to as angular dispersion. The angular dispersion is equal D
,
(6.3.5)
In order to obtain an expression for angular dispersion, we should find a derivative d /d from eq. (6.3.4) and change further the differentials into finite increments (neglecting the minus sign). At small angles, eq. (6.3.4) can be rewritten as b ≈ m and then m . b
D
(6.3.6)
Linear dispersion Dl is the value numerically equal to the ratio of the linear distance in the spectrum between the spectral lines l to the difference corresponding to those lines Dl
l ;
(6.3.7)
m , e
(6.3.8)
and at small Dl f
where f is the focal length of lens L (Figure 6.13). From the above formulas, it can be seen that the dispersion (both angular and linear) is larger for higher order of the spectrum m.
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Another device that permits the decomposition of incident radiation in the spectrum is a prism. Decomposition in the spectrum by a prism is due to the dependence of the angle of refraction on the wavelength. The corresponding formula for the prism can be obtained using expression (6.1.2). Notice that the sign of the derivative / for the diffraction grating and for the prism is different. On wavelength resolution (resolving power) of the spectral device, there is a minimal distance at which two close spectral lines are accepted as being two instead of seeing them as one single widened line (Figure 6.18). Rayleigh has offered a criterion by which the spectral lines are considered as resolved if the middle of the maximum position of one line coincides with the edge of the adjacent line (Figure 6.18, center). The resolving power R of a spectral device is a dimensionless reversed value of the wavelength difference of the resolved neighboring lines to the wavelength of one of them: the value is R
.
(6.3.9)
Using the Rayleigh criterion, we arrive at the expression R mN .
(6.3.10)
It can be seen that the resolving power is larger when a longer grating length and higher order reflections are used. In Figure 6.19, two spectral lines obtained with three different diffraction gratings are presented. Gratings I and II are characterized by identical resolution (lines have identical half-widths) but provide a different dispersion, whereas gratings II and III have different resolution (maxima have different half-widths at identical dispersion).
Rayleigh’s criterion
0 (a)
0 (b)
0 (c)
Figure 6.18 Image of two distant point objects formed by a converging lens; (a) the angular separation of the objects is so small that the images are not resolved, (b) the objects are farther apart and the images obey Rayleigh’s criterion of resolution, (c) the objects are still farther apart and the images are well resolved.
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2 I
1
2 II
1
2 III
Figure 6.19 The intensity patterns of light with wavelength 1 and 2 incident on the different gratings; grating II has the same resolution as I however higher dispersion, grating III – the same dispersion but lower resolution.
6.3.5
X-ray diffraction
X-rays, discovered by W.K. Röntgen in 1895, as well as visible light are both electromagnetic waves; however the X-ray wavelength is 103–104 times shorter (about 1010 m, i.e., 0.1 nm). This circumstance defines their high penetrating ability, which the great majority of mankind has experienced during medical inspections. Our interest here is in X-ray diffraction (XRD) in crystals. For experimental observation of diffraction, the radiation wavelength should be of the same order of magnitude as the diffraction grating period. This follows from eq. (6.3.4): in order to measure diffraction angle the ratio mb/ should have the order of unity, i.e., b should be commensurable to . Therefore, to observe XRD using diffraction gratings is extremely difficult in practice. At the same time, a diffraction grating with a period of about 1 Å has the nature that the interatomic distances in crystals are about this size. As the interatomic distances are approximately 1010 m and the size of even the smallest crystal is 107 m (repetition is 103 in the majority of cases), the crystal can be considered infinite. If a beam of X-rays falls on a crystal, under the action of an electromagnetic wave the atoms’ electrons begin to oscillate and scatter secondary radiation of the same wavelength in all directions (compare with Compton-effect, Section 6.6). As the atoms in a crystal are ordered, these secondary waves are coherent and interfere; this defines the diffraction effect. The diffraction problem of X-rays in crystals “on transmission” has been solved by M. Laue. However, a more evident picture has been given by W.L. Bragg and also independently by G.V. Wulf. Formalizing the picture described above, they reduced the scattering of secondary waves to the X-ray reflection from so-called crystallographic planes (see Section 9.1). (These are planes drawn through the nodes of a crystal lattice.)
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(a)
(b) 1 2
A
C
B
d
AC+CB= 2d sin
Figure 6.20 The Bragg’s law: (a) the ordered atom’s array and two arbitrary crystallographic planes, (b) an incident X-ray beam scattered by the entire family of crystallographic planes, the X-ray’s paths difference ACB is marked.
The atoms of a crystal and two most rational crystallographic planes are shown in Figure 6.20a. An X-ray beam falls on a crystallographic plane at an incident angle . Because of its high penetrating ability, the X-ray radiation passes into the crystal without refraction (the refraction index n 1). Therefore, the difference in the lengths traveled by waves 1 and 2 can be easily counted, making 2d sin , where d is the distance between the nearest parallel planes (interplanar distance). The maximum of intensity will be observed if this difference is equal to an integer of wavelengths (refer to Section 6.2): 2 d sin m,
(6.3.11)
where m is the reflection order. This formula is referred to as the Bragg formula. Knowing the arrangement of atoms in a crystal, it is easy to calculate the intensity of X-ray reflection. More difficult, however, is the problem of calculating the arrangement of atoms in a crystal from an experimentally measured diffraction picture. This problem is the essence of modern X-ray crystal structure analysis for which M. von Laue (1914) and W.L. and W.H. Bragg (1915) were awarded Nobel Prizes.
6.4 6.4.1
POLARIZATION
Polarized light: definitions
An important feature of a wave beam is its polarization. A wavetrain has two mutually perpendicular planes in which oscillations of the vectors of E and H takes place (Figure 5.45). It has already been mentioned that the action of an electromagnetic wave is defined mainly by the vector E. Therefore, vector H in many cases is neglected in drawings, whereas the plane of E vector oscillation is referred to as the plane of oscillations. The wavetrain is, therefore, linearly polarized, i.e., it possesses a single plane of oscillation (Section 2.8.1; Figures 6.2a and 6.21b and c).
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Because all atoms of a source emit electromagnetic waves independently, the beam consists of large numbers of independent wavetrains; their planes of oscillations are not correlated, such light being referred to as nonpolarized or natural. In this case, axial symmetry of the oscillation planes disposition takes place (Figure 6.2b and 6.21b and c). The direction of the axis of symmetry coincides with the direction of the wave propagation. If there is a partial infringement of the axial symmetry, the light beam is partly polarized (Figure 6.2b and 6.21b and c). The following designations are accepted in the schematic representation of light polarization in physics literature (Figure 6.21). The plane of oscillations of the vector E is set by arrows. The polarized beam is represented accordingly by a number of parallel arrows. If the plane of oscillations is perpendicular to the drawing plane, arrows are projected in points. A nonpolarized beam is represented by alternate points and arrows. 6.4.2
Malus law
There are devices called polarizers, sensitive to polarization of a light beam. These devices freely transmit the incident electromagnetic waves with a plane of oscillation parallel to the plane of the polarizer, and completely absorb oscillations perpendicular to this plane. Hence, behind the polarizer the natural light becomes polarized with the plane of oscillation parallel to the plane of the polarizer. If, in the way of this secondary beam, a second polarizer is installed with a plane perpendicular to the first, it will detain the first polarized beam completely. This second polarizer is in the position of an analyzer; it is sensitive to the degree of polarization of the light beam. What will happen if the plane of oscillations in the beam makes an angle with the plane of the polarizer? Let the plane polarized beam falls on the analyzer with the oscillation plane oriented at an angle relative to this plane and the plane of the polarizer. Separate the E0 vector into two components: parallel and perpendicular to the polarizer planes (Figure 6.22). The perpendicular component will be completely absorbed by the polarizer, whereas the component of the electric field in the parallel position will be equal to E0 cos , and the corresponding intensity will be I ( ) E 20 cos2 I 0 cos2 .
(6.4.1)
(a) Natural light
(b) Partly polarized light
(c) Plane polarized light
Figure 6.21 Schematic representation of light polarization: (a) natural light, (b) partly polarized light and (c) completely polarized light.
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2 I ( ) E0
1 I0
Figure 6.22 Illustration of Malus’ law: (1) the polarization device and (2) the plane of polarization.
This equation is called Malus’ law. It can be seen that if the wave’s plane of oscillation is parallel to the polarizer axis, the beam will pass through with no intensity loss ( 0, cos 1). In contrast, at /2 cos 0 and light will be absorbed completely. If natural light falls on a polarizer the intensity of the passed light is proportional to average value of cos2 ; since in an interval 0 /2 the value cos2 is equal to 1/2, Ipol. Inat/2: intensity of light passed through the polarizer is a half of that of the incident natural light.
6.4.3
Polarization at reflection: Brewster’s law
If natural light falls on the border surface of two media the reflected and refracted beams are partly polarized. This occurs because of the fact that from a dielectric surface only the component of the E vector which is parallel to the border surface (perpendicular to the incidence plane) is reflected. Then, in the reflected light the oscillations perpendicular to the plane of incidence will predominate, whereas in the refracted beam the oscillations parallel to the plane of incidence will prevail. It has been experimentally established that when reflected from a dielectric surface light is completely polarized if there is a certain relationship between the incidence angle and the refraction index: tan B n.
(6.4.2)
Here the angle B is referred to as the Brewster angle and the given reflection is known as Brewster’s law. However, the refracted beam is polarized only in part. When the beam falls on the two-phase border at the Brewster angle, the angle between the reflecting and refracting beams is equal to /2 (Figure 6.23). Indeed, as tan (sin /cos ) and (sin /sin ) n to satisfy Brewster’s law (tan B n) it turns out that cos B sin , which is possible only when B /2.
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Incident natural ray
Reflected planepolarized ray
B =
Refracted partly-polarized ray
Figure 6.23 Illustration of Brewster’s law.
EXAMPLE E6.6 A beam of natural light falls on a polished glass surface plate submerged in a liquid. The beam of light reflected from the plate is at an angle of 97° to the incident beam. Define the refraction index of the liquid if the reflected light is completely polarized (Figure E6.6).
1
1'
n1
n2
2'
Solution: According to Brewster’s law, when reflected from a dielectric light is completely polarized if the tangent of the incidence angle is equal to tan 1 (n2/n1) n21 where n21 is the relative index of the second body (glass) relative the first (liquid). The relative refractive index n21 is the ratio of absolute indexes, i.e., tan 12 (n2/n1). The reflected beam makes an angle of 2 and, consequently, tan ( /2) (n2/n1). Therefore, n1 n2/tan ( /2) and the refraction index n1 is 1.33.
6.4.4
Rotation of the polarization plane
When passing plane-polarized light through some substances, the plane of polarization can change its position in space, namely, it rotates around the light wave vector k. Substances
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possessing such properties are referred to as being optically active. Among optically active substances there are many anisotropic crystal (i.e., whose structure does not relate to cubic and hexagonal systems, see Section 9.1) and liquids (e.g., turpentine, nicotine, solutions of many organic and inorganic substances in inactive solvents, etc.). Experience shows that the angle of rotation of the polarization plane around the wave vector k in optically active media is proportional to the length l, traveled by a beam in a sample l
(6.4.3)
The coefficient , generally dependent on the wavelength, is referred to a rotation constant and is expressed in angular degrees on millimeters of distance run. In solutions of optically active substances the angle of polarization plane rotation is proportional to the length traveled l and concentration of the active substance c: [] c l,
(6.4.4)
where [] is the specific rotation constant. So, knowing [] and having measured l, it is possible to define the concentration of an active substance in a solution. The direction of rotation of the polarization plane depends on the substance: if the plane of polarization turns clockwise in relation to k, the substance is referred to as a right-hand (or dextrorotatory); if it turns anticlockwise, the substance is a left-hand (or laevorotatory) substance. Thus the direction (the wave vector k) and the beam direction of rotation in a dextrorotatory substance forms a left-hand system, and in a laevorotatory substance forms a right-hand system. For an explanation of the rotation of the polarization plane it is supposed that plane polarization in inactive substances is the superposition of two oppositely directed circular polarizations with identical amplitude and angular velocity. In Figure 6.24, a scheme explaining this supposition is given. On the left, vectors E1 and E2 rotate around the k vector in opposite directions with equal angular velocities, therefore the total vector E oscillates in the vertical plane. If the angular velocities differ, the plane of oscillations turns around k (Figure 6.24, on the right). The angular velocity’s characteristic in optically active substances is caused by an asymmetric arrangement of atoms in molecules and crystals. In Figure 6.25, an example of a hypothetical tetrahedron in two various enantiomorphous forms is depicted. In the center of the tetrahedron is an atom (e.g., carbon as a complexation atom, not shown in the picture), and in the vertexes various atoms are arranged A, X, Y and Z. If the tetrahedron is looked at from above at a detour alternation XYZ (Figure 6.25a), a clockwise motion takes place. The tetrahedron in Figure 6.25b is a mirror image of the one Figure 6.25a. Such molecules are referred to as enantiomorphic. Therefore, if a substance with tetrahedrons of a-type in the structure is, for instance, dextrorotatory, an isomer with tetrahedrons of b-type is a left-handed isomer. Research into the effects of rotation of the polarization plane is one of the methods of structural chemistry.
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P'
P E E1
P
E E2
E2 E
Figure 6.24 The polarization plane rotation.
A
A
Y
X
Y
X (b)
(a) Z
Z
Figure 6.25 Enantiomorphic atomic arrangement.
6.4.5
Birefringence: a Nichol prism
Let us consider now a phenomenon known as double refraction in anisotropic crystals. In the XVIIth century, Huygens discovered that light passing through some crystals is split into two beams (Figure 6.26). One passes through the crystal in strict conformity with the laws of geometrical optics and is referred to as an ordinary beam (marked on the figure by the letter “o”). The other beam is called an extraordinary beam (marked on figure by the letter “e”); it passes the crystal’s surfaces with an infringement of the law of refraction: i.e., it cannot lie in one plane with an incident beam and a normal to an interface. The important thing is that both beams are completely polarized in mutually perpendicular planes. This is the basic, and most practically important, property of birefringent crystals. In the crystals described there are one or two directions along which the double refraction does not occur. These directions are referred to as the optical axes of a crystal (in Figure 6.26 and further defined by line MN). Certainly, they are determined by the atomic structure of a crystal. If the crystal has one such direction it is referred to as a single-axis crystal; there are also biaxial crystals with two such directions. Any plane which runs through the crystal’s optical axis is referred to as the main section or the main plane. Most interesting is the main section containing the light beam. The plane of the vector E oscillations in an ordinary beam is perpendicular to the main section and in extraordinary beam lies in the main plane.
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The properties of isotropic media (including optical properties) are identical in all directions. Nearly all gases and liquids and highly symmetric crystals with cubic and, in part, hexagonal structure (see Chapter 9.1) are referred to as isotropic ones. With anisotropic crystals light interacts differently than with isotropic media. Remember that the refraction index n defines the light speed in a medium c/n. For weakly magnetic substances (M ≈ 1), it is connected to the dielectric permeability of the medium by an equation n . In anisotropic crystals the dielectric permeability depends on direction. In particular, in the optic single-axis crystals the dielectric permeability in the direction of the optical axis and perpendicular to it have different values, || and ⊥ respectively. In other directions has intermediate values. If we draw a sketch of values in a single-axial crystal for different directions by segments from an origin, the ends of these segments form a rotation ellipsoid. Its axis of symmetry will coincide with the crystal optical axis. In Figure 6.27, an ellipsoid of the dielectric permeability of a single-axis crystal is presented. Because the light speed in a substance depends on the dielectric permeability , the given scheme also represents a diagram of the dependence of on the crystal direction. In this case this figure is called an indicatrix of speeds. Since the light speed does not depend on direction in isotropic media, the indicatrix is represented by a spherical surface. In anisotropic crystals, the properties of which depend on direction, the indicatrix differs from a spherical one. Moreover, they can differ for ordinary and extraordinary
M e 0
N
Figure 6.26 The birefraction of natural light by a single crystal of spar CaCO3: MN is the optical axis, o is the ordinary ray (o-ray), e is the extraordinary ray (e-ray). y g
S
x x
z
Figure 6.27 Huygens wave surfaces generated by a point source S embedded in calcite.
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beams. Therefore, two indicatrixes exist: spherical for ordinary beams and as rotational ellipsoids for extraordinary beams. They appear to be “inserted” in each other. Both indicatrixes touch in the direction of an optical axis because in this direction they have an identical light velocity. In a perpendicular direction, both indicatrixes differ maximally. In single-axis crystals there are two opportunities: in optically positive crystals the velocity of extraordinary beam e is less than that of ordinary 0, in optically negative crystals e 0 (Figure 6.28). By taking into account the difference in the optical properties of the crystal, it is possible to find the refraction of all rays in all directions graphically. Nichol suggested a relatively simple method of making a completely polarized beam; the method is based on arranging two split pieces of a crystal of Iceland spar (CaCO3) in such a way that the beam transmitted through it is polarized. Such a device is now referred to as a Nichol (Figure 6.29). In order to obtain such a polarizer the single crystal of Iceland spar should be cut first into two pieces of proper orientation and then be stuck together by a special glue substance. This substance should have the refraction index n, lying in an interval between indexes n0 and ne of an initial crystal (n0 n ne); it is the Canadian balm. The angle of fall onto the plane of pasting is selected in order to make the ordinary beam undergo a total internal reflection (refer to Section 6.1) and then be absorbed by the frame of the prism. The extraordinary beam freely passes through the thin layer of balm and leaves the Nichol. Accordingly, it is completely polarized and can be used in optical measurements. υe
υo υo
υe
M
N
O
M
(a)
N
O
(b)
Figure 6.28 Ellipsoid index of the light wave velocity in a mono-axis crystal: (a) single axis positive and (b) negative. MN is the optical axis.
M 48°
B
C
22
68°
S
c A N
76.5°
o D
Figure 6.29 Schematic presentation of a Nichol. BD—the Canadian balm layer.
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A similar Nichol can be used as an analyzer. If two Nichols are one after another, one as polarizer and the other as analyzer, their rotation around the beam axis essentially influences the transmission of light through this double system. If both Nichols are installed identically the light transmission has maximal intensity, the rotation of the second Nichol (the analyzer) around the beam at a right angle (crossed Nichol) completely extinguishes light. If, however, an optically active substance is placed between the Nichol, the sample becomes visible in crossed Nichols. To determine the angle of polarization plane rotation it is necessary to turn the Nichol-analyzer at a certain angle to achieve extinction again. The angle of rotation will be equal to angle (Section 6.4.2). This fact relates to optical methods of substances research.
EXAMPLE E6.7 A plate of quartz of thickness d1 1 mm cut out perpendicularly to the optical axis of a crystal, turns a polarization plane of monochromatic light of a certain wavelength at an angle of 20°. Define: (1) what the thickness of the quartz plate placed between two parallel Nichols should be in order to extinguish the light completely; (2) what lengths a tube l with a solution of sugar of mass concentration C 0.4 kg/l should be placed between Nichols in order to obtain the same effect? The specific rotation of the sugar solution is 0.665°/(m.kg.m3) (refer to Section 6.6.4). Solution: (1) An optically active medium rotates a polarization plane at an angle d*. Therefore we can present the thickness of the quartz plate as d2 ( 2/)**, where 2 is an angle totally extinguishing the light ( 2 90°). The rotation constants can be found from * formula ( 1/d1). Substituting this expression into **, we obtain d2 ( 2/ 1) d1. Executing the calculations, we obtain d2 4.5 mm. (2) The length of the tube with the sugar solution can be found from the expression 2 []Cd which defines the sugar solution turning angle of the polarization plane 艎 2[]C. Substituting all known data we obtain l 0.38 m.
EXAMPLE E6.8 A parallel beam of light with a wavelength 0.5 m falls normal to a diffraction grating. A lens with a focal length l 1 m is behind the diffraction grating to project a diffraction picture on a screen (Figure E6.8). The distance between two first-order maxima is l 20.2 cm. Determine (1) the diffraction grating constant, (2) the specific number of grating grooves (on 1 cm), (3) the limiting diffraction maxima and their total amount N and (4) the angle max corresponding to this maximum.
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Diffraction grating
L
II
I
0 l
I
II
Screen
Solution: (1) The diffraction grating formula is d sin k. In our case k 1. sin ≈ tan (because L p l) and 2 tan l/L. Therefore, d·艎/2L . From this equation it follows that d 2L/艎. Executing calculations, d 4.95 m. (2) The specific number (on 1cm) of grating grooves is n (1/d) 2.02 103 cm1. (3) In order to find the number of diffraction maxima we need to calculate the kmax corresponding to kmax d/(sin max)*. kmax corresponds to max sin . Therefore, kmax 9.9; however for this number sin is larger than 1, therefore, kmax is 9. This value allows us to find the general number of diffraction maximums N. The obvious relation exists N 2kmax 1. Therefore, N 19. (4) max can be found from star relation: max arcsin(kmax/d) 65.4°. 6.5
DISPERSION OF LIGHT
All phenomena caused by the dependence of the refraction index on frequency (or on wavelength) are united under the name light dispersion. Dispersion is referred to as normal if the refraction index steadily falls with an increase of wavelength (dn/d 0) or, grows with an increase of frequency ((dn/d ) 0); otherwise the dispersion is referred to as being anomalous ((dn/d ) 0). A typical picture of dependence n() for a normal dispersion is given in Figure 6.30.
n
Figure 6.30 Normal dispersion: the relation between refraction index and .
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For a better understanding of the presence of abnormalities in the dependence n(), it is useful to consider the process of light interaction with atoms in a substance. In Section 4.2.4 in the analysis of electronic types of dielectric polarization, it was shown that the internal electric field in an atom linearly depends on displacement and creates a force that returns the charge to its initial position. This means that an electron, being forced out of its equilibrium position, begins to make harmonic oscillations with a frequency of its natural frequency 0. The strength of an electric field in a light wave acts with its own frequency . Under the action of an electric field, an electron starts to oscillate with frequency causing secondary radiation. The process of forced oscillation was considered in Chapter 2.7. There it was shown that the amplitude of the forced oscillations of a system A( ) depends inversely on the difference of squares of natural frequency and the frequency of the driving force ( 02 – 2). When 2 approaches 02, a sharp increase in oscillation amplitude can be observed (eq. (2.7.4) and Figure 2.15). Accordingly, the resonance is accompanied by an additional absorption of the incident wave; which is expressed in more or less sharp lines in spectra of absorption and emission. The refraction index n is expressed through dielectric permeability and magnetic susceptibility by an equation n ()1/2. In Section 4.2.5, it was mentioned that only electron polarization will be exhibited in light with a frequency ≈ 1015 sec1. Accordingly for substances with ≈ 1 (which is characteristic of the overwhelming number of chemical compounds) n2 . Taking into account that 1 and ℜ/0E (refer to Section 4.2.2, eq. (4.2.8)), we obtain n2 1(ℜ/0E). For dielectrics in this frequency range the polarization ℜ can be presented as: ℜ np (eq. (4.2.2)). In its turn an induced electric moment p, according to definition, is the product of the charge ⏐e⏐ and shoulder x of the induced dipole; then
n2 1 e
nx o E
(6.5.1)
The displacement value x can be found by solving a differential equation of forced oscillations. In this case the electric field strength is acting as the force: Ffrc ⏐e⏐E(t) ⏐e⏐E0cos t, a restoring force can be considered as an elastic force Frst x m 20x. Therefore, the differential equation for x determination has the form: ⎛ d2 x ⎞ m ⎜ 2 ⎟ m 20 x e E0 cos t. ⎝ dt ⎠
(6.5.2)
The solution of such an equation is given in Chapter 2.7 where it was shown that:
x
e E0 cos t m( 20 2 )
e E (t ) m( 20 2 )
.
(6.5.3)
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Substituting the result into eq. (6.5.1) we obtain
n2 ( ) 1
ne2 . o m( 20 2 )
(6.5.4)
At ^ 0 or p 0, the value n2 does not depend on and is near to 1. If the light frequency approaches 0, resonance phenomena appear and the refraction index rises significantly (Figure 6.31). In reality the picture, n2( ) is like that presented in Figure 6.32. It is important that the position of abnormal dispersion ((dn/d ) 0) coincides with the light absorption line in a substance.
n2
1
0
Figure 6.31 Graphic representation of the relationship of the square refraction index and the light frequency (as follows from eq. (6.5.4)). n2
1
0
Figure 6.32 The experimentally measured dependence of n2 versus the incident light angular frequency . The position of the anomaly dispersion maximum coincides with the absorption line position.
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The figure at which n 1, the light speed in a substance becomes larger than the speed of light c, should not confuse the reader. In this case the phase speed is considered, i.e., the speed of distribution of the given, constant phase. The wave energy and momentum are transferred by group speed which cannot be more than the speed of light (see Section 2.9.4).
6.6 6.6.1
THE QUANTUM-OPTICAL PHENOMENA
Experimental laws of an ideal black body radiation
The radiation of electromagnetic waves by a heated body refers to as thermal radiation. Any body radiates at any temperature, however at medium temperature range its intensity cannot be measured by an ordinary device at sure. For the quantitative characteristic of thermal radiation the concept of an emittance R is used: the emittance is referred as to the energy that is emitted by a unit surface of a heated body in all directions within a solid angle 2 (a half of full solid angle, i.e., one side of a plane) in a unit time in a whole interval of frequencies (wavelengths). Thermal radiation basically contains waves of all frequencies. When allocate an interval of frequencies d at temperature T, part of emittance dR corresponds to it: the wider d the higher dR. However the ratio between them is not linear, it depends on frequency (wavelength ). The value r, connecting dR with d depends also on radiation frequency and referred to as spectral density of emittance, i.e. dR(T ) r ( , T )d , or dR( , T ) r ( , T ). d
(6.6.1)
The total emittance R can be obtained by integration of function r( ,T ) over the whole interval of frequencies, therefore R does not depend on frequency and is entirely defined by temperature:
0
0
R(T ) ∫ dR( , T ) ∫ r ( , T )d .
(6.6.2)
Function r( ,T ) also describes of a body’s ability to emit thermal radiation. Basically, the curve of dependence r ( ) at the fixed temperature T can be experimentally measured (Figure 6.33). The figure shows: a muffle furnace with a radiating body inside, aperture in the furnace door and in a screen cutting out the desirable stream of
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heated body
screen
spectral unit
intensity final curve wavelength
Figure 6.33 The experimental scheme of the measurement of the spectral composition of heat radiation; furnace to heat a body, screen protector, spectral device to measure the radiation intensity; the experimental graph coordinates are shown.
the thermal energy directed further to a spectral device. A purpose of the spectral device is to decompose the whole falling radiation into a spectrum on frequency. Consequently the spectrum of thermal radiation is obtained which should be investigated and explained. To measure such curve represents significant difficulty as beams of different wavelength demand various techniques. In a result, the various parts of a general curve are “tailored”. Figure 6.34 schematically depicts the emission spectra of bodies heated to various temperatures, from almost room temperature to temperature of the sun surface. In the same figure the curve of luminosity presents (see also Figure 6.1) which enables one to know radiation at what emitter temperature is radiation perceived by eye as light. Part of a spectrum of thermal radiation can be felt as heat on the skin of the human – this is mainly IR radiation. UV radiation can also be felt by the skin – this we think of as sunburn. All this is an insignificant part of the general thermal radiation. Perhaps, only the sun radiation contains in its spectrum all the wavelengths that humans perceives with almost all their sensory organs. The curves can be schemed in coordinates r( ) and r() keeping in mind the relation 2c ,
(6.6.3)
In order to transfer from r() to r( ) one should to compare equivalent peaces of the graphs areas of both functions r ( )d r ( )d .
(6.6.4)
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r pick of 180 cm height
6000 K
3000 K , m
2000 K 1000 K
B
r
vertical scale drawn up in 250 times
V R
1000 K 800 K 400 K , m
VR
Figure 6.34 Spectral intensity distribution of the heat radiation at different emitter temperature: boiling water (400 K), electric heater (800 K), red incandescence (1000 K), blowtorch (2000 K), voltage arc (3000 K) and the sun radiation (6000 K).
Derivation of the eq. (6.6.3) gives d
2c 2 d d . 2c 2
(6.6.5)
(The minus sign in this expression specifies only that with increase one value another one decreases. Therefore the sign minus further will be omitted). Changing in the equation (6.6.4) d on d according to (6.6.5) we can obtain r ( )d r ()(2πc/ 2)d r()(2/2c)d , or, finally: r ( ) r ( )
2c 2 r ( ) . 2c 2
(6.6.6)
The radiation power flux d ( ) is the thermal radiation energy emitted in the unit of time from the surface dS of the radiated body d ( ) R( )dS.
(6.6.7)
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Falling on any surface, this flux will be partly absorbed (this part we will call the absorbed flux of energy d ( )), and partly reflected. The dimensionless value
a( , T )
d ( , T ) d ( , T )
(6.6.8)
is referred to as the absorbing capacity of a body a. In general cases, absorbing capacity a depends both on frequency and on temperature of the radiator. For a body that completely absorbs all the thermal radiation falling on it in all ranges of frequencies and at any temperature, the value a is constant and equal to 1; such a body is called a perfect black body (PBB). If the body’s absorbing capacity depends on frequency and/or on temperature, but is constant (less than 1), the body is called gray. The majority of bodies, however, are not PBB or gray, their absorbing capacity is less than 1 and depends on frequency and temperature; these are arbitrary bodies. There is a certain connection between the emitting and absorbing capacity of a body namely the Kirchhoff law: the ratio of the body’s emitting and absorbing capacity does not depend on the nature of the body but is identical, i.e., universal, for all bodies’ functions of frequency and temperatures (function f( ,T ) r ( , T ) f ( , T ). a( , T )
(6.6.9)
This function is referred to as Kirchhoff function. The Kirchhoff law defines one of the most important properties of thermal radiation, distinguishing it from other types of radiation (fluorescence, luminescent, etc.): thermal radiation is an equilibrium one. From eq. (6.6.9), it follows that the more a body absorbs, the more it radiates. Hence, in an isolated system of bodies their temperature will eventually be equalized, becoming identical. If a body absorbs more, it also radiates more. The values r( ,T) and a( ,T) can differ, but their ratio is identical. The analysis of the curves, similar to those presented in Figure 6.34, allows one to understand and formulate some laws of thermal radiation. Thus it has been experimentally established that emittance is proportional to the fourth degree of absolute temperature: R(T ) T 4 . For a gray body the given ratio can be rewritten: R(T ) aT 4 .
(6.6.10)
This is the Stefan–Boltzmann law. The value of constant in this law is experimentally established: 5.7 108 W/(m2 K4). The expression (6.6.2) shows that the area under the curves in Figure. 6.34 is proportional to the fourth degree of absolute temperature.
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From the same figure it can be seen that, depending on temperature, the radiation spectrum is shifted to the shorter wavelengths. This law can be written as max
b T
(6.6.11)
where max represents the wavelength corresponding to the maximum of radiation spectral distribution. This dependence is referred to as Wien’s law of displacement, constant b is equal to 2.90 103 mK. All the above-mentioned experimental facts and laws require a theoretical explanation. 6.6.2 Theory of radiation of an ideal black body from the point of view of wave theory: Rayleigh–Jeans formula Before examining the theory of thermal radiation, one should suggest the model of an ideal black body (IBB), i.e., a body that absorbs all falling radiation. The most simple, yet successful, is a model representing an almost completely closed cavity with a small aperture (Figure 6.35): all beams that get inside the cavity lose their intensity after consecutive reflections, and do not leave the cavity. Because the heated-up walls of the cavity are a source of thermal radiation and only an insignificant part of it leaves, a certain equilibrium density of radiation is established in the cavity. Standing waves (such as in a string (see Section 2.9.3) are produced in the cavity, the wavelength of which is defined by eq. (2.9.8). As for standing waves in a string, the maximum wavelength of a standing wave is determined by the size of the cavity. Also, the minimum length of a wave is determined by the discrete character of the material from which the walls are made. Therefore, the density of standing waves in the cavity is finite. After detailed consideration, one can obtain 2/42c2. Within the framework of classical physics, an oscillator with certain frequency can be put in conformity to each standing wave. According to the law of uniform distribution of energy on degrees of freedom (see Chapter 3), to every oscillator an average energy T can be attributed (because two (½)T goes to the oscillation degree of
Figure 6.35 Model of an ideal black body (IBB).
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freedom connected to kinetic and potential energy). Hence, the total energy of radiation of an IBB is r ( , T )
2 T . 4 2 c 2
(6.6.12)
or in terms of wavelengths
r ( )
2c T . 4
(6.6.12)
This formula was first suggested by Rayleigh and Jeans, and was irreproachable from the point of view of the wave nature of light. It equally concerns experimentally measured dependence of heat radiation of an ideal black body. At the same time, the theory and experiment are in a glaring contradiction with one another, mainly in the area of small wavelengths (large frequencies). From Figure 6.36b, it can be seen that at small wavelengths (or at increase of frequency, in other figures) the theoretical curve soars sharply upwards, whereas the curve achieved by experiment (Figures 6.34 and 6.36a) goes downwards. This also leads to the incorrect conclusion that the luminosity R (6.6.2) becomes senselessly infinite. Because of the area where the divergence takes place, the discrepancy between theory and experiment has been called an “UV accident,” thus recognizing the inability of the theory to explain the laws of radiation within the framework of wave theory as it existed at that time. A completely different approach to the theory of radiation was, therefore, necessary. One was proposed by W. Wien (awarded the Nobel Prize in 1911) who, on the basis of the laws of thermodynamics, obtained a “bell-shaped” theoretical curve. A revolutionary approach was suggested by Planck (1900), which resulted in the full agreement between theory and experiment.
r
r
r
h too small
"catastrophe"
λ
λ (a)
h=6.64×10−34 J.sec
(b)
h too big
λ
(c)
Figure 6.36 The spectral density of heat radiation r versus the wavelength : (a) experimental curve, (b) “ultraviolet catastrophe,” (c) the correspondence of the theory and experiment in the framework of Planck theory.
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EXAMPLE E6.9 The maximum of spectral density of the sun’s radiation emission corresponds to the wavelength 500 nm. Assuming that the sun radiates like an IBB, determine: (1) emittance of sun R* (R* means the emittance of IBB), (2) energy flux of sun radiation , (3) total mass m of electromagnetic radiation irradiated by the sun in one sec (refer to Section 6.6.1). Solution: (1) The Stefan–Boltzmann law describes the radiation emittance R* of an IBB R* T 4, where equals to 5.67 108 W/(m2K4). The sun’s surface temperature can be determined using the Wien’s law max b/dT, where b 2.90 103 mK. Combining these two formulas, we obtain 4
⎛ b ⎞ Rⴱ ⎜ . ⎝ max ⎟⎠ Executing calculations, we arrive at R* 6.4 107 W/m2. (2) The energy flux radiated by the sun is the product of radiation emittance and the sun’s surface area RS or 4r2R* where r is the sun’s radius. Substituting all data into the last formula, we arrive at 3.9 1026 W. (3) The total mass of electromagnetic radiation emitted by the sun in 1 sec can be determined using the correspondence mass and energy E mc2. The energy of the electromagnetic radiation in the time t is equal to the product of the flux and time E t. Therefore t mc2 and further m (t/c2). Executing calculations, we arrive at m 4.3 109 kg. 6.6.3 Planck’s formula: a hypothesis of quanta—intensity of light from wave and quantum points of view Analysis of the state of theory and experiment concerning IBB radiation and the mathematical description of the phenomenon led Planck to recall Newton’s hypothesis that light is a stream of particles (corpuscles); this had been rejected on the basis of successive works on interference and diffraction. Planck suggested a revolutionary idea: that each particle of radiation is a corpuscle or quantum, i.e., a particle bears a portion of energy h where h is a certain constant, and h/2. Then the total radiation energy flux should be expressed by the total number of quanta, , i.e., E Ni1ni i, where is the energy of a single quantum, ni is the amount of such quanta and N is their total number. The distribution of quanta on energy is set by the Boltzmann factor. According to the statistical method of average values calculation (see eq. (3.2.11)), we can write a similar expression for the average energy of a quantum oscillator having replaced integrals by sums. Therefore, following Planck, we arrive at: N
⎡ n ⎤ ⎥⎦ [exp(x ) 2 exp(2 x ) ] , 1 exp(x ) exp(2 x ) ⎡ n ⎤ exp ⎢ ⎣ T ⎥⎦
∑ n0 n exp ⎢⎣ T ∑ n0 N
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where x ( /T). In the last expression a fraction can be rewritten as d log[1 exp x exp 2 x ] dx d 1 log dx 1 exp x exp(x ) 1
from which the sought value is derived:
. ⎛ ⎞ exp ⎜ 1 ⎝ T ⎟⎠
(6.6.13)
Leaving unchanged the part of the calculation in which the oscillation density in the cavity of an IBB was counted, and attributing to every oscillator the above-mentioned average energy , Planck came to the formula of function r( ,T) for an IBB
r ( , T )
2 4 2 c 2
. ⎛ ⎞ exp ⎜ 1 ⎝ T ⎟⎠
(6.6.14)
The formula obtained not only correctly reflected an agreement between the theory and experiment, but also allowed the determination of the h value; in Figure 6.36c the theoretical results are “adjusted” to the experimental, from which the value of the constant h ( 6.626 1034 J sec) has been determined. This value was later named after M. Planck; it is typical that Planck’s constant has the dimension of the momentum or quantum of action (i.e., the product of energy and time of its action). Proceeding further to corresponding functions from wavelength ((6.6.1) and (6.6.6)), it is possible to obtain the dependences r and f on :
r (, T ) f (, T )
4 2 c 2 5
1 exp ⎛⎜⎝ 2c ⎞⎟⎠ 1 T
.
(6.6.15)
Planck’s formula well describes limiting transitions. So, at ^ T the exhibitor in the denominator of function (6.6.14) can be decomposed in a series and be limited by two terms. This leads to Rayleigh–Jeans formula which describes very well experiment in this area of frequencies. In contrast, at p T the unit in the denominator can be neglected and functions r and f fall according to the exponent that is found by experiment.
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The Planck formula suggests how to find numerical values of constants in Stefan– Boltzmann and Wien laws. In particular on integration of Kirchhoff’s law on the whole frequency range one can arrive at the Stefan–Boltzmann formula. The constant in Wien’s law b can be found by derivation of the Kirchhoff’s function on frequency and equalizing it to zero. We hope that readers can carry out these calculations themselves. Thus in the phenomena described, radiation is represented by a flow of corpuscles, quanta of energy which have been called photons. The energy of each photon is defined by product . Though for a long time it was known that in other experiments the same radiation manifests itself as a flow of waves. There is a problem, which has occupied the minds of many physicists, which is now known as particle-wave dualism and to which we shall pay more attention. Wave and quantum theory lead to completely different representations of the intensity of light. We should remember that intensity is understood as energy falling normally on a unit of area in a unit of time. Within the framework of wave theory, the intensity of a monochromatic beam of light is defined by the square of the wave amplitude and does not depend on frequency, i.e., I ~ A2. In quantum theory at a fixed wavelength (and, correspondingly, frequency) intensity is defined by the number of quanta, i.e., I ~ N . As will be shown below, a number of experiments have excellently confirmed Planck’s quantum hypothesis. Let us emphasize once again that the theory of thermal radiation became the starting point for quantum mechanics, which has subsequently received confirmation in many areas of physics. Laws of thermal radiation are widely used in technology to initiate and support chemical processes and to measure the temperature of bodies, contact with which is either impossible or complicated (e.g., measuring the temperature of the stars, the heated up gases during the launch of missiles, etc). These laws form the basis of optical pyrometry. It can be seen in Figure 6.34 that measurement of an integral of luminosity R (the area under the curves) can be a measure of body’s temperature. There are several kinds of pyrometry: one is based on color and another on brightness. The first is based on the position of the curve maximum and the second on the ordinate at fixed wavelength. Certainly, all the laws used here are only fair for IBB, however there are ways to account for the uncertainty arising in experiments. The stated theory of thermal radiation also allows an explanation of a phenomenon that has an influence on life on earth. This is the so-called green-house effect. The sun’s radiation (the spectrum is depicted in Figure 6.34), passes through open space, and reaches the external layers of the earth’s atmosphere, naturally with a loss of intensity, but without a special change of spectral composition. In the atmosphere there is selective absorption of the sun’s radiation by natural and industrial gases. This selectivity is defined by the structure of molecules, by their concentration and properties. It is natural also, that absorption of radiation depends on humidity, dust content and other properties of the atmospheric layers close to the surface of the earth. The sun’s radiation reaches the surface of the earth and heats it up and, together with the internal heat of the planet, defines the temperature of its parts (depending on geographical place). At the same time, the earth’s surface also radiates thermal energy. The temperature of the “radiator” in this case is essentially less than the sun’s temperature, accordingly the entire spectrum, under Wien’s law, is shifted to the long-wavelength region area (see Figure 6.34).
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This radiation is directed from the earth surface and should again penetrate through the atmosphere in the opposite direction. However, being long wavelength the earth’s radiation is absorbed by the atmosphere differently than the radiation from the sun. The transmission ability of long-wavelength radiation is less than solar radiation, the earth’s radiation is appreciably “absorbed” in the atmosphere, heating it up. This is the green-house effect. Taking billions of years to establish, the thermal equilibrium in the solar system defines life on earth. Every large-scale action can affect the established balance to some extent, displacing it in one way or another. In particular, the industrial activity of mankind leads to a change in the chemical composition of the atmosphere, increasing the concentration of industrial waste products. This change influences the absorption of the radiation falling to earth and leaving it. However, to a much greater degree, it concerns the radiation of the earth rather than that of the sun. All these events cause “over-warming” of the atmosphere and disturbance of the equilibrium. The effect is probably not so large: it is estimated at approximately 1–2°. The results, however, can be catastrophic. One example is the appreciable effect on the people living in those European countries that are below sea-level and protected from the sea by dams (e.g., Denmark, The Netherlands). The increase in atmospheric temperature can melt much more ice than would normally maintain the existing balance. The consequences are dangerous for large cities such as Venice, Saint Petersburg and will affect the climate of Florida and many other pearls of human civilization. In this connection it is also worth mentioning the so-called “ozone holes”—the local destruction of a centuries-old balance in the composition of the atmosphere resulting from the products of industrial activity (e.g., chlorofluorocarbons—freons) which create areas (holes) in the atmosphere that are transparent to short-wave UV radiations. These holes in the ozone layer are making the affected areas dangerous to live in because of the excess of UV radiation, which is harmful to life on earth. It is also probable that short wavelengths can cause undesirable mutations in living organisms. 6.6.4
Another quantum-optical phenomena
Planck’s hypothesis was confirmed and developed by Einstein’s theory of an external photo-effect. The photo-effect consists of knocking electrons out of the surface layers of some metals and oxides on their irradiation by quanta of electromagnetic radiation. The scheme of an experiment on the photo-effect is presented in Figure 6.37. The main part of the experimental equipment is a vacuum bulb with two electrodes C and A with a window allowing the irradiation of electrode C. The interaction of electromagnetic radiation of definite frequency and amplitude A results in knocking the electrons out of the surface of electrode C. A voltage difference is applied to the electrodes (Figure 6.38). First a negative pole is applied to K accelerating the kicked-out electrons. As the voltage increases, all the electrons reach electrode A, and saturation takes place (the horizontal line in the scheme). However, a device permits the signs of the electrodes to be changed; when a decelerating voltage difference is applied, part of the electrons are not able to reach electrode A. When the difference is zero only those electrons that have their own high enough kinetic energy
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can reach A. Furthermore, even at a coercive force some of the energetically active electrons can still reach A. Only a locked-out voltage can stop the current through the bulb. If the light intensity (i.e., the amplitude of the incident light, A) is increased and the same is kept, the saturation increases but the locked-in voltage remains the same. This means that it is not the light electric field amplitude (intensity) but the frequency which is responsible for kicking electrons out from the electrode and locking them. In fact, if the light frequency is increased, the locking in voltage also increases. The result obtained in the experiment is in agreement with the supposition that, in a given phenomenon in the photo-effect, light behaves as a flux of particles (photons). The results obtained regarding the volt–ampere characteristic shown in Figure 6.38 are deceptive. These results can be explained as follows. The energy of the falling quantum is transferred to an electron in photocathode producing the work A. Part of the energy is expended for the work A1 of moving the electron from the deep layers of the photocathode up to its surface, then in overcoming electron binding to the photocathode body A2; the remaining energy is left to the
Light
-
C
A µA V
R
−
+
B1 −
0
B2 −
+
+
Figure 6.37 Diagram of a device for the photo-effect measurements: an electric device permits to change an electric field polarity.
I
I
I
1
saturation
2
1= 2
1= 2 Locking voltage (lv)
V (a)
lv
V (b)
lv2 lv1
V (c)
Figure 6.38 The V–A characteristics of the photo-effect at different light frequencies: (a) typical V–A characteristic, (b) the same frequency and different light intensity, (c) the same intensity of light however different frequencies.
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photoelectron as its kinetic energy. According to the law of energy conservation we can write:
A1 A2
my 2 . 2
(6.6.16)
In this expression, work A1 is undetermined since it is not known at precisely what point of the cathode body the collision took place. The work A2 is a characteristic of every metal and oxide and is referred to as work function. If we exclude the unknown term from this line, i.e., remove the term A1, the equation become valid only for those electrons which, at the moment of collision, were on the surface of electrode A; they have the highest possible kinetic energy. Therefore:
A2
2 m y max . 2
(6.6.17)
It is possible to determine the kinetic energy of the photoelectron using the experimental value of the locking out voltage. Then we can write:
eU loc
2 m y max . 2
(6.6.18)
It follows from this equation that there is a limiting frequency k below which the photo-effect in a given photocathode disappears completely. In fact, the quantum k does not have enough energy to tear an electron out of the surface of electrode C. This happens when is lower than the work function. The so-called photo-electric threshold takes place at k A2. In addition to the theory of an IBB, it has been proved in quite another experiment that, in some circumstances, light behaves as a particle flux rather than a wave. Albert Einstein was awarded a Nobel Prize in 1921 for his outstanding contribution to physical science in general and especially for the photo-effect theory, which belonged mainly to Einstein and provided convincing confirmation of Planck’s hypothesis energy quanta not only in the theory of heat radiation but also in some other physical events. Einstein was also responsible for the concept of a photon which is widely used in modern physics. A short-wave border of X-ray radiation is another phenomenon, which supports the quantum idea. Discovered by W.C. Röntgen in 1895 and referred to as X-rays, this is the electromagnetic radiation with a wavelength of the order 1010 m (see Chapter 5, Table 5.3), arising on the electron transition in atoms and also on electron movement with acceleration (linear or centripetal). In the majority of countries this radiation is referred to as X-rays as it was called by Röntgen himself, but in Germany and Russia the term “Röntgen rays” is used. The generator of such radiation is the X-ray tube, the principle scheme of which is given in Figure 6.39. There are two electrodes in vacuum glass cylinder. The cathode represents a heated string and the anode is a massive metal cylinder, compulsorily cooled from inside by
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flowing water. The cathode’s task is to emit electrons. A potential difference from tens to hundreds of keV (and more) is applied between the cathode and the anode. This electric field accelerates all emitted electrons up to an energy of 10–100 keV. The main part of the electron energy, allocated in the anode as heat, is taken away by flowing water. The remaining energy is used in the excitation of X-rays. Two kinds of X-ray radiation are known. Characteristic radiation results from the return electron’ transitions from excited to ground state levels in atoms. This radiation has a linear spectrum and is widely used in modern science and technology for the analysis of chemical structures (refer to 7.6.4). Bremschtralung radiation arises at the instant of the electron stopping in the anode substance. According to classical theory, the distribution of a frequency (wavelength), arising due to electron stopping X-ray radiation should cover a wide range of spectrum from zero to infinity (as in thermal radiation spectra). Experimental results contradict this supposition: in Figure 6.40 the X-ray intensity versus their wavelength is plotted, the curves sharply terminating at the shortest wavelengths. An explanation of this fact can be found in the quantum theory of radiation. The law of energy conservation in this case can be written as: N
eU ∑ i .
(6.6.19)
i1
The electron energy before impact with an anode is written on the left-hand side of this expression; on right-hand side is the sum of all the photons’ energies, which appeared on collision. Since the process of electron braking is uncontrollable in this process, photons of all energies are produced, and the spectrum contains all wavelengths (so-called “white” spectrum). However, a limiting case exists when an electron gives all its energy to produce only one single unique photon. In this experiment this photon possesses the largest energy. It defines the boundary value of the wavelengths; photons of larger energy (smaller wavelength) in the bremschtralung X-ray spectrum cannot appear. For such a photon, expression (6.6.19) becomes simpler eU max
2c , min
(6.6.20)
cathode anode cooling water device to heat cathode X-rays high voltage transformer ground
Figure 6.39 Scheme of an X-ray tube source.
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10 50×8 8 40×8 I
6
4 30×8 2 20×8
r 0.02
0.04
0.06 .Å
0.08
0.10
Figure 6.40 The relation of the X-ray bremschtralung short-range limit versus wavelength .
whence min
hc . eU
(6.6.21)
The expression excellently coincides with experiment. The presence of a short-wave limit of X-ray radiation in the X-ray tube spectrum is a fact that cannot be explained by wave theory; in this experimental arrangement, the photon with larger energy cannot appear under the law of energy conservation: all the electron energy has already been given to the single X-ray quantum, the photon with smaller wavelength (greater energy) simply has no electron energy to appear; The Compton effect is another phenomenon contradicting classical wave theory. This effect arises on X-rays scattering by electrons weakly bonded to atoms. The scheme of the experiment is given in Figure 6.41. A beam of monochromatic X-rays (with wavelength 0) falls through a collimator onto a sample and is scattered. A special device investigates the intensities of both incident and secondary radiation scattered at an angle . Proceeding from wave theory, it follows that the scattered radiation should contain only one wavelength: the one that falls on the sample, i.e., 0. In fact, the electric field of an electromagnetic wave in the X-ray range should oscillate the electrons, which in turn should radiate secondary waves of the same wavelength. However, in the scattered radiation, experiment reveals that besides one unshifted 0, there is one more component, referred to as a shifted component with wavelength greater than 0 (Figure 6.42). It is experimentally established that the value of displacement (shift) 0 does not depend on the sample material and that this displacement is greater, the larger the scattering angle , namely,
0 (1 cos ) 2 0 sin 2 . 2
(6.6.22)
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Pe X-ray source
ricoil electron
sample
hk
diafragms spectral unit hk'
Figure 6.41 The Compton scattering experiment.
It is possible to explain all the features of the Compton effect if we consider them as a process of elastic collisions of X-ray photons with peripheral atomic electrons (in terms of the theory of particle collisions, see Section 1.4.5). We recall that, in elastic collision, both conservation laws (kinetic energy and momentum) are valid. Since a feedback electron can have a speed commensurable with the speed of light, it is more appropriate to use the relativistic theory (refer to Chapter 1.6) for the analysis. At an initially rested electron, weakly bonded to an atom (with its kinetic energy and momentum practically equal to zero), the photon falls with energy and a momentum k. In this case, the above-mentioned conservation laws in this case look like: h h (m m0 )c 2
(6.6.23)
⎛ ⎞ hc hc 1 m0 c 2 ⎜ 1⎟ . ⎜⎝ 1 2 ⎟⎠
(6.6.24)
the energy conservation law, and
the momentum conservation law, /c. For the feedback electron, having lost its bonding to the atom, the momentum relativistic expression pe can be given as: pe
m0 y 1 2
,
(6.6.25)
where is its speed. Using a vector diagram (Figure 6.43), projecting electron and photon momentums on x-axis, we obtain h h cos pe cos ,
(6.6.26)
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= 0.7 Å
= 90°
= 0°
ns
ns
sh
= 35° ns
= 135°
sh
ns
sh
Figure 6.42 Results of the measurement of the Compton effect versus the scattering angle : ns, nonshifted component; sh, shifted component.
and on the y-axis
0
h sin pe sin .
(6.6.27)
Exclude from the last expressions the electron parameters and we arrive at the final expression:
h (1 cos ) (1 cos ). m0 c
(6.6.28)
The expression h/m0c is referred to as Compton wavelength and is denoted as Λ. The same results can be obtained in the framework of nonrelativistic physics though the Compton effect belongs to the relativistic case. Both arrive at the same result, but the nonrelativistic derivation is simpler. However, when examining the feedback electron, it is necessary to use relativistic theory. For his discovery and explanation of the effect, A. Compton was awarded the Nobel Prize in 1927. At present this effect is used for the study of atomic valence electrons in the structure of the chemical compounds.
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m
hk θ hk'
Figure 6.43 Vector diagram of Compton scattering.
EXAMPLE E6.10 An electron runs accelerating voltage 104 V in an X-ray tube. Determine the wavelength corresponding to the short-wave limit of the bremschtralung spectrum of the X-ray radiation. Solution: A simplified construction of an X-ray generating device is given in Chapter 6.6. The accelerating electrons in the X-ray tube knock on an anode, X-rays being emitted in this process. In this example, a continuous spectrum is of interest. The shortest wavelength limit in this bremschtralung spectrum appears. It corresponds to the case when the whole electron energy transfers to a single X-ray quantum. Therefore, for this point the energy conservation law is valid eU hc/. Therefore, min
hc . eU
Substituting all the values we arrive at
min
o 6.63 1034 3 108 1.24 A. 19 4 1.6 10 10
EXAMPLE E6.11 Determine the cesium photoelectric threshold 0 if at its surface irradiation by violet light 400 nm a maximal speed of photoelectrons max is equal to 0.65 106 m/sec (refer to Section 6.6.4). Solution: The threshold corresponds to the situation where both the speed and energy of photoelectrons are equal to zero. Therefore, Einstein’s equation is
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A where A is the photoelectric work function and is the electron kinetic energy. We can obtain the expression 2c 2c A or 0 = ⴱ. 0 A The photoelectric work function can be determined using Einstein’s equation
A K max
2 2c m y max . 2
To execute calculations we should express all the values in the SI system: 1.05 1034 J sec, c 3 108 m/sec, 400 nm 4 107 m, m 9.11 1031 kg, max 6.5 105 m/sec. Calculations give us A 3.05 1019 J. To define photoelectric threshold 0, we should substitute the already known data and obtain 0 651 nm. EXAMPLE E6.12 A photon of energy 0.75 MeV is scattered by a nearly free electron at an angle 60°. Assuming that the electron’s kinetic energy and momentum before the collision were negligible small, define: (1) the energy ′ of the scattered phonon, (1) the kinetic energy K of the recoil electron and (3) the direction of its movement. Solution: (1) According to the Compton formula (refer to Section 6.6.4):
2 (1 cos ), mo c
we can express and ′ using energy of photons and ′: 2c 2c 2 (1 cos ) or m0 c 1 1 (1 cos ) . m0 c 2 Solving this equation regarding we obtain:
(1 cos ) 1 m0 c 2
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Executing calculations we arrive at 0.43 MeV. (2) The kinetic energy can be found from the energy conservation law K – 0.32 MeV. (3) The direction of the electron recoil motion can be found by applying the momentum conservation law (see Figure 6.43) p p m0. From a triangle OCD we can find CD
CA sin or OD OA CA cos sin p sin si n . tan p p p cos cos cos p tan
.0 (We use here the general equation p /c). Let us express (1 cos ) 1 m0 c 2
tan through the given data; therefore we find the ratio (1 cos ) 1. mo c 2 Hence tan
sin . ⎛ ⎞ 1 (1 cos ) ⎜⎝ m c 2 ⎟⎠ o
Taking into account some trigonometric relation we arrive at 2 . tan 1 m0 c 2 c tan
Executing calculations we obtain tan 0.701 and correspondingly 35. 6.7
THE BOHR MODEL OF A HYDROGEN ATOM
In previous sections, experiments which do not have explanations within the framework of Newtonian physics have been described. It is also known that reliably measured linear
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spectra of atoms, primarily hydrogen, also required the development of a completely new approach. This approach was suggested by Bohr, who formulated the theory of the hydrogen atom (Nobel Prize, 1922). Today it is considered to be semi-quantitative although it has not lost its significance. If we accept that atoms radiate electromagnetic waves with an energy h, it is necessary to establish where this energy originates from. We can equate energy h to the loss of energy E2 - E1 but it is then necessary to explain the nature of these energies. It is tempting to accept Rutherford’s planetary atomic model, but this seems impossible since it is known that the movement of a charged particle on a curvilinear trajectory undergoes a continuous loss of energy and the electron will inevitably fall onto a nucleus. However, in order to connect energy E2 and E1 with orbital movement, it is necessary to understand the stability of their orbits. The answers to all these questions were given by Niels Bohr in his theory of hydrogen atom. Bohr’s planetary model of the atom states that electrons in a hydrogen atom move in a circular orbit of radius r around a proton. The proton is so heavy in comparison with the electron that the center of mass of this system coincides with the position of the nucleus. Following Bohr, calculate the total electron energy. According to Newton’s second law 2 e2 , m r 4 0 r 2
(6.7.1)
where a Coulomb force electron–proton interaction equates to ma, a being centripetal acceleration. The kinetic energy K can be derived from this equation: K
m2 e2 , 2 8 0 r
(6.7.2)
whereas the potential energy of a negative charge in a field of the positive nucleus can also be found U (e)
e2 . 4 0 r
(6.7.3)
where is the nucleus electrostatic potential at distance r (refer to eqs. (4.1.21) and (4.1.22)). The total energy is then
E K U
e2 . 8 0 r
(6.7.4)
Since the orbit radius can apparently take on any value, so can the energy E. The problem of E quantization reduces to quantization of r.
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All the values listed above are unequivocally connected with r, other values are expressed with its help as well. In particular, the linear electron speed can be expressed as e2 . 4 0 mr
(6.7.5)
The frequency is
e2 . 2r 163 0 mr 2
(6.7.6)
The linear momentum p is p my
me2 . 4 0 r
(6.7.7)
L pr
me2 r . 4 0
(6.7.8)
And the angular momentum L is
Thus if r is known the orbit parameters K, U, E, , , p and L are also known. If any one of them are quantized all others must be also. Up to this point only classical physics has been used. Here, Bohr suggested that the necessary quantization of the orbit’s parameters shows up most simply when applied to the angular momentum and that, specifically, L can take on only values given by L n n
h , 2
(6.7.9)
where n can accept integer values 1, 2, 3, etc. The Planck constant appears again in a fundamental way. Combining eqs. (6.7.8) and (6.7.9) leads to r
h2o me2
n2
(6.7.10)
and E
i.e., to the energy quantization.
me 4 1 8 20 h 2 n2
(6.7.11)
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If we accept n 1, the well-known radius of the first orbit in H-atom can be obtained r0
h2o me2
.
(6.7.12)
Further, the atom radiates or absorbs energy only when an electron passes from one stationary orbit to another. This portion of radiation has been referred to as a quantum (of energy). As the energy is connected to the frequency of the quantum and, accordingly, to wavelength (in vacuum) the frequency of the quantum can be expressed through the quantum numbers corresponding to two orbits (j and k being their quantum numbers) (compare with (7.5.33)):
1⎞ me 4 ⎛ 1 2⎟. 2 3 ⎜ 2 8 0 h ⎝ j k ⎠
This is the famous serial formula which allows calculation of all the spectral lines in the hydrogen atom spectrum. An expression me4/802h3 is referred to as a Rydberg constant. The quantum mechanical theory of the hydrogen atom is given below (see Chapter 7.5) The Bohr model of the hydrogen atom is a transition from purely classical presentations to quantum mechanical ones: the motion of electrons along the orbits is accepted; however not all orbits are permitted, the angular momentum is accepted, though its values and orientations are subject to strict limitation. One can consider the Bohr model as the transition from classical mechanics to quantum mechanics with the preservation of many its attributes. As a result, many of the ideas of the Bohr model will often be met in order to simplify the students’ understanding. A typical quantum mechanical object such as an atom possesses some classical characteristics unexplainable within the framework of generally accepted presentations (no orbital motion, yet the existence of angular momentum; no rotation of an electron around its own axis, yet intrinsic angular and magnetic moments, i.e., spin, etc.). As a result, these terms are used irrespective of their classical sense.
EXAMPLE E6.13 In the framework of the Bohr model of the hydrogen atom (refer to Section 6.6.7) calculate the radius of the first electron orbit r and the linear electron speed . Solution: From Section 6.6.7, we know that r and are united in equation mr n (in our case n 1). In order to find two values one more relation is needed. For the rotation of an electron around a nucleus we can write the Newton’s second law equation
m
1 e2 y2 . 4 0 r 2 r
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or
my 2
1 e2 . 4 o r
Solving all equation we can find
r
4 o 2 me2
.
Substituting all known values, we arrive at r1 ao 5.29 1011 m and /mr 2.18 106 m/sec.
EXAMPLE E6.14 It is measured experimentally that the CuK X-ray beam ( 1.542 Å) being diffracted by a corundum single crystal deviates from its initial direction at an angle 41.66° (compare to Figure 6.20). The diffraction takes place from the crystallographic plane (600). Find the deviation angle from the same crystallographic plane using the MoK radiation ( 0.710 Å).
d
A
B
C
Solution: Bragg’s law should be used to solve this problem (refer to Section 6.3.5 and eq. (6.3.11), Figure 6.20). In the equation mentioned, is the incident angle and is the deflection angle. The picture is repeated here to make the situation clearer (Figure E6.14; an atomic arrangement is not depicted substituted by two reflected planes with interplanar distance d600). It is seen in the picture that 2. The crystallographic plane’s index is (600), in fact there is no such plane in the crystal; this should be understood as a 6th order of reflection from the plane (100) (i.e., at distance
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ACB in Figure 6.20 six wavelengths stack). Therefore, sin( /2) 6/2d600; this equation is valid for both wavelengths. Dividing two equations for two wavelengths we obtain (sin( Mo/2) sin( Cu/2)(Mo/Cu). Executing calculations we arrive at sin( Mo/2) 0.164, therefore, Mo is equal to 4.72°.
PROBLEMS/TASKS 6.1. The surfaces of a glass wedge form an angle 0.2. On the wedge perpendicular to its surfaces a beam of monochromatic light with wavelength 0.55 m falls. Determine the width of interference strips b (the distance between the adjacent maxima). 6.2. The diameters of two light Newton rings are d1 4.0 and d2 4.8 mm. It is known that three light rings settle between the two measured rings. The rings were observed in reflected light. Find the curvature radius R of the plane-convex lens. 6.3. In the experiment with Newton rings, a liquid oil was poured between a lens and a sample stage table, with its refraction index less than that of glass. The radius of the eighth dark ring is d8 2 mm ( 700 nm) whereas the radius R of the planeconvex lens is 1 m. Find the refraction index n of oil. 6.4. On illuminating diffraction gratings by a white light, the spectra of second- and third-orders partly overlap. On what wavelength in the second-order spectrum, does ultraviolet (UV) of the third order ( 0.4 m) fall. 6.5. A monochromatic light with wavelength 600 nm falls on diffraction gratings with the period d 10 m at an angle 30°. Find the diffraction angle corresponding to the second main maximum. 6.6. The energy flux radiated through a muffle’s sight hole is 34 W. Assume that the muffle radiates as an IBB and find its temperature T if its area is S 6 cm2. 6.7. Assume that the sun radiates as an IBB; calculate its emittance R and surface temperature T. The solar constant (the energy radiated by the sun per unit area measured on the outer surface of earth’s atmosphere) is C 1.4 kJ/m2 s. Assume the sun–earth distance to be d 1.49 1011 m. 6.8. Because of the change of temperature of an IBB, the position of the maximum spectral density emittance shifts from 1 2.4 m to 2 0.8 µm. How many times are the emittance R and the maximum spectral density of emittance r changed? 6.9. The temperature T of an IBB is T 2000 K. Calculate: (1) the spectral density of emittance r() for the wavelength 600 nm, (2) emittance R in an interval of wavelength from 1 590 nm to 2 610 nm. Assume that the averaged spectral density of emittance in this interval is equal to that for 600 nm. 6.10. Compton scattering of X-ray 55.8 pm occurs by graphite plate. Find the wavelength ′ of light scattered at an angle 60° 6.11. A photon ( 1 pm) is scattered by a free electron at an angle 90°. What part of its energy W does the photon transmit to the electron? 6.12. The spectral density maximum rmax of the emittance () of the bright star Arcturus corresponds to 580 nm. Assuming that the surface of this star emits as an IBB, determine its temperature T.
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6.13. Find the wavelength of a 1 MeV photon. Compare it with resting electron mass m0. 6.14. A photon’s wavelength is equal to Compton length C. Determine the photon energy and momentum p.
ANSWERS
6.1. 6.2. 6.3. 6.4. 6.5. 6.6. 6.7. 6.8. 6.9. 6.10. 6.11. 6.12. 6.13. 6.14.
b /(2n) 3.15 mm. R 880 mm. n 1.4. 0.6 m. arcsin (sin m/d) 38.3°. T 1 kK. R 64.7 MW/m2; T 5.8 kK. Increases R 81 and r 243 times. r() 30 MW/m2 mm, R 600 W/m2 . 57 nm. 0.70. T 4.98 kK. 1.24 pm, mph 1.8 1030 kg, pph 5.3.10 22 kg m/sec, mph 2moe. 0.511 MeV, p 2.7 1022 kg m/sec.
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–7– Elements of Quantum Mechanics
7.1 7.1.1
PARTICLE-WAVE DUALITY
De Broglie hypothesis
After the experiments and theoretical explanations that led to the revival of the corpuscular theory of light outlined in Section 6.6, it was logical to take the following step: restoring symmetry and assigning wave characteristics to those micro-objects that are known to us as particles. This step was taken in 1924 by the French theoretical physicist L. de Broglie (Nobel Prize, 1929). His idea consisted of the following. If quanta of electromagnetic radiation—photons—possess corpuscular characteristics, i.e., possess a mass (m E/c2), momentum (p = hk) and energy (E = h), the converse must also be true, i.e., a wave process must be related to a moving particle. De Broglie suggested a formula for the wavelength of such a process:
h , m
(7.1.1)
where h is the Planck’s constant, m is the particle mass and is its velocity. These waves are called “de Broglie waves.” It must be said that no one can imagine what lies behind this statement. We will see here a case that is quite typical in quantum mechanics: our normal imagination is not adequate to conceive of the main principles of this science. There is certainly an explanation for this effect. Indeed, let us imagine a 10 g bullet moving at a speed of 1000 m/sec. According the formula (7.1.1), the corresponding wave process (de Broglie wave) has the wavelength ~ (1034/(102 103)) ~ 1035 m . In order to reveal the character of such a wave, we must find a grating with a comparable period. Certainly, there are no such gratings. Generalizing, it can be said that in the everyday world there is no way to prove that de Broglie waves corresponding to a moving body can be visualized. Therefore, it is purely academic. This is why the referee of de Broglie’s thesis, the well-known French physicist P. Langevin, was not able to evaluate de Broglie’s formula and stated “de Broglie’s idea is either the fruit of his morbid imagination or, conversely, a brilliant idea, though in both cases it deserves high recognition.” The Nobel Prize was the relevant award. 423
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Quite a different situation occurs in the micro-world. Here, because of small particle masses, the de Broglie wavelength is commensurate with the interatomic distances in crystals; therefore, the corresponding experiment can be realized: a crystal can serve as a diffraction grating (refer to Section 6.3.5). In fact, according to formula (7.1.1) and the conservation law we can obtain
h h 150 ⬃ 1010 m, 1 2 m (2 meU ) E
(7.1.2)
i.e., a voltage of 150 V is needed to produce an electron beam with a wavelength of 1010 m and to observe diffraction. It is quite natural to ask the question: what do the microparticles (electrons, neutrons, etc.), on one side, and the phonons on the other represent? The modern development of science enables it to be stated that all micro-objects simultaneously possess a set of properties, among which both wave and corpuscular properties are equally present, and they reveal them depending on the conditions to which an experimenter subjects them. This statement is the essence of corpuscular-wave dualism, which in turn is the basis of quantum mechanics. 7.1.2
Electron and neutron diffraction
Only 2 years were needed to prove electron diffraction experimentally. In 1927, simultaneously in several laboratories in different countries, attempts were made to observe electron diffraction. Davisson and Germer in America, Thomson in England and, several years later, Tartakovsky in Russia observed electron diffraction after the interaction of electrons with single crystals (C.J. Davisson and G.P. Thomson, Nobel Prize 1937) and polycrystalline films. In a simplified representation, the principles of an electron diffraction device are depicted in Figure 7.1. The main part of the instrument is the vacuum column, in which all the elements of the device are contained. The heating filament 1 (cathode) emits electrons by thermoemission, which then speed in an electrostatic field with the potential difference U. Passing through the diaphragm, monochromatic electrons (i.e., electrons with constant wavelength, refer to formula (7.1.2)) fall onto the polycrystalline film sample 3. The polycrystalline sample contains an enormous number of small microcrystals, absolutely chaotically oriented in space. From the whole set of microcrystals, some are oriented with their crystallographic planes p1, with interplanar distance d1 with respect to the incident beam of electrons at an angle 1, for which the Bragg condition is met (refer to 6.3.5). Planes p1 of this crystal will give a reflected beam with an angle 21 with respect to the direction of the primary electron beam. In a polycrystalline sample, an ensemble of other microcrystals can always be found which have the planes p1 identically oriented with the same reflecting angle ; all the p1 planes look like they are rotated around the primary beam direction. Therefore, all the reflected beams are situated on a cone around the direction of the primary beam with a cone opening angle 41. Each electron colliding with the fluorescence screen 5 produces a flash; all the flashes create a diffraction ring.
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P1
1 2
Θ1
3 2Θ1
4Θ1
4 5
Figure 7.1 An electron diffraction investigation scheme.
Under known instrument parameters (distance from the sample to the screen L, applied voltage U, etc.), the diffraction circle diameter D is connected with the interplanar distance d (refer to Sections 6.3.5 and 9.1) by the formula
nh 2 meU
2 d sin ⬵ 2 d
dD , 2L
and, accordingly, d
2 Lh D 2 meU
,
(7.1.3)
where m and e are the mass and charge of the electron. There are other crystallographic planes p2, p3 and p4 in the polycrystalline sample too, with other diffraction cone opening angles 42, 43 and 44, respectively. As a result, the typical electron diffraction pattern looks like a set of concentric circles, all circles having a different intensity. The set of interplanar distances di and the corresponding intensity Ii of each circle characterize the material analyzed. The diffraction pictures of MgO films obtained by means of the diffraction of X-rays (usually accepted as waves) and electrons (usually considered as particles) are compared in Figure 7.2; since they were taken using different wavelengths, for the best comparison both pictures were brought to the same scale. (Spots on the electron diffraction rings are caused by the presence of partial crystalline order in the MgO film.) The similarity of the diffraction images made with electrons and X-rays can clearly be seen in this picture. In 1932, the English scientist J. Chadwick discovered a neutral elementary particle—the neutron with a mass practically equal to the proton mass (Nobel Prize, 1935). Experiments to prove neutron diffraction immediately followed. In 1936, such experiments gave a positive result. However, in the modern view as it is currently used, neutron diffraction as a method of structure investigation became possible only after nuclear reactors were set up. Neutron diffraction became one of the most powerful methods for the analysis of
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Figure 7.2 A MgO powder diffraction patterns obtained by X-ray (left) and electrons (right).
solids. In 1995, the American physicist C. Shull, who contributed most to the development of the method of neutron diffraction in physics and chemistry, was awarded a Nobel Prize. Along with neutron diffraction, experiments of inelastic neutron scattering (where the neutron behaves as a material particle interacting with phonons (refer to Chapter 7.3), according to collision theory (refer to Section 1.5.5)), were also realized. For the discovery of inelastic neutron scattering and establishing a new method of solid state investigations, the Canadian physicist B. Brockhause was awarded the Nobel Prize (1995). Thereby, the de Broglie hypothesis has been widely accepted as a scientific law. To reiterate, particle-wave dualism is one of the cornerstones of quantum mechanics.
EXAMPLE E7.1 An electron acquires a kinetic energy K under an accelerating voltage U. Determine its de Broglie wavelength for two cases U1 51 V and U2 510 kV. Solution: The electron wavelength depends on its momentum p: (2/p)*. In the case of classic mechanics the momentum depends on energy p 兹2苶m 苶苶 0K ; however, for a relativistic case this relationship is p (1/c)(兹(2 E K )K ). The rela苶苶0苶苶苶苶 tion * for two cases mentioned can accept the forms 2/(兹2苶m 苶苶 0K ), and correspondingly 2/冢(1/c) 兹苶2苶 E苶 苶 K 苶 )K 苶 冣. 0 Compare the electron’s kinetic energy for two cases and compare them with the electron’s rest mass; the kinetic energy is 51 eV and 0.51 MeV; the electron’s rest mass is just 0.51 MeV. Therefore, the second case corresponds to a relativistic one. The wavelengths for these cases are: 1 2/eU1 172 and 2 2/ 兹苶3m0c 1.4 pm.
EXAMPLE E7.2 Assume that on a narrow slit of a width a 1 m, a plane-parallel electron beam of very low intensity, having speed 36.5 106 m/sec, is directed (Figure E7.2,
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refer also to Figure 6.15). Taking into account the electron wave properties, define the distance between two first minima of intensity of the first order (that is width of a zero maximum) in a diffraction spectrum. Assume the distance from the slit to the screen to be L 10 cm.
a
L
Ltg x
Solution: Assume that according to de Broglie hypothesis, a wave of wavelength 2/m* is attributed to a particle of mass m moving with a speed . Therefore, diffraction should be observed. The diffraction minimum in the spectrum can be found according to the relation a sin (2k1)(/2) (refer to Section 6.3.2), where k indicates the diffraction order (in our case k 0; Figure E7.2). Besides diffraction angle is small, sin ⬇ tan⬇ and therefore, a (3/2). The sought distance x on the screen is . x 2 L tan ⬇ 2 L (3 2)(a ) 3L a. Using the equation * we obtain x
6L a1 . m
Executing calculations, we obtain x 6 .105 m. This means that a spectator can find a particular electron in this range, i.e., the uncertainty to locate the electron’s position is x.
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EXAMPLE E7.3 A plane-parallel incident electron beam falls on a crystallographic plane of a nickel single crystal at an angle of 64°. Diffraction takes place. Accepting the interplanar distance of the given crystallographic planes as d 200 pm, define the de Broglie electron wavelength and their speed . Solution: Diffraction of X-rays is described in Section 6.6.4; electrons exhibit the same property. The Bragg’s equation is applicable to this case 2d sin n, n being the diffraction order, n 1 in our case. From this equation we can find the wavelength 2d sin 360 pm. From the de Broglie relation the electron’s speed can be found:
2
7.2
2 Mms. m
HEISENBERG’S UNCERTAINTY PRINCIPLE
Particle-wave duality entails important consequences. The question is, can a micro-object simultaneously possess precise values of its coordinate and momentum? Indeed, a certain internal self-contradiction exists between the characteristics of a material particle that can be localized in space with arbitrary accuracy and the monochromatic de Broglie wave, which according to its nature is extended from ∞ to ∞ and is completely delocalized in space. However, it is exactly the latter that possesses a certain, exact wavelength and, accordingly, fixed momentum. Quantitative analysis allowed W. Heisenberg in 1927 to suggest the principle (Nobel Prize, 1932), which nowadays is given as follows: there exists no state in which the coordinates of a microparticle and its momentum have precise values. If a micro-object is traveling along the x-axis, one can characterize the uncertainty of the coordinate and component of momentum by values x and px; then the Heisenberg principle (for coordinate and momentum) has the form x p x h,
(7.2.1)
i.e., the product of the uncertainties in the coordinate and corresponding component of momentum cannot be less than h. It is possible to apply another interpretation to the Heisenberg uncertainty principle. It is well-known that a wave can be characterized by the precise value of the wavelength, when it spreads from ∞ to ∞. It is also known that such a wave is a mathematical abstraction; any wave has an origin and an end. Accordingly, this model corresponds to the precise values of the wavelength (and wave vector k) and, consequently, momentum p. It means that in this case, uncertainty in the momentum px is zero (Figure 7.3, on the left). As a result, we are unable to assign any position to the particle; its uncertainty is equal to infinity. Certainly, the product of x and px can be less than h. Such values of uncertainties save an equitable correlation of Heisenberg uncertainties.
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∆px ∆px −∞
∆x
p +∞
∆x
Figure 7.3. An uncertainty principle for x and px: the higher the particle localization, the lower the determination of its momentum.
However, if we try to reduce the uncertainty in the particle’s position and put it in a state when x becomes any finite quantity, say a value L (Figure 7.3, middle), the product L 0 cannot be larger than h. In other words, this will bring about the appearance of an uncertainty in the momentum, which is displayed in Figure 7.3 (on the right) in the form of a curve, the maximum width of which is x (which can be evaluated as h/L). The Heisenberg uncertainty principle imposes essential restrictions on some laws of classical mechanics. In particular, these affect a very important notion such as trajectory. As an example, let us consider a hydrogen atom within the framework of the Bohr model: an electron revolves around a proton along a circular orbit. From the known electron mass and charge, as well as the electrical constant, within the framework of classical electrodynamics one can define (at the order of magnitude) the linear electron velocity, which turns out to be approximately 106 m/sec. Then the uncertainty in the coordinate x is x
h h 1034 30 ⬇ 1010 m, p m 10 106
i.e., it coincides with the atomic size. One can conclude, therefore, that the notion “trajectory” in this case (and in quantum mechanics, in general) loses it’s meaning: the uncertainty in the electron coordinate is comparable with the object size. It is clear that a new approach to describing micro-objects is necessary. The principle of uncertainty itself allows one, in some cases, to arrive at the decision not to solve the problem exactly. As an example, one can consider the state of a particle limited in its motion in space (i.e., existing in a potential well, refer to Section 1.5.4) by the width L. Let us pose a question: can the particle energy accept any values or an undetermined one in this case? Can a particle “settle down to the bottom” (i.e., possess exact (zero) energy and, accordingly, exactly determined momentum)? In order to decide, let us choose an uncertainty in the momentum: let this uncertainty be equal to 100%, i.e., it will accept pp. Bearing in mind the relationship of energy E with the momentum, we can write: pp 兹苶2苶 m苶 E. The uncertainty in the coordinate x in this case is the width of well L: we know that the particle is in the potential well, but do not know precisely at what point. As a result, the uncertainty principle looks like x px (兹苶2苶 m苶 E) L h, whence E
h2 2mL2
.
(7.2.2)
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This shows that the answer to the question above is as follows: a particle cannot occupy a position on the bottom of a potential well of finite width, and the expression derived presents the lowest permitted value of energy. This solution has been reached only by the uncertainty principle, without using the main attributes of quantum mechanics. As we will see below, this conclusion complies well with the result of the exact solution of this problem. The uncertainty principle also offers the energy E of a micro-object and the lifetime of a system in this state: the product of uncertainty in energy E and in the time the system exists in this state cannot be less than h (7.2.3)
E h.
For the ground state of a micro-object that can exist in this state infinitely long (t∞), the uncertainty in energy E is zero, i.e., the energy of the ground state can have an absolutely precise value. Though for excited states with a limited lifetime of, say 108 sec, the uncertainty in energy is: E ⬇ 1034/108 1026 J⬇107eV. This is a very small value: however, in some cases it plays an important role in the physical phenomenon. In Figure 7.4, the illustration of broadening of a spectral line on account of uncertainty principle (for energy and time) is depicted. The linewidth caused solely by the energy level broadening because of the Heisenberg uncertainty principle (i.e., not subjected to the influence of an instrumental imperfection), is called a natural spectral linewidth. It is necessary to note that the uncertainty principle does not impose any restrictions on the possibility of the simultaneous existence of precise values of coordinates and momentums along different coordinate axis. In other words, products y px and x py can be equal to zero, i.e., the corresponding values of coordinates and component of momentums can be determined precisely.
E1
E1
E0
E0
I Γ = 2∆E/h
Figure 7.4 An uncertainty principle for energy and time of life. Two energy levels are presented (left) without uncertainty principle accounting (the both are infinitely narrow and a spectral line is infinitely narrow too); taking the principle mentioned into account leads to the broadening of the width of both an excited energy level E1 and spectral line ( is the natural width of a spectral line).
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EXAMPLE E7.4 An electron’s kinetic energy in a hydrogen atom is of the order 10 eV. Using the Heisenberg uncertainty principle, determine the linear size of the hydrogen atom. Solution: The momentum and coordinate uncertainties are related by eq. (7.2.1). Let the atom have a linear size l, then the electron will be somewhere in the limits xl/2. The uncertainty principle can then be written as:
2 . p
Suppose that the physical reasonableness p should be less than p: p p. Therefore, we can assume p ⬇ 2 mK
and min
2 2 mK
.
Substituting the known and given values, we arrive at lmin 124 pm, i.e., commensurable to the atom’s size.
EXAMPLE E7.5 Using the uncertainty principle of energy and time, define the natural width of a spectral line in the excited state and at its transition from the excited state to the ground state. Define also the corresponding . Assume equal to 108 sec and wavelength 600 nm. Solution: The natural width of the energy level E and the time of its life are related by the Heisenberg ratio . Therefore the natural width is / , i.e., ⬇
1034 1026 J. 108
Since the photon energy and the wavelength are related as
2c ,
the uncertainty relation can be found by
2c 2
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(the minus sign is ignored). Hereafter,
2 . 2c
Substituting the given value, we arrive at 2 1014 m
7.3 7.3.1
WAVEFUNCTION AND THE SCHRÖDINGER EQUATION
A wavefunction
What we have discussed above makes it necessary to develop another approach to the description of micro-objects, different from that in classical mechanics. This was done by the Austrian physicist E. Schrödinger (Nobel Prize, 1933), who (together with P. Dirac and W.K. Heisenberg) suggested the idea of the wavefunction ψ(x, y, z, t) as well as, an equation that this wavefunction should obey. Accordingly, a microparticle state in quantum mechanics is defined by the wavefunction ψ(x, y, z, t). Knowing the wavefunction, for instance for electrons in an atom, one can define their behavior when changing the chemical bonding between atoms, the probability of forming one or other molecular structures, judging the strength of interatomic bonding in molecules, etc. Thereby, the wavefunction of electrons is a key to deciding many principle problems of chemistry. However, function itself has no physical sense; it cannot be measured itself. The wavefunction presents a certain mathematical expression by means of which it is possible to find the probability of one or other real physical features of electrons in atoms and molecules. In order to calculate these probabilities, we need to use a value *, where * is a complex conjugate with , since in general can be a complex function. In a particular case, when a wavefunction is real, * 2. We will restrict ourselves to some limitations: we will only deal with those problems that do not depend on time; an electron spin will be incorporated further, see Section 7.5.5; besides, we have considered so far only nonrelativistic problems. The physical meaning of ⱍ ⱍ2 consists in the following: ⱍ (x, y, z)ⱍ2 is proportional to the probability dw(x, y, z) of finding a particle in the elementary volume dV in the vicinity of a point with coordinates x, y, z: 2
2
dw( x, y, z ) ( x, y, z ) , dV ( x, y, z ) dxdydz.
(7.3.1)
The square of the wavefunction | (x, y, z)|2 is hence the probability density (i.e., the probability related to the unit volume) of finding a particle in a point x, y, z: dw( xyz ) 冷 ( x, y, z ) 冨2 dV
(7.3.2)
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Certainly, wavefunction can be dependent on time, because physics knows many problems that are time dependent. However, according to our task, we will consider only stationary processes, not depending on time: the force fields in which particles move are stationary (refer to (1.4.4), i.e., they do not depend on time). It is possible to show that in this case the wavefunction (x, y, z, t) disintegrates into two factors, one of which depends on coordinates only and the other cyclically depends on time: ( x, y, z, t ) ei t ( x, y, z ) e(i ) Et ( x, y, z ),
(7.3.3)
where is frequency, and E is the total particle energy. In this case ( x, y, z, t ) ⴱ ei t e i t ⴱ ( x, y, z ) , 2
2
(7.3.4)
i.e., the wavefunction square is also time independent. 7.3.2
The Schrödinger equation
The wavefunction is the solution of a certain equation that was introduced by E. Schrödinger—the Schrödinger equation. This is the main equation in quantum mechanics. In general, it cannot be derived theoretically; however, its validity is proved in practice: the results obtained in solving this equation are confirmed in numerous experiments. Here it plays the same role as the second Newtonian law in classical physics (refer to Section 1.3.3). For stationary nonrelativistic problems, the Schrödinger equation can be written as follows:
( x, y, z )
2m [ E U ( x, y, z )] ( x, y, z ) 0. 2
(7.3.5)
Besides the well-known notions, a new one is presented in the equation: the Laplacian operator that is the sum of the second partial coordinate derivatives acting on the function that follows it (in our case on wavefunction (x, y, z)): ⎛ 2
2
2 ⎞ ⎜ 2 2 2⎟.
y
z ⎠ ⎝ x
(7.3.6)
If a one-dimensional problem is being solved, only the first term of eq. (7.3.6) is used. The term U(x, y, z) is the particular particle potential energy in the force field; the information on the particular type of problem is concentrated exactly in this term. In order to solve a quantum mechanical problem, we should substitute an analytical expression for the particle potential energy in the given force field, U(x, y, z), in eq. (7.3.5); find the values of parameter E at which the Schrödinger equation allows solutions and calculate the wavefunction (x, y, z) as a solution of this equation. The parameter E stands out as the particle energy.
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Standard requirements that the wavefunction should obey
In all cases, the wavefunction (x, y, z) as the solution of the Schrödinger equation should possess some properties that themselves play an important role in quantum mechanics. Before describing the methods for solving Schrödinger equation, we have to analyze the standard requirements (conditions) that a wavefunction should obey; these requirements result from the type of equation and the physical sense of the functions. This especially concerns cases when the problem should be divided into pieces because of the complexity of the potential distribution. So, the wavefunction (x, y, z), in any case, must be a single-defined (univocal) function of coordinates, finite and continuous in the whole range of variable coordinates x, y, z, including infinity. In the case of the potential energy partition, the most important is that (x, y, z) must join smoothly at the boundaries. The wavefunction must be finite, i.e., we require that (x)t0 at xt -. As far as the Schrödinger equation contains the coordinate second derivatives, the first coordinate derivative (x, y, z) should be continuous also. All of these are obvious from physical and mathematical considerations. Certainly, | (x, y, z)|2, and, consequently, (x, y, z) , must have no more than a single value under fixed coordinates x, y and z or under such operations which return a particle into the former point of space (requirement of univocacy). Otherwise the solution would be ambiguous, which would make no sense. Furthermore, as we know that the micro-object really exists in the whole region, the wavefunction (x, y, z) should obey the normalization condition. The total probability of finding a particle in the whole range of variable must be unity, i.e.,
∫ dw ∫∫∫ dw ∫∫∫ ( x, y, z)
2
dxdydz 1.
(7.3.7)
The properties mentioned are called the standard requirements (conditions) that the wavefunction should obey. They result from the type of equation and physical sense of these functions. The requirements imposed on the wavefunction have the result that the Schrödinger equation, as a second-order differential equation in partial derivatives under the given U(x, y, z), can have solutions satisfying them only at definite values of E, values which play the role of parameters in the equation. Values of E, under which the Schrödinger equation has solutions satisfying the standard conditions, are called eigenvalues of energy. The set of eigenvalues E forms an energy spectrum. If eigenvalues form a set of definite values E1, E2, E3, etc., the energy spectrum is called a discontinuous one. If energy E can accept any values, the energy spectrum is called a continuous one. Below we will see that a discrete energy spectrum appears when a particle motion is restricted in space (motion is finite). In the case of infinite motion, the energy spectrum is continuous (energy can change continuously). The wavefunctions N that correspond to parameters EN are called the eigenfunctions.
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MOST GENERAL PROBLEMS OF A SINGLE-PARTICLE QUANTUM MECHANICS A free particle
Let us consider first the simplest case: free particle motion. Free motion means the case with U ⬅ 0, i.e., the motion of a particle on which no irrelevant forces are exerted. Such a particle is moving uniformly and in a straight line. Let the x-axis be directed along the particle motion. The Schrödinger equation in this case has the form 2m d2 ( x ) 2 E ( x ) 0. 2 dx
(7.4.1)
As here depends on only one variable, the partial derivatives are replaced onto the full derivative. The energy in this case is kinetic energy, because potential energy for free particles is zero. Therefore, E
p2 , 2m
(7.4.2)
where p is particle momentum. Let us introduce the definition
2 mE k2. 2
(7.4.3)
( x ) k 2 ( x ) 0.
(7.4.4)
Then the equation accepts the form
The solution of such an equation is well-known: ( x ) A exp(ikx ),
(7.4.5)
where k p/ (refer to Section 1.6) and, correspondingly, k 2/ (eq. (2.8.4)). Because of the fact that we are dealing with the uniform Schrödinger equation and the derivation from the wavefunction is taken on the coordinate x, this solution can be multiplied by any coefficient not depending on the coordinate. Let this coefficient be ei t. Then the solution accepts the form (x,t) A exp(ikx) exp(i t) A exp(i tikx). The real part of it is therefore the wave running in x direction ( x, t ) Acos(i t ikx ),
(7.4.6)
(refer to 2.8.2). It is of use to express this expression via the energy and momentum 1 ( x, t ) A cos ( px x Et ).
(7.4.7)
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The calculation of the probability density to find a particle on any point of the x-axis gives dw( x ) ⴱ A exp(i t ikx ) A exp(i t ikx ) A2 , dx
(7.4.8)
i.e., the probability density is uniform along the whole x-axis. Thereby, if a particle momentum is strictly defined (as in our case), its position in space is indefinite: there is no priority to find it on any point of the x-axis, the particle is as if it is smashed along the x-axis. It is worth noting that this result completely corresponds to the uncertainty principle (refer to Section 7.2). Correlation (7.4.6) allows us to express carefully the statement (not attempting any of the serious generalization), that the wavefunction (x, t) in the form of waves can be called the wave of probability, although this statement certainly does not make the physical picture clearer. The real physical sense has only * 2. The solution obtained satisfies a standard condition: the energy E and particle momentum can accept any magnitude (the spectrum of energy is continuous). Motion is infinite and energy, as in classical physics, can accept any value. We must pay attention to one more important property of the Schrödinger equation (7.4.1), apparent from its uniformity: any superposition of solutions is also a solution of this equation. Such a superposition property often leads to what in chemistry is called hybridization.
7.4.2
A particle in a potential box
Let us consider now the state of a particle placed in a one-dimensional square potential box. This problem is not a far-fetched abstraction, but presents a model of a bound molecule particle. In Section 1.5.4, the Lennard-Jones potential was considered (and below the Morse potential will be used). It was shown that under the total energy E lower than the depths of the potential well, a particle makes an oscillatory motion (whether it is harmonic or anharmonic plays no essential role at present) near its equilibrium position. Exactly such a potential function has to be substituted in the Schrödinger equation to decide this problem. To solve such an equation in analytical form, however, is impossible at the moment. Therefore, the potential of a complicated profile is approximated by a model, the consideration of which, certainly, does not solve the problem precisely, but nevertheless allows U
r
Figure 7.5 A Lennard-Jones’ and a model’s potentials.
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U
I
II 0
III L
x
Figure 7.6 A one-dimensional infinitely deep potential box.
us to make a number of important conclusions. In Figure 7.5, the Lennard-Jones potential is approximated by a rectangular, one-dimensional potential box. In Figure 7.6, this box is provided in the form in which it will be used below. First consider an infinitely deep potential box of the width L. Beyond the borders of the box, potential energy U is infinitely large, hence the particle cannot exist there. Inside, the potential U can be taken as zero. As a result a particle in the potential box is considered, the latter being assigned as follows (Figure 7.6): U(x) ∞ at x 0 and at x L, U(x) 0 at 0 x L. Accordingly, three areas are presented in the scheme: I, II and III. In areas I and III, the particle cannot exist, the wavefunctions there are zero: I (x) III(x) 0. In area II the potential energy is zero (U 0) and the Schrödinger equation gains the type II ( x ) kII2 II ( x ) 0, similar to eq.(7.4.4). Solution of the equation in this area gains the form II ( x ) A sin kx B cos kx.
(7.4.9)
From standard conditions for wavefunction continuity, it follows that at the point x 0, the wavefunctions on the left and on the right sides are zero: I(0) II(0). It follows that 0 A 0 B 1, hence B 0. At the point x L the same condition requires A sin kL to be zero. This signifies that kIIL n, where n is an integer. Taking the expression (7.4.3) into account this equality can be rewritten as En
2 2 2 n . 2 mL2
(7.4.10)
The fundamental result has been obtained: the particle energy in an infinitely deep potential well can accept only discrete values! In other words the bound particle energy is quantized. The integer n, incorporated earlier as an arbitrary integer, now stands out as a certain parameter that determines the value of energy. This integer is called the quantum number. The position of the energy levels of a particle in the infinitely deep potential well is given in Figure 7.7.
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E
0
n
n2
5
25
4
16
3
9
2 1
4 1
Figure 7.7 The derived energy levels of a particle in a one-dimensional infinitely deep potential box.
It is worth noting that the quantization of energy in this instance has resulted not from mathematical deduction but by using one of the standard conditions of wavefunction continuity. A result of the same significance (7.2.2) was obtained earlier in the analysis of the problem by means of the uncertainty principle; the particular expression differs from eq. (7.4.10) by only a constant multiplier nearer to unity. We want to emphasize this logical unity and intercoupling of different approaches in quantum mechanics, including the standard conditions the wavefunction must obey. Each of these aspects can be used in different cases when deciding a particular problem. In order to obtain the total final expression for the particle wavefunction in the infinitely deep potential box, one has to define the normalizing multiplier. The normalizing condition (7.3.7) can be used here. In this instance it looks like L
∫ 0
2
L
( x ) dx A2 ∫ sin 2 kx dx = 1.
(7.4.11)
0
Replacing the integrand according to the trigonometry formula sin2kx (1/2)(1/2) cos2kx, we can arrive at A2 2/L. Finally, the expression for the wavefunction of a particle inside the rectangular infinitely deep potential box is given as II ( x )
2 n sin x. L L
(7.4.12)
The graphs of the wavefunctions (a) and their squares (b) of the particle in the infinitely deep potential well in different quantum states are depicted in Figure 7.8. As a matter of convenience, they are distributed along the ordinate axis corresponding to the quantum numbers. The specific question is, can n be equal to zero? This is equivalent to the question already discussed in Section 7.2: whether a particle can “lie” on the bottom of a well; to which we have already obtained a negative answer from the uncertainty principle.
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U
U
439
n=3
n=3
*
n=2
n=2
n=1 n=1
0
x
L
0
x
L
Figure 7.8 Wavefunctions of a particle in a one-dimensional infinitely deep potential box (a) and their squares (b).
Graphs of n(x) are extremely reminiscent of graphs of standing waves and, particularly, the string oscillations (refer to 2.9.3). There is nothing amazing in this fact: moving in the potential box, particles are reflected from potential barriers on the right and on the left so that their wavefunctions interfere, forming standing waves. Half of the standing wavelength is equal to L/n (refer to formula (2.9.8)). Only in such a case is a state stable! This example is suitable to illustrate the Bohr correspondence principle. It states that all regularities of quantum mechanics turn into the regularities of classical mechanics under the increasing quantum numbers. It is well known that the different levels of physical approximations are characteristic to certain areas of this science. Transformation from one area to another occurs not abruptly, but gradually. So, Newtonian mechanics becomes less and less exact when the velocity of particle motion increases, transforming into the relativistic one. We are interested here in the transition from quantum mechanics (in which quantization plays a fundamental role) to classical (in which the energy levels discontinuity is not observed). Let us start from the formula for energy (7.4.10). From this formula and Figure 7.7, one can see that the distance between adjacent levels increases with increasing the quantum number n. However, if one analyzes not the absolute but the relative value of energy, this fraction with increasing n decreases. In fact, (En1En)/En E/E and E (n 1)2 n2 2 n 1 1 2 ⬇ 0, E n n2 n
(7.4.13)
manifestation of quantization decreases rapidly. Consider now (on the qualitative level) what will occur if the depth of a potential box becomes finite; let it be denoted U0 (Figure 7.9). Suppose herewith, that the total particle
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U
U0 (x)
ekx
sin kx
e−kx
E
0
L
x
Figure 7.9 A wavefunction of a particle in a one-dimensional potential box of a definite depth.
energy E is less than U0, i.e., the particle remains bounded. In this case the potential U in the Schrödinger equation (7.3.5) will not be equal to infinity and the wavefunction in areas I and II will not be zero. Moreover, the expression in brackets (E – U) is negative. In order to avoid the operation with complex qualities let E and U change their places in the brackets ( x ) [U 0 ] ( x ) ( x ) k 2 ( x ) 0.
(7.4.14)
This problem can be solved exactly, but we will restrict ourselves to qualitative consideration. From the given equation, one can see that the second derivative from the wavefunctions on the coordinate in areas I and III must have the same sign as the function itself. An exponent AIexp( kIx) satisfies this condition. In the area I (∞ x 0), the requirement for the wavefunction to be finite corresponds to a sign “” in the exponent. Therefore, the solution for I(x) must be of the form AI exp(kIx). In area III (L x ∞), for the preservation of the wavefunction to be finite, the sign in the exponent must be negative: III(x) AIII exp(–kIIIx). Constants AI and AIII are to be determined from boundary conditions I(0) II(0) and II(L) III(L). The solution for wavefunction expression and conditions of quantization are, certainly, changed, though not in principle. The graph of relationship 2(x) for n 1 is schematically depicted in Figure 7.9. The most important point is that the particle wavefunction is not zero even in those areas of space where E U0, i.e., where the total energy is less than the potential or where the kinetic energy is formally negative. In classical physics, this cannot be the case: total energy must always be higher than potential since kinetic energy cannot be negative; the particle cannot exist where this condition is not fulfilled. The explanation of this apparent confusion is contained in the uncertainty principle! The fact is, potential energy depends on the particle positions, but kinetic energy is a function of momentum. Therefore, a particle in quantum mechanics cannot simultaneously possess precise values of potential and kinetic
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energy. Quantum mechanics operates mainly with total and potential energy, and only in some cases considers kinetic energy when a system does not possess potential energy at all.
EXAMPLE E7.6 An electron is in a one-dimensional potential box of infinite depth and width of l. Calculate the probability to find the electron in the excited state (n 2) in the middle third of the box (see Section 7.4.2) and Figure E7.6).
n=2
Solution: The probability to find a particle in an interval x1 x x2 is defined by an integral x2
w ∫ n ( x ) dx, 2
x1
where (x) is normalized wavefunction corresponding to a given state. In our case this function is 2 n n ( x ) sin x (n 2) (eq. (7.4.12)). Substitute all given values into this equation, we obtain x
w
2 2 2 2 sin xdx. x∫ 1
In our case x1 (l/3) and x2 (2l/3). Using a trigonometric equation, sin x 2
2 1⎛ 4 ⎞ x ⎜ 1 cos x 2⎝ ⎟⎠
we arrive at w 0.195. 7.4.3
A potential step
The results we obtained in preceding sections can help us to analyze quantitatively some other quantum mechanical problems, not having analogues in classical physics. We will consider further a potential step and a potential barrier.
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U(x) |(x)|2 U0
A2I
A2Iexp(-2k IIx) E
I
II
x
Figure 7.10 A rectilinear potential step.
The graph of potential energy as a function on the distance (along the axis of particle motion x) in quantum mechanics is referred to as a potential step: U(x) 0 at x 0 and U(x) U0 at x 0 (Figure 7.10). This problem can be solved at two total energy E: higher and lower the heights of the potential step U0. The latter case E U0 corresponds to an infinite particle motion; consequently, the energy E can accept a continuous spectrum of values. We will move our attention to another case with E U0. The problem can be solved analytically in the framework of Schrödinger’s equation and the standard conditions for (x), in particular using the requirement to be continuous at the step boundaries (at x 0). As in preceding cases, divide the problem into two parts: x 0 (area I) and x 0 (area II). In area I the particle motion is infinite; it can be described by the periodic wavefunction I(x) (7.4.2) with constant probability of finding a particle in any point of this area. Let us denote the amplitude of the de Broglie wave AI, then ⏐ I(x)⏐2 AI2. In area II the energy E U0, therefore all that has been said in Section 7.4.2 about the wavefunction behavior outside a potential box of finite depth (see above) is applicable to the particle falling onto the potential step (Section 7.4.2). The Schrödinger equation is the same as above (eq. 7.4.14). The solution of this equation is analogous to the case of exponential decrease of the probability of finding the particle while moving away from the boundary, i.e., AII exp(-kIIx). This wavefunction corresponds to the probability reduction of finding a particle under the step at x 0. From the condition of the wavefunctions continuity, it follows that AI AII. In Figure 7.10 a graph of the wavefunction⏐ II(x)⏐2 AI2 exp(–2kIIx) is presented. It can be seen that the probability of finding a particle under the potential step is exponentially decreased when moving away from the border. Considering this problem, we ignore the possibility of a wave reflection from the potential step boundary: this does not change the essence of the solution and is not important for us at the moment. 7.4.4
A potential barrier: a tunnel effect
Let us apply the results obtained to the consideration of the problem of a potential barrier depicted in Figure 7.11. Consider the case when the total particle energy E is lower than
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U ||2 A2Ie−2kIIx
U0 A2I
A2Ie−2kII d E x I
0 II d
III
Figure 7.11 A rectilinear potential barrier.
the height of the barrier U0. As previously, we subdivide the problem into three areas: area I (−∞ x), area II (0 x d) and area III (d x). The Schrödinger equation may be written as follows: ( x ) k 2 ( x ) 0,
(7.4.15)
for areas I and III with k2I,III (2mE/2), and ( x ) k 2 ( x ) 0
(7.4.16)
for area II with kII2
2m(U 0 E ) 2
Solutions should be sought in the following form: In area I: I(x) AI exp(–ikIx) BI exp(ikI x); In area II: II(x) AII exp (–kIIx) BII exp(kIIx); (7.4.17) In area III: III(x) AIII exp(ikIIIx). In these expressions, the amplitudes marked by the letter A correspond to the wavefunction propagation from left to right and amplitudes B describe a reflection from external and internal edges of the potential barrier. We shall not take reflected waves into account and, accordingly, shall not consider the waves containing the amplitudes marked by B (the exact solution using the boundary conditions and the wavefunction’s properties does not present a particular difficulty though they have no special interest). We denote the incident wave amplitude as AI and will express all other qualities using this value. A graph of the wavefunction’s amplitudes is given in Figure 7.11. In area I, the value of the square of the wavefunction amplitude AI2 is depicted. As was shown earlier for free particles, this value does not depend on coordinates and, within the given area, is constant. In area II the solution is not periodic but exponential (as was obtained when solving a potential
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step problem). The boundary conditions consideration gives: AI(0) AII(0). So the solution for the second area is given by the expression: II ( x ) AII exp(kII x ), which is also depicted in Figure 7.11. In the third area a motion once again is infinite and, accordingly, wavefunction is presented by periodic function and constant probability density. Its absolute magnitude is defined by one of the boundary conditions (condition of continuity). It links AIII with AI by the correlation 2 | III ( x ) |2 AIII AI2 exp(2 kII d ).
(7.4.18)
Since kI kIII, the de Broglie wavelengths in areas I and III are the same, whereas the amplitudes are different. The result also shows that, regardless of the energy of the falling particles, they can penetrate the barrier from area I to area III. Let us estimate what part of the particles falling onto the barrier penetrates the barrier or, what amounts to the same thing, how probable is it that a single particle shall penetrate it. We can call this value the barrier transparency and denote it by the letter D. Because of the fact that squares of amplitudes describe the “intensity” of de Broglie waves in a particular area, the barrier transparency can be defined as the ratio of their “intensities,” or
D
III (d ) I (0 )
2
2
AI2 exp(2 kII d ) AI2
⎛ 2d ⎞ 2 m(U 0 E) ⎟ . exp(2 kII d ) exp ⎜ ⎝ ⎠ (7.4.19)
One can see that the transparency of the rectangular potential barrier depends exponentially upon several factors: on the particle mass, on the barrier width and on the difference (U0 – E). The table below illustrates the barrier transparency for electrons upon the barrier width. d (10–10 m) D
1.0 0.1
1.5 0.03
2.0 0.008
2.5 10–7
If the barrier is not rectangular and its form is described by the function U(x) (Figure 7.12), the barrier transparency can be given by the expression
⎛ 2b ⎞ D exp ⎜ ∫ 2 m [U ( x ) E ] dx ⎟ . ⎝ a ⎠
(7.4.20)
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U
E
0
x
L
Figure 7.12 A potential barrier of an arbitrary form.
EXAMPLE E7.7 An electron with an energy E 4.9 eV moves in the positive direction of the x-axis (see Figure 7.11). The height of the potential barrier is U0 is 5 eV. At what width of the barrier will the probability to penetrate it be w 0.2? Solution: The penetration probability (the barrier transparency) is given by the eq. (7.4.19) ⎛ 2d ⎞ w⎜ 2 m (U 0 E )⎟ ⎝ ⎠
and hereafter, nw
2d 2 m(U 0 E ).
It is possible now to find the main equation:
d
n(1w) 2 2 m(U 0 E )
.
Calculation gives d 4.95 1010 m 0.495 nm. 7.4.5
Tunnel effect in chemistry
An approximate formula for the temperature dependence of the reaction rate coefficient K is given by the Arrhenius equation: ⎛ Q⎞ K exp ⎜ ⎟ , ⎝ T ⎠
(7.4.21)
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where Q is an activation energy and is the Boltzmann constant. If the activation energy is constant, a plot of ln K versus 1/T gives a straight line. This equation is often given in so-called Arrhenius coordinate (Figure 7.13): ln K
Q . T
(7.4.22)
In order to give a demonstrative presentation of the activation energy, let us imagine some group of atoms having two equilibrium states in two potential wells. Let them be characterized by two potential curves as depicted in Figure 7.14 (refer to Section 1.5.4). Such a state can be characteristic, for instance, for two atoms creating a molecule after their collision, or for two independent molecules forming a more complex molecule, or monomer molecules joining in polymer chains, and others. Certainly, different states are characterized by potential wells of a different form and depth: the more stable states have a deeper potential well. Suppose that the energy of an initially stable (ground) state in one potential well is characterized by the energy level noted in Figure 7.14 by a horizontal line. In order for the reaction to occur, the system should overcome the potential barrier Q, noted in the same scheme. This can occur either if the system has enough energy and can overcome the potential barrier Q (concentration of such active particles was calculated in eq. (3.3.7), Figure 3.6) or by means of penetration of the potential barrier. The overwhelming majority of chemical reactions really complies with the Arrhenius law and, in the logarithmic scale, is schematically expressed by a straight line (Figure 7.13). However, not long ago, the effect of the limitation of chemical reactions’ ratio at low temperatures was observed, the ratio became constant (Figure.7.13), in contradiction to Arrhenius theory. An explanation of this behavior can be given within the framework of the tunnel mechanism. In Figure 7.14, it can be seen that a system can overcome the potential barrier, not by climbing over the barrier, but having penetrated the barrier from one potential well into another one (in the manner of using an underpass in a mountainous area). From eq. (7.4.19)
ln K
l/T
Figure. 7.13 A chemical reaction ratio K versus reciprocal temperature 1/T.
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U
Q
Figure 7.14 A Scheme of activation (over a potential barrier with activation energy Q) and tunnel (penetration of a barrier) mechanisms of chemical reactions.
it can be seen that the barrier transparency indeed does not depend on temperatures. The penetration of the barrier is purely a quantum mechanical effect. As an example, we will describe the polymerization reaction of formaldehyde. In Figure 7.15, the potential curves of formaldehyde both in monomer and polymer states are shown. From a comparison of the curves it can be seen that the polymeric state is more stable than the mixture of monomers. The scheme above shows chain linkage in monomers and in polymer. In order for a monomer molecule to join the polymer chain, an electron from the double CO bond should abandon it and “penetrate” into the space between the atoms of carbon and oxygen of the adjacent molecule with the formation of a single bond C–O (this is shown in the middle of the scheme above). Certainly, herewith a redistribution of electron density occurs, developing and changing the interatomic distances: the density of polymer is higher than of the monomer mixture (the corresponding distances df, di and d are shown in the scheme). Since the polymerization process brings about greater energy of bonding (the potential well is deeper for the polymer) and the allocation of heat. The time of joining of the next monomer molecule to the already created polymeric chain, measured by very sensitive equipment, proves to be 102 sec, whereas computed from the Arrhenius equation gives a value near 1030 years. The effect of tunneling enables us to take quite a new look at some physical, chemical and, particularly, biological processes in respect of the organization and behavior of biologically active systems, accompanying all natural processes—from the formation of planets to the most complex particularities of biosyntheses. The scales of these phenomena are inconceivable though it is presently possible to assert that “tunneling effects” play a very important role in many processes of vital activity.
7.5
THE HYDROGEN ATOM
Consider now a more complex problem, which is very important for chemistry: the motion of a charged particle in a spherically symmetric electric field. In this case a particle’s
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polymer O df
C H
H
monomer O
O C
C H
H
H
O
O ∆d
H H
U(d) 2
di
C H
O
C H
C H
H
H
-CH2-Opolymer unit
1
CH2=O monomer molecule Q
df di
∆d
distance between adjacent molecules d
Figure 7.15 The tunnel effect of formaldehyde polymerization (after V.I. Goldanski et al.).
energy depends only on the distance from the center but not on direction (refer to expression (1.4.22)). The simplest problem of this kind is the motion of an electron in the field of a positively charged, dimensionless, heavy (in comparison with the electron) nucleus (proton), i.e., the problem of a hydrogen atom (and a singly ionized helium atom). The solution to this problem has an exceptionally important role in quantum mechanics and especially in quantum chemistry. Firstly, this is a problem that can be solved analytically (though some special mathematical functions must be used). Secondly, the solution is of great importance for chemistry where the electronic orbits arise from the solution; moreover the theory of the chemical bond has been worked out using the results. Thirdly, an empirically modified hydrogen atom’s orbits are widely used generally for heavier atoms because there are no other ways of achieving results. 7.5.1
The Schrödinger equation for the hydrogen atom
Let us choose the nucleus (proton) as the origin. Assume the nucleus is point-like. Because the proton’s electric field is spherically symmetric, we have to change from a Cartesian (x, y, z) to a spherical (r, , ) coordinate system. Figure 7.16 illustrates this transform: x r sin sin , y r sin cos , z r cos .
(7.5.1)
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z
r
y
x
Figure 7.16 Spherical coordinate system.
Note, that the z-axis is formally distinguished from the other axes. It can be said that this axis is selected: it is specially distinguished only geometrically, though it is always specially distinguished if there is an outer influence (electric, magnetic, etc.). It is said that z is the distinguished axis. The Schrödinger equation for hydrogen atom has the form (refer to (7.3.5))
(x,y,z )
2m [E U (x,y,z )] (x,y,z ) 0, 2
(7.5.2)
in which U(x, y, z) is the electron’s potential energy in the proton’s Coulomb field
U ( x, y, z ) U (r )
1 e2 , 4 0 r
(7.5.3)
(refer to Section 4.1.4), where r 兹苶 x2 苶苶 y2苶 苶z2苶. This expression has to be substituted into the Schrödinger equation and solved. As was noted earlier, this problem can be solved analytically; however, it takes much effort and a long time. Therefore, we will use a reasonable mathematical simplification paying more attention to the physical meaning. The Laplace operator (7.3.6) in spherical coordinates has the form: (r , , )
1 r2
⎧ ⎛ 2 ⎞ 1 ⎛
⎞ 1
2 ⎫ sin ⎨ ⎜⎝ r ⎬. ⎟ ⎜ ⎟
r ⎠ sin ⎝
⎠ sin 2 2 ⎭ ⎩ r
(7.5.4)
The Schrödinger equation in this case can be expressed as ⎧1 ⎡ 2
⎡
1 ⎤ ⎤ sin (r , , ) ⎥ r (r , , ) ⎥ 2 ⎨ 2 ⎢ ⎢
⎦ r sin ⎣ ⎦ ⎩ r r ⎣ r 2 1
⎪⎫ 2 m 2 2 (r , , )⎬ + 2 [E U (r )] (r , , ) 0. 2 r sin ∂ ⎭⎪
(7.5.5)
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Let us represent the wavefunction (r, , ) as the product of three independent functions where each function depends only on one argument: either on a radial coordinate or on the angles and : (r , q, ) R(r )()() R(r ) Y (, ),
(7.5.6)
Y (, ) ()()
(7.5.7)
where, is the so-called angular part of the wavefunction, and the function R(r) is the radial part. Consider first the electron’s motion along a spherical surface at fixed radial coordinate. This corresponds to the expressions r const. and Y/ r 0. This permits us to find the electron’s probability distribution on a sphere of fixed radius. The first term in eq. (7.5.5) becomes zero and therefore, the equation for the angular part of the wavefunction will be of the form 1 r2
⎧⎪ 1 ⎡ ⎫⎪ 1
2 ⎤ sin Y ( , ) Y (, ) ⎬ ⎨ ⎢ ⎥ 2 2
⎦ sin ⎩⎪ sin ⎣ ⎭⎪ 2 2r + 2 [E U (r )]Y (, ) 0.
(7.5.8)
The problem has been reduced to the investigation of the motion of a body along a sphere with fixed radius. Such a problem is called the rigid rotator; the classic version of which was considered in Section 1.3.9 (Figure 1.17). Recall that the electron’s angular momentum L in the framework of the Bohr model is L [rp] m[r].
(7.5.9)
It is perpendicular to the two vectors r and p m. If the angle between the two vectors r and p is 90°, the magnitude of L is L mr
(7.5.10)
It was shown in Section 1.3.9 that the rotation of two masses with distance d between them around a motionless center of mass O can be represented by the planar rotation of one mass (m1m2)/(m1m2)(the reduced mass) around the axis that passes through the center of mass. The result relates both to rotation of a diatomic molecule relative to its center of mass and to the rotation of the electron relative to the nucleus. The moment of inertia is called the reduced moment of inertia Ie I d 2 ,
(7.5.11)
where d being the interparticle distance. It is worth to notice that the reduced mass for hydrogen atom is half of electron mass whereas for hydrogen molecule is half of hydrogen atom mass. The kinetic energy of rotation is K
L2 L2 . 2 I 2d 2
(7.5.12)
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One can see that the same approach applies both to diatomic molecular rotation and to the hydrogen atom. 7.5.2.
The eigenvalues of the electron angular moment projection L z
The kinetic energy of a free particle forms a continuous spectrum, i.e., can possess any value. Within the framework of quantum mechanics we can ask: is the spectrum of values of the kinetic energy of a molecule’s free rotation (within the framework of the rigid rotator model) as well as an electron in the hydrogen atom either discrete or continuous? The answer is not a priori obvious. To answer this question we need first to solve the Schrödinger equation (7.5.8) with boundary conditions imposed on the wavefunction. For the rotation of the electron in the hydrogen atom, it is necessary to find a solution for the function Φ(). With reference to Figure 7.16, it will be noticed that this function describes the electron’s rotation in a xy plane, which is perpendicular to the z-axis. Consequently, it describes the behavior of the projection of the angular moment L onto the z-axis, i.e., Lz. It was shown above that free particle translational motion is described by the wavefunction (x) a exp(ipxx/) (time dependence is not important here). If we remember that the formulas of translational and rotational motion have the same structure (Section 1.3.9, Table 1.1), it is easy to write down the wavefunction for rotation by replacing px on Lz and x by . Therefore, we can write ⎛i ⎞ () a exp ⎜ Lz ⎟ . ⎝ ⎠
(7.5.13)
The condition of a single-valued solution has the form () ( 2).
(7.5.14)
Certainly, one revolution brings a system back to the initial state. Combining eqs. (7.5.13) and (7.5.14), and using the Euler formula we obtain: ⎛i ⎞ ⎛i ⎞ ⎛i ⎞ a exp ⎜ Lz ⎟ a exp ⎜ Lz ⎟ exp ⎜ Lz 2⎟ , ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ and further, ⎛ L 2 ⎞ ⎛ L 2 ⎞ cos ⎜ z ⎟ i sin ⎜ z ⎟ 1. ⎝ ⎠ ⎝ ⎠
It follows from this equation that (LZ/)2 m(2), where m is integer. And finally, Lz m.
(7.5.15)
Thus, without solving the Schrödinger equation, starting only from the boundary condition of uniqueness of the wavefunction, an important conclusion has been obtained: the
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projection of the angular momentum onto the distinguished z-axis can take only integer values of . The integer m, equal to 0, ±1, ±2, ±3, etc., is a quantum number that defines the value of Lz projection. The wavefunction Φ() is, consequently, expressed as m () a exp( im).
(7.5.16)
The magnitude a can be found from the normalization condition: 2
2
2 im im d a 2 2 1, ∫ () d a ∫ e e 2
0
0
i.e., a 2 (1 2) and finally ()
1 im e . 2
(7.5.17)
(7.5.18)
Thus the angular part Φ() of the wavefunction has been specified. 7.5.3.
Angular momentum and magnetic moment of a one-electron atom
Eq.(7.5.8) is modified by taking into account that E – U is the kinetic energy K because of the fact that in this particular case the potential energy of free rotation is zero (K E), and m is the reduced mass of the rigid rotator. Since the angular part of the wavefunction is the product of two functions Y(, ) Φ()Θ(), we can obtain () ⎡
() ⎤ () 2() 2r 2 sin 2 ( E U )()() 0. (7.5.19) ⎢ sin ⎣
⎥⎦ sin 2 2
Multiplying the equation by sin2/Φ()Θ() we can obtain 2 sin ⎡
() ⎤ 1 2() 2 2r sin sin E 0. () ⎢⎣
⎥⎦ () 2 2
(7.5.20)
Note that in the last equation each term depends either on or on . They can be grouped in the following manner: 2 sin ⎡
() ⎤ 1 2() 2 2r sin sin E . Θ() ⎢⎣
⎥⎦ () 2 2
(7.5.21)
A function Θ() is on the left-hand side and a function Φ() is on the right-hand side. Therefore, this equality can be satisfied only if both sides are equal to a constant value. Since the function Φ() (7.5.18) is already known, we can find that constant. In fact, this function is equal to:
1 2() m2 . () 2
(7.5.22)
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It means that 2
sin ⎡
() ⎤ 2r sin sin 2 2 E m 2 . () ⎢⎣
⎥⎦
(7.5.23)
This illustrates the previous statement that the wavefunction of the hydrogen atom can be presented as a product (7.5.6). The eq. (7.5.23) is a Legendre equation. It follows from the theory of Legendre equations that they have solution only if the coefficient of expression at sin2 is equal to 2r 2 E ( 1), 2
(7.5.24)
m .
(7.5.25)
where l is an integer with
Taking into account eq. (7.5.12) one can write E
2 ( 1). 2I
(7.5.26)
Therefore, we have found the energy eigenvalues of hydrogen atom and rigid rotator energy. The angular eigenvalues will be given later in Section 7.5.7. Thus the answer to the question that we posed earlier is: the rotational energy can take only definite discrete values (i.e., it is quantized). Using eq. (7.5.12), we can obtain also L2 2l(l1) or L ( 1),
(7.5.27)
i.e., the quantization of the kinetic energy of the rigid rotator originates from the quantization of the absolute value of angular momentum. The expression (7.5.27) and a gyromagnetic ratio (see Section 5.2.1) allow us to determine an orbital magnetic moment of the one-electron atom. Because the gyromagnetic ratio is /L ge/2m (where g 1), the magnetic moment is as:
e 2m
( 1) B ( 1).
(7.5.28)
The value B (e)/(2m) is known as the Bohr magneton. The general solution of eq. (7.5.23) for the function Θ() can be written with Legendre polynomials. The expressions for Y(, ) for the quantum number l 0,1,2 are given below in Table 7.1. Two important conclusions can be drawn from the solution obtained. One concerns the quantization of the rigid rotator energy; the other describes the properties of angular momentum in quantum mechanics. We will use the first one later in Section 7.8.2 in the description of rotational spectroscopy; here, we consider the properties of the angular momentums in quantum mechanics. As has already been noted, if one has to select a particular axis it is usually identified with the z-axis. It follows from the results described by eqs. (7.5.15), (7.5.25) and (7.5.27), that the
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Table 7.1 Wavefunctions for s-, p- and d-electrons n
l
ml
1
0
0
1s
0
⎛Z⎞ 2s 4 2 ⎜⎝ a0 ⎟⎠
2
2
2
3
3
3
3
0
1
1
0
1
1
2
1 ⎛Z⎞ ⎜ ⎟ ⎝ a0 ⎠ 1
3 2
e 32
(2 )e( 2 ) 32
2pz
⎛Z⎞ 4 2 ⎜⎝ a0 ⎟⎠
32
2p x
⎛Z⎞ 4 2 ⎜⎝ a0 ⎟⎠
32
2p y
⎛Z⎞ 4 2 ⎜⎝ a0 ⎟⎠
32
0
⎛Z⎞ 3s 81 3 ⎜⎝ a0 ⎟⎠
32
0
3pz
2 ⎛Z⎞ 81 ⎜⎝ a0 ⎟⎠
32
3p x
2 ⎛Z⎞ 81 ⎜⎝ a0 ⎟⎠
32
3p x
2 ⎛Z⎞ 81 ⎜⎝ a0 ⎟⎠
0
±1
±1
0
1
e( 2 ) cos
1
e( 2 ) sin cos
1
1
e( 2 ) sin sin
(27 18 2 2 )e( 3)
(6 − )e( 3) cos
(6 )e( 3) sin cos
(6 )e( 3) sin sin
⎛Z⎞ 3d z 2 81 6 ⎜⎝ a0 ⎟⎠ 1
3d xz
2 ⎛Z⎞ 81 ⎜⎝ a0 ⎟⎠
32
2 e( 3) (3cos2 1)
32
2 e( 3) sin cos cos
(Continued )
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Table 7.1 (Continued ) n
l
ml
2 ⎛ Z⎞ 81 ⎜⎝ 0 ⎟⎠
32
3
2
±1
3d yz
3
2
±2
3d 2
3d xy
⎛Z⎞ 81 2 ⎜⎝ a0 ⎟⎠
x y2
2 e( 3) sin cos sin
⎛Z⎞ 81 2 ⎜⎝ a0 ⎟⎠ 1
1
32
2 e( 3) sin 2 cos2
32
2 e( 3) sin 2 sin2
Note: The following units are accepted in the table: (Z/a0)兹2苶r苶; a0 (h2/4e2)
absolute value of angular momentum and its projection on the z-axis satisfy certain conditions; in particular they signify that the absolute value of the angular momentum vector is always longer than the length of any (even the longest) of its projections (Figure 7.17). Indeed, mmax l and l 兹苶 l(苶 l苶 1) 兹苶 l 2苶 苶l and , therefore Lz is always smaller than |L| (with one exception when l 0). Therefore, the vector L is fixed in the space within angle . It follows also from the results presented that the vector L can never lie on the z-axis. This law concerning angular momentum vectors is called spatial quantization. Figure 7.18 shows the spatial quantization for several l values. Moreover, the theory gives no way of finding the other projections (Lx and Ly); they are completely nondeterminated. This fact can conditionally be added to uncertainty principles. Note that this is not a lack of theory but is the law of nature. Generally speaking, there is no need to give visual evidence of a given fact: this is the nature of things. However, attempts to present a situation as a precession of the vector L around the axis z are sometimes met. According to our understanding, this interpretation is unlikely to be correct: with precession a certain frequency and, further, energy must be bound though no one has observed such an additional energy. More acceptable is a model of “uniform smearing” of the L vector upon a conical surface at a fixed angle, however this model also does not give any reliable ideas for further development. At a given quantum number the quantum number ml can accept 2l1 values. Therefore, the angular momentum vector can be situated on one of 2l1 cones (Figure 7.18). The opening angle can be found as: cos
LZ m L ( 1)
(7.5.29)
where is an angle between the vector L and z-axis and m is one of the quantum numbers connected to l. Complications appear also when we have to sum up the angular momentums vectors of several rotators (electron orbits). All restrictions described above are valid in this case too.
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z LZ |L |
0
Figure 7.17 A space quantization of the angular momentum.
Suppose, for example, that it is necessary to sum up two vectors L1 and L2 in order to obtain the third (resulting) vector L. One has to remember that the vectors being added cannot be exactly parallel or antiparallel to each other. Really, since 兹苶 (苶 )(苶 1苶 苶 2苶 苶 1) 兹苶 苶2苶 (苶 1苶 )苶1兹苶苶 苶 1苶), the length 1 1苶 2(苶 2苶 of the resulting vector L is always less than the sum of the absolute values of composite vectors L1 and L2 (兩L兩 兩L1兩兩L2兩). Formally, one can write L L1 L2, where L1 兹苶 1苶 (苶 1) and L2 兹苶苶 苶 1苶), 1苶 2(苶 2苶 the resulting vector having magnitude L 兹 苶 1). All the vectors have z-projections 苶(苶苶 which are quantized: L1z m1, L2z m2, Lz m, where mmax for all three vectors. Since projections are scalar values, projection of the resulting vector will be found as an algebraic sum of the projections of the individual vectors. The maximum value of the projection of the vector L is Lz , where l2, whereas the shortest length of vector L will be if projections have opposite signs, L l2. Thereby, when adding the angular momentum vectors, the resulting vector L can take all values from maximum to minimum with quantum numbers , running over all values from 1 2 to 1 2 in integer steps. Figure 7.19, illustrate this summation. One should remember that the resulting vector L in quantized as well relative the new axis z. Since the magnetic moment vector is tightly connected with the angular momentum, everything that has been said applies to them both. The main conclusion from the whole consideration is that the summation of angular momentum vectors in quantum mechanics is accomplished according to a certain scheme. The procedure L1 L2 L must be understood as a summation of two quantum mechanical angular momentum vectors giving the third vector, complying to the same rules. An attempt to give a primitive image is presented in Figure 7.19. However, it appears that there is no need to carry out such a huge drawing especially in the case of several vectors, reducing the geometric summation to the combination of quantum numbers. Sometimes, the prescription given above for vector summation is reduced to the symbolic sum l 2, suggesting combining quantum numbers 1 and 2 according to the general rule presented above. After introducing electron spin, we have to deal with the set of quantum numbers related to orbit and spin.
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/2
+½ (a) B −½
) ×2
(1 (1 ×2
×3
)×
l=½ 2l + 1 = 2
3/
2
1× 2
+1
1×2
(b) B
2 1×
l=1 2l + 1 = 3
−1
(3/
+½
2)× (5/
2)
+
(c) B
(3/ 2)× (
5/2
)
l= 2l + 1 = 4
2)
(5/ 2)×
(3/ −½
)
5/2 )×( (3/2
−
Figure 7.18 Permitted projections of the angular momentum vector. (a) 1/2, (b) 1, (c) 3/2.
7.5.4 A Schrödinger equation for the radial part of the wavefunction; electron energy quantization Assume that the values of angles and are constant. Thus we set all partial derivatives of these arguments to zero and look at the behavior of the radial part of the wavefunction. The Schrödinger equation (7.5.5) in these circumstances will take the form
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z′ L
L2
L1
Figure 7.19 Summation of two angular momentum vectors.
1 ⎡ 2 R(r ) ⎤ 2 m r [E U (r )]R(r ) 0,
r ⎥⎦ 2 r 2 r ⎢⎣
(7.5.30)
where instead of the reduced mass the electron mass m is again substituted. This equation has solutions at the definite values of the eigenvalues E, and the wavefunctions R(r) are expressed via special functions, the adjoined Laguerre functions. The eigenvalues of energy can be expressed in terms of the fundamental physical constants: En
e4 m 1 13.56 2 eV. 2 2 2 n 32 0 n
(7.5.31)
In this expression the energy E is given in eV in order to avoid a source of errors caused by the fact that in different systems of units, the constants are written differently. The integer n in this expression is the principal quantum number; it determines the electron energy E in the hydrogen atom. It can take any positive nonzero value. Eq. (7.5.31) describes the discrete spectrum of electron energies (Figure 7.20). From this drawing, one can see that value –13.56 eV is the energy of the ground state of the hydrogen atom. Figure 7.20 also depicts the values of some excited states with n 1–7. Excitation of an electron up to the energy E 0 transfers it to a state with a continuous energy spectrum and infinite motion. This corresponds to the removal of the electron from the atom or, in other words, to the ionization of the atom. The first ionization potential is the lowest energy that should be given to an atom from outside to tear the ground state electron away. Consequently, the first ionization potential of a hydrogen atom is 13.56 eV. If the atom is already excited, the ionization energy is less. Accordingly, one can distinguish the second (3.4 eV), third (1.5 eV), etc., ionization potentials of the hydrogen atom. The transitions between electron energy levels, associated with absorption or emission of a quantum of electromagnetic radiation, are not limited by selection rules, all transitions are permitted. The main equation is: En En ,
(7.5.32)
where n and n are principal quantum numbers corresponding to the initial and final energy levels. Equation (7.5.32) corresponds to serial formula (6.7.13) (see Section 6.7).
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E(eV) 0
n ∞
7 6
5
4 3 2
−13.56
1
∞
Figure 7.20 The hydrogen atom energy levels; the visible part of hydrogen atom radiation (Balmer series) is presented below: wavelengths —656, —486, γ—434 and δ—410 m. A spectrum border is 365 m.
Different series of spectral lines are associated with the quantum number n. According to the quantum number n, the so-called series is distinguished: Lyman series (n 1), Balmer series (n 2), Pashen series (n 3), etc. The Balmer series lies in the visible region of the spectrum. The eigenfunctions R(r) are presented in Section 7.5.7 and in Figure 7.22. As has already been mentioned above, the Schrödinger equation can be solved exactly for several particular problems only, including the hydrogen atom and the helium ion He. Even for the neutral helium atom with its two s-electrons, the equation is rather complicated because the potential energy U must take into account not only the interaction of each electron with the nucleus, but the electrons’ interaction with each other as well. The expression for the total potential energy takes the form
U (r1 , r2 ) ⬃
1 ⎡1 1 1 ⎤ , 4 0 ⎢⎣ r1 r2 r12 ⎥⎦
(7.5.33)
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r1 and r2 being the radial coordinates of each electron, and the last term takes into account the mutual interaction of the electrons. One can imagine to what extent the Schrödinger equation becomes complicated. This equation can be solved only approximately, though modern methods of quantum chemistry enable a solution with good enough accuracy to be obtained. However, cases exist when the solution for the hydrogen atom can be successfully used for more complicated atoms. These are the so-called hydrogen-like atoms, in which a particular electron is rather more distant from the nucleus than the other electrons. The latter form a closed shell of inner electrons, and the residual electron is situated in the complex electric field. In this case the changes in the energy are taken into account by atomic number Z:
E
13.56 2 Z eV. n2
(7.5.34)
It must be emphasized however, that the value Z in this case is not an exact atomic number because an external, valence electron is presented in the effective field, created both by the nucleus and by inner electrons. This effect is termed “screening” because the inner electrons make a negative shield around the positive nucleus. Therefore, the effective Z value differs from the ideal atomic number and is often called the effective atomic number. 7.5.5
Spin of an electron
Besides an orbital angular momentum, the electron has an inherent angular momentum that has been called spin. This is often explained as the rotation of the electron around its own axis. However, neither orbital nor spin “motions” have been considered as motion, but as a state, characterized by definite quantum numbers. It was found by Einstein and deHaas (1915) and Stern and Gerlach (1922) in classic experiments that the number of possible projections of the electron angular momentum vector on the selected axis is two. So in accordance with eq. (7.5.3) one can obtain s ½ (because 2 2s1), s being the spin quantum number. This signifies that the length of electron intrinsic angular momentum vector LS according to the general rule for the magnitude of the spin angular momentum vector is 兹苶 s(苶 s苶1苶), therefore, LS
3 3 . 4 2
(7.5.35)
According to the quantization rule this means that the spin projections onto the selected z-axis can have two values, namely, 1 L m z s 2 where mS is a quantum number of spin projection.
(7.5.36)
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The quantum number mS does not influence the electron wavefunction significantly; however, it influences the electron distribution among the energy levels and quantum cells to a great extent. The Pauli principle is a consequence of these quantum mechanical rules. Because of the fact that not the absolute value of the angular momentums but their projections are measurable quantities, it can often be said that the electron spin is (1/2) or simply (1/2). The spin spatial quantization is depicted in Figure 7.21. In full analogy with the orbital state, the spin magnetic moment can be determined through the gyromagnetic ratio g (Section 5.2.1). For spin g 2 (in e/2 m units), the absolute value of spin magnetic moment is therefore equal to M 2
1 1 3 e ( 1) 2 B , 2m 2 2 4
(7.5.37)
The projection of the magnetic moment on the z-axis is M z B .
(7.5.38)
When chemists look at atomic and/or ionic electron states, and distribute electrons over the energy levels and/or fill in corresponding quantum cells by arrows in accordance with the Pauli exclusion principle (not more that two electrons into one quantum cell) and the Hund’s rule, they bear in mind exactly these two projections of spin magnetic moment. Note that arrows are represented by schematically signs and , since projections themselves are not vector but algebraic values. 7.5.6
Atomic orbits: hydrogen atom quantum numbers
Now it is desirable to generalize the information that we have obtained following the analyses of the Schrödinger equation for the hydrogen atom and introducing the electron spin. We mean the systematization of the values characterizing an atom’s state. Herewith
+½
)
1½
½(
B −½
½(
1½
)
Figure 7.21 Spin space quantization.
S=½ 2S + 1 = 2
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we will try to analyze the approaches of physicists and chemists, which in some respect look different. A typical quantum mechanical object such as an atom possesses some classic characteristics that are unexplainable within the framework of generally accepted presentations (refer to Section 6.7) (no orbital motion, yet the existence of angular momentum; no rotation of an electron around its own axis, yet intrinsic angular and magnetic moments, i.e., spin, etc.). As a result, these terms are used irrespective of their classical sense. Moreover, the Bohr model is a transition from the purely classical presentations to the quantum mechanical ones: the motion of electrons along the orbits is accepted, but not all orbits are permitted; the angular momentum is accepted though its values, and orientations are the subject of strict limitation. It is possible to consider the Bohr model as a transition from the classical mechanics to quantum, with the preservation of many of its attributes. As a result, many of Bohr model notions will often be met in order to simplify the students’ understanding. Quantum mechanical angular momentums and its projections, as well as the electron energy, are defined now by quantum numbers: the electron energy is defined by the principal quantum number n, the angular momentum vector length is defined by quantum number l, its projection on the selected z-axis is given by a quantum number ml. Each of the quantum numbers enumerated is included into a particular wavefunction, and their product gives total wavefunction describing an electron distribution in the language of probabilities. Such an image of wavefunctions and their squares brings about the atomic orbits. For chemists, it is precisely the most valuable result. The quantum number mS is characterized by the z projection of the electron spin angular momentum. In first approximation, it does not influence the energy and wavefunction shape, but influences significantly the electron distribution among the energy levels. Consequently, the total wavefunction is the product of all three parts (refer to eq. 7.5.6): (r , , ) R(r )()() R(r )Y (, ),
(7.5.39)
the R(r) function depends on n and Y(, ) on l and ml. 7.5.7
Atomic orbits
The general expressions for R(r), Θ() and Φ() can be written using the special mathematical functions. They are presented in special literature on mathematics, quantum mechanics and quantum chemistry. We restrict ourselves to giving here the description of the main physical concepts of electron orbits as the basis for the theory of chemical bonding. The state of an electron in an atom is much more sophisticated than can be expected from the Bohr theory. Quantum mechanics shows that an electron can be found in any point of space, but the probability of its presence changes from point to point. The notion of an electron orbit appears more productive than electron clouds. Under the electron orbit, the physicist often understands the mathematical expression of the wavefunction itself corresponding to definite quantum numbers. In chemistry, the orbit is understood as a set of electron positions around a nucleus taking the probabilities into account. This probability is defined by wavefunctions R, Θ, Φ.
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Expressions for s-, p- and d-electrons in the analytical form in the spherical coordinate system (Figure 7.16) are given in Table 7.1. In the radial part of the wavefunction R(r) argument r is given in atomic units, i.e., in the unit of the first Bohr atomic radius a0 0.5292 1010m. Graphs of the R(r) functions (a) and the probability density curves, i.e., the probability density of finding an electron in the spherical layer with a thickness dr(dw/dr 4r2R2(r))(b) as dependent on r are depicted in Figure 7.22. It should be noted that R(r) at the point r 0 (i.e., on the nucleus) has the maximum value. However, this does not contradict common sense because the probability of finding an electron at point r 0 (Figure 7.22b) is equal to zero. A scheme for drawing graphs of the angular part of the wavefunction Y(, ) and its square Y2(, ) (particularly for pz orbit) is given in Figure 7.23a and b, respectively. The value of Y for the given is proportional to a line OM. It is worth noting that the function Y() is presented as spheres, whereas the Y 2() is presented as the elongated dumbbells more popular in chemistry. The wavefunctions in Table 7.1 (above) are presented for n 1, 2 and 3. In the first line the data for the 1s-state is given, in this case R(r) has a maximum at r 0 and falls down with increasing r. The Y(, ) function depends neither on , nor on , therefore the ||2 graph is spherically symmetric. The same is true for the 2s- and 3s-states. The analytical expression for n 2, and 1, ml 0 and 1 are given in the next three lines. It can be seen that the pz-orbit solution has the simplest look than the two others (px and py); such inequality is the result of the spherical coordinate system; moreover, the last ones have imaginary form. In order to obtain the expression in a real form, one should compose a linear combination of particular solutions, i.e., carry out the hybridization of orbits (since any combination of the solutions of the Schrödinger equation is also the equitable solution). We must use here the Euler formula and compose the linear combination of Y1,1 and Y1,−1 orbits: 1 1 3 Ypx (Y1,1 Y1,1 ) sin [cos i sin cos i sin ] 2 2 8 3 sin cos 8
(7.5.40)
and, Yp y
=
1 ⎛ 1⎞ (Y1,1 Y1,1 ) ⎜ ⎟ ⎝ 2i ⎠ 2i 3 sin sin . 8
3 sin [cos i sin cos i sin ] 8 (7.5.41)
The angular parts in the real form for electron d-states were also obtained in this way. In this form, they are widely used in chemistry. Having determined all parts of the wavefunction in any point, the r(r, , ), the total wavefunction can be derived by multiplication of all its parts. In the abstract case of the absence of any external influence when there are no arguments for the choice of quantization of axis z, all solutions of the Schrödinger equation and
0.8 4 0.4 n = 1, l = 0 0 0.8
2 1
0.4
9:45 PM
n = 2, l = 0
n = 1, l =0 0 1
4πr 2[R(r)] 2
0 n = 2, l =0
0 1
n = 3, l =0 0
0.4
n = 3, l = 0
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5
0 0.8
1
0.4 n = 2, l = 1
0 1
0 0.4
n = 3, l =1
0
n = 3, l = 1
0 0
1
2
3
4
5 (a) r
6
7
8
9
10
0
1
2
3
4 5 (b) r
6
7
8
Figure 7.22 A wavefunction’s radial parts R (a) and corresponding values 4πr2R2 (b) for some electron states.
9
10
7. Elements of Quantum Mechanics
n = 2, l =1
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z
z M
M
0
(a)
0
(b)
Figure 7.23 How to draw the wavefunction’s angular parts Y(, ) (a) and Y2(, ) (b).
all their linear combinations can virtually exist. However, there is no physical possibility of finding either all or some of them, because any attempt to select a quantization axis destroys the atom itself. This fact reveals a property of quantum mechanics: an instrument of investigation destroys the state of an object. If the atom considered falls into the orbit of other atoms, the mutual chemical influence makes essential changes to their state. In different circumstances, it can appear that other linear combinations can be more advantageous, for example, the well-known s-p and s-p-d hybrid orbits are created (see Tables 7.1). Note that the probability density of finding an electron in different points of an orbit graph is different. To depict the total electron distribution in three-dimensional form and to make them understandable to the reader is a very difficult task. Nevertheless, some efforts have been made to create a suitable image. The most recent is the presentation of electron density as a graph of the charge distribution dq(r, , ) |e|||2(r, , )dV as a spatial pattern in the form of isolines and three-dimensional pictures. The results of X-ray diffraction investigation are at present given in such a form. Graphs in the form of closed surfaces are often used in chemistry; inside a closed volume, a definite amount of atomic electrons are contained (more often 90%). Such a picture is presented in Figure 7.24 showing the orbits of different electron states in the hydrogen atom. Note that the orbits do not touch the origin (the nucleus). This is because of the fact that in this area the probability density of finding an electron is very small due to the radial part of the wavefunction (the argument r is too small). Therefore the total density is also small. It turns out that, even for hydrogen-like atoms, atomic orbits appear vastly more complex. Regrettably, it is impossible to obtain the exact solution even for these atoms. So in quantum chemistry different kinds of approximation are used, more or less successfully, to describe one system or another, and one atomic area or another. For instance, the factor is introduced as a multiplier in the exponent order of the radial wavefunctions to describe an orbit’s compression-expansion (Slater multiplier). Sometimes, not one but two, or even several, multipliers are used, each of which is better for describing the electron density near the nucleus or far from it. These empirical modifications for different atoms are given in quantum chemistry.
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z
1s
+
x y z
z
z
+
−
2px
−
+
x
y
2pz
2py
x
x
+
y
y z
z
+
−
+
−
+
−
x
x +
y
3dx2-y2
y
−
3dz2
z
z −
+ −
−
+
x
+
z −
−
x
+ y
y
y
+
+
3dxy
3dxz
x
− 3dyz
Figure 7.24 Representation of a wavefunction’s angular parts in different electron states.
EXAMPLE 7.8 A hydrogen atom is in a ground 1s state. Determine the probability of finding an electron w in a sphere of radius r 0.1a. Solution: The probability of finding an electron in a spherically symmetric 1s state is given by the normalized wavefunction 100(r)(1/兹苶 苶 a3)exp((r/a)) the volume ele2 ment being dV 4r dr. Therefore
dw
2
4 ⎛r⎞ ⎛ 2r ⎞ exp ⎜ ⎟ 4r 2 dr 3 exp ⎜ ⎟ r 2 dr. 3 ⎝ a⎠ ⎝ a⎠ a a 1
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It is convenient to move to the atomic units further r/a; therefore r2 2a2, dr ad and dw 4 exp(2)2d. The probability can be found by integrating dw in limits r1 0 to r2 0.1a (or from 1 0 to 2 0.1) 0.1
w 4 ∫ 2 exp(2)d . 0
Decompose the exponential into a MacLoren series exp(2) ⬇12 .. and limit by two terms; we can present the integral as 0.1
0.1
0.1
0
0
w 4 ∫ (1 2)2 d 4 ∫ 2 d 8 ∫ 3 d 0
4 3 3
4 0.1 0 8 4
4 −3 −3 0.1 0 10 0.2 10 . 3
Therefore the final result is w 1.53.103. It is useful to compare this result with Figure 7.22.
7.5.8
A spin–orbit interaction (fine interaction)
The above scheme of the Schrödinger equation solution did not take two circumstances into account: firstly, relativistic effects and, secondly, electron spin. The relativistic effects appear when a particle possesses high energy and consequently moves at a speed close to the light velocity. In an atomic planetary model, the inner electrons are nearest to a nucleus (E is negative and great in the absolute value, refer to eq. (7.5.34) at Z 30), are precisely the relativistic particle. In this case, one should take relativism into account; we will not, however, discuss this further. Consideration of electron spin raises the question of the so-called spin–orbital interaction influencing the electron energy values. The mechanism of this interaction consists of the fact that the electron orbital “movement” produces the magnetic field acting on its own spin. In order to make the picture more understandable, let us place the origin in an electron, the nucleus in this case appears moving along the circular orbit around the electron under consideration, creating a magnetic field at its position. Interaction of the magnetic moment of the electron spin with the magnetic field produced by a nucleus is called the spin–orbit interaction. The energy of this interaction must be taken into account when estimating the atomic total electron energy. Fortunately, such a complex procedure can be considerably simplified using the procedure of substitution of the vectorial summation by the combination of quantum numbers (see Section 7.5.3). To account for the spin–orbit interaction, the additional quantum number j is introduced, j being the sum of quantum numbers l and s j s 12
(7.5.42)
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Instead of the single energy level, characterized by quantum number n, two levels appear with quantum number l equal to l (1/2). (At l0 quantum number j accepts only one value j(1/2).). Accounting for the spin–orbit interaction, the electron energy becomes dependent not only on the principal quantum number n but on the quantum number j as well. Exact calculations give ⎛ ⎞ 13.56 2 13.56 Z 4 2 ⎜ 1 3⎟ En, j 2 Z ⎜ 1 4n ⎟ n n3 ⎜ j ⎟ ⎝ ⎠ 2
(7.5.43)
(as in expression (7.5.30), the energy is given here in eV units). Estimation of En,j shows that it is in the order 104 eV. The dimensionless value (e2/c) with good accuracy is equal to 1/137; it is called the constant of the fine structure. Its rational value has been the reason for a great deal of speculation about the rational correlations between atomic characteristics and, accordingly, the creation of the united theory of elementary particles. However, all of these came to nothing and were personal tragedies for a number of scientists. In spite of the fact that the energy of fine interactions is small, analysis of the last two formulas shows that because of fine interaction all energy levels with l0 split into two. This is developed, for instance, in the splitting of some spectral lines of atomic electron spectrums. So the well-known transition p→s in alkaline metal atoms instead of one spectral line contains two reliably measurable lines.
EXAMPLE E7.9 An excited electron is in a 3p-state. Find the change in the orbital magnetic moment of the atom’s transformation to a ground state. Solution: We have to find the difference MlMl2Ml1. The orbital magnetic moment depends only on orbital quantum number: Ml B 兹苶( 苶 苶1苶)苶. Therefore, in the ground state l 0 and l2 0. In the excited state, l 1 and Ml1 B兹2苶. Hence the difference is Ml B兹2苶. Execute the calculation remember that B 0.927 1023 J/ T, we arrive at the numerical value Ml –1.31 1023 J/ T.
7.6
A MANY-ELECTRON ATOM
As we already know, it is impossible to solve analytically the Schrödinger equation for atoms with two and more electrons. This prompted the development of new quantum mechanical methods of approximate solutions or modified solutions, equitable for the hydrogen atom, by introducing empirical adjustments. In this chapter we will consider intra-atomic interactions between electrons and the complication that this interaction causes. (Here we must emphasize that any complication in the theory forces the development of more sophisticated experimental methods of investigation,
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opening new possibilities of studying more precise questions relating to the chemical structure.) The wavefunctions of many-electron atoms will not be studied in this book; these are the subject of quantum chemistry. 7.6.1
Types of electron’s coupling in many-electron atoms
If there are several electrons in an atom, the total angular momentum (sometimes it is called a mechanical moment unlike magnetic) is obtained as a sum of mechanical moments of atomic electrons. Depending on the electron interaction energies there exists two ways to combine all mechanical moments (spin and orbital) into the total one: either to take into account the fine spin-electron coupling first (refer to Section 7.5.8), obtain the quantum numbers j’s and corresponding LJ or primarily unite all mechanical moments of all electrons, orbital and spin separately, obtain total L and Ls and finally obtain LJ. Such coupling is referred as Russell-Saunders binding. Consequently, to obtain the total angular momentum of a manyelectron atom one should first to add to each other the orbital moments of all electrons LL, then sum all spin moments into Ls and only afterwards to sum them both into the total one ( and s being orbital and spin quantum numbers of a single electron and L and S the quantum numbers of the total angular momentum of the whole atom). Herewith there is no need to combine vectors with provision for rules of the angular momentum summation in the quantum mechanics, it is easier do this by combination of the quantum numbers. Let us do the summation of the two electrons with quantum numbers l1 and l2, s1 and s2. Summation of orbital moments gives L 1 2 , 1 2 1, 冨 1 2 冨 .
(7.6.1)
Consequently, the total angular momentum LL, can accept as much values as many terms contain the series (7.6.1). Their values can be determined according the general rule (refer to 7.5.3) L L L ( L 1).
(7.6.2)
We can operate quite analogously with spin angular momentums. As s 1/2 and N is the number of electrons, the total spin angular momentums S can accept the values S N 12 , N 12 1, ..., 0,
(7.6.3)
S N 12 , N 12 1, ... , 12 ,
(7.6.4)
when N is even and
when N is odd. The total spin angular momentum is expressed by analogy with the previous expression L S S (S 1).
(7.6.5)
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In order to obtain the total (atomic) angular momentum (mechanical moment), we must put the quantum numbers together J L S, L S 1, ... , L S ,
(7.6.6)
where J is the total atomic quantum number; it determines the atomic mechanical moments L J J ( J 1). Its z-projections are LJ,z mJ where
(7.6.7) (7.6.8)
mJ J , J 1, J 2, ... ,J ,
(7.6.9)
overall 2J 1 values. The graphic summation will be given below. When the individual electron spin–orbit interaction energy prevails over the spin–orbit interactions of different electrons, we first have to sum the orbital and spin moments of every electron (to obtain all j’), then sum these j’ and further obtain the total atomic angular momentum quantum numbers and finally the overall quantum number J, characterizing the total angular momentum of the atom. This type of binding is called a j–j bond. Since the spin–orbit interaction is proportional to Z 4, this type of binding occurs mainly in heavy atoms. We will not consider further this type of coupling. Sometimes all states with the same quantum number J have the same energy, because different mI’ does not often influence it. In these cases, states are called degenerated ones. However, there are often cases when the energy with definite J and different mI’ have different energy. These are cases without degeneration. Sometimes, this only partly concerns a state. For instance, in a state with J 2 corresponding to 2J 1 values of mJ (2, 1, 0, –1, –2), all of them can possess the same energy. Such states are called fivefold degenerate ones. An external influence (for instance, a crystal field) can split one degenerated fivefold electron level into twofold and triple fold split levels (this situation is often encountered in the d-state of d-element complexes). Such a process is called lifting (or removing, or waiving,) the degeneration. In this case the degeneration is removed only partly. Which quantum number the degeneration lifting relates to is usually indicated. 7.6.2 Magnetic moments and a vector model of a many-electron atom. The Lande factor In Section 5.2.1 the gyromagnetic ratio g, the ratio of the magnetic moment to the angular momentum, was introduced. This ratio is given in the unit (e/2m) and allows the determination of the magnetic moment knowing the angular (mechanical) moment. The g value for the orbit (g 1) can be calculated theoretically, whereas for spin (g 2) it is determined experimentally. In this section, we shall consider the problem of what to do if, in the many-electron atom with a Russell-Saunders coupling scheme, both orbit and spin participate in producing the magnetic moment. The vector model of an atom will help us.
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Formally, one can write the expression for the magnetic moment MJ
g B J ( J 1),
(7.6.10)
where B is the Bohr magneton and J is the total quantum number of atomic mechanical moments. Our task is to determine the g-value when both orbit (quantum number L) and spin (quantum number S) participate in magnetic moment creation. The mutual disposition of all total atomic mechanical and magnetic moments is presented in Figure 7.25. The vectors LL and LS, obtained according to the rule described above for Russell–Saunders coupling can be seen in the picture. Their sum gives the general mechanical moment LJ (all of them in the scale of ). The magnetic moments (in the scale B) correspond to each mechanical moment; all of them are directed oppositely to their counterparts because of the negative electron charge. The length of magnetic moments takes into account the corresponding g-factor: g 1 for orbit and g 2 for spin. Therefore, their sum J* is directed not strictly antiparallel to its counterpart J , but at some angle (denoted as η) to the line of LJ. (In an atom, as in any other rotating system, the main vector is the angular (mechanical) momentum vector; it specifies the main system axis). Therefore, the total atomic magnetic moment is not J*, but J that is a projection of the first on an axis specified by the LJ vector. It is seen in Figure 7.25 that M J M J
ⴱ
cos
(7.6.11)
The total vector J length is a sum M J M L cos M S cos .
(7.6.12)
As ML 1(e/2m)兹苶 L苶 (L苶 苶1苶) 1 B兹苶 L苶 (L苶 苶1苶) and MS 2(e/2m)兹苶S苶 (S苶 苶1苶)2B 兹苶S苶 (S苶 苶1苶) we have to determine cos α and cos . Let us consider a triangle with sides LL, LS and LJ. According the cosine theorem: L2S L2L L2J 2 L L L J cos and L2L L2S L2J 2 L S L J cos . Then, cos
L2L L2J L2S L ( L 1) J ( J 1) S (S 1) 2L L L J 2 L ( L1) J ( J 1)
and cos
L2S L2J L2L S (S1)J ( J1) L ( L +1) . 2L S L J 2 S (S +1) J ( J +1)
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B,z
LJ
LS
LL
-
ML
MJZ MS
MJ
MJ*
Figure 7.25 The vector diagram of a multielectron atom.
Substituting these values into eq. (7.6.12), we arrive at ⎛
M J B ⎜ L ( L 1)
L ( L 1) J ( J 1) S (S 1)
2 L ( L 1) J ( J 1) S (S 1) J ( J 1) L ( L 1) ⎞ 2 S (S 1) ⎟. 2 S (S 1) J ( J 1) ⎠ ⎝
This expression can be re-written as: ⎛ ⎝
M J B ⎜ 1
J ( J 1) S (S 1) L ( L 1) ⎞
2 J ( J 1)
⎟⎠ J ( J + 1).
(7.6.13)
Comparing expressions (7.8.10) and (7.8.13), we obtain g 1
J ( J 1) S (S 1) L ( L 1) . 2 J ( J 1)
(7.6.14)
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Expression (7.6.16) determines the so-called Lande factor (multiplier). It permits us to write the expression of the magnetic moment of the multi-electron Russell–Saunders atom in a compact form: MJ B g
J ( J 1)
(7.6.15)
In the particular cases at S 0, J L and the Lande factor is, certainly, equal to 1; at L 0, J S g 2, which correspond to purely spin magnetic moment. Expressions that determine the magnetic moments of multielectron atom are obtained in full. 7.6.3
The atomic terms
It is obvious now that an atomic state has to be characterized not only by the electron distribution between energy levels (with Pauli exclusion principle validity), but also by moments: total LJ, orbital LL and spin LS, their mutual positions being taken into account. The range of quantum numbers J, L and S comprise the so-called atomic term. The symbolic presentation of the atomic term is given as: βAJ : A is characterized by the quantum number L according to the following scheme: L A
0 S
1 P
2 D
3 F
The number is called the term multiplicity; it characterizes the number of acceptable states at fixed quantum number L. The multiplicity is equal to 2S 1 at S < L and 2L 1 at S L. For example, for states with L 1 and S 21 multiplicity is equal 2 12 12 (term 2P1/2, 2P3/2); for states with L 1 and S 2, 2 1 1 3 (terms 3P3 , 3P2 and 3P1), etc. The definite energy corresponds to each term. However, systems can often be degenerated. The external action can remove the degeneration, i.e., make each term have its own energy. 7.6.4
Characteristic X-rays: Moseley’s law
As we saw in Section 6.6.4, an electro-magnetic radiation with a wavelength of 10–10–10–11 m is regarded as an X-ray. Two types of X-rays are known; both of them are produced in an X-ray tube as the result of the interaction of primary radiation (X-rays or electrons) with an anode material. X-rays of the first type are produced as a result of intraatomic energy transitions and are called characteristic X-rays; their line spectrum is specified by the anode material’s atoms. The second are bremsstrahlung X-rays emitted while an electron moves with high deceleration in the surface layers of the anode material. This spectrum is white, i.e., it contains a continuous range of wavelengths (see Section 6.6.4). In both cases a great amount of heat is emitted and then removed by a special water flow inside the anode.
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White radiation is used in material X-ray radiography (translucence, human body). Its properties were mentioned in Section 6.6.4. We will restrict ourselves in this section to a consideration of characteristic X-rays. The primary electrons in the X-ray tube (see Figure 6.39) are emitted by a heated cathode by thermo-emission and accelerated by an electrostatic field between anode and cathode with the potential difference 103–105 V. In the nonexcited atoms of the anode, all lower levels are occupied in accordance with the Pauli exclusion principle. Rapidly moving electrons strike a metal target (refer to Section 6.6.4). Colliding with the atom, the accelerated electron kicks the inner, lower-level electron out producing free places (holes) in the low-lying (ground) levels. An upper-level electron drops down to fill the hole. The transition is accompanied by photon emission, its wavelength being determined by the energy difference: 2c/(E – E), where (E – E) is the energy difference. In Section 7.5.4, the spectrum of the hydrogen atom was discussed. Equation (7.5.34) shows that the energy levels, at least for the hydrogen-like atoms, go down (in a negative region) proportional to Z 2: the potential well deepens though the relative positions of the levels do not change significantly. At the above-mentioned voltage difference and Z~10–40, the emitted photons enter the X-ray range. In other types of atoms, the mutual positions of levels are changed: now the energy level depends not on the principal quantum number n only but on the orbit L, spin S and total J numbers as well. A spectral fine structure appears (Figure 7.26). It can be seen that the X-ray spectrum has a line feature with a relatively small (in comparison with optic molecular spectra) number of spectral lines. The X-ray tube radiation usually contains low intensity white radiation background and strong lines of characteristic spectral lines rise above it (Figure 7.27). The most important fact is that the appearance
2D s/2 2
D3/2
2P
M-shell
3/2
2P
1/2
2S
1/2
M series
2P
3/2
2P
L-shell
1/2
2
S1/2
L series 1 2 1 2 1 2 2S
K-shell
1/2
K Series
Figure 7.26 Electron levels of a multielectron atom.
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I K
K
0.4
0.6
0.8
(Å)
Figure 7.27 The tube’s X-ray spectrum; high-intensity characteristic spectral lines over the background of a white bremsstrahlung can be seen.
of the spectrum is very specific: each element has its own spectrum and it can be used to reveal the material that has emitted from the spectrum. The X-rays resulting from electron transition from any higher level to a lower level with the principal quantum number n 1 are known as K-series; transition to a level with n 2 are known as L-series; and series M, N... describe with n 3, 4..., etc. Because of the fact that every level after K is split, the transition L → K in particular produces three closely set lines, the most intense being denoted as Kα1. A Z-dependence of an emitted X-ray radiation frequency ν is given by Moseley’s law (1913). The British scientist H.G.J. Moseley found that the most intense short-wavelength line in the characteristic K-series X-ray spectrum from a particular target (anode) element varied evenly with the element atomic number Z (Figure 7.26). He also found that this relationship could be expressed in terms of X-ray frequency by a simple formula v 2.48( Z 1)2 1015 Hz
(7.6.17)
The Z1 value is the so-called effective atomic number. The point is that one electron kicked out of the K level left a second one untouched. The latter takes part in the screening of the positive nuclear charge by the negative electron charge. The above formula only applies to an L→K transition. An analogous expression describes the other series. An illustration of Moseley’s law for K-series lines of different elements is given in Figure 7.28. The X-ray spectra of all elements are tabulated. Moseley’s discovery is closely connected with the Bohr atomic model, both being announced in the same year (1913). The spectra discussed are produced by inner transitions and are not affected by the valence state and chemically bonded atoms. Therefore the characteristic spectra are the probe of the chemical composition of the material investigated. Analysis of the specimen
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24 Zr
√·10-s (Hz1/2)
Y Cu
16
Cr Cr Fe
Ti Cl 8
Al
Ni
Zn
V K
Si
0 1
8
16
24
32
40
Z
Figure 7.28 A graph of versus periodic number of a chemical element Z (Moseley law).
chemical composition on the basis of Moseley’s law is the essence of the chemical X-ray fluorescent method. For excitation, a corresponding device can use either the primary X-ray beam or primary electron beam described above (refer to Section 7.2). In the latter case, special magnetic coils (lenses) can squeeze the electron beam down to microns and carry the experiment out to the chemical element distribution point by point. Such devices are called quantometers. In order to see the composition of secondary X-ray radiation, X-ray diffraction analyzers are used (refer to Section 6.3.5): an analyzing crystal with a known structure and, hence, with known crystal interplanar distance enables the measuring of the reflecting angle and then the wavelength of the secondary radiation. By consulting tables of the characteristic X-rays, one can find the sample chemical composition in every point, in crystallites and intercrystallite junctions, and in amorphous samples, etc. As an example, the characteristic spectrum of a BaTiO3 ceramic is presented in Figure 7.29. The wavelength is plotted along the abscissa and the relative intensity along the ordinate. The numbers below are the tabulated data of corresponding elements and series of Ti, Ba and O. The line intensity is proportional to the relative amount of a particular element in the sample (at the point irradiated). The spectral lines of Ti-Kα and Ba-Lα are located near each other, whereas the O-Kα line is at a great distance from the first ones. Moreover, such a wavelength is the subject of intense absorption by the air molecules. Therefore, a high vacuum device is needed for such experiments. The method is successfully used with relatively heavy substances (from Be up to U). The analysis of light element (C, S, O, etc.) is rather troublesome because their fluorescent radiation is of a long wavelength and is strongly absorbed by air. Therefore, the precision and sensitivity increases with the increase in the atomic mass of the analyzing materials; however, the possibility does exist to analyze elements with Z 2. Its application ranges from on-line industrial analysis and in-field inspection of geological samples to ultra-trace analysis of semiconductor surfaces.
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I
TiK BaL
27.48 27.75
OK
235.7
(Å)
Figure 7.29 Results of X-ray investigation of a chemical composition of the ferroelectric ceramic BaTiO3: in an arbitrary scale as a graph of the X-ray intensities versus the wavelength is presented (after Yu. Ya. Tomashpolsky et al.).
7.7
AN ATOM IN THE MAGNETIC FIELD: THE ZEEMAN EFFECT
The Zeeman effect consists of atomic energy level splitting and, accordingly, the splitting of the spectral lines of a sample when an external magnetic field is imposed on a sample. If a multi-electron atom is placed in an external magnetic field, depending on the magnitude of the magnetic field induction B, two cases can occur. In the first case, the weak magnetic field is unable to tear out the Russell–Saunders coupling; this is the case of a weak magnetic field or anomalous Zeeman effect. In this particular case, Larmor precession definitely takes place, indeed all cones in Figure 7.25 are precession cones: the whole “construction” takes part in the precession around the direction of this field (z-axis) and mechanic vectors. In the second case, corresponding to a strong magnetic field, the Russell–Saunders coupling is thrown out, and the orbital and spin angular momentums participate in precession independently (normal Zeeman effect). Consider first the case of a weak magnetic field when the bonding of the orbit and spin angular momentums remains unbroken. It was shown in Section 5.1.5 (eq. (5.1.35)) that magnetic moment in the magnetic field with induction B depending on the angle α between vectors M and B acquires additional energy (MB) MB cos . This energy can be expressed in the form E (M J B) M J B cos( ) M J B cos ,
(7.7.1)
(refer to Figure 7.25). ( − ) is an angle between vectors M and B. The product M J cos is the projection of the magnetic moment onto the z-axis, eq. (7.6.15), i.e., M J*. Then, E Mⴱj B g B m J B.
(7.7.2)
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The value E is, accordingly, that energy which an atom admits by the interaction of its magnetic moment with the external magnetic field (with induction B) and which is added to the main electronic energy levels. As a consequence of ∆E depending on the quantum number mJ, and taking 2J 1 values, the subsidiary energy also takes such values. This signifies that the main electron energy level splits into 2J 1 sublevels. If, before the imposition of the magnetic field, the states with different mJ had the same energy, the magnetic field has brought about the levels splitting, i.e., lifting the degeneration on this quantum number mJ. The number of sublevels depends on the quantum numbers, i.e., splitting depends on the atomic term. In Figure 7.30 an example of the splitting of the atomic S- and P-levels in an external magnetic field is presented. Before the field was imposed, each term was degenerated, the S-level was double, and the P-level was sixfold degenerated; both levels can be regarded as singular (Figure 7.30b), i.e., the energy of all sublevels is the same. The energy transition between them is defined by the energy difference EP - ES E, which corresponds to the quantum emission with the frequency 0
E .
(7.7.3)
At the same time, the spin–orbit interaction, even in the absence of an external field (B 0), brings about the splitting of the P-level into two (one of them with J 3/2 and the other with J 1/2, Figure 7.30c). This corresponds to the lifting of the degeneration on the quantum number J. The S-level remains singular. The value of splitting of the P-level is defined by the expression (7.7.2). In the experimentally measured spectrum, provided the resolution is sufficient, instead of one line with the frequency ω0, two lines (a doublet) appear, whereas the central line disappears (as shown in the scheme below). The shift of the lines from the position of the central (absent) line ω0 is defined by the value of energy gB divided by .
mJ 3/2 1/2 −1/2 −3/2 1/2 −1/2
B=O +1
B=O
2
P
0
P3/2
B=O
B=O
2
P1/2
−1
2
S1/2
S
0
gmJ 2 2/3 −2/3 −2 1/3 −1/3
1/2 1 1/2 −1
0
0
0
0
0
(a)
(b)
(c)
(d)
(e)
Figure 7.30 Zeeman effect: (a) normal Zeeman effect, (b) energy levels before imposing a magnetic field, (c) transition P→S (without the magnetic field), splitting because of spin–orbit interaction, (d) and (e) an anomalous Zeeman effect.
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Further splitting occurs when the external field is imposed (B 0, Figures 7.30d and e). The degeneration on the quantum number mJ is lifted according to eq. (7.7.2): at J 3/4 four levels appear, at J 1/2 there are two. In Figures 7.30d and e, the splitting scheme is presented specified by values mJ and gmJ for all sublevels. Transitions can occur between levels (and sublevels). Selection rules limit their number. Strictly speaking, selection rules follow from the analysis of quantum mechanical transition probabilities, however, a qualitative explanation of these rules can be suggested; the law of spin conservation can be attracted. The point is that an electromagnetic radiation quantum (photon) carries away from a system (from an atom) the spin equal to its own spin. The latter equals l . The spin angular momentum can be oriented in a triple way regarding the photon wave vector: perpendicular to it (upward ( ) and downward ( )) and along it (0). Accordingly, mJ can accept three values mJ 0, 1
(7.7.4)
Transitions with mJ 0 are referred to as and with mJ 1 as σ transition; photons are differently polarized, correspondingly. Thus, at transitions P1/2→S1/2, four spectral lines occur whereas at transition P3/2→S1/2 there are six of them (Figures 7.30d and e). The line’s shift relative ω0 is defined as:
B B (mJ g − mJ g),
(7.7.5)
where two strokes characterize the top and one stroke characterizes the bottom levels. This description of splitting of the spectral lines, corresponding to an abnormal Zeeman effect, is given only by quantum physics and cannot be explained within the framework of classical physics. Quite a different picture appears when a high magnetic field is imposed; the Russell– Saunders coupling is torn out, and the orbital and spin moments participate separately in
Z B LL LS
Figure 7.31 A Russell–Saunders coupling breakdown in a high magnetic field.
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precession around the direction of an external field (Figure 7.31). The additional energy E consists of the contribution of both the orbit and spin accounting for the gyromagnetic ratio anomaly: ∆E B BmL 2 B BmS B B(mL 2 mS ).
(7.7.6)
Because transitions with a change of spin quantum number are forbidden (as a quantum emission is connected only with the orbital atomic state change), the spin term in the sum (7.7.6) is excluded and only the degeneration lifting on mL is considered. The S-level remains single- and doublefold degenerated, whereas the P-level (with L 1) splits into three doublefold degenerated levels (Figure 7.30a) (i.e., the degeneration on mL is lifted). The spectral triplet turns out to be exhibited. The shift of each spectral line from the central one is defined as
B B ,
(7.7.7)
In spite of the fact that D-, F- and the other subsequent levels split into a greater number of sublevels because the g-factor is equal to unity in all level splitting, and the selection rule is still valid here as well, the normal Zeeman effect is exhibited as a spectral triplet regardless of the transitions. The exhibiting of a triplet is referred to as a normal Zeeman effect. A normal Zeeman effect can be explained within the framework of classical physics. Displacement of both the energy levels and the spectral line (7.7.7) corresponds to the Larmor precession frequency (refer to eq. (5.2.19))
eB , 2m
(7.7.8)
This coincides with the previous statement that it is possible to treat the additional energy obtained by the magnetic moment in the external magnetic field either as a shift of the energy level on the value (M B), or as the energy of the Larmor precession (7.7.8). In conclusion, we have established that the precession gives a gain of additional energy. Therefore, to treat spatial quantization of the angular momentum (7.5.3) as precession is incorrect from our point of view, because this effect does not produce any specific energy.
7.8 7.8.1
A QUANTUM OSCILLATOR AND A QUANTUM ROTATOR
Definitions
Like translational motion, all molecular motions accomplish rotational and oscillatory motions, too. In Section 2.4.5, an example was given of a diatomic molecule with an interatomic distance d rotating around a stationary axis z passing through the molecular center of inertia. It was shown that rotation of these two masses can be substituted by the rotation
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of one reduced mass accomplishing rotation around the same axis at the same distance d. Remember that any molecule rotated around a stationary axis is called a rotator; if the molecular intra-atomic distance remains constant the rotator is referred to as a rigid one. The expression for the reduced moment of inertia Ie d2 and kinetic energy KIe 2/2 was also given there. No requirement on the value of rotational kinetic energy is imposed in Newtonian physics. Since, in this case, the potential energy is accepted as zero, the kinetic energy is the total one. In Section 1.5.4, a potential curve for classic harmonic oscillations in parabolic form (refer to Figure 1.33), as well as a potential curve describing anharmonic oscillations (Lennard-Jones potential “6–12”, Figure 1.31) were presented. In both cases, the total particle energy in a potential well can have a continuous range of values. Also remember that any oscillating system is called an oscillator. The quantum mechanical consideration of molecular rotation and intramolecular oscillation is of significant interest since it is the basis of optical methods of investigation, both scientific and technological.
7.8.2
Quantum oscillators: harmonic and anharmonic
The potential energy of a system accomplishing small oscillations around an equilibrium position was considered in Section 2.4. The condition “small” signifies that the restoring force is linearly dependent on displacements; oscillations then behave according to the harmonic law (sine or cosine). For classical harmonic oscillation, alongside the continuous spectrum of energy, is the distinctive fact that the probability of finding a system beyond the amplitude displacements is zero. Because we consider here a conservative process, the total mechanical energy is preserved (remember that in a conservative system the energy does not dissipate). In accordance with Section 7.3.2, in order to solve the quantum harmonic oscillator problem, we have to write the potential energy expression in the form x2/2, to substitute it into the Schrödinger equation (7.3.5), find the wavefunction satisfying the standard condition and then find the spectrum of the energy whether it is continuous or discrete. This particular problem can be solved in analytical form. However, while not solving the Schrödinger equations, we will give here the essence of the results. Firstly, the equation for the quantum harmonic oscillator shows that the energy can accept only definite values of energy equal to E 冸 v 12 冹,
(7.8.1)
where v is the oscillation quantum number. The corresponding spectrum of allowed energies is depicted in Figure 7.32 on the background of the classical potential curve. It consists of the equidistant levels with the interlevel distance E ⎡⎣冸 v 1 12 冹 冸 v 12 冹 ⎤⎦ h .
(7.8.2)
Secondly, the number of levels is not limited; the quantum number v can accept any value.
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Thirdly, calculations of the quantum mechanical probability of the energy transitions showed that the quantum number can be changed in increments of 1 only (v 1); the transitions are allowed only between the adjacent levels. In other words, since the levels are equidistant, only one spectral line can be emitted (or absorbed) regardless of which levels the transition takes place between. The emitted/absorbed quantum energy is in any case, always ω. Fourthly, at T 0 K (v 0), the oscillations do not come to an end. The so-called zero oscillations are preserved even at absolute zero temperature. The zero point oscillation energy is equal to ω/2. Finally, the eigenvalues of the wavefunctions can be expressed by Hermite special polynomials. Some of these, together with heir squares, are depicted in Figure 7.32. The principle difference from the classic case is that the probability of finding the oscillator beyond the amplitude values is not zero; it vanishes as the deviation is increased. Of course, we are interested mostly in the oscillation of atoms in molecules. For the sake of simplicity we will deal with the diatomic molecule. As a rule they are the anharmonic ones, i.e., described by a nonsymmetrical potential curve. One with the prevailing potential of such a type in physical chemistry is the Morse potential presented in Figure 7.33: U (r ) U 0 [1 e(rr0 ) ]2 .
(7.8.3)
The potential curve U(r) is equal to zero at r r0; this is an equilibrium interatomic distance [U(r) U(r0) 0]. In the case of small oscillations near r r0, the exponent can be decomposed into a series 1–α(r–r0)±… and we arrive at the quadratic harmonic law
U ψ(x)
X (a)
|ψ(x)|2
V=3
V=3
V=2
V=2
V=1
V=1
V=0
V=0 (b)
(c)
Figure 7.32 A linear harmonic quantum oscillator: (a) classic potential energy curve, (b) wavefunctions and (c) their squares.
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U(r) ⬇ 2(r r0)2. At large values of r the exponent approaches zero and U(∞) → U0. Therefore U0 is the depth of the potential well. Notice that the Lennard-Jones and Morse potentials resemble each other; they differ in the choice of origin of the ordinate axis (one is lifted relative to another by a value of U0; this is insignificant because the potential energy is determined accurate to the constant, U0 in this case; refer to Section 1.5.4). To solve the quantum mechanical problem of anharmonic atomic oscillations in a diatomic molecule it is necessary to accomplish the standard procedure: substitute the Morse potential expression into the Schrödinger equation and solve it, the eigenvalues of energy and eigenwave functions can be found. The given problem cannot be solved in analytical form. We will use only eigenvalues of energy E(v). This is given by the approximate expression E(v) [(v 12 ) (v 12 )2 ].
(7.8.4)
In this expression is the anharmonicity coefficient that characterizes the peculiarities of the interatomic interactions. For some simple diatomic molecules lies in the limit 0.01–0.07; it can be found in the reference books for many molecules. The Morse potential curve is depicted in Figure 7.34 with energy levels of anharmonic oscillations. At small v the second term in eq. (7.8.4) is small and the energy levels are approximately the same as for the harmonic oscillator. While increasing the quantum number v, the negative contribution of the second term increases. As a result, the curve E(v) has a maximum and further increase of the v value brings about a decrease of energy (Figure 7.35): this reduction has no physical sense. So the number of levels of the inharmonic oscillator is limited, i.e., there exists a maximum value of v (v vmax). There are two ways to find the value of this limiting quantum number: either find the extreme value of the functions E(v) from an equation dE/dx0, or equate ∆E to zero at v vmax. In the latter case one obtains 0 E(vmax ) ⎡⎣冸 vmax 1 12 冹 v冸 vmax 1 12 冹2 冸 vmax 12 冹 v冸 vmax 12 冹2 ⎤⎦ , (7.8.5) U U0
r0
r
Figures 7.33 Potential Morse for unharmonic oscillator.
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U U0 3 2
D
1 0 r0
r
Figure 7.34 Potential Morse, the energy levels and the dissociation energy D.
therefore, [1 (vmax 1)] 0,
(7.8.6)
and further, vmax
1 1. 2
(7.8.7)
1 . 2
(7.8.8)
As far as 1, one can obtain vmax ⬇
Substituting this value into the general expression for energy (7.8.4), we obtain the maximum energy and, consequently, the depth of the potential well Emax U 0
. 4
(7.8.9)
In order to tear out the interatomic bond, one has to excite a molecule into the state with E U0, i.e., into the state when an infinite motion of one atom relative to another is realized. If the molecule is in the lower energy quantum state, this energy is E U 0 E0
. 4 2
The energy D required for tearing out the chemical bond of the molecule being in the lower state with v 0 is called the dissociation energy. Accordingly, it is equal to ∆E, i.e., D
(1 2). 4 2 4
(7.8.10)
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E Emax
max
Figure 7.35 Energy of unharmonic oscillator versus the oscillation quantum number v.
The coefficient can be derived from the last expression:
. 4D2
(7.8.11)
As can be seen, the dissociation process is tightly connected with the molecule’s anharmonic oscillation. Within the framework of the harmonic model, it is impossible to explain the dissociation process. The frequency , appearing in all the last equations, is the frequency of natural vibrations, i.e., a frequency corresponding to the lower molecule energy level E /2; the anharmonic term brings a negligible contribution to this energy. When a molecule turns from one energy level into another, emission or absorption of energy in the form of quantum of the electromagnetic radiation occurs. The energy of such quanta is equal to the energy difference. The quantum mechanical calculations show that all transitions are allowed, i.e., there are no selection rules in this case. The frequency of a photon emitted at transitions between adjacent levels is equal: ph 1(Ev1Ev). The corresponding wavelength is
2c . Ev1 Ev
(7.8.12)
An arbitrary quantum transition has been shown in Figure 7.34. EXAMPLE E7.10 The natural frequency of HCl molecule vibration is 5.63 1014 sec1, its anharmonicity coefficient is 0.0201. Find (1) the energy ∆E2,1 molecule transition from the second to the first energy levels (in eV) of the oscillatory spectra; (2) the highest quantum number vmax; (3) the maximum oscillatory energy Emax; (4) the dissociating energy D. Solution: (The theory of molecular anharmonic oscillations is presented in Section 7.8.2 (eq. (7.8.4))).
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U
U0
3 2 =2h
D 1 0
r
r0
(1) The energy difference at transition is Ev+1,v [12(v1)] the quantum number v being the lowest level. Executing calculations, we obtain E2,1 1.09 1019 J 0,682 eV. (2) The vmax can be found according to eq. (7.8.8) Vmax 1/(2)23. (3) The Emax can be obtained if we substitute vmax into the expression for oscillation energy (7.8.4). We can obtain (Emax /4). Substituting all values already obtained we arrive at Emax U0 7.38 1019 J 4.61 eV. (4) The dissociation energy is given by eq. (7.8.10) D U0(1-2). Calculation shows that D 4.43 eV. The corresponding drawings is presented in Figure E7.10. 7.8.3
A rigid quantum rotator
The rotation of micro-objects around a motionless axis was analyzed in Section 7.5.3. The orbital motion of an electron was used as an example. It was found that in this case the rotational energy can accept only discrete values defined by eq. (7.5.26). Since potential energy in free rotation is accepted to be equal to zero, the total energy is kinetic. One of the important characteristics of such movement is the rotational constant B:
B
2 . 2I e
(7.8.13)
The Ie value is called the reduced moment of inertia d2. The quantum number in the optical rotational spectroscopy is defined by another letter (letter j instead of l). Therefore the rotational energy can be written as E Bj ( j 1),
(7.8.14)
(compare with eq. (7.5.26)). No restrictions in quantum number are present here. Besides, j can accept the value j 0, corresponding to zero rotational energy. The rotational energy levels E(j) are depicted in Figure 7.36. The selection rule in the case of rotational spectra
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j
1
22
33
j( j+1)
3
12
2
6
1
2
0
0
44
Figure 7.36 The rigid quantum rotator energy levels.
is the same as in harmonic oscillations, the quantum number difference can accept values j 1, i.e., transitions between adjacent levels only are allowed. It is easy to calculate the energy distance between rotational levels. Naturally, E E j1 E j B[( j 1)( j 2) j ( j 1)] 2 B( j 1).
(7.8.15)
The absolute distance between levels increases at increasing j; however, the relative values, E/E vice versa, decrease. This corresponds to the Bohr correspondence principle.
EXAMPLE E7.11 For an HCl molecule determine: (1) the moment of inertia; interatomic distance is d 91.7 pm, relative atomic masses are correspondingly 1 and 19 a.m.u.; (2) the rotational constant; (3) the energy necessary to excite the molecule from the ground state to the first rotational level (j 1) (see Section 1.3.9, Figure 1.17 and Section 7.8.2, Figure 7.36.) Solution: (1). The moment of inertia can be found according to eq. (1.3.50) I 1.33 10–47 kg m2. (2) The rotational constant is given by eq. (7.8.13). Calculation gives B 4.37 10–23 J 2.73 meV. (3) The energy in question can be calculated according to eq. (7.8.15) E1,0 2B 5.46 meV.
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EXAMPLE E7.12 A multielectron atom is in a state 3P. For this term draw a vector diagram for the maximal atomic orbital angular momentum L (Figure E7.12) and determine the angle between this vector and the orbital angular momentum LL (refer to Figure 7.19)
α LL
LJ LS
Solution: The atomic term 3P is characterized by a set of quantum numbers (refer to Sections 7.5 and 7.6.3): L 1 and S 1. One can draw three vector diagrams like that depicted in Figure 7.19 for different J (which can take on values from L S to |L - S|, i.e., 2, 1, 0. It was mentioned in the problem situation that the maximum value of L is supposed (J 2). Hence the vector diagram is presented in Figure E7.12 according to LS兹苶2, LL兹苶2 and LJ兹苶6 (see eqs. (7.6.5), (7.6.2) and (7.6.7)). The angle sought can be found according to cosine law (Figure 7.25), angle cos
L ( L 1) J ( J 1) S (S 1) 2 L ( L 1) J ( J +1)
3 . 2
Hence arccos 兹3苶/2(/6)30°. EXAMPLE E7.13 A beam of neutrons is thermolyzed by a room temperature moderator (T 300 K) in a nuclear reactor. It then passes a special collimator in a hole in the reactor wall and falls on a graphite single crystal. A diffracted plane of the first order n 1 from the graphite base with interplanar spacing d 33.5 pm was measured at an angle 2 25.5o. Find neutron wavelength , its velocity υ and mass m. Solution: Produced in chain nuclear reactions, fast neutrons should be slowed down in order to participate in further reaction (refer to Section 1.5.5 and Example E1.25). Besides, slow neutrons are widely used in crystal structure investigation and solid state physics (refer Section 7.1.2). Therefore, this example has a practical interest. According to Bragg’s law (eq. (6.3.11) and Figure 6.20) at n 1 we can write 2dsin . Knowing d and angle 2 we can find the neutron wavelength . 2 3.35 1010 sin 12.75 1.46 1010 m.
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The neutron mass can be found from de Broglie formula (7.7.1) and rms thermolyzed neutron velocity (eq. (3.3.7″/)) m
h2 43.96 1027 1.66 1027 m. 3T 2.13 9 1.38 2
Dimension checking gives
J 2 s2 m. m (JK)K 2
EXAMPLE E7.14 Define into how many sublevels will energy levels 2S, 3P, 4D, 9F split in atoms with Russell-Saunders bonds mode owing to spin–orbit interaction and write down the terms describing each sublevel. Solution. In light atoms, a Russell-Saunders type of electron interaction usually take place. In Section 7.6 the method is described in detail. The atomic energy state, in this case, is characterized by a set of quantum numbers: the total orbital quantum number L, total spin quantum number S and total internal quantum number J. Thus quantum number J can accept values from Jmax L S up to Lmin |L –- S |, changing on unit. A certain energy state of atom (spin–orbital interaction) corresponds to each value J, i.e., an energy sublevel appears. The number of sublevels or number of possible mutual orientations of vectors LL and LS at is defined by the ratio 2S 1 referred as multiplicity. At L < S the number of sublevels is defined by another ratio 2L 1 (see Section 7.6.3). Accordingly, for a term 2S we have: multiplicity 2S 1 2, since S ½; L 0 corresponds to symbol S (to not be confused with S-state). Hence, condition L < S and the number of sublevels determined by expression 2L 1 1 is valid. This means that the given energy level is not split and is characterized by quantum number J L S ½ and accordingly term 2S1/2 is created. For term 3P multiplicity is 2S 1 3, therefore S 1; to symbol P corresponds L 1 and L S; the number of sublevels determined by expression 2S 1 3 coincides with term multiplicity, i.e., it is equal 3. Thus, to each sublevel there corresponds the term with various J values. As Jmax L S 2 and Jmin |L – S | 0. J accepts values 0, 1, 2 and then corresponding terms will be as 3P0, 3P1, 3P2. The term 4D corresponds to S 3/2, since multiplicity is 2S 1 4, L 2 (D state). As condition L S is valid J accepts values from L S 2 3/27/2 up to (L – S) (2 – 3/2) 1/2, i.e.,1/2, 3/2, 5/2, 7/2. For this set of sublevels the quantum numbers is equal to multiplicity (i.e., 4); the corresponding terms are 4D1/2, 4 D3/2, 4D5/2, 4D7/2. The term 9F has S 4 (i.e. 2S 1 9); therefore, the F-state meets L 3. Thus, inequality L S and the number of possible mutual orientations of vectors LL and LS and consequently the number of sublevels is defined by a ratio 2L 1 23 1 7 is fulfilled. Thus quantum number J will accept the following values; 1, 2, 3, 4, 5, 6, 7 and each sublevel will be characterized by a term: 9F1, 9F2, 9F3, 9F4, 9F5, 9F6, 9F7.
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EXAMPLE E7.15 In the Stern–Gerlach experiment a narrow beam of sodium atoms in the ground state was passed through a highly nonuniform magnetic field (Figure E7.15). What nonuniformity of the magnetic field ∂B/∂z should be provided so that the distance between the components of the sodium atoms split beam fixed on the screen will be equal to 6 mm? The installation dimensions are l1 10 cm, l2 15 cm, speed of atoms is equal to υ0 400 m/sec. The whole installation is placed in a hermetic vacuum shield. z sample
S
S = 2( 1+ 2) υ0
heater N
N
υ0
υz
1 2
x screen
l2
l1 z
(a)
(b)
Solution. Figure E7.15a shows the main way of producing a highly nonuniform magnetic field due to tailor-made poles and Figure E7.15b describes the path the atom travels. We have to consider some details. The basis for the decision of this problem is formula (5.1.32): a force working on a magnetic dipole depends on both the nonuniformity of the magnetic field and the orientation of the atomic magnetic moments relative to the quantization axis z. If the atomic magnetic moments submit to laws of classical physics, there could be any values of angles and the experiment would result in a fuzzy maximum. However, in quantum mechanics this is not a case! The force working on an atom depends on atomic state, degree of nonuniformity of the magnetic field in the device and a set of magnetic quantum numbers. Eq. (5.1.32) is still valid in quantum physics though the angle depends here on the set of quantum numbers mentioned: M z /Mcos mJ/兹苶 J(苶 J苶1苶). The quantum number mJ can accept an amount of the discrete values dependent on quantum number J, that is in all 2J 1 values. Since to each value mJ corresponds the force as in eq. (5.1.32), the atomic beam is split on some component; there will be a system of strips on the screen. To each strip there corresponds a certain number of mJ . The sodium atom in the ground state has one 3s electron in M shell. Therefore, the set of quantum numbers are: L 0, S1/2 and J1/2, this corresponds to mJ 1/2. So we know that there should be two strips on the screen. In order to calculate the distance between them and find the magnetic field nonuniformity, we have to solve the problem qualitatively. There is no need to calculate g-factor Lande since we know that only spin contributes to the sodium atomic magnetic moment, i.e., g 2 (see Section 7.6.2). In Figure E7.15, a beam of atoms from a heater pass some of the slits shown. Entering the nonuniform magnetic field, a beam splits into two parts, which move
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along curvilinear (parabolic) trajectories. The acceleration inside the magnet can be calculated according to the second Newtonian law azFz/m. Then the shift is azt12 az12 1 where t1 1 and 1 . The movement along l2 is assumes 0 2 202 uniform. In time t2 the atom acquires the velocity υz azt2. The shift gained is 12 2zt2azt1t2az . The total relative shift is then 2(12) and 02 az1 ( 22). Taking into consideration all calculations we obtain 02 1
⎛ B ⎞ g B ⎜ ⎟ mJ ⎝ z ⎠ m02
1 ( 1 2 2 ).
Therefore, we arrive at
A02
B
z B 1 ( 1 2 2 ) N A (keeping in mind that g|mJ| 1). Executing calculations we obtain
B 6 103 23 103 4002 J T
z 9.27 1024 0.1 (0.1 2 0.15) 6.02 1023 98.9 JT ⬇ 1 Tcm. 7.8.4
Principles of molecular spectroscopy
The electromagnetic radiation emitted or absorbed by the substances under investigation is the subject of the oscillation–rotation (molecular) spectroscopy. In this section, the main attention is concentrated on molecular spectroscopy near and within the optical range of frequencies. Emission of the radiation quanta or their absorption is defined by energy transitions. Therefore, the basis of all spectroscopy methods is the discontinuity of energy spectrum and the mutual position of energy levels. Of course, this is only a general treatment: many factors playing an important role in spectroscopy remain outside our discussion. Naturally, the best effect in the interaction of radiation with matter can be reached in the case when the energy of radiation quanta used coincides in order of values with energy transitions. Therefore, we will consider that range of quanta energy complies with that of energy transition on the basis of the data given in Table 5.3. The total molecule energy can be written as a sum: E En Ev E j ,
(7.8.16)
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where quantum numbers n, v and j denote electronic, oscillatory and rotational contributions. Let us evaluate each part separately. The electronic energy levels are assigned by eqs. (7.5.31) and (7.5.34). Depending on the nuclear charge Z the ground state energy varies from 10 eV up to 100 keV. Therefore, only transitions among the high lying atomic levels enter the optical frequency range (for example, the Balmer series with v 2). In order to evaluate the oscillatory energy levels, it is necessary to know their intrinsic frequencies. Experiments show that they are 10121014 Hz. Correspondingly, the emission frequency belongs to the IR radiation region and the wavelengths are 104–106 m. The purely rotational frequencies lie in the frequency range dependent on the molecule’s rotational constant (7.8.13). Estimations show that B ⬇ (1068/1047) ⬇ 1021 J or 102103 eV. This corresponds to the wavelength ⬇ 103 m, which corresponds to a microwave frequency far from the optical range. Important conclusions follow from these consideration. First, emission and/or absorption due to electron transition, lying far from the optical range in the short wavelength side, are not considered in molecular spectroscopy. Second, the same applies to purely rotationary spectra, though the shift is in the long wavelength region. The main interest in optical molecular spectroscopy concerns oscillatory-rotationary transitions, presented by the last two terms of the sum (7.8.16). The energy transitions in this case correspond precisely to the visible and adjoining wavelength regions (IR and UV). An arrangement of the electronic, oscilationary and rotationary levels of energy of a hypothetical molecule is submitted in Figure 7.37. The distances between oscillatory levels only, determined by (eq. (7.8.2)), are larger than distances in the rotationary levels (eq. (7.8.14)). Therefore, rotary levels settle down between oscillatory levels (according to the sum (7.8.16)), however, the number of rotary levels between the next oscillatory levels is limited to the distance between the last. If E(v) accepts a certain value, the number of rotationary levels is defined by their highest value (in an accepted interval E(v)). In other words, the difference E Bjmax ( jmax 1) should be less than or equal to E. Transitions are carried out between levels with two sets of quantum numbers (v, j). Let us consider in general the scheme of a spectroscopic experiment. We usually distinguish emission spectroscopy (Figure 7.38a) and adsorption spectroscopy (Figure 7.38b). In the first case, the sample under investigation is excited in order to force it to emit quanta of radiation (e.g., by heating the sample). Radiation is directed onto the spectral device with a prism or diffraction grating (see 6.3.4), decomposing the radiation in a spectrum along wavelengths. The investigator writes down the results on a paper (screen) as the dependence of intensity on wavelength (wavenumber). In adsorption spectroscopy the light source emits a “white” spectrum (containing the entire wavelength in a certain interval), which then goes through the sample. The falling radiation is absorbed by the substance that cuts out a definite characteristic wavelength from the incident beam. The corresponding spectral device gives the results as strips with the “cut out” frequencies. As an example, the absorption spectrum of chloroform is presented in Figure 7.39. Traditionally in molecular spectroscopy energy is measured in wavenumbers 1/, i.e., it is expressed in cm1. It is simple to derive the relation between wavenumbers and energy:
2c hc. v
(7.8.17)
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E n
V
j
2
−3
1
−2 −1 0
0 En" 2
1
−3
0
−2 −1 0
En'
Figures 7.37 The overall electronic, oscillator and rotator electron levels (not to scale).
I K Sp S 1/ (a)
I I
K
1/
S Sp
E (b)
1/
Figure 7.38 Molecular spectroscopy schemes: (a) emission, (b) absorption; are samples, K collimation slits, Sp spectral recorder, E source of radiation, I(1/) type of specter’s measured. Incertion: corresponding source spectra.
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Cl C
H
Cl
Cl 3
5
C-Cl 7
9
11
13 16
25 ν
Figure 7.39 Chloroform absorption optical spectrum.
In conclusion, we should note that when translating from the SI system to reciprocal centimeters (cm–1), it is necessary to divide the result by 100.
PROBLEMS/TASKS 7.1. A narrow beam of sodium atoms in the ground state passes through a Stern-Gerlach device with a nonuniformly magnetic field ( B/ z1 T/cm) and l 10 cm in length. Determine the distance between the components of a split beam at the outlet of the magnet. The sodium atom’s speed is 300 m/sec (refer to Figure E7.15). 7.2 An atomic state is characterized by the two spectral terms 1D and 1P. Find all possible quantum numbers J for these terms and draw a scheme of level splitting in a weak magnetic field. 7.3. Determine the minimum energy of ⏐E⏐min (in meV) of an atom in the state 2F in a uniform magnetic field B 0.8 T. 7.4. An atom is in a state characterized by the term 4D. Calculate the minimum value of the total angular momentum LJ and draw the corresponding vector model (refer to Figure 7.19). Determine the angle between the spin LS and total LJ angular momentums. 7.5. An electron of energy E 0.5 U0 moves in the positive direction of an axis x. Estimate the probability that the electron will penetrate through the potential barrier of height U0 10 eV and width d 0.1 nm. It is useful to make a corresponding drawing. 7.6. An electron of energy E 9 eV moves in the positive direction of an axis x. At what width of potential barrier d will the transparency factor be equal to D 0.1 if the height of the barrier U0 is 10 eV? It is useful to make a corresponding drawing. 7.7. A monochromatic electron beam meets in its motion a potential barrier d 1.5 nm in width. At what energy difference (U0 – E) (in eV) will the barrier penetrate 0.001 of incident electrons. It is useful to make a corresponding drawing. 7.8. An electron meets in its motion a potential barrier of the height U0 10 eV and in width d 0.2 nm. At what electron energy E (in eV) will the barrier transmission probability w be w 0.01. It is useful to make a corresponding drawing.
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7.9. A particle is in an infinitely deep potential box of width L. The particle state corresponds to wavenumber k (/L). What is the probability of finding the particle in an area (L/2) x (L/4)? It is useful to make a corresponding drawing. 7.10. A particle is in an infinitely deep potential box of width L. The particle state corresponds to the quantum number n 2. What is the probability of finding the particle in an area (L/3) x (2L/3)? It is useful to make a corresponding drawing. 7.11. A vanadium atom in the state 4F3/2 passes through a Stern-Gerlach device (refer to Figure E7.15b). The atom’s velocity is 400 m/sec. Determine the distance between the upper and lower component of the split beams spots if l1 l2 10 cm and B/ z3 T/cm. 7.12. A silver atom’s beam is passed through a Stern-Gerlach device (refer to Figure E7.15b). The atom’s velocity is 300 m/sec. Determine the magnetic field gradient B/ z if the distance between the ends component of the split beam is 2 mm, l1 10 cm and l2 0. Silver atoms are in the ground state (L 0, J 1/2, g 2). 7.13. Find the number N of vibration energy levels for HBr molecule if the anharmonicity coefficient is 0.0201. 7.14. Knowing the natural angular frequency of a CO molecule (ω 4.08 .1014 sec–1), find its rigidity (quasi-elastic) coefficient . 7.15. Determine the dissociation energy D (in eV) of a CO molecule if the natural frequency is ω 4.08 1014 sec–1 and the anharmonicity coefficient is 5.83 10–3. 7.16. For an O2 molecule, find (1) the reduced mass , (2) the internuclear distance d, if the rotational constant B 0.178 meV, (3) the angular velocity ω if the molecule is on the first rotational energy level. 7.17. Find the angular momentum of an O2 molecule if its rotational energy EJ is 2.16 meV. 7.18. Can monochromatic electromagnetic radiation with wavelength 3 µm excite vibrational and rotational energies of the HF molecules if it is in a ground state? 7.19. Determine the multiplicity of the energy levels of a diatomic molecule with quantum number J. 7.20. Calculate the internuclear distance d in CH molecules, if the V interval in the purely rotational emission spectrum is 29 cm1.
ANSWERS 7.1. B gJl ( B z ) 2.70 mm. m2 2
7.2. For the term 1P J 1 and mJ 0, 1; for the term 1D J 2 and mJ 0, 1, 2. 7.3. ⏐E⏐min Bg ⏐mJ⏐min, B 0.0199 meV (g (6/7) at J (5/2); ⏐mJ⏐min (1/2). 7.4.
LJ ,min
3 ; cos 0.447; 166.5 2
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⎡ 2d ⎤ 2 m(U 0 E ) ⎥ 0.1 7.5. w exp ⎢ ⎣ ⎦ 7.6.
d
n(1D) 2 2 m(U 0 E )
0.22 nm. 2
7.7.
⎡ n 冸1w 冹 ⎤⎦ (U 0 E ) ⎣ 0.20 eV. 8md 2 2
7.8.
E U0
[n(1w)] 5.0 eV. 8md 2
1 1 7.9. w 0.409. 4 2 1 3 7.10. w 0.196. 3 4
7.11.
7.12.
3 B Jgl 2 m2
B
z 3.7 mm.
B m2 116 Tm.
z 3 B gJ 2
7.13. N Vmax (1/2) ⬇ 24. 7.14. 2 1.91 kN/m. 7.15. D
1 2 11.4 eV. 4
7.16. (1) 1.33 .10–26 kg , (2) d 121 pm, (3) ω 7.61 .1011 sec–1. 7.17. L 3.66 .10–34 J sec. 7.18. Only rotational motion occurs. 7.19. 2J 1. 7.20. d 112 pm.
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8.1
SELECTED ATOMIC NUCLEI CHARACTERISTICS
So far we have considered the atomic shell and the point (dimensionless) nucleus with charge Ze interacting with electrons by Coulomb forces according to the laws of quantum mechanics. The account of the nuclei characteristics with a definite dimension raises many important and interesting phenomena. They are the basis of modern methods of investigation of atomic and molecular structure of chemical substances. The aim of the present chapter is to give relevant information on the atomic nucleus. Taking into account the general aim of the book, we will only touch on those nuclei properties, which have a direct bearing on the description of methods, so this information cannot be considered as comprehensive. Among the modern chemical and physical methods of investigation, spectroscopic methods are widely used (they are often called instrumental methods). These methods allow quantitative information on composition and structure to be obtained effectively and rapidly, and also enable analytical and other problems to be solved without the destruction of the materials tested. Further, these methods are in the process of development and improvement. Very popular among these methods are the nuclear resonance (spectroscopic) methods, which are based on the fundamental discoveries made in quantum mechanics and radioengineering in the second part of the 20th century, namely, nuclear magnetic resonance (NMR), nuclear quadrupole resonance (NQR), -resonance (Mössbauer) spectroscopy (RS), electron paramagnetic resonance (EPR), and others. As these resonance methods are based on electron–nuclear interactions, we must provide some information on the nuclear physics required to understand the subject. 8.1.1
A nucleon model of nuclei
A nucleus possesses an internal structure. It consists of many elementary particles— nucleons—the main of which are protons and neutrons. They move in the field of nuclear forces and continuously change their state from that of charged particle (proton) to neutral particle (neutron) and back. However, on an average in a given nucleus, a certain number 497
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of protons, Z (which defines the given element’s number in the periodic system) and number of neutrons, N (which defines the particular isotope of a given element) are always present. Since the mass of protons and neutrons are approximately 2000 times greater than the mass of electrons, the total mass of an atom is practically comprised in its nucleus. The total number of nucleons, A, is called the mass number, i.e., A⫽N⫹Z. The symbol AZN is used to identify a nucleus, where Z is the chemical symbol of a given element. Mass number A defines the mass of a nucleus. Thus, there are 92 protons and 143 neutrons in the uranium nucleus 235U92. The number of nucleons Z is often omitted because the relevant information is contained in its chemical symbol. Thus, 235U is the usual symbol. In the text below, a proton mass is marked by mp. Mass can be measured in kilograms (mp⫽1673⫻10⫺27 kg), or can be denominated in atomic units of mass (1 a.m.u.⫽ 1.66⫻10⫺27 kg). It is sometimes also expressed in energy units (mp⫽938.2 MeV). Each nucleon possesses its own angular momentum, i.e., spin. The quantum numbers of neutron and proton spins are the same as an electron spin, i.e., 1/2. In general, a complex nucleus also possesses a spin angular momentum. As a result of the partial or complete compensation of nucleonic spins, the total spin can have values of 0, 1/2, 1, 3/2, etc. Herewith, in full analogy with the electronic shell, the angular moment (mechanical moment) of a nucleus is: LI ⫽ I ( I ⫹1),
(8.1.1)
where I is the spin quantum number of a particular nucleus. Analogously with electron characteristics, the projection LI on the selected axis z can have the values: Lz ⫽ mI ,
(8.1.2)
where quantum number mI ⫽ I, I⫺1, … 0, … ⫺I can have 2I ⫹ 1 values. From eqs. (8.1.1) and (8.1.2), it can be seen that the value of electron and nucleon angular momentums have the same order (⬃). Because the nucleus spin is stipulated by the behavior of both neutrons and protons, the existence of nuclear magnetic momentums can be expected as well. The vectors of mechanical and magnetic moments are tightly bonded to each other. Nuclear physics was developed much later than atomic physics and this fact has influenced the approach to the description of nuclear properties. So, eq. (5.2.2), describing the electron gyromagnetic ratio of magnetic to mechanical moments, is applied to the nuclei as well. As a result, the gyromagnetic ratio for a spin-possessing nuclei was formally transcribed as: 冏e冟 M ⫽ gN LI 2 mp
(8.1.3)
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(in SI), however, instead of the mass of electron me in the denominator, the proton’s mass mp is present. Hence, the magnetic moment of a nucleus can be derived as:
M I ⫽ gN
冷e冨 I ( I ⫹1) 2 mp
(8.1.4)
The value e ⫽ N 2 mp
(8.1.5)
is referred to as the nuclear magneton and denoted as N (N ⫽ 5.05 ⫻ 10⫺27 J/T). Since in nuclear physics the mutual orientation of the angular and magnetic moments can be both parallel or antiparallel, the gyromagnetic ratio can also be positive (parallel orientation of these two vectors) and negative (antiparallel orientation as for orbital electron state). Therefore, the nuclear g-factor (gN) (in units e/2mp) can have both positive and negative signs. They cannot be calculated but are defined from experiment only. Therefore, one can see that the gN sign is historically casual. For proton gp is ⫹5.5851, for neutron gN⫽⫺1.9103. It can be seen from eqs. (8.1.4) and (8.1.5) that the magnitudes of the nuclear magnetic moment are approximately 2000 times lower than the corresponding electron values. Therefore, special instruments are necessary for observing them. The z-projection of MI is: M Iz ⫽ gN N mI
(8.1.6)
Because MI and LI are tightly bonded, nuclei precession in the magnetic field also takes place (refer to Section 5.2.4). The magnetic field created by electron shell can in this case be regarded as external to the nucleus. It is necessary to note that the nature of the nucleon’s magnetism is still not sufficiently clear. For instance, it is not understood why such an electrically neutral particle as a neutron nevertheless possesses a magnetic moment. 8.1.2
Nuclear energy levels
One of the distinctive signs of any quantum mechanical system, including nuclei, is the discontinuity of its energy states. The permitted values of nuclear energy depend on its particular structure, the nature of the nuclear forces, and more. Unfortunately, at present, nuclear physics does not allow us to calculate theoretically the arrangement of energy levels; these are experimentally found values. Nuclear ground states E0 are stable, not changing their energy in the course of time. However, by one or other external influence, they can be excited to states E1, E2, etc. All
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excited states are unstable: in the course of time, nuclei spontaneously transform to the lower or ground state, emitting in the surrounding space an excess of energy in the form of rays. The time during which the number of originally existing nuclei decreases in e times is referred to as the nuclear lifetime . For a given nucleus in different excited states, can have widely different values, however, a quantum mechanical relationship always exists between the value and uncertainty of energy E in the given energy state (refer to Section 7.2). E ⱖ h.
(8.1.7)
For instance, if ⫽⬁ (ground state), energy E0 can have a precise value (E⫽0). If ⬃10⫺8 sec, then E⬃10⫺7 eV. Usually, relative to E0, the first excited energy levels of many nuclei are E1⫺E0⬃104 eV. Therefore, the ratio E/(E1⫺E0) is of 10⫺11 in the order of value and seemingly can be regarded as negligible. However, as we will see below, this is not so. 8.1.3
Nuclear charge and mass distribution
The distribution of neutrons and protons in a nucleus can be different. In the majority of experiments, a nucleus behaves as a system possessing a center of symmetry and axis of symmetry of an endless order. Consequently, a nucleus also possesses a mirror plane of symmetry perpendicular to the symmetry axis passing through the center of symmetry. The charge distribution can be described by the function of the charge density (x,y,z). The normalization of the charge to the total nuclear charge gives:
∫ (r )d ⫽ Ze,
(8.1.8)
V
The distinctive features can be mentioned in this description. Firstly, the average density function can differ for charge and mass distribution. If, in particular, protons are distributed at the nucleus periphery and neutrons are concentrated nearer the origin, the characteristic radii of the two can be different. They can differ even for the two energy states of one and the same nucleus. Secondly, the nuclear forces are of short-range nature. Therefore, nuclei are sharply outlined in space. There is an empirical expression for spheri cal nuclei size: rЯ ⬇ 1.25 ⫻10⫺15 3 A m,
(8.1.9)
where A is the mass number. Thirdly, the nucleus can deviate from the spherical symmetry toward the ellipsoid of revolution with three mutually perpendicular second-order symmetry axes. Such an ellipsoid is of two radii: a is directed along z-axis and b is perpendicular to it. The ellipsoids of revolution can be flattened (a⬍b) or elongated (a⬎b) (Figure 8.1).
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z a z
z
x
b y
y
y
x
x (a)
(b)
(c)
Figure 8.1 Models of different nuclear forms: (a) spherical, (b) elongated and (c) flattened.
8.1.4
Nuclear quadrupole electrical moment
Point charge and dipole electrostatic fields have been described in Chapter 4. There are more complex electrical systems created by irregularly distributed charges (refer to Appendix 3). It is shown in Appendix 3 that such a system can produce an electrostatic field, which can be described by the sum of terms differently dependent on the distance r (Table 8.1). Being charged, the nucleus has nonzero monopole contribution. Since nuclei possess the definite symmetry elements listed above, it cannot possess a dipole moment; a dipole moment as a vector cannot coexist with the symmetry operations mentioned. This signifies that the nucleus does not possess a dipole moment. However, this does not prohibit a nucleus from having an electrical multipole moment of higher order: quadrupole, in particular. In other words, the deflection of the nucleus from a spherical form brings about the existence of the quadrupole moment eQ (refer to Appendix 3). Value eQ is measured in C.m2 units (in SI). The mathematical description of the quadrupole moment can be given as:
∫∫∫ ( x, y, z)(3z
2
⫺ r 2 )dV ⫽ eQ
(8.1.10)
V
where r2⫽x2⫹y2⫹z2, dV ⫽ dx dy dz. Integration is subjected over the whole nucleus volume V. Bearing in mind that in the reference system connected with the nucleus (refer to Figure 7.16), this expression can be presented in trigonometric form: eQ ⫽ ∫∫∫ ( x, y, z )r 2 (3 cos2 ⫺1)dV .
(8.1.11)
V
As shown in Appendix 3, the charged system elongated up to the z axes possesses a positive quadrupole moment, though for the flattened nucleus it is negative. The deflection of the nucleus from a spherical form and the nucleus spin are interrelated. Therefore, one can expect that the experimental manifestation of a quadrupole moment and nuclear spin are also correlated. This is exhibited in the following. The uncertainty in the angular momentum orientation (see Figure 7.17) is present in the quadrupole moment
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Table 8.1 Properties of a number of electric systems Potential as dependent on distance r
Charges system
Electric field E as dependent on distance r
Energy in an external electric field
Monopole
Ec ⬃
q r2
c ⬃
q r
U ⬃ q
Dipole
Ed ⬃
p r3
q ⬃
p r2
U⬃p
Quadrupole
Eq ⬃
qa 2 r4
q ⬃
qa 2 r3
U ⬃ qa 2
d dr d 2 dr 2
too. The measured value is not the eQ itself but the quantity eQz that can be expressed through the nuclear spin I as: eQz ⫽
I ⫺ (1Ⲑ 2) eQ. I ⫹1
(8.1.12)
This means that at I⫽0, the quadrupole term is meaningless; at I⫽1/2, it cannot be measured and it is reliably measured at I⬎1/2. 8.2 8.2.1
INTRAATOMIC ELECTRON–NUCLEAR INTERACTIONS
General considerations
If we consider a nucleus being not a point but a volumetric nucleus, additional effects in the electron–nucleus interaction appear. They play a very important role in the physical description of an atom. These additional effects are exceedingly small in comparison with the main Coulomb and even with fine interactions (refer to Section 7.5.4). So, they refer to the number of intraatomic “superfine interactions.” Remember that the energy of the electrostatic interaction of two electric charges is defined by product q (refer to Section 4.1.4) in which is a potential created by the electron’s charge in the point where the proton’s charge is located. Therefore, the additional interaction energy appears. (Certainly one can consider that the field is created by a volumetric nucleus and an electron is in this field; this effect will also be considered further.) The most general expression for the electrostatic (marked by letter E) interaction energy can be written as:
EE ⫽ ∫∫∫ ( x1 , x2 , x3 )( x1 , x2 , x3 )dx1 dx2 dx3 . V
(8.2.1)
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For convenience in this expression, instead of Cartesian coordinates x,y,z we can choose another form of coordinates x1;x, x2;y, x3;z. The function that describes a charge distribution in a nucleus is (x1,x2,x3), the potential created by the electron shell at the nucleus is denoted (x1,x2,x3), and the integral is taken upon the whole nucleus volume V. Decomposing the function (x1,x2, x3) in the MacLoren series near the origin: 3 ⎛ ⎞ x ( x1 , x2 , x3 ) ⫽ (0,0,0) ⫹ ∑ ⎜ x ⎟⎠ 0 ⫽1 ⎝
⫹
1 3 ⎛ 2 ⎞ 2 1 3 3 ⎛ 2 ⎞ x ⫹ ∑ ∑ ⎜ x x ∑ 2 ⫽1 ⎜⎝ x 2 ⎟⎠ 0 2 ⫽1 ⫽1 ⎝ x x ⎟⎠ 0
(8.2.2)
where (⭸/⭸x)0 are electric field components along the axis x, (⭸2/⭸x2)0 are the electric potential gradient along the same axis. Because of the small dimension of nuclei, it is possible to limit the number of terms in expression (8.2.2) by the second order. The electric field gradient possesses the symmetry axis along the x3 axis. Therefore, all crossed terms of the type (⭸2)/(⭸x⭸x) with ⫽ are equal to zero. Substituting the expression (8.2.2) into the electric energy (8.2.1) we have: 3 ⎧⎪ ⎛ ⎞ 1 3 ⎛ 2 ⎞ 2 ⎫⎪ EE ⫽ ∫∫∫ ( x1 , x2 , x3 ) ⎨(0,0,0) ⫹ ∑ ⎜ x ⫹ ⎟ 2 ∑ ⎜⎝ x 2 ⎟⎠ x ⎬ dx1 dx2 dx3 V ⫽1 ⎝ x ⎠ 0 ⫽1 0 ⎩⎪ ⎭⎪
(8.2.3) Opening the brackets and carrying out elementary transformation we obtain: 3 ⎛ ⎞ EE ⫽ ∫ ( x1 , x2 , x3 )(0, 0, 0)dV ⫹ ∑ ⎜ ⎟ ⭈ ∫ ( x1 , x2 , x3 ) x dV ⫽1 ⎝ x ⎠ 0
⫹
1 3 ⎛ 2 ⎞ ⭈ ( x1 , x2 , x3 ) x2 dV ⫹... ∑ 2 ⫽1 ⎜⎝ x2 ⎟⎠ 0 ∫
(8.2.4)
For our convenience, let us write every term separately presenting eq. (8.2.4) as a sum EE⫽EE1⫹EE2⫹EE3 and then: EE1 ⫽ ∫ ( x1 , x2 , x3 )(0, 0, 0)dV Because (0,0,0) is the electric potential at the origin, it is the number and can be taken out from the integral. The rest is the nucleus charge, Ze. Therefore, EE1 ⫽ eZ (0, 0, 0),
(8.2.5)
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The first term in eq. (8.2.5) EE1 is the interaction energy of atomic electrons with the nucleus, the charge of which is contracted to point; the nucleus becomes dimensionless. Integrals of the type EE2⫽ (x1,x2,x3)xdV in (8.2.4) are equal to zero because of the fact that according to their mathematical structure, they describe nuclear electric dipole moments though we already know that the nucleus cannnot have one. Therefore, eq. (8.2.4) reduces to two terms, the first and the third: EE ⫽ eZ (0, 0, 0) ⫹
1 3 ⎛ 2 ⎞ ( x1 , x2 , x3 ) x2 dV , ∑ 2 ⫽1 ⎜⎝ x2 ⎟⎠ 0 ∫
(8.2.6)
By some transformations, the term EE3 can be transformed into the sum: EE′ 2 ⫽ 23 e2 Z 冷 (0) 冨2 冬r 2 冭
(8.2.7)
1 3m 2 ⫺ I ( I ⫹1) . EE′′3 ⫽ zz [eQ] 2 4 3I ⫺ I ( I ⫹1)
(8.2.8)
and
Consider all terms separately. We will refer here to the energy level scheme in Figure 8.2. 8.2.2 Coulomb interaction of an electron shell with dimensionless nucleus The first term in (8.2.4) (EE1 in eq. (8.2.5)) describes the electrostatic interaction of an electron shell with the point nucleus charge at the origin (Figure 8.2a). This term corresponds fully to the Coulomb energy, which was discussed in Section 7.5; hence, the quantization of the energy is just the same as was accepted above. In other words, E1 is the energy that describes the electron shell state in the absence of superfine interactions. Evaluation of the energy has already been made in Section 7.5.4. It shows that depending on Z, the energy EE1 is 101⫺104 eV. Remember that the addition to EE1, the energy of spin–orbit fine interactions is Enj⬃10⫺1/10⫺3 eV. Other types of interactions have made their own particular contribution to the energy levels. 8.2.3 Coulomb interaction of an electron shell with a nucleus of finite size: the chemical shift Let us consider the next term eq. (8.2.7). Two multipliers exist in this term: 冓r2冔 is the mean-square nuclear charge radius (determined according to the general rule (eq. (3.1.6⬘)) 冓r2冔⫽( (x1,x2,x3)r2dv)/( (x1,x2,x3)dv) and 兩 (0)兩2 is the probability density of finding an electron at the origin. Notice that this term accounts for the nonzero nuclear size.
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(a)
(b)
505
(d)
(c)
(e) +3/2
∆2
+3/2
3/4 1/4
+1/2
+1/2 –3/2 –1/2
–1/2 –3/2
+1/2 1/4
∆1
–1/2
∆ 0
0
∆ 0
0
Figure 8.2 The particular effects influences on the energy state of an atom: (a) a stripped nucleus, (b) the chemical shift (refer to Section 8.2.3), (c) the fine electron-nucleus interaction (refer to Section 7.5.8), (d) and (e) an atom in an external magnetic field (in (d) influence of both quadrupole and magnetic splitting is very complicate), (e) the influence of the magnetic field produced by the electron shell on the nucleus’s magnetic moment position (nuclear Zeeman effect, see Section 7.7). The shift of the spectral line relative 0 is presented in the lower part of the Figure. The energy transitions scale is not followed. The long arrows represent R transitions, the double arrow in the figure represent the NMR transitions.
It follows from eq. (8.2.7) that this energy contribution characterizes the electron cloud interaction with the nuclear charge. This type influences the shifting of the nuclear energy levels but does not split them. The energy shift depends on a product: the electronic parameters [e兩 (0)兩2] and the nuclear properties (Ze冓r2冔) (Figure 8.2b). Because nucleus excitation can lead to a change of its mean-square radius 冓r2冔, this value can differ for two levels—the ground and the excited ones. Therefore, the energy difference can be changed. This phenomenon is referred to as the chemical shift. The experimental spectral lines are schematically shown under the energy transitions. Analysis of the electron radial curves (Figure 7.22 and Table 7.1) shows that only s-electrons can influence the 兩 (0)兩2 value. In the isolated atom, the energy E⬘E2 could change the line position only a little. This is schematically shown in Figure 8.2b. The E⬘E2 value is often 10⫺4 eV and has a positive sign. The difference between the energy of two levels is significantly smaller than the energy itself and is usually 10⫺7–10⫺8 eV. As a result, the line position on the experimental spectrum is shifted as well by , as shown in the figure. For a given atom in different atomic groups and/or molecules, the E⬘E2 value can be changed. When an atom enters the chemical bonding with other atoms, it produces new molecules and the whole electron system undergoes a change (s-electrons including). This last influences the E⬘E2 value. Thus, the chemical shift is a very sensitive measure of the chemical bonding.
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It is impossible at present to calculate the E⬘E2 value theoretically. However, the change in the energy transition can be found experimentally. In order to give this value a quantitative character, measurements are carried out relative to a specially chosen reference substance. In this case, one can measure the quality: EE′ 2 ⫽ Ze2 冬r 2 冭冦冷 (0, 0, 0) 冨2 ⫺ 冷 (0, 0, 0) 冨2refr 冧,
(8.2.9)
where 冓r2冔⫽{冓r2excited冔⫺冓r2ground冔} is the change of the mean-square nuclear charge radius in the transition from the excited to the ground state. As soon as the reference substance is chosen (the second term in the difference in the figure bracket), the E⬘E2 quality is called the chemical shift relative to a definite compound. In modern chemistry, the chemical shift is one of the main, widely used characteristics of an atom in a molecule. Measurement of the chemical shift can be achieved by means of NMR and R. It should be noticed, however, that the specificity and individual development of these two methods are the reason that different data and different reference substances determine the difference in results, although the physical content of both methods remains the same. 8.2.4 The nuclear quadrupole moment and the electric field gradient interaction A second term in eq. (8.2.6) corresponds to the so-called quadrupole effects related to electron–nuclear superfine interactions. Unlike the preceding, it brings about the splitting of nuclear energy levels into sublevels, lying both above and below the main level because the energy of this interaction can have both positive and negative signs. As well as chemical shift, the quadrupole interaction expression consists of two terms: one describes the nucleus feature, i.e., the nuclear quadrupole moment, and the others present the electrical field gradient in the point where the nucleus is located (⭸2/⭸x2)0⫽(⭸E/⭸z)0 The nuclear quadrupole moment was considered in the previous chapter (refer to formulas (8.1.10) and (8.1.11)). It depends on the quantum numbers I and mI: depending on these numbers, the originally singular level splits into several sublevels (degeneration on the quantum number mI is partly lifted). The scheme of quadrupole splitting with the nuclear spin I⫽3/2 is given in Figure 8.2c. Since in the expression for the energy, the square of quantum number mI is entered, the four values of mI (⫾1/2 and ⫾3/2) correspond only to two energy levels. When a system turns from the upper (split) onto the ground (singular) level, two lines appear (as shown in Figure 8.2c). The distance between sublevels depends on the electrical field gradient at the nucleus. These are called quadrupole splitting. In order for quadrupole splitting to be developed two conditions must be fulfilled. First, the nucleus must possess a quadrupole electric moment (refer to eq. 8.1.10). Second, it is necessary for the nucleus be in the point where the electric field gradient exists. This appears in those cases where an atom is in an asymmetric environment. Thus, a relationship of the quadrupole splitting and the chemical structure appears. Unlike the chemical shift, which depends only on s-electrons, the gradient of an electrical field can be created by p-, d- and f -and other electrons.
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The evaluation of the value developed in this splitting is usually in the order of 10⫺7⫺10⫺8 eV. It is possible to observe nuclear quadrupole splitting using R and NQR methods. 8.2.5
Interaction of a nuclear magnetic moment with an electron shell
In addition to the electrostatic interactions of a nuclear charge with an atomic electron shell, there is also the interaction of a nuclear magnetic moment with a magnetic field created by an electronic shell at the point of nucleus location. Two phenomena appear simultaneously: the electron shell creates a magnetic field at the point of nucleus location, and, on the other hand, the nuclear magnetic moment creates a magnetic field influencing the electron cloud. The interaction of the nuclear magnetic moment with the electron magnetic field can be described by the Zeeman effect (refer to Section 7.7). The Zeeman splitting in this instance is: EM ⫽ gN mI N B(0, 0, 0),
(8.2.10)
where B(0,0,0) is the magnetic field induction produced by electrons at the origin. In magnetically disordered materials, as a result of the thermal chaotic motion of atoms, the averaged nuclear magnetic field is zero and the effect is nonobservable. In magnetically ordered materials, the chaotic motion does not occur in full measure and a nuclear Zeeman effect takes place. In this case, the degeneration upon mI is lifted (refer to Section 7.7), i.e., the nuclear energy level with a certain I is split into 2I⫹1 sublevels. Figure 8.2e presents such spectral lines splitting with I⫽3/2: this level will form six equidistant superfine magnetic splitting lines. The splitting energy is defined by an expression close to eq. (7.7.5). It can be seen from expression (8.2.10) that the additional energy of the magnetic interaction depends on the magnetic field imposed on the nucleus. Because the nucleus magnetism is 1840 times lower than the electronic, the Zeeman nuclear splitting is significantly less than the electronic. However, owing to the large magnitude of the magnetic field created by electron shell at the nucleus position, this splitting can be experimentally observed, for instance, by the -resonance method. The same happens when the nuclear magnetic moment acts on the atomic electron’s shell. This makes it possible to investigate the changes in the shell caused by the chemical interactions. 8.2.6
Atomic level energy and the scale of electromagnetic waves
Let us analyze the results. A general scheme of nuclear energy levels with consideration of all electronic–nucleus interactions is given in Figure 8.2. A scheme corresponding to idealized models of stripped nucleus (without electronic shell) is given in Figure 8.2a. As an example, the energy states—ground and excited—were accepted with quantum numbers I ⫽ 1/2 and 3/2, accordingly. The excitation energy for the majority of cases is of the order 104 eV.
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When a nucleus undergoes transformation from an excited to a ground state, a -quantum is emitted, the frequency of which is ⫽(E3/2⫺E1/2)/. For stripped nucleus, we denote this frequency as 0. If we put the originally taken nucleus of nonzero size into the usual electronic shell and observe the interactions, we should take the Coulomb forces of the nucleus with electrons into account (eq. (8.2.5)) (Figure 8.2b). This event shifts both levels. The displacement value is in the order of 10⫺4 eV. (Attention should be paid to the fact that it is impossible to draw the levels and splitting in Figure 8.2 to the right scale: on the background of main transition 104 eV, we must depict details of 10⫺7⫺10⫺8 eV, the difference is 11 orders of value less.) However, it is important that the nucleus in ground and excited states can have different radii (different root mean square radii 冓r2冔) and displacements of those two levels are different (however, 兩(0,0,0)兩2) in both cases is the same). Measurements show that this difference is of the order 10⫺7 eV. So, transition frequency decreases by the value ⬃10⫺7/10⫺15⬃108 Hz (in comparison with 1019 Hz). The next step is an account of quadrupole interaction (eq. 8.2.8) provided the nucleus possesses a quadrupole moment and is in a nonuniform electric field. If both these conditions are fulfilled, the energy levels split according to eq. (8.2.8) (see Section 8.2.4). When quantum number mI present in this expression is in square, the top level splits into two sublevels (mI2 ⫽ 9/4 and 1/4) and the lower level does not split at all (mI2 ⫽ 1/4). Therefore, one existing spectral line splits in two. The experimental evaluations show that the usual value of quadrupole splitting is about 10⫺7 eV. If one takes into consideration the magnetic field created by the electronic shell at the nucleus position (Figure 8.2d), the scheme of energy transitions becomes more complicated (see Section 8.2.5). The degeneration on mI is lifted. The value of splitting is defined by the particular magnitudes of the terms in eq. (8.2.4). If the magnetic and quadrupole contributions are presented simultaneously, the spectral picture gets more and more complex (not shown in Figure 8.2d). If we consider magnetic splitting alone, the spectral picture gets comparatively simple (Figure 8.2e). Here, we should keep in mind that the selection rules permit only the transition for which quantum number mI is changed to unity or is not changed at all. If we take I⫽(3/2), the spectrum will split into six lines. The magnitude of splitting depends on the nuclear g-factor and on magnetic field induction at the nucleus position. In addition to the transition between sublevels of excited and ground states, other transitions are also possible, in particular between two sublevels of a single ground state (the double arrow in Figure 8.2e). The single arrows correspond to RS and the double arrows describe the NMR. 8.3 8.3.1
-RESONANCE (MÖSSBAUER) SPECTROSCOPY
Principles of resonance absorption
In certain circumstances, a substance is capable of absorbing electromagnetic energy, particularly, powerfully and selectively. This happens when the energy of a falling quantum coincides with the electron and/or nuclear energy level difference. Then, so-called resonance absorption of the electromagnetic radiation occurs.
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Suppose that the sample contain atoms with two intrinsic energy levels E1 and E0. The resonance condition can be written as: E1 ⫺ E0 ⫽ res ⫽ gN N B,
(8.3.1)
(remember that the change in quantum number mI is ⫾1). Here, B is the magnetic induction at the nucleus position. An experimental scheme is depicted in Figure 8.3. In Figure 8.3a, the main parts of the experimental equipment are shown: an emitter (1), a sample (2) and a detector (3). The resonance is observed when condition (8.3.1) is met. In Figure 8.3b, the absorption curve is
1
2 –V
3 V (a)
Emitter
Detector
Sample
Absorption
Absorption
(b)
Intensity
Ι
(c)
Γ
Derivation
dI d
(d)
Figure 8.3 An experimental arrangement of the -resonance experiment: (a) experimental arrangement, (b) absorption curve, (c) transmission curve and (d) dI/d curve. is the width of a bell-like curve on the half-height (half-width).
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presented, whereas in (c) the transmission is depicted. Only a single frequency can be found in experiments. In order to measure the whole curve and reveal reliably resonance absorption, a displacement from the resonance frequency should be accomplished. The frequency can be changed although it is easier to move the emitter (or sample) relative to each other using the Doppler effect. In some cases, the derivative curve is measured (Figure 8.3d). The point is that if both emitter and detector are motionless, the device would register quanta with just the emitted frequency. If either the emitter or detector begins to move along the connecting straight line with relative velocity V, the harmony is destroyed. Depending on V, the detector will register another frequency ⬘. The difference
⫺ ⬘⫽ can be found according to the Doppler effect (refer to Section 2.8.4 and to eq. (2.8.23)), namely, V ⎛ V⎞
′ ⬇ 0 ⎜ 1⫹ ⎟ ⫽ 0 ⫹ 0 , ⎝ ⎠ c c
(8.3.2)
The resonance curve is characterized by several parameters: position, width on half of the height (half-width) and intensity (height of maximum or absorption “depth”). 8.3.2
Resonance absorption of -rays: Mössbauer effect
Assume that there are two samples with exactly the same structure and nuclei. This signifies that the positions of two energy levels in both samples are exactly equal. Assume further that there is a way to initiate the nuclei of the first sample to emit quanta. The emitted spectral line E1⫺E0⫽E will exhibit a frequency, . One can evaluate the value of the natural width of this spectral line (i.e., the width as defined by quantum physics (refer to Chapter 7.2) but not by imperfection of the experimental equipment). The uncertainty principle gives: 10⫺15 c ⫽ ⫽ ⫺7 ⫽ 10⫺8 eV 10 c
(8.3.3)
where is the natural half-width of the spectral line and is the average time-of-life of a nucleus in an excited state with the energy E1. The ratio of the natural half-width to the absolute energy value will then be /(E1⫺E0)⬇(10⫺8 eV/10⫺4 eV)⬇10⫺12. It can be seen that this relative width is very narrow. If this radiation is directed onto a second quite similar sample, resonance should occur, i.e., the resonance absorption should be observed. Indeed, the -quantum energy is precisely equal to the energy difference E1⫺E0. The absorption of the electromagnetic waves with -quantum energy just equal to the energy difference in a sample is caused by the resonance absorption, i.e.:
res ⫽
E1 ⫺ E0
(8.3.4)
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If it is possible to change smoothly the electromagnetic radiation frequency, then the resonance radiation absorption can be observed in the background of the nonresonance processes. Such an experimental scheme is depicted in general form in Figure 8.3. However, there are at least two factors that destroy the resonance. The first factor is recoil, which obtains free nucleus when emitting the quantum. Really, the law of momentum conservation requires that the total system momentum, being equal to zero before emitting the quantum (0⫽p⫹pn), should be kept at zero after disintegration, so 0 ⫽ p ⫹ pn
or
p ⫽ ⫺pn ,
(8.3.5)
where p and pn are the p⫺ and pn⫺nuclear momentums. The law of energy conservation in this case has the form E2⫺E1⫽E⫹En, where: E ⫽ ( E2 ⫺ E1 ) ⫺ En
(8.3.6)
It can be seen that the -quantum energy E is less than the ideal value E1⫺E0, on the value of nucleus recoil energy En (denoted usually as R). From eqs. (8.3.5) and (8.3.6), we can find values EnR. The energy of a photon is correlated with its momentum by the equation (see Chapter 1.6): E ⫽ p c
(8.3.7)
where c is the speed of light. Energy of recoil, R, from the above given equations is R⫽
p2 E2 pN2 ⬃ 10⫺3 ⫺10⫺2 eV ⫽ ⫽ 2 mN 2 mN 2 mN c 2
(8.3.8)
This signifies that the emitting line is displaced along the energy axis on R to a side of its reduction. This shift itself is small, particularly in comparison with the quantum energy (104 eV); however, in comparison with the natural width of a spectral line (10⫺8 eV), it is large. Similarly, the spectral absorption line will be displaced as well, but in the opposite direction. Two lines, the natural width of which are 10⫺8 eV, move apart by 10⫺3 eV (Figure 8.4). Thereby, no overlap of the spectral lines takes place, and, consequently, no resonance absorption occurs. The second factor is chaotic thermal atomic motion: different nuclei are emitted in different chaotic states of movement. As a result of the Doppler effect, the broadening D of both spectral lines, emitting and absorbing, occurs (Figure 8.4); moreover, at room temperature, this broadening is of many orders of magnitude larger than the natural line width. As a result, only the tails of the spectral lines overlap; absorption will reach a miserable value from the expected effect. Quite another picture is observed if both nuclei are confined in a crystal. In this case, the whole crystal needs to be considered as a closed system. The theory of this effect (at energies
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I D
E R
R
Figure 8.4 An infringement of the resonance.
of quanta smaller than the binding energy of atoms in the crystal) shows that, while emitting, two possibilities can arise. The first is the creation of an elastic wave in the crystal, i.e., phonon (refer to Section 9.3.1), which would carry away the excess of the energy of a quantum. This is inelastically scattered nonresonance quantum. The other possibility is an elastically scattered quantum, when the recoil energy is transmitted to the whole crystal. Herewith, in the formula for the determination of R, instead of the nuclear mass, mN, we should substitute the macroscopic crystal mass; then the recoil energy reduces practically to zero, and the -quantum energy becomes almost precisely equal to the energy difference E1⫺E0. The Doppler broadening from the thermal motion also vanishes. As a result, the emitting and absorbing lines narrow down to the natural width and coincide with each other. The resonance requirement becomes satisfied. The theory of this effect was worked out by R. Mössbauer (Nobel Prize, 1961) and the effect itself is referred to as the Mössbauer effect. It consists of the recoilless emission and resonance absorption of quanta by nuclei. The ratio of the number of elastically emitted quanta to their total number is called the Mössbauer effect probability f and is proportional to: ⎧⎪ 冬 x 2 冭 2 ⎫⎪ f ⬃ exp ⎨⫺ ⎬, c 2 ⎪⎭ ⎪⎩
(8.3.9)
where 冓x2冔 is a mean-square displacement of nuclei from their equilibrium position in a crystal as a result of their thermal motion (in the direction of quantum emission). The probability is proportional to the atomic mobility in a crystal. How can -radiation resonance absorption be seen and measured? Assume that the emitter and absorber materials are quite similar and are in the same state. The maximum value of resonance absorption must be observed, when emitter and absorber are motionless (V⫽0) (eq. 8.3.2). When one begins to move, e.g., the emitter regarding the absorber, this resonance absorption is destroyed and the experiment reveals an absorption curve; a very small relative velocity is needed to “separate” the spectral lines (emitted and absorbed). V ⬃ ⬃ 10⫺12 E c
(8.3.10)
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The relative velocity needed to destroy the resonance can be found from this equation. A striking result appears: despite the fact that light travels very fast, the speed needed to destroy the resonance is very small V⫽c⫻10⫺12⫽10⫺4 m/sec! Therefore, the simplest way to observe resonance absorption is to measure it as a function of the relative speed, V. A schematic diagram of a R experiment is presented in Figure 8.5a. 8.3.3
-Resonance (Mössbauer) spectroscopy in chemistry
As far as the volume of information that can be obtained on chemical and physical properties of chemical compounds, RS occupies a leading position. Nuclear transitions correspond to energy difference 104 eV but the effect to be measured relates to 10⫺7–10⫺8 eV; however, Mössbauer spectroscopy resolution is so high that it successfully solves such problems. In addition, this method can find nearly all the effects of nuclear interactions with an electronic shell; it carries extremely valuable information on molecular or crystal structure. One of the main parameters of a resonance curve is its position on the velocity axis. In Figure 8.5a, a resonance curve is presented schematically when the emitter and absorber are one and the same motionless substance: the maximum of absorption falls on the zero value of relative velocity. Intensity of absorption (depth of curve minimum) is proportional to the probability of the resonance absorption, f. So, such measurements enable the measurement of a change in the atomic mobility in a sample under different treatments
(a)
(b)
(c)
(d)
0
∆ 0
∆
0
2E/h
0
Figure 8.5 Exhibiting different effects in R: (a) primary position of the absorption spectral line on the velocity axis by a stripped nucleus 0, (b) chemical shift (c) quadrupole splitting and (d) magnetic SF interaction development.
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connected with such chemical problems as catalysts, chemisorption, polymerization and others. If the source substance and absorber are different, the spectral absorption line is shifted along the velocity axis; that is caused by the additional energy EE2 (8.2.9). The energy EE2, denominated in units of velocity, is called in RS the chemical shift (Figure 8.5b). The sodium nitroprusside Na2[Fe(CN)(NO)5]⭈2H2O is mostly used as the standard (reference) substance in Mössbauer spectroscopy. Indication of the reference substance in scientific publications is obligatory. If some atoms in a substance are in an asymmetric environment and their nuclei possess a quadrupole moment, quadrupole splitting can occur (Figure 8.5c). Knowing spin I, the splitting allows the electric field gradient value of the nuclei that is dependent on the structure’s peculiarities to be found. If the substance under investigation is in a magnetically ordered state (ferro-, antiferro- or ferri-magnetic, refer to Section 5.3), a nuclear Zeeman effect takes place: energy levels split according to eq. (8.2.10). In Figure 8.5d, the case of a nucleus with I⫽3/2 in symmetrical surroundings in a magnetically ordered substance (e.g., antiferromagnetic, -Fe2O3) is given. The number of lines allows the nuclear spin I to be found, whereas the distance between lines enables us to find the magnetic field induction (strength), created by the electronic shell, at the nucleus position B(0 ,0 ,0) (or H(0,0,0)). The latter is a feature of electronic shell and differs for different atomic states. Let us look at some examples. A particular interest for chemical investigations is the relative change of the chemical shift in a number of various isostructural absorbers at a fixed (motionless) source. For example, one can judge the valence state of a Mössbauer atom in various compounds by measuring the chemical shifts. So, all bivalent tin compounds have positive chemical shifts relative to grey tin (the source of the radiation moves to the absorber), whereas all tetravalent compounds have a negative shift relative to grey tin (the source of the radiation goes from the absorber). The chemical shift value is proportional to the ionicity of a Mössbauer atom in the compound under investigation. In Figure 8.6, the spectra of a series of isostructural tin tetra-halogenides are presented with tetrahedrons SnI4, SnBr4 and SnCl4, confirming this belief. However, from the same figure, it can be seen that there are no tetrahedrons with equivalent Sn–F bonds in the SnF4 compound; a couple of resonance lines are present. Theoretical analysis of the spectrum obtained shows that there is a quadrupole electric super-fine interaction in this compound; this is explained by the reticular coordination polymer structure with nonequivalent Sn-F bonds (Figure. 8.6). If a nucleus with energy E possesses a magnetic moment M, which is in a magnetic field B, the energy of the nuclear state (8.2.10) will change by a magnitude E. In R experiments, the transitions between two different nuclear states are observed. As EM is proportional to the first order of mI, the (2I⫹1)-fold degeneration is lifted in the magnetic field: to every mI, there corresponds its own value EM. Distinction is measured in RS effect at the transitions between the sublevels of two various nuclear states (Figure 8.2c). Thus, the number of lines of radiation/absorption spectra is defined by selection rules mI⫽0, ⫾1. For example, for nucleus 57Fe and 119Sn (Iex⫽3/2, Igr ⫽ 1/2) from eight formally possible transitions, only six are realized because of the selection rule.
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0 0.5
515
SnI4
v
2Γ
1.0
F
Absorption
v
SnBr4
SnF4
2Γ
F v
SnCl4 2Γ
v
SnF4 2Γ
∆
2Γ
Figure 8.6 Results of a RS investigation of tin tetrahalogenides. Quadrupole splitting is seen for SnF4 (after E.F. Makarov et al.).
For chemical researchers studying magnetic fields acting on the nucleus of Mössbauer atoms, this is of essential interest as the magnetic field on the nucleus is caused by spin, radial and angular distributions of electronic density in an atom. 8.3.4
Superfine interactions of a magnetic nature
If an atom with a nucleus with a magnetic moment M appears in an inner field with magnetic induction B, the nuclear energy E (8.2.20) increases: EM ⫽⫺(M B),
(8.3.11)
The transition between the two nuclear states is measured. As EM is proportional to the first degree of mI, (2I⫹1)-fold degeneration is lifted, and a separate value of the energy level EM corresponds to each quantum number mI. This value is measured in RS as the energy of the transitions between the sublevels of the two different nuclear states (Figure 8.2e). The number of spectral lines is the subject of limitation by selection rules mI⫽0, ⫾1 and depends on the number of energy lines in both ground and excited states. It can be seen from eq. (8.3.31) that experimentally measured EM allows calculation of the magnetic field strength H (or induction B), which electrons of the same atom create on the nucleus. We will not discuss this event in more detail, but would like to underline that these fields appeared to be rather high (up to 105 A/m and even higher). In Figure 8.7, the
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temperature dependence of RS in ferromagnetic iron metal measurements is depicted: at elevated temperatures, all the SF spectral lines collapse into a single one; the crystal undergoes transformation to a paramagnetic state. The interline distance in spectra gives the magnetic field mentioned above.
8.4 8.4.1
NUCLEAR MAGNETIC RESONANCE (NMR)
Introduction
NMR consists of the resonance absorption of electromagnetic radiation by a system of nuclear magnetic moments of the substance under investigation. NMR was discovered in 1946 by E.M. Parcel and F. Bloch (Nobel Prize, 1952). At present, the method is widely used in chemistry because of its remarkable ability to solve many chemical problems. There are several approaches to nuclear resonance descriptions. In this section, we will try to develop the energy approach, which is simpler for understanding. In Chapter 7.7 the behavior of a system of atomic magnetic moments in an external magnetic field was considered. Additional energy appeared; being applied to nuclei, this energy is E⫽gNNmIB0, where gN denotes the nuclear g-factor, N is the nuclear magneton, mI is
Absorption, arbitrary units
773°
744°
444°
22°
V mm/s
Figure 8.7 Magnetic R experimental curves of iron heated to above TC (see Section 5.3.1): all the SF lines collapse at T ⬎ TC (after E.F. Makarov et al.).
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the nuclear quantum number that determines the projection of the magnetic moment vector onto the quantization axis and B0 is the induction of the permanent external magnetic field applied to the substance. The splitting of the ground state of the nuclear energy levels is the result of the nuclear Zeeman effect (Figure 8.2). This effect lifted the degeneracy of the ground state energy making it dependent on the quantum number, mI. This means that the system that previously had one degenerated level, has 2I⫹1 nondegenerate sublevels in the external magnetic field. For proton I⫽1/2, the number of sublevels is two with mI ⫽⫾(1/2). Therefore, the ground state level will admit either positive or negative energy; in other words, it will be split, the splitting gap being E ⫽ N gN B0
(8.4.1)
(the quantum number mI is absent in this equation because the transition difference corresponds to mI⫽⫾1). This is depicted by a grey double arrow in Figure 8.2e. An alternating electric field with definite frequency is additionally applied onto the sample. This alternating field will be absorbed more intensively by the sample if the following resonance condition is achieved: res ⫽ E ⫽ gN B B0
(8.4.2)
There is a special coil to measure the absorption of an electromagnetic field. There are many types of NMR apparatus available. Figures 8.8 depicted two main schemes: an old one (a) and greatly advanced one (b). Since a very stable magnetic field is needed, the superconducting magnetic coil (SMC) is constructed, merged into liquid helium. The special devices permit one to change samples without warming the main vessel. Moreover, they enable one to change the frequency range as well. It can be seen from eq. (8.4.1) that the resonance condition can be met either by changing the frequency of the secondary alternating magnetic field at constant magnetic field B, or the induction variation B at constant frequency. Both possibilities can be realized in modern NMR apparatus. Sometimes it is easier to change the field (by means of sweep coils “c” in Figure 8.8a) and use the constant frequency. The majority of devices work according to these principles; actually, results do not depend on the choice of measurements. The frequency ranges used are presented in Table 5.3. So, the nuclear paramagnetic resonance NMR is the selective absorption of either the alternating electromagnetic energy in the presence of a permanent magnetic field by a substance containing atoms whose nuclei have nonzero spins, or vice versa, the absorption of magnetic induction B at constant frequency of the alternating field. 8.4.2
Use of nuclear magnetic resonance in chemistry
In order to realize the resonance absorption of an alternating magnetic field, one should have a sample with nuclei that possess nonzero magnetic moments, i.e., nonzero spin quantum number. The nuclei 1H (protons), 11B, 13C, 19F and some others have such nonzero
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(a)
(b)
z −1 a x
N2
b
N2 He
He
b
SCC
d S
N
y
−2
SCC 4 −3 CM
g c
c f
h
probe
e
Figures 8.8 Schemes of NMR spectrometers. (a) An original old spectrometer scheme of the continuous wave (a: sample tube; b: magnet; c: sweep coil; d: receiver coil; e: transmitter; f: amplifier; g: register unit; h: recorder), the electromagnet poles are also seen. (b) Section of a more advanced NMR spectrometer, the nitrogen and helium vessels with the superconducting magnetic coil are depicted. A sample is introduced from above; however, the device (probe), inserted from below, realizes the connection of sample measurement coils with the register system. CM: position of the sample investigated.
spins. It can be seen that many atoms dealt with in modern chemistry are present in this list. Proton magnetic resonance (PMR) is most popular in chemistry, but resonance on 19F and 13C is also often used. However, the information quality is so great that using the spectra of these atoms, one can find the behavior of other atoms of the mother molecule. Consider the behavior of hydrogen nuclei (protons) in atoms in a certain molecule in the external magnetic field, B0. The proton spin quantum number is 1/2, therefore, its lower energy level is split (Figure 8.2e). The purpose of the constant magnetic field is to split the single level in two due to the nuclear Zeeman effect. The alternative electromagnetic field B( ) facilitates transitions between sublevels. The third coil is to measure absorption. The principle difference between NMR and RS is as follows: on account of its enormous resolution, resonance allows one to define the superfine structure investigating the transition between different nuclei states (levels) with different I, whereas NMR allows one to measure the transition
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between sublevels of one and the same ground level. So, in spite of the difference between the methods, the results obtained often have the same nature. Consider the PMR ability using the example of the ethyl alcohol molecule CH3–CH2–OH. It contains six protons in three different atomic groups. However, only hydrogen protons give a contribution to the spectrum in the given frequency range. The PMR spectrum of ethyl alcohol obtained by a spectrometer with a very low resolution would contain only one spectral line to which all the protons contribute simultaneously. The information that can be obtained from such a spectrum is restricted by the statement that in fact there are hydrogen atoms in the sample. Substitution of hydrogen by, say, fluorine atoms would remove the signal completely. By improving the spectrometer resolution, the amount of information can be significantly increased. Consequently, it can be found that protons of different atomic groups absorb an electromagnetic wave under a different external magnetic field (Figure 8.9a). The amount of information in such a spectrum increases: the presence of three maxima shows that there are three types of hydrogen atoms in a compound, the resonance for different functional groups CH3, CH2 and OH takes place at the three different experimental conditions. This is because of the fact that, as well as the outer magnetic field, an additional magnetic field acts on each proton. The origin of these additional fields is the diamagnetic response of the given atom to the external field. Every particular atom has its
(a) Thu Jan 12 15:47:14 2006: (untitled) W1: 1H Axis = ppm Scale = 25.39 Hz/cm
5.500
5.000
4.500
4.000
3.500
3.000
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2.000
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1.000
0.500
0.000
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(b) Thu Jan 12 15:49:45 2006: (untitled) W1: 1H Axis = ppm Scale = 25.39 Hz/cm
5.500
5.000
4.500
4.000
3.500
Figure 8.9 The NMR spectra of ethyl alcohol obtained at intermediate (a) and higher (b) resolution. Zero position signal corresponds to TMS.
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own electron cloud subject to some individual changes under the influence of the chemical bonding. This diamagnetic field is local, i.e., different for different atomic groups. How small the differences in the electron shell of hydrogen atoms in different positions, this difference is exhibited in the diamagnetic screening of the given nucleus. This signifies that not only will the external magnetic field with the induction B0 act on the given proton, but an additional local field as well, specific only to this particular atom. It was shown in Section 5.2 that the diamagnetic field is proportional to the external magnetic field and is in the opposite direction to it. Thereby, BA ⫽ B0 ⫺ A B0 ⫽ B0 (1⫺ A ), BB ⫽ B0 ⫺ B B0 ⫽ B0 (1⫺ B ), BC ⫽ B0 ⫺ C B0 ⫽ B0 (1⫺ C ),
(8.4.3)
where A, B and C show the structure-nonequivalent atoms of hydrogen in the molecule. Values are screening constants; they depend upon the electron structure, which, in turn, depends on more distant effects of the influence of the neighboring atoms and atomic groups. Dimensionless values correspond to atomic diamagnetic susceptibility; they are of the order 10⫺6 (this value’s order dictated the high requirement of magnetic field stability). On the background of unity, this is a very small value; nevertheless, it is reliably measured in an experiment. This is because of the fact that being in the outer magnetic field B0, protons of the hydrogen atoms “fill” not that field but fields BA, BB, BC, etc. Consequently, resonance occurs under some field other than B0 (at a fixed frequency). Thereby, the NMR spectrum consists of as many spectral lines as the structure-nonequivalent hydrogen atoms present in the molecule: for each of them, resonance will be observed under the different induction of the magnetic field. Such displacement of the position of the NMR signal along the magnetic induction axis depending on the atomic electron shell properties is referred to as the chemical shift. NMR spectrum of the same ethyl alcohol molecules is given in Figure 8.9b measured by a high-resolution NMR spectrometer. It can be seen that the proton resonance of the different functional groups CH3, CH2 and OH takes place at different values of the magnetic field (on account of the difference in ). There follows, in particular, an important conclusion that the number of NMR signals on the spectrum is equal to the number of structurenonequivalent protons in the molecule. (For instance, in benzene there will be one signal, in mono-substituted benzene there will be three, in ortho-di-fluorine-benzene—one, in para-di-fluorine-benzine—one, in meta-di-fluorine-benzine—two, etc). It would be useful to impart a quantitative meaning to the notion of the “chemical shift” in NMR. For this purpose, for instance, it was possible to obtain an NMR signal for a “stripped” proton and measure the shift from its signal. However, it is too complicated to obtain a signal from a “necked” proton, particularly for every sample. It was agreed, therefore, to choose a standard “reference” substance. At present, tetramethilsilan Si(CH3)4 (TMS) has been chosen as the most widespread standard substance. This substance has many advantages. Firstly, there is a relatively large amount of hydrogen in this compound.
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Secondly, all protons of methyl groups are structure-equivalent, so TMS exhibits itself by only one NMR signal. Thirdly, in solutions the CH3 groups revolve around the Si–C bonds and, therefore, the magnetic fields are averaged and the TMS signal is very narrow. Fourthly, the screening constant of the methyl group is minimal with respect to the majority of other hydrogen-containing substances. Consequently, the TMS reference line lies nearly always on the right side on all other signals on the axis of the magnetic field induction. From this line, the chemical shift is always positive (measured to the left side). To compare chemical shifts obtained on different instruments with different working frequencies, it was accepted to express the chemical shift quantitatively in relative units (in millionth units, or in m.u.) (not dependent neither on B0 nor 0): i ⫽
B0 i ⫺ B0 ( TMS) B0 ( TMS)
6 ⫻ 10 ,
(8.4.4)
where B0i and B0 (TMS) are the induction of external magnetic fields corresponding to the resonance absorption at the principle frequency of the group of equivalent nuclei (i) of molecules under investigation and in TMS, correspondingly. Thereby, i is defined by the relative displacement of the NMR signal of a given molecule on the spectrum to the left from the TMS lines. The evaluation shows that, for the hydrogen atom, removing one selectron brings about a chemical shift of 20⫻10⫺6 or in 20 m.u. Therefore, i for protons in the majority of different atomic groups have values from 0 to 10 m.u. There exist other, less often used, scales of chemical shifts. The generalization of NMR spectra allows experimental results on chemical shifts to be exhibited in the form depicted in Figure 8.10. Since the protons in similar atomic groups have somewhat different , the chemical shift values are plotted on the abscissa axis in the form of segments (not points) of i (in m.u.); the corresponding functional atomic groups are plotted on different levels on the ordinate axis. The black strips correspond to those cases, which are met with more often. 14
12
10
8
6
4
2
0 TMS
– COOH C(O)H C–NOH N = CH ArH RC(O)NHR(H) ArOH C = CH
C = C – NH C = OH RNH SH CH CH2 CH3
Figure 8.10 Generalization of the proton chemical shift in different atomic groups.
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Having obtained a PMR spectrum, the information can be used to gain a representation of whatever substance is being dealt with. Similar tables are accumulated for other nuclei as well. The picture described, however, is equitable only in some respects. There actually exist other finer effects influencing the spectrum. The mutual interaction of the proton magnetic moments of different functional groups can be mentioned in the first order. This interaction turns out to be stronger than was assumed, keeping in mind the interaction between two nuclear magnetic moments (magnetic dipole–dipole interaction). This is because magnetic moments of protons superpose magnetization onto electrons, participating in the chemical bonding in the molecule (particularly, electrons participating in chemical coupling), which, in turn, creates a magnetic field on the neighboring proton. This additional chemical bond magnetization noticeably enlarges the hydrogen proton chemical shift, as well as enlarging the interaction between proton containing groups, and does its more long-range acting. Such a type of interaction is called a spin–spin interaction. Spin–spin interaction is exhibited in the high-resolution spectrum. The spectrum gets more complicated whereas the information on the molecular structure increases. Let us consider this phenomenon using the example of the same ethyl alcohol, measured on the high-resolution spectrometer (Figure 8.9b). As has already been noted, three types of hydrogen groups manifest themselves in three line series, belonging to protons of different functional groups. The one right utmost line is a mark of resonance in TMS (⫽0). The group of lines at ⬇1.2 m.u. corresponds to protons of the CH3, the group of lines at ⬇3.5 m.u. corresponds to the group CH2, and at ⬇4.5 m.u. to the OH group. Let us look at the spin–spin interaction between protons of three functional groups in the molecule of ethyl alcohol. The proton can occupy two positions in the external field with projections defined by quantum numbers mI⫽⫾(1/2). Denote the state with mI⫽⫹1/2 by the letter and with mI⫽⫺1/2 by the letter . Analyze first the action of protons of group CH2 on the nearest groups, CH3 and OH. Two protons of CH2-groups can accept three total quantum numbers, S: ᎑ spin ⫽⫹1 ᎑ and ᎑ spin ⫽ 0 ᎑ spin ⫽ −1
S ⫽ 1, S ⫽ 0, S ⫽⫺1,
the total number of states with S⫽0 being twice as large as the other S values. Correspondingly, the neighboring lines of the CH3 and OH groups split into three with relative intensity 1:2:1 (Figure 8.9b). Evaluate now the action of CH3 on the protons of their neighbors. The number of states of the group of three protons can have four values of the total S: ᎑ spin ⫽ 23 , , ᎑ spin ⫽ 12 , , ᎑ spin ⫽⫺ 12 ᎑ spin ⫽ ⫺ 23
S ⫽ 23 , S ⫽ 12 , S ⫽⫺ 12 , S ⫽⫺ 23 ,
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with the relative number of states 1:3:3:1, accordingly. The SF dipole–dipole splitting is depicted in Figure 8.11 (without following the scale on). Consequently, the influence of group CH3 on group CH2 will be exhibited in the splitting of the signal. The action of the hydroxyl group in a favorite case obeys the general rule herewith each of the four lines must else be split into two. So, the high-resolution signal of the group CH2 in the ethyl alcohol is usually split into eight components. The general rule for nuclei with I⫽1/2 corresponds to this example: the number of spin–spin splitting caused by spin–spin interaction, is n⫹1, where n is the number of protons in the adjacent atomic group whose influence is considered (in the general case with I⬎½, this number is 2nI⫹1). This phenomenon can be described quantitatively by the spin–spin interaction constant J, proportional to the distance between the lines of the spin–spin splitting in the spectrum. Unlike the chemical shift, J does not depend on inductions of magnetic field B0 and so it is expressed right in Hz. For many atomic groups, J is well known and is reported in reference books. If, in the molecule in the main atomic group alongside the protons, there are other atoms with magnetic nuclei they can participate in spin–spin interaction not exhibiting their own magnetic moments. In this case, the splitting character is more complicated, though the amount of information on the substance increases as well. The above-considered example of the PMR spectrum of ethyl alcohol is equitable only for very pure material. If a small admixture of water is present in the sample, the OH group will not be exhibited in the spectrum. This is because the proton of the hydroxyl group
Hz
, m.u.
4.0
200
3.0
Figure 8.11 The scheme of a SF nuclear dipole–dipole splitting (scale not followed).
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sometimes participates in the rapid exchange between the alcohol and water molecules. The time of exchange influences the width of the PMR signal. Depending on the rate of this exchange, a signal can be significantly broadened or disappear completely. This also illustrates the possibilities for PMR in chemical applications. In the last decade, some methods have been offered that are more complex and significantly more informative. In particular, a number of methods have greatly increased the ability of NMR (R.R. Ernst, Nobel Prize, 1991). Ernst found that the ability of NMR, previously limited by the small number of nuclei, could be greatly extended by substituting a slow variation of the magnetic field by strong short pulses, decomposed by means of Fourier transformations in a certain number of harmonics. Thereby, for one pulse, a field runs the whole spectrum of frequencies. The inverse Fourier summations permit a normal NMR spectrum to be obtained. The method allows greatly enlarged sensitivity of the NMR spectroscopy and involves a greater number of nuclei. The method allows influence upon the processes of spin–spin interaction, enlarging or reducing their intensity. A method of two-dimensional scan has also been worked out with the participation of R. Ernst. This approach permits one to influence the spin–spin interactions, enlarging and/or suppressing them. This enables one to untangle complex spectra of polymeric and biological objects. A mechanism of spin–lattice and spin–spin relaxations was also worked out at the same time. The point is that the relative levels occupancy N/N, between which a transition occurs, is defined by the Boltzmann factor. The probability of transition between two levels is greater the larger the distance between them and, consequently, the more N/N differs from 1. In -gamma resonance, the energy difference is of the order of 104 eV, so the upper level is always nearly free and resonance absorption of quantum is nearly always possible. In NMR resonance, the difference E⫺E is extraordinarily small and factor N/N is close to 1. Under such conditions, the occupancy of both levels is the same and the probability of transition should be very small; and, basically, paramagnetic resonance must not be observed. However, it turns out that another mechanism of removing the excitation from the upper level exists. Two such processes are the excitation of either thermal crystal lattice oscillations (spin–lattice relaxation) and/or excitement of oscillation in the nuclear magnetic system (spin–spin relaxation). Both processes are characterized by constants T1 and T 2: T1 is the spin–lattice relaxation time and T 2 is the spin–spin relaxation time. Values of T1 and T2 render an essential influence upon the form of resonance curves. On the other hand, the relaxation processes depend on the mobility of one atomic group or another in the substance. So, experimental measurement of the relaxation processes serves as a method of measurement of the dynamics of chemical conversions depending on time, temperature, chemical conversion particularity and other factors. New, rather complex instruments—tomographs—have been constructed on the basis of NMR. They can be successfully applied in the medical diagnosis of normal and pathological human tissues. The general principle of tomography consists of the fact that the amplitude of the NMR response is proportional to the proton concentration in the sample investigated. In the human body, hydrogen is present in water and in adipose tissue; in the latter, its percentage being higher than in water. So, in picture imaging the distribution of hydrogen, one can see a distinct contrast between adipose and other tissue. To increase contrast, another characteristic of the NMR signal can be used: the strong dependence of the relaxation times on the chemical composition of the medium.
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Being practically harmless, NMR is used for the diagnosis of pathology in the human body, particularly for the hard-to-reach parts of the body such as the heart, liver and brain.
8.5
ABILITIES OF NUCLEAR QUADRUPOLE RESONANCE
NQR is a method based on the resonance absorption of electromagnetic energy in a substance owing to quantum transitions between nuclear energy levels, created by a nuclear quadrupole moment interaction with the gradient of the intracrystal electric field (Section 8.2.4, Figure 8.2c). NQR has been developed on account of the interaction of the electrical field gradient of the shell electrons with the gradient of electrical field appearing because of nonspherical electron density in atoms (refer to Appendix 3). Quadrupole interaction brings about a change in the energy states, corresponding to nuclear spin different spatial orientation regarding the crystallographic axes (formula (8.2.10)). An alternating electromagnetic field causes a magnetic dipole transition between sublevels of the main level, which is found as resonance absorption of electromagnetic energy. Since the energy of quadrupole interaction changes are in rather broad limits depending on the nuclear characteristics and on crystal structures, so the NQR frequencies lie in a range from hundreds kHz to thousands MHz. At present, NQR is not used as widely in science and technology as NMR. Nevertheless, many questions can be solved more quickly and easily with NQR than by NMR, not least because this method does not require an external magnetic field and availability of single crystals. NQR measurements are characterized by a high spectral resolution, selectivity and rate. In particular, of current importance is the remote control of air on the presence of different nitrogen compounds by means of NQR. Nuclear charge quadrupole moment eQ interacts with the gradient of electrical field, e2Qq, defined by an asymmetry parameter, ; it contains information on the molecular structure. These parameters can be determined from experiment. The NQR method can also be used in solids and, predominantly, crystals. There are four areas of successful investigation: electron density studies in molecules, in particular, changes in the orbital occupancy at complexation and substitutions, study of molecular dynamics, in particular, reorientation, rotation of atomic groups, hindered rotation, phase transformation study, revealing and studying defects and mixed crystals investigation. The NQR method is used in nuclear physics for the determination of nuclear quadrupole moments. In crystal chemistry, the method gives information on symmetry and structure, macromolecules ordering degree, their motion and nature of chemical bonding. The method permits the definition of a number of structure-nonequivalent atoms even with nonzero quadrupole moments. While with the NMR method, the structure of a molecule manifests itself only as broadening and splitting of lines; with NQR measurements, the crystal structure is defined by resonance frequencies themselves. Since in NQR, there is a distinctively strong dependence of the spectral line width from defect concentration in crystals; the line-width measurement allows the study of internal stresses in the crystal, the presence of admixtures and ordering processes. The parameter of asymmetry strongly depends on structures; so, in particular, in an ion NC2H2⫹, when substituting two hydrogen atoms on carbon, it is changed from 0 to 1, which is easy to observe in an experiment.
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EXAMPLE E8.1 Calculate the relative emitter/absorber velocity V requires to compensate recoil energy and observe resonance on a free 197Au isotope (wavelength is ⫽0.016 nm). Solution. In order to destroy the resonance, one can move the resonance line on its width (eq. (8.3.10)) taking into account the fact that the recoil process relates both to emitter and to absorber. Therefore, to compensate for this effect, it is necessary to move one of them with velocity V, which can be determined according the equation:
V ⫽ 2 R. c
Therefore, V⫽
( )2 c 2 Rc 2 h . ⫽2 ⫽2 ⫽ 2 2 mN mN 2 mN c
Substituting all known values into this equation, we arrive at V ⫽ 126 m/sec. This value is not small, however, in comparison with quantum speed (V/ c)⬃(0.1/108) ⫽ 10⫺9, which seems negligible.
EXAMPLE E8.2 The spin quantum number of a nucleus of 14N is I⫽1. Therefore, the nucleus has three energy levels with frequencies 0, ⫹ and ⫹ (for mI⫽⫹1, 0, ⫺1, correspondingly). Two frequencies of 14N in sodium nitrate (NaNO2) are ⫹⫽4.640 MHz and –⫽3.600 MHz. Calculate (1) NQR frequency 0, (2) asymmetry parameter and (3) quadrupole coupling constant e2qzzQ/h. Solution. (1) A simple average v0 can be easily calculated v0⫽v⫹⫺v⫺⫽1.040 MHz. (2) The quadrupole moment is v⫹⫽3/4(e2qzzQ/h)(1⫹/3), v⫺⫽3/4 (e2qzzQ/h) (1⫺/3): the solution of these two equations give: ⫽3(v⫹⫺v⫺)(v⫹⫹v⫺)⫽0.38. (3) The quadrupole moment of 14N is: {(e2qzzQ/h)⫽2(v⫹⫺v⫺)/]⫽5.474 MHz
8.6
ELECTRON PARAMAGNETIC RESONANCE (EPR)
The effect of resonant absorption of electromagnetic waves by electron paramagnetic centers in a substance in a permanent magnetic field is referred to as electronic paramagnetic resonance (EPR). In the EPR method, the resonance condition for electrons looks like that for nuclei (refer to eq. (8.3.1): (8.6.1) E1 ⫺ E0 ⫽ res ⫽ gE B B,
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though instead of nuclear, electron characteristics are included. A comparison of resonance in nuclear and electronic subsystems (see Section 8.3.1) shows that the frequency of the EPR-resonance used is higher than that in NMR, i.e., it is approximately 109 Hz, which corresponds to the wavelength 10⫺1 m (several centimeters) lying in the microwave range (see Table 5.3). In EPR, an external permanent magnetic field influences electron spin energy level split; however, a high-frequency field throws the spins from one state to another. In other words, the external permanent magnetic field removes degeneration from electron energy states on quantum number mS. The basic EPR circuit is similar to that for NMR though the technique of EPR measurements differs significantly from that of NMR. Intraatomic superfine interaction is described differently depending on the aggregative state of the substance investigated. In particular, it has the same nature in liquids as that described in Section 8.3 and is referred to as contact interaction; it is described by the term:
a⬵
8 M I 冷 (0, 0, 0) 冨2 3 I
(8.6.2)
It can be seen that contact interaction is caused mostly by s-electrons, because only for s-electrons is the wave function on the nucleus distinct from zero. However, contact interaction of the same character can be observed in some cases in the absence of a noncoupled s-electron; it can arise to the account of -electron of aromatic hydrocarbons belonging, for example, to anion-radicals: contact interaction of a nucleus with s-electrons is rather sensitive even to small electron excitation at hybridization. The interaction of noncoupled electron spins with the nucleus can cause SF EPR line splitting EPR signal. The character of the splitting, or SF splitting of the EPR signal, is defined by a nucleus spin interaction with the number of specific atomic configuration in the nearest neighborhood. For a proton, the rule n⫹1 (where n is the number of nonequivalent atoms) is valid in this case as well, since I⫽(1/2), like electron spin. The simplest case is atomic hydrogen where SF splitting is defined by electron interaction with a nucleus (proton) belonging to the same atom. Notice that the reorientation time of a nucleus in an alternating magnetic field is much longer than the corresponding electron time. This means that the nuclear spin assigns the quantization axis and electron spin reorients alone (transition from mS⫽⫹1/2 to mS⫽–1/2). Quantum numbers for an electron spin and proton are identical and equal to 1/2. This means that the number of possible states is equal to two (spin of an electron and nucleus are parallel, the coupled spin is 1; or antiparallel, the coupled spin is zero). Therefore, the EPR spectrum of an atomic hydrogen consists of two SF signals (Figure 8.12a) split on 502 Oe (remember that the size of splitting does not depend either on the external field or on frequency). A deuteron’s spin quantum number is equal to 1. Therefore, there are three possible mutual orientations of the deuteron and electron spins resulting in 1, 0 and –1. Hence, the EPR signal of atomic deuterium consists of a three-line SF structure divided by a field 78 Oe (Figure 8.12b). The splitting is proportional to a; this allows one to determine the magnitude 兩 (0,0,0)兩2 for hydrogen (deuterium) atoms. Notice that the rule n⫹1 for a deuteron is no longer valid, since I⬎1/2.
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(a) 502 oe
(b)
78 oe
Figure 8.12 EPR spectra of atomic H (a) and D (b).
The EPR spectrum of each atom, ion or radical, has a specific shape (due to SF electron–nuclear interaction) well known to experts and given in the specialist reference literature. The most successful fields of application of the EPR method are: atoms and radicals (stable and/or unstable) with odd number of electrons, short-lived particles as elements of intermediate stages of the chemical reactions, ions of transition elements with partly builtup electron shells, crystal defects, etc. Integral signal intensity is proportional to paramagnetic center number; hence, the EPR spectrum contains important characteristics of a substance, such as the concentration of free radicals, number of defects of a crystal lattice, etc. The position of a signal allows one to determine the g-factor of a paramagnetic atom. Therefore, one can find the state of the paramagnetic center (refer to Section 7.6.2). Notice that, in view of relativistic effects, the spin g-factor is not precisely to 2, but is 2.0023. PROBLEMS/TASKS 8.1 A free 40K nucleus emits a -photon with the energy ⫽30 keV. Find the relative displacement / of the spectral line due to nucleus recoil. 8.2 A 67Zn nucleus with the excitement energy E⫽93 keV emits a -photon transferring from the excited to the ground state. Determine the relative change of (/) of the -quantum energy because of the recoil of the initial nucleus.
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8.3 Calculate an atom’s recoil energy R after emission photon in (1) visible range energies (⫽500 nm), (2) in X-ray range (⫽0.5 nm) and (3) in range (⫽5⫻10⫺3 nm). Consider one and the same nucleus of mass m⫽100 a.u.m. 8.4. The atomic bonding energy Eb in a crystal is Eb ⫽ 20 eV; atom mass is 20 a.m.u. Find the minimum quantum’s energy, which kicked the atom out of the crystal. 8.5. Simulate the R spectrum of ilmenite (FeTiO3) if iron’s chemical shift EI relative to stainless steel is 1.3 mm/sec and the quadrupole splitting EQ⫽0.5 mm/sec. The Fe excitation energy is 14.4 keV. Express EI and EQ in eV units. ANSWERS 8.1. ⫽1.7⫻10⫺17; (/)⫽(/2mNc2)⫽4⫻10⫺7 m. 8.2. /⫽(/2mNc2)⫽7.45⫻10⫺7. 2 ⎛ ⎞ ; (1) 33 peV; (2) 33 meV; and (3) 0.33 eV. m A ⎜⎝ ⎟⎠
8.3.
R⫽
8.4.
⫽ c 2 mN Eb ⫽ 8.6 ⫻10 −5 eV .
8.5. EI ⫽ 6.24 ⫻ 10⫺8 eV, EQ ⫽ 2.4 ⫻ 10⫺8 eV.
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–9– Solid State Physics
Solids are mostly subdivided into crystal and amorphous substances. In this chapter the crystalline state is predominantly considered although some features of the amorphous and liquid states will also be briefly touched on.
9.1
CRYSTAL STRUCTURE, CRYSTAL LATTICE
A crystal is characterized by a three-dimensional, regular, periodic array of particles— atoms and/or molecules. Modern experimental techniques allow us to see the structure of large molecules in crystals using an electron microscope. Figure 9.1 is an electron photograph of a crystal of tobacco mosaic virus, showing the regular packing of the molecules in a crystal. Remember that by “molecule” we usually mean the smallest part of a substance that can exist alone and retain the characteristics of that substance. Imagine that we can divide a crystal until it becomes the “brick,” which retains the main (but, certainly, not all) characteristics of the whole crystal. The contents and form of this “brick” defines the crystal structure: the relative amount and mutual disposition of atoms (molecules), the chemical composition of crystalline material, the interatomic distances and valence angles (i.e., chemical bonding), etc. The smallest part of the crystal that retains the specified characteristics (the “brick”) is referred to as a unit cell. Using the property of periodicity one can build the whole crystal by regular repetition of unit cells along coordinate axes, as shown in Figure 9.2. A crystal can be characterized both by the unit cell (the carrier of the chemical composition and atomic structure), and by three translation vectors a, b and c; the latter can be used to build the whole crystal from original cells. Each of these three vectors corresponds to a symmetry operation, since these operations superpose a crystal with itself: each atom (supposed to be a point) moves over to a similar atom in the nearby unit cell. Any point (atom, including) can be transferred to another crystal point, identical to the first, by operation of a translation t, which is t manb pc,
531
(9.1.1)
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Figure 9.1 Image of the tobacco mosaic crystal as seen in the electron microscope (size of molecules is approximately 25 nm).
z
y
>
c
>
>
(a,b) = (a,c)= (b,c)=
b
x 0
a
Figure 9.2 Crystal lattice: a unit cell based on a, b, c vectors and the crystal build-up.
where m, n and p are integers which number the unit cell to which the translation is taking place. Unlike the symmetry elements considered in Section 1.3.7, translation is an attribute of the crystalline state, because it translates a given atom to a similar one in the neighboring unit cell, but does not combine an atom with itself. It is important to note that the crystal
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is here considered to be of infinite size; this is a good approximation, since crystal size is usually many orders of value larger than the unit cell. The system of translations forms a so-called crystal lattice; this presents a mathematical abstraction: describing the translational characteristics of a given crystal. In Figure 9.2 this position is illustrated. Vectors a, b and c represent a set of three translations in three dimensions, which together with angles , and , define the form and size of the unit cell. The crystallographic axes x, y and z are usually directed along the main translations. Note that the choice of three main translations is ambiguous, there exist certain rules to make this choice but this is not important here. The cross points of the axes are called lattice nodes. It should once again be emphasized that while the crystal structure is the arrangement of atoms in the crystal, the crystal lattice is a system of translations, describing the translational properties of the crystal. The form of cells, the location of atoms in them and, accordingly, the crystal’s physical properties are defined by the symmetry laws. All these comprise the subject of crystal physics and crystal chemistry. Correlations between the translation vector length and the angles between them define possible crystal classes or syngony, resulting from the unit cell form and the crystal lattice. There are seven crystal classes (or syngony), plotted in Table 9.1. The position of origin is chosen from considerations of rationality. So, in Figure 9.3 the atomic structure and crystal lattice of a Cl2 crystal is depicted: the origin is chosen not in the position of an atom, but in the CM of the Cl2 molecule. (In this instance a crystal node is in “emptiness.”) Planes drawn through the nodes (but not in general through atoms) are called crystallographic planes (see Section 6.3.5). A lattice node is defined by its coordinates (in the units of the vector length), which are placed in double square parentheses (for instance, [[001]]). All parallel directions in a crystal are equivalent. So a straight crystal line (or simply line) is conducted through the origin and its indexes are defined by the coordinate of the first node, lying on this line; a line’s indexes are enclosed in brackets (so, direction [001] complies with the direction to axis z, since the first node on it has the coordinates [[001]]).
Table 9.1 The crystal classes (syngony) Syngony
Relations between lattice periods
Relations between angles
Triclinic Monoclinic Orthorhombic Tetragonal Trigonal (rhombohedral) Hexagonal Cubic
abc ab c abc abc abc abc abc
90° 90° 90° 90° 90°, 120° 90°
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Figure 9.3 Crystal unit cell of Cl2: there is an arbitrary rule in the choice on the unit cell.
z 2
1 0
1
2 y
1 2 3 x
Figure 9.4 An order of the Miller indexes determination.
The family of parallel crystallographic planes is defined by three indexes h, k, l called the Miller indexes. The indexes are inversely proportional to the segments cut by the given plane on the axes, provided it is the nearest plane to the origin; they are bracketed in parentheses. Let a plane cut definite segments on the axes. In Figure 9.4 such lengths (in units a, b and c) are 3,2,2. (These lengths in unit cell periods can be whole or fractional numbers.) They must be “turned over” and multiplied by a single whole number (here the number is 6) to make them whole numbers
冤冢
冣
冥
1 1 1 → (2,3,3) . The three whole numbers obtained (2,3,3) are 3,2,2
the Miller indexes of this plane. If a plane cuts an axis on the negative side of the origin, the corresponding index is negative indicated by placing a minus sign above the corresponding index (e.g., h ,k, l). In Figure 9.5 an example of some crystallographic planes in a cubic lattice is presented. Index zero amongst the Miller indexes corresponds to the fact that a plane cuts on the corresponding axis a segment equal to infinity, i.e., this plane is parallel to a given axis. If there are two zeroes amongst the indexes it means that a given plane is parallel to the corresponding unit cell edge.
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z
535
z
(110)
z
(100)
y x
y
x z
y x
z
z
y x
y
x (111)
(110)
y
x (100)
(200)
Figure 9.5 Selected crystallographic planes in a cubic crystal.
The correlation presented describes the so-called primitive cells in which there is only one atom (each atom in the nodes belongs simultaneously to eight cells, which gives for one unit cell 8 (1/8) 1 atom). The unit cell of each crystallographic class is characterized by a definite symmetry corresponding to the crystal space symmetry. It is possible to place an additional atom in a primitive cell in such a manner that the cell symmetry does not change. For instance, in the cubic primitive cell one can introduce an additional atom into the cell’s center
冤冤 12 , 12 , 21 冥冥 without destroying its symmetry. A body-centered
lattice (BCC) with two atoms to the cell is obtained. Iron crystals, for instance, possess such a structure. It is possible to place atoms at the centers of the edges of a primitive cell
冤冤 2 2 0冥冥, 冤冤 2 0 2 冥冥 and 冤冤0 2 2 冥冥 preserving the cubic symmetry. In a similar manner, 1 1
1
1
1 1
the so-called face-centered cubic (FCC) with four atoms in the unit cell is obtained. Copper, for instance, has a FCC structure. Such cells are called cells with basis (or Bravais lattices). Crystallographic directions are characterized by identity distances (periods), i.e., the shortest distance between identical atoms in a given direction. Parallel crystallographic planes are also identical to each other. The distance between two adjacent planes with indexes h, k, l, or what amounts to the same things, the distance from the origin to the nearest crystallographic plane with the same indexes are called interplanar spacing dh,k,l. Exactly this value falls into the Bragg equation (refer to Sections 6.3.5). Periods of lattices a, b, c together with angles , , , plane indexes (h, k, l) and interplanar spacing d are bound by a so-called quadratic form. For the simplest case of a cubic crystal, the quadratic form is presented by the equation 1 h2 k 2 l 2 d2 a2
.
(9.1.2)
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Structure and crystal lattice define the positions of atomic CM in the unit cell. In addition, atoms in crystals perform thermal oscillations, significantly influencing the crystal’s physical properties. These oscillations are characterized by root mean square displacements. The main method of crystal structure determination is X-ray diffraction analysis (refer to Section 6.3.5). Experiments that use small single crystals and/or polycrystalline samples allow one to determine the mutual location of atoms in crystals of rather complex chemical compounds (for instance, a thousand or more atoms in the unit cell) together with the nature and parameters of atomic thermal vibrations (their root mean square displacement values). For investigation of special questions, methods of neutron and electron diffraction are also used (refer to Chapter 7.2). EXAMPLE E9.1 Knowing the density of a Ca single crystal 1.55 103 kg/m3 determine: (1) the lattice period a; and (2) the closest interatomic distance d. The Ca crystal is of the FCC structure type. Solution: The unit cell volume V of a cubic crystal can be bounded with the lattice period a by a simple expression V a3. On the other side, it can be expressed as a ratio of the molar volume to the number of unit cells in one mole of the crystal a3 Vm/Zm*. The molar volume Vm is Vm M/. The number of the unit cells in one mole is Zm NA/n where n is the number of atoms in the unit cell. Substituting these 3 values into * we obtain a3 nM / NA wherefrom a 兹苶 (n苶 M苶/苶 苶 N苶 A). Taking into account the number of atoms in the FCC unit cell (Chapter 9.1) we arrive at a 556 pm. (2). The closest interatomic distance in the FCC lattice is the face diagonal d a兹2苶 / 2. Therefore, d can be found d 393 pm, where the atomic radius can be calculated from r (d/2) 1.81 Å. EXAMPLE E9.2 Write down the indexes of crystallographic directions presented in Figure E9.2 by a bold line (refer to Chapter 9.1).
z
y x
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Solution: The crystallographic direction does not cross the origin (axes symbols without asterisks). However, we know that all the nodes of a crystal lattice are equivalent. Therefore, we can shift the origin in the point [[100]] in the figure. This point is the new origin. Then the line will cross the node [[ 101]]. These are the indexes of the direction [ 101].
EXAMPLE E9.3 In Figure E9.3 a crystal lattice is presented. Write down indexes of the crystallographic plane crossing three nodes indicated in the figure.
x
y
z
Solution: It can be seen from the Figure E9.3 that points cut on coordinate axes (in periods units) are x 3, y 1 and z 2. According the rule we should obtain 1 3
冢
1 2
冣
the reciprocal values; they are ,1, . Multiply the three value on 6 we arrive at
( 2, 6, 3 ). Note that indexes can be multiplied by a constant value, including negative. All planes obtained will be identical.
9.2 9.2.1
ELECTRONS IN CRYSTALS
Energy band formation
In Chapter 7 the electron structure of free atoms, i.e., not subjected to any external influences, was considered. In a crystalline state the distance between the atoms is comparable with their size; so each of them appears strongly influenced by its neighbors. The interatomic interaction and periodic character of a crystal field render an extremely strong
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influence on the atomic ensemble. As a result, crystals possess a complex of the properties strongly distinguishing them from ensembles of free atoms. Let us consider the essence of the process using the example of the hydrogen-like atom of sodium with 11 electrons in its shell in a free state. In Figure 9.6 (above) a potential electron energy curve of two neighboring atoms in a field of their nuclei is presented. All levels of energy up to 2p are occupied completely though there is one noncoupled electron in the outer 3s level. The potential curves go up to U 0 but are not overlapping. Each electron is fixed near its nucleus. The same picture is presented in Figure 9.6 (below); however, here the distance between atoms corresponds to the shortest interatomic distance in sodium metal (a 4.3 Å). Firstly, the potential curves in a crystal form a unique periodic potential with the maxima located appreciably below the zero level of energy. Secondly, an upper occupied 3s-electron level has risen above the potential barriers and a corresponding electron appears capable of moving along the whole crystal. It is usually said that collectivization of valence electrons has occurred. Such collectivized electrons form an ensemble of quasi-particles that has lost part of their initial properties (see Section 9.2.2). The overlapping of the outer 3s orbits has occurred. On approach, changes also touch the inner electrons. The mutual influence of atoms transforms the very “thin” electron’s levels into a band of final width. In Figure 9.7a scheme of such level expansion is given for several crystals. In Figure 9.7a the 1s level of a lithium atom is expanded insignificantly whereas the 2s level has formed a rather wide
E
E
+ r
r
E
E E=0
Figure 9.6 Interaction of two sodium atoms. above – the initial state, the potential curves of the two free atoms not interacting to each other, below – overlapping of the upper electron shells at the interatomic distance became closer reduced to that in metallic sodium: this give rise to Bloch’s free electron formation.
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E
539
E
E
E
E
2p 2s
2s
1s
0
(a)
Ec
0
P S
Ev
1s
r
a
E
r
r
a
0 (b)
a (c)
Figure 9.7 Broadening of the electron’s energy levels and formation of energy bands: (a) lithium crystal (in the 2s band part of the level is occupied, this is shaded in a section, whereas other part remains free), (b) bands in beryllium, (c) bands in diamond-like crystals.
2s band. In a beryllium crystal (Figure 9.7b), 2s and 2p levels have formed an overlapping band. This mixed band is called a hybrid band. In diamond-like crystals with strongly pronounced covalency the bands were split in two (Figure 9.7c); each of them contains four states for one atom: one s-state and three p-states. These bands are subdivided by the forbidden energy band. The lower occupied band is referred to as a valence band, and the upper band as a conductivity band. The picture described corresponds to the so-called strong-binding approximation. In a weak-binding approximation the electron behavior in a periodic crystal field is considered. For the analysis of the latter case it is necessary to solve the Schrödinger equation for a particle moving in a periodic field F. Bloch has offered a function consisting of the product of two functions: U ( x ) u( x ) e ikx ,
(9.2.1)
The function u(x) describes the electronic potential of a single atom and the other (exponential) ensures the periodicity u(x a) u(x). This function is called the Bloch function. Such potential should be substituted into the Schrödinger equation and the allowed values of energy can be derived. Instead of a smooth parabolic function E (h ¯ 2 / 2m)k 2 a curve with breaks is derived. Some representation of the physical origin of such breaks can be given using diffraction electron properties (see Section 7.1): breaks occur at values k, corresponding to back reflection according to Bragg’s equation (6.3.11). Back reflection in this case can take place when an angle is 90° and, accordingly, n 2d (n is an integer designating the order of reflection). Remembering that 2/k the equation can be presented as
k
n , d
(9.2.2)
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E
2 − a
− a
a
0
2 a
k
Figure 9.8 Electron energy E as a function of the wavenumber k. The same curve for a free atom is presented by a broken curve.
where d is the interplanar distance. If the planes (h00) are considered, the interplanar distance is a and eq. (9.2.2) appears in the form
k
n . a
(9.2.3)
As a result the curve E(k) looks like that given in Figure 9.8: on a background of classical parabolic dependence in the places determined by expression (9.2.3) the curve has breaks. The forbidden energy gap, which we met above, is formed. We are unable now to call the particles as electrons; we must call them quasi-particles (quasi-electrons). The properties of a quasi-particle with a wavenumber k close to n/a (at the bottom or near the top of a band), differ appreciably from the properties of really free electrons. In particular, their effective mass m* differs at the “top” and “bottom” of the energy band. It follows from this fact that m* is defined by the second derivative of an energy E on wave vector k, i.e., m* (d2E/dk2); from the curve it can be seen that m* depends on k and can even change sign. 9.2.2
Elements of quantum statistics
Because of mutual influence, some atomic electrons are generalized forming a “gas” of quasi-electrons in crystals. These electrons preserve some properties of free electrons (e.g., each of them possesses a classical momentum) but, at the same time, also possess properties that distinguish them from really free particles (e.g., they have a mass different from the classical electron mass). Some well-known electric properties of metals are caused by this “gas.” However, it appears that in a model of free “electron gas” theoretical calculations strongly overestimate experimentally known characteristics; only a small part of the generalized electrons can take part in the formation of these properties.
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Are these electrons indeed the gas of free electrons? The answer to this question has already been obtained: no, there are no free electrons in metals; electrons form energy bands that can either be overlapping, or be separated by the forbidden energy gap. Besides, moving in a periodic crystal field, electrons are symmetry-dependent. It is necessary to answer one more question: how are electrons distributed among these bands and how are they distributed inside the band? In classical Newtonian physics the elementar volume of a configurational space cell is infinitesimal (it looks like Planck’s constant ¯h is accepted to be zero); the electron distribution upon their energy is given by Maxwell–Boltzmann statistics: there are large amounts of particles, all of which tend to occupy the state with the lowest energy, though chaotic temperature motion, on the other hand, scatters them on different energies. This process is described by the Boltzmann factor. In quantum physics the volume of a phase space cell is no smaller than h3 because of the uncertainty principle; the number of cells is limited accordingly and this leads to the limited amount of energy levels in the bands. The problem of particle distribution among cells (and energy levels) is not apparent. Regulation is given by quantum statistics. The energy spectrum is presented by energy bands with limited “capacities.” Consider the energy structure of a valence band. The example of sodium atom was given above; every atom from its 11 electrons delivers only 1 to the valence band; besides, every atom “brings” one level to each energy band; levels regularly fill the band, they cannot overlap since the uncertainty principle and exclusion principle do not permit overlapping. If we assume that sodium crystal comprises N atoms, the conduction band is half-filled because of the spin degeneration. How are these electrons distributed on the levels? Figure 9.9 presents such a half-filled band. The upper occupied level at 0 K is referred to as the Fermi-level (EF); energy counts up from the bottom of the band. The same is presented in Figure 10a; the graph f(E) forms a step, all levels under EF are occupied whereas all levels over EF are empty. It is suggested in quantum statistics that particles with half-integer spin obey Fermi–Dirac statistics; such particles are called fermions. Electrons have a spin quantum number equal to 1/2 and therefore are fermions and must obey Fermi–Dirac statistics. The mathematical description of Fermi–Dirac statistics is given by the Fermi distribution function f ( E ) 冤e( EEF ) (T ) 1冥1 .
(9.2.4)
The magnitude of this function depends on the sign of difference E EF. At 0 K and E EF the difference is negative and f(E) 1. At E EF the difference is positive and f(E) 0. This corresponds to the fact that up to Fermi energy, all levels are occupied by electrons according to Pauli’s exclusion principle. When the temperature increases, only those electrons close to the EF level transmit to higher levels (Figures 9.9 and 9.10b) and take part in excitation. The number of such electrons is proportional to T, as is shown in the figure. The point is that only such electrons can participate in electric properties. At T EF function f(E) corresponds to nondegenerated gas and Fermi–Dirac distribution transforms to Boltzmann distribution. However, this can happen only at very high temperatures when the crystal hardly exists.
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At absolute zero temperature all levels are occupied by electrons up to EF. When temperature increases some of the electrons are excited and occupy free levels of energy. These electrons define the electron conductivity of metals. The average energy width of this “excitation band” is equal to T, as is marked in Figures 9.9 and 9.10.The opposite case is presented by particles with integer spin, in particular photons with a spin quantum number 1. Such particles are referred to as bozons (Bose particle). They obey Bose–Einstein statistics. A distribution function f(E) for bozons is expressed as f ( E ) 冤 e ( E ) (T ) 1冥1 .
(9.2.5)
The value in this equation is the so-called chemical potential, i.e., the thermodynamic function of a system, which is determined by the system energy change at the change of the particle number by 1. A feature of photon gas is the fact that, at first, photon gas cannot exist by itself: a source of photons as heated body surface is necessary for its maintenance. Further, all photons in vacuum move at an identical speed equal to the speed of light c and possess energy ¯h , momentum ¯h /c and mass E/c2 (refer to Section 1.6). They cannot stop; their resting mass is zero. Moreover, photons do not collide with each other and their equilibrium is achieved only on interaction with heated surfaces; therefore their quantity is not fixed, it is established by the equilibrium with the heated body.
E
0
{
T
E EF
zone bottom
Figure 9.9 An energy band structure. E is the total depth of the band, EF is the Fermi energy count from the band bottom. Schematically the dark field presents the fully occupied by electrons band at zero temperature, non-occupied levels space is white, a variable part of a band of T in width is active, i.e., can be exited and participate in conductivity. The energy levels are very close to each other and therefore are not shown in the diagram.
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f 1 T=0
(a)
EF
E
f 1 T>0
(b)
E
T f 1 T
(c)
∞
E
Figure 9.10 The graph of Fermi-Dirac distribution: (a) at 0 K it looks like a step, (b) at intermediate temperature, the border near Fermi energy is fuzzy. (c) at T → ⴥ the distribution transforms to Boltzmann law.
An electron with half-integer spin is a fermion, i.e., it has to obey Fermi–Dirac statistics. All the usual metal electroconductivity properties are in agreement with these statistics. However, in some cases two electrons can produce the so-called Cooper’s pairs with compensated spin, the spin of the Cooper pairs is zero. Therefore, such pairs transform electrons from fermions to bozons. There are no Pauli’s exclusion principles for bozons; they can be “condensed,” i.e., they occupy all the levels, and nearly all participate in electroconductivity. Superconductivity can then take place. This suggestion theoretically explains the metal and intermetallic compound superconductivity phenomenon (J. Bardin, L.N. Cooper and R. Schrieffer, Nobel Prize 1972), discovered earlier by H. Kamerling-Onnes (Nobel Prize, 1913). This phenomenon occurs only at very low temperatures (⬇20 K). However, superconductivity has been discovered in nonmetallic, oxide-type chemical compounds with critical points of superconductivity up to ⬇140 K (in liquid nitrogen region) (so-called high-temperature super conductors, HTSC) (J.G. Bednorz and K.A. Muller, Nobel Prize, 1987). Intensive attempts to synthesize new materials of this kind are in progress.
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Band theory of solids
Band theory relates to the electrical properties of solids and the character of their energy bands. The corresponding relation is given in Figure 9.11. The case of metallic sodium already examined above is presented in Figure 9.11a. Electrons fully occupy the 2p band, but the conducting 3s band is half-filled. Therefore, 2p electrons do not participate in conductivity but the electrons lying near EF in the 3s band can easily be excited by an external electrical field and take part in an electrical current. Consequently, sodium, like all other alkali metals, is a conductor. This situation is shown in Figure 9.11b, where a scheme of another case of a conductivity band created both by 3p- and 3s-electrons is depicted. Another picture of energy band filling is shown in Figure 9.11c. It corresponds to full occupation of a valence band (EF equals the band top energy) with the presence of the forbidden energy gap Eg between the valence band and the band of conductivity. Electrons of the valence band cannot be so easily excited now by the action of an inner field to take part in conductivity (for this, electrons have to be transmitted to the conduction band, overcoming a broad energy gap); the corresponding materials possess a dielectric property. If the gap is narrow in comparison to T, the material is an intrinsic semiconductor (Figure 9.11d). In addition, there exist two great classes of semiconductors characterized by the presence of local energy levels (to the account of different admixtures) in the forbidden energy gap. If these levels lie close to the top of the valence band (Figure 9.12a) (this is called the acceptor level), electrons of the valence band occupy them and release some levels in this band. The so-called hole conduction appears (on account of vacancies near the top of the valence band). Such materials are semiconductors of p-type. Germanium crystals with indium admixture can be numbered amongst them. If occupied local electron energy levels are near the bottom of an empty conduction band they can be excited directly to this band (Figure 9.12b). Such levels are called donor levels and the material is a semiconductor of n-type (germanium with the admixture of arsenic). In intrinsic semiconductors (for instance, pure germanium and silicon) the local levels are absent; however electrical conductivity appears because of the narrowness of the forbidden gap and at moderately high temperature an excitation of electrons from the valence band directly to the conduction band is possible (Figure 9.11d).
3s
3p
2p
3p
Eg
3s
Eg
Eg
Eg
2p
2p
(a)
(b)
2s (c)
3s (d)
Figure 9.11 Band structure of different crystal types: (a) and (b) conductors, (c) dielectric, (d) semiconductors.
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Conduction zone
545
Conduction zone EA
Ec
Ec
Eg
Eg
Donor levels
Ea
Accepter levels
(a) Valence zone
(b) Valence zone
Ev
Figure 9.12 A band structure of doped semiconductors: (a) n-type, (b) p-type.
9.3 9.3.1
LATTICE DYNAMICS AND HEAT CAPACITY OF CRYSTALS
The Born–Karman model and dispersion curves
By crystal dynamics one usually understands the theory describing atomic oscillations around their equilibrium positions and those features of the properties that depend on these oscillations. Only harmonious oscillations occurring because of the action of quasi-elastic forces will be considered here (refer to Chapters 2.2 and 2.4). Being independent, oscillations of all atoms can be described by a system of running waves with certain frequencies, polarization and wave vectors (see Section 2.9.5). According to the general principles of particle-wave duality (refer to Chapter 7.1), each wave associates with a particle called a phonon with energy ¯h S, polarization s (s 1, 2, 3) and momentum p ¯h k. Since phonons exist only within a crystal they are related to the category of quasi-particles. Nevertheless, interaction of phonons with other particles (electrons, neutrons, etc.) occurs according to classical laws of collision (refer to Section 1.4.5). Consider the simplest model of a crystal as a one-dimensional chain of identical atoms. Direct the chain along an axis x. Denote a period along the chain by the letter a, and let l be the number of the atom counted from an arbitrary chosen atom (Figure 9.13). Value xl represents instant coordinate of atom l. In this model one more simplification is entered, namely, the nearest neighboring atoms are considered to be interacting among themselves only. Such a model is called the Born–Karman model. The equation of movement (Newton’s second law) can be written for the atom l as ⎤ ⎡ m l [( l l1 ) ( l1 l )] 2 ⎢ l l1 l1 ⎥ , 2 ⎣ ⎦
(9.3.1)
where is a force constant and the expression in square brackets is the shift of the atom l from its equilibrium position. This system of equations can be solved, the solution being sought in the form of a running wave l 0 exp[i( t kla)].
(9.3.2)
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a
a
1
2
3
Figure 9.13 A one-dimensional chain of similar atoms with inter-atomic distance a ( is atom’s deviation from the equilibrium position), is a force constant of the inter-atomic bonds.
The connection between and k is 2 m (e ika eika 2) (e( ika 2 ) e( ika 2 ) )2 4 sin 2
ka , 2
(9.3.3)
wherefrom
4 ka sin . 2 m
(9.3.4)
A dispersion curve is a curve of allowed -values as a function of k. Within the framework of the Born–Karman model this dependence is represented by eq. (9.3.4) (Figure 9.14). Let us look at its main points. Firstly, this equation is nonlinear; the dispersion of waves (refer to Section 2.9.4) is taking place. Secondly, this dependence is periodic; in space of wave vectors an independent (repeating) part is lasting from zero to /a. This area is referred to as the Brillouin zone and a value /a is the border (in this case –one-dimensional border) of this zone: outside this zone the function (k) repeats periodically. Besides, the number of waves with various wavelengths is limited: from below by the period of a chain a, from above by the length of all chain L. At small k((ka / 2) ^ 1) sin(ka/2) in eq. (9.3.4) can be substituted by its argument. Then
m
ka
(9.3.5)
and the dependence (k) is getting linear. The wave phase speed can be derived from eq. (9.3.5) (refer to eq. (2.8.12)); this is an acoustic dispersion wave curve. The derivative (d / dk) is zero at the zone boundary. The dispersion curve is referred to as an acoustic branch. A more complex model is a diatomic chain consisting of alternating atoms of different masses m1 and m2 which are located at equal distances from each other (which still equal a) and between which the same elastic forces operate (Figure 9.15). “The unit cell” of such a “crystal” is twice as large as the previous one. Two equations for movement should be written and solved together. The solution in the form of running
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ka
( / m)
547
–π/a
π/a
0
k
Figure 9.14 Dispersion curve (k) for Borman–Karman model. a
1
a
m1
m2
2
Figure 9.15 A one-dimensional model consisted of two different atoms with the same inter-atomic distance. A period is equal to 2a.
waves has the same appearance, but frequency depends on two masses and has two solutions: 2
2
2 ⎛ 1 ⎛ m m2 ⎞ 1 ⎞ sin ka ⎜ ⎟ ⎜ 1 . 4 ⎟ m1 m2 ⎝ m1 m2 ⎠ ⎝ m1m2 ⎠
(9.3.6)
Accordingly, a graphic representation of the (k) function has two branches (Figure 9.16). It is typical that one solution ( ) behaves in the same way as in the previous case. However another solution ( ) essentially differs: firstly, at k 0 frequency aspires to a nonzero value, secondly, at k ( / 2a) the curve comes to different point on the Brillouin zone border. (The factor 2 in the denominator appeared because the identity period in the model presented in Figure 9.15 has changed.) The difference between frequencies on the Brillouin zone border is proportional to the distinction of masses of the two kinds of atoms. This second branch is referred to as an optical branch. In a real three-dimensional crystal there are similar waves extending in all directions. Taking into account that in the crystal the existence of waves of three polarizations is possible (one being longitudinal L and two transverse wave T1 and T2, refer to Section 2.8.2),
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A
–/2a
/2a k
0
Figure 9.16 Graphical representation of dispersion curves (k) for the two-atomic chains. Acoustic (A) and optic (O) branches are presented.
L
T1
T2
0
k
/2
Figure 9.17 A general case of dispersion curves of three different kinds: L—longitudinal wave, T—two transverse waves.
three branches (k) can occur (Figure 9.17). In some crystals their degeneration is probable: two or all three branches can merge into one. Similarly to the one-dimensional chain, the borders within the framework of which function (k) is independent are outlined. Such borders form the three-dimensional Brillouin zone. The picture of the dispersion curves dependent on direction and from wave polarization, becomes complex and, frequently, confusing. Figure 9.18 shows an example of experimentally measured dispersion curves in an aluminum crystal for the different directions specified in the figure; Figure 9.19 presents an example of the oscillation frequency distribution function g( ). Note that the acoustic branch corresponds to the oscillation of the crystal unit cells relative to each other, whereas the optical branches describe the oscillation of atoms relative to each other within the volume of a single unit cell. Since the elastic waves in crystals are caused by atomic oscillations and are interconnected, waves are developed as collective excitations. Hence, in modern terminology,
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[110]
[111]
arb. units
[100]
549
0
1.0
0
0.5
k/k max
Figure 9.18 Dispersion curves for different crystallographic directions (shown in the scheme) in aluminum obtained by the neutron inelastic scattering method.
g
2
1
0
3
6 /2, TΓµ
D
Figure 9.19 The experimental frequency density distribution g( ) for aluminum ( D is the Debye frequency).
lattice heat capacity is a development of collective excitation in solids, i.e., the heat energy goes to phonon excitation (see below). Participation of electron excitations in heat capacity is limited by quantum statistics laws (refer to Section 9.2.2). As only a small part of the electrons takes part in excitation, the electronic heat capacity usually makes a very small addition in comparison with the lattice one.
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In a polycrystalline sample consisting of a large number of microcrystals, ideally— randomly dispersed in space—the resulting picture is averaged. As a result, the function g( ), presents the projection of all dispersion branches to the axis of frequencies (phonon spectrum). It describes the distribution of oscillations on frequencies: product g( )d
gives the number of oscillations dz falling on an interval d . In Figure 9.19 the experimentally measured function g( ) for aluminum is given as an example. The fluctuation distribution function determines the internal energy of a crystal:
U ∫ g( )d ,
(9.3.7)
where is an averaged oscillation energy. The dispersion curves and g( ) function can be obtained by neutron scattering methods. 9.3.2
The heat capacity of crystals
Heat capacity was introduced in Section 3.4.3 as the heat energy capable of heating a body by 1 grad. For an ideal gas in molecular kinetic theory the mole heat capacity appeared to be equal to CV
ief i 2 R and CP ef R, 2 2
where ief is the effective number of a gas molecule’s degrees of freedom. In this case ief turns out to be the sum of translational, rotational and oscillation movements; in the latter ief, osc 2 because oscillation simultaneously possesses potential and kinetic energy. Apply these representations to a solid. For this purpose we should first analyze the character of possible movements of the molecules in it. Clearly, translational motion is excluded in this case. The rotation of molecules in a crystal is basically possible. For example, in crystal NH4Cl a group NH4 at different temperatures can make rotations around axes of different symmetry and sometimes exhibit a free rotation. However, the contribution of rotation to the thermal capacity of solids is appreciably less than oscillatory degrees of freedom. As a result only oscillations of atoms basically define the crystal thermal capacity. Furthermore, the three-dimensional character of a crystal should be taken into account. The molar internal energy of a crystal U will then be given as a product
U 3
ief,osc RT 3RT . 2
(9.3.8)
The derivative from internal energy on temperature gives a mole thermal crystal heat capacity C. As the volume of a crystal does not vary appreciably with temperature, we exclude index V. We shall then obtain C 3 R,
(9.3.9)
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i.e., the molar heat capacity of a crystal does not depend on temperature and is numerically equal to 3R (⬇ 25 J/(mol K)). This result is referred to as the Dulong–Petit law. Experiment shows that at moderate temperatures this law is carried out well enough for crystals with rather simple structure (Na, Al, Fe, Cu, Sn, etc.). However, crystals with more complex structures fail the Dulong–Petit law. So the atomic thermal capacity of a boron crystal is 14.2 and diamond is 5.7 J/(mol K). It has also been experimentally established that in the region of low temperatures the heat capacity falls as T 3, coming nearer to zero at T → 0 K. The experimental facts therefore show that the simple classical theory considered is applicable only in a narrow temperature interval and for crystals with a simple structure. The first quantum theory of heat capacity was suggested by Einstein. Einstein’s model assumed that each atom in a crystal is an independent quantum oscillator (refer to Section 7.8.1). The energy spectrum of such an oscillator is presented in Figure 7.32. Because of the selection rule, the spectrum of absorption/emission contains only one frequency. A level’s population is defined by the Boltzmann factor E ⬃ exp(E/T ), i.e., the population gets less when the energy increases. In Section 6.6.3 it was shown that the average oscillator’s energy is (h ¯ ) / (exp (h ¯ / T ) 1) (eq. (6.6.13)). Accordingly, the internal energy U of one mole of a substance is
U = E N A = N A
+U , exp(
T ) -1 0
(9.3.10)
where U0 is the zero oscillation energy. The latter is of no significance in the heat capacity of crystals. The heat capacity C is obtained as 2
C
exp(
T ) dU ⎛ ⎞ NA ⎜ . ⎝ T ⎟⎠ (exp(
T ) 1)2 dT
(9.3.11)
Denoting the ratio (h ¯ /) by , the last expression can be rewritten as 2
exp( T ) ⎛ ⎞ . C R⎜ ⎟ ⎝ T ⎠ (exp( T ) 1)2
(9.3.12)
The value is called the characteristic temperature. If one derives a function of the heat capacity upon T/ for a series of simple substances, it will be represented by a single curve (Figure 9.20). It appears that Einstein’s characteristic temperature defines the border behind which an essential deviation from the Dulong–Petit law takes place for every element. If we substitute the frequency 兹 苶苶/m 苶 (see eq. (2.4.5)) into the expression for the value, we can arrive at
, m
(9.3.13)
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6
C/n
5 4
Ag Al C(grafite) Al2O3 KCl
3 2 1 0
0
0.5
1.0 T/
1.5
2.0
Figure 9.20 A reduced temperature dependence of heat capacity for some simple substances.
i.e., the value is inversely proportional to the square root from atomics’ mass. Let us consider some extreme cases. At T the exponent can be decomposed into a series (ex ⬇ 1x ...), and we can limit ourselves to the first two terms. For a system of one-dimensional oscillators we can arrive at ⎛ ⎞ ⎛ ⎞ 1 冢 T 冣 C ⬵ R⎜ ⎟ R ⎜ 1 ⎟ ⬇ R ⎝ T⎠ ⎝ T ⎠ 冢 2 T 2 冣 2
or for a three-dimensional crystal C 3R, i.e., the Dulong–Petit law. At T the exponent in eq. (9.3.12) is much larger than 1. It gives C ⬵ 3R ( / T )2 ( / T ) e , i.e., heat capacity is decreases with the decrease in temperature (because the exponent influence dominates over the term ( / T)2). Apparently, Einstein’s model gives good conformity with Dulong–Petit at high temperature and explains the decrease of thermal capacity at low temperatures. However, it contradicts the law of approaching absolute zero: according to experiment, the corresponding curve should look like a cubic parabola T 3, whereas according to Einstein this dependence is described by the law e(1/T). P. Debye modified the Einstein’s model by introduction of inter-atomic forces in a crystal model. This is equivalent as to take phonons into account (refer to Section 9.3.1). To each elastic wave (phonon) the Born-Karman atomic chain was attracted spread out in a three-dimensional array (Figure 9.13 and 9.15). As a result of reflection from external crystal borders, standing waves with various values and k (refer to Section 2.9.2 and 2.9.3) are formed. There is the certain relationship between the wavelength of standing waves and the size of the crystal L expressed by the eqn (2.9.8). Phase speed of running
, therefore elastic wave Ph is related to and k by expression (2.8.3) Ph k
kPh
2Ph T Ph n, L
(9.3.14)
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where n is the number of the harmonic wave. At the crystal volume V a large but limited number of standing waves are formed. The wavelengths of these standing waves are limited by the period of the crystal lattice from the short wavelength size ( a) and by the crystallite size L ( L) from the large one. Accordingly,
min
2ph max
2ph 2L
ph L
(9.3.15)
and
max
2ph min
ph a
.
(9.3.16)
These values limit the range of frequencies. The number of standing waves dz is proportional to the elementary volume 4 2d . For a cubic crystal with volume V L3 the number of harmonics (normal oscillations) in the limits from 0 to is equal to z (2L /n)3 8 (V / n3) or more precisely
z
4V V 2 3 2 . 3n 2 ph
(9.3.17)
Differentiating z by we can obtain the frequency distribution g( ) as
g( )
dz 3V
2 . 3 d 2 2ph
(9.3.18)
The Debye function g( ) is depicted in Figure 9.21.
g
0
Figure 9.21 The frequency density distribution g( ) in the framework of the Debye model ( D is the Debye frequency).
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The maximum spectrum frequency is called the Debye characteristic frequency D. It can be determined from the normalization to the total number of normal frequencies (the number of standing waves).
D
V 3
∫ g( )d 22D3
3N ,
ph
0
from which
⎛ 6N ⎞
D ph ⎜ ⎝ V ⎟⎠
1 3
.
The notion of “characteristic temperature” acquires a new meaning:
D . T
(9.3.19)
According to eq. (9.3.7) the general expression for U in a given case is:
D
U
∫ 0
D
g( )d
∫ 0
9N 3 3 d . exp(
T 1) max
(9.3.20)
Differentiating over temperature, we can obtain the crystal heat capacity
C
⎡ ⎛ T ⎞ 3 T ⎛ x 3 dx dU 3( T ) ⎞ ⎤ 3 R ⎢12 ⎜ ⎟ ∫ ⎜ x x ⎟ dx ⎥ . dT ⎢⎣ ⎝ ⎠ 0 ⎝ e 1 e ( T ) 1⎠ ⎥⎦
(9.3.21)
The expression in square brackets is the Debye function. This function is tabulated. We shall find extreme values of heat capacity: at T p D the Debye function aspires to 1 and C → 3R; at T ^ D the integral in the Debye function is close to 1 and C → (T / )3. Good consent with experiment is achieved. Figure 9.22 shows the results of comparative approaches for three models of crystal dynamic properties. In conclusion, we give in Table 9.2, the Debye temperature for some substances. Despite the success of the theory, it is necessary to ascertain whether its use in practice meets many difficulties. This is visible even from a comparison of the curves of dependence (k), obtained theoretically and experimentally (with the help of inelastic scattering of thermal neutrons) (Figures 9.19 and 9.21). At concurrence of some features of the curves (square dependence at small phonon wave vectors k, sharp decrease at certain values of frequencies, etc.) a definite difference can also be found. The theory allows one to
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D
B-K
E
/a
k
Figure 9.22 Comparison of the g( ) function for different models: Born–Karman (B–K), Einstein (E) and Debye (D). Table 9.2 Debye temperature values for some simple substances Substance
Debye temperature (K)
C (diamond) Be Si Cr Pb
1910 1160 658 420 95
obtain only estimated or comparative values of heat capacity; therefore experimental data are often used in serious work. A promising direction in chemical technology developments in the last 10 years is nanotechnology, i.e., the syntheses of materials consisting of microscopically small particles (clusters), small in number (N ⬃ 103 atoms), being comprised on the “nanoscale” L ⬃ 109 m. This technology gives us a good assurance that this chapter is of use in the book. In practice, this border is somehow degraded, because in technology by “nanoparticle” one usually implies an atomic or molecular object, the internal energy of which complies in order of magnitude with its surface energy. More important is the fact that the mechanical, electrical, magnetic and other characteristics of nanoparticles are vastly distinct from similar characteristics of the same bulk material. Some of thesxe characteristics are directly connected with the nanoparticle’s size, so this opens the way to adjust them to a goal-directed image. This fact permits one to produce new materials with predetermined properties, e.g., semiconductors that are required for creating more reliable generations of computers.
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Let us consider as an example the well-known Debye theory of the heat capacity of crystals. The internal energy of a bulk crystal minus zero energy is given in this theory by an equation 9 N T 4 U 3D
D T
∫ 0
x3 dx, e x 1
where x (h ¯ / T ) A nanoparticle is distinguished from a bulk crystal since its phonon spectrum is truncated not only from the site of high frequencies but from low frequencies as well (refer to Example E9.5). Let us consider the averaged solid nanoparticle model with number of atoms N, crystal lattice period a, primitive cubic lattice with the unit cell volume V and linear size L. The physicochemical features, however, and in particular the nanoparticle’s heat capacity in the implicit description can depend on their forms, for instance, by the way it fastens to the substrate and the nature of interaction with it, i.e., from border conditions. A detailed consideration of all these factors forms the subject for specialized future studies. Therefore, the further evaluations are based on the assumption of the independence of the nanoparticle’s thermal characteristics from the particular type of border conditions (similar to that of the classical theories of Einstein and Debye). The longest wavelength corresponds to the lowest frequency of the nanoparticle that can be put in correspondence to the temperature, i.e., N
N 2 , N
therefore,
N
2 ⬇ . L 2 N 1 3 a N 1 3 a
Since the Debye temperature is assumed to be D (h ¯. /a), the ratio (N/D) N 1/3 3 or (N / D) ⬇1/10 with N ⬇ 10 . Therefore, nanoparticles have two characteristic temperatures that distinguish them from that of the bulk materials: the usual Debye temperature D and the temperature N both depend on the particle size. For further simplification, let us restrict ourselves to the limit that corresponds to the condition D T, because of the fact that this condition corresponds to the lowest value of x (h ¯ / T ) (N T ) 1 . This permits us to simplify the integral expression, i.e., assume ex p 1. In this case the internal energy of the nanoparticle in the low temperature approximation is equal to
UN
9 N T 4 3D
∫
N T
x3ex dx
9 N T 4 3 ( x 3 x 2 6 x 6) 3D
N T
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For diamond dust (D 2230 K), L ⬃ 10a, N ⬃ 103, N (D /10) ⬃ 223 K, i.e., the condition (N / T ) 2 is fulfilled at temperatures T 111 K. At ex p 1 the expression for the nanoparticle heat capacity is changed as well. The Debye equation at T ^ D, i.e., ⎛T ⎞ C 9 Nk ⎜ ⎟ ⎝ D ⎠
3
x 4 e x dx
∫ (e x 1)2 0
can be simplified by decomposition to
⎛T ⎞ CN 9 Nk ⎜ ⎟ ⎝ D ⎠
3
∫
N T
3
⎛T ⎞ x e dx 9 Nk ⎜ ⎟ ( x 4 4 x312 x 2 24 x24)ex ⎝ D ⎠ 4 x
N T
.
The well-known classical Debye law for bulk crystals Cⴥ⬃ T 3 is disobeyed, and, as the calculations show, rather significantly. Once more, this confirms the fact that nanoparticles really are new materials, with physicochemical characteristics greatly different from the characteristics of materials with the same chemical composition but in the bulk state.
EXAMPLE E9.4 Determine the amount of heat Q for NaCl of mass m 20 g heated at T 2 K in two cases (1) from T1 D and (2) T2 2 K. Accept the characteristic Debye temperature D to be equal to 320 K. Solution: In general, the amount of heat Q needed to heat a system from 1 to
冕
2 can be calculated according to the expression Q
2
1
CdT where C is the heat
capacity of the system. The heat capacity of a body is related to molar heat capacity C (m/M)Cm where m is body mass and M is molar mass(refer to Section 3.4.3). m
Substituting into the integral gives Q
冕
2
1
CMd T *. In a general case, CM
depends on T and therefore it is not allowed to take it out of the integral sign. However, in the first case we can neglect the CM change and consider it to be constant, throughout the interval T, equal to CM(T1). Therefore, the integral takes the form Q (m/M)CM(T1)T**. The molar heat capacity in Debye theory is given by eq. (9.3.21). In the first case calculations give CM 2.87 R. Substituting this result into ** we obtain Q 2.87(m/M)RT and executing calculations we arrive at Q 16.3 J. In the second case (T ^ D) calculation of Q is simplified by the fact that we can use the limited property of the Debye law where the heat capacity is
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proportional to T 3: we cannot use the previous approximation but have a mathematical expression for 3
12 4 ⎛ T ⎞ CM R⎜ ⎟ . 5 ⎝ D ⎠
Q
12 4 m R 5 M 3D
T2 T
∫
T 3 dT .
T2
Executing integration we obtain
Q
12 4 m R ⎡ (T2 T )4 T24 ⎤ ⎥. ⎢ 5 M 3D ⎣ 4 4 ⎦
Since T2 T 2T2 this equation takes the form
Q
3 4 m R m R 4 15T24 or 9 4 T2 . 5 M 3D M 3D
Substituting all known values and executing calculations we obtain Q 1.22 mJ.
EXAMPLE E9.5 How much does the amount of heat required differ for heating a nanoparticle consisting of N atoms from temperature T1 0 K to T2 D/50, compared to the amount of heat required to heat the same amount of particles in the form of a bulk crystal. Accept the nanoparticle’s size L 10a, a being the period of the crystal lattice; D is the Debye temperature).
L
a
k
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Solution: The frequency of a nanoparticle’s dependence on the wavenumber k is depicted in the Figure E9.5. From the problem’s conditions, we have D 1 . N L 10 The internal energy U of a bulk crystal consisting of N atoms at T1 T2 is equal to
UN
9 N T 3D
D 4 T2
∫ 0
x3 dx . e x 1
The upper limit in the task is (D / T2) 50 1, therefore one can accept it, as usual, to be equal to ⴥ. The tabular point D 50 1, T2
x3 ∫ e x 1 15 6.5. 0
For T T2 the internal energy is
UN
9 N T24 3D
∫
x3ex dx,
N
T2
where N 50 5. T2 10 The zero energy as independent of temperature cannot be taken into account. The integral in the last expression for UN can be calculated in parts:
∫x e
3 x
dx ( x 3 3 x 2 6 x 6) ex
5
1.57
5
Then the ratio (Uⴥ) /(UN) (6.5 /1.57) ⬇ 4.15, i.e., the internal energy difference for a bulk crystal and a nanoparticle with equal amounts of atoms is really very significant.
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EXAMPLE E9.6 Using a free electron model calculate for potassium at 0 K: (1) Fermi energy F, (2) Fermi temperature TF, (3) the ratio of averaged potential energy 具U典 of the two adjacent electrons in the electron gas to their averaged kinetic energy 具典. Assume the potassium density 860 kg/m3 and atomic mass A = 39.1 103 kg/mol. Solution. Although in general in the main text we can come to the conclusion that the free-electron model is applicable to crystals, we can use in some cases, with certain limitations, the classical approach to the evaluation of electron gas properties. (1). At T 0 K, the Fermi energy (refer to Section 9.2.2) depends only on the electron concentration n F (h ¯ 2 / 2me)(32n)2/3. Assume that every potassium atom gives one electron to the free electron gas state. Therefore, the concentration of free electrons is equal to the potassium atom concentration, namely n nat (/A)NA. In order to avoid very large expressions and calculation we will deviate from our recommendation and calculate each quantity separately. Thus, we can calculate the electron concentration accordingly:
n
860 6.02 1023 m3 1.32 1028 m3 39.1103
We can calculate the Fermi energy according to the mentioned formula as F
(1.05 1034 )2 冸3 2 1.32 1028 冹2 3 J 3.23 1019 J or 2.02 eV 2. 9.111031
(2) The Fermi temperature can be calculated from the equation F T: T F /. Executing calculations we obtain: TF
3.23 1019 2.34 10 4 K 23.4 K 1.38 1023
(3) In order to find the zero temperature electron gas pressure we can use the expression p (2/3)n具典. Here we will assume 具典 the averaged value of kinetic energy at T 0 K. Find this expression using the distribution of the free electron upon energies dn() (1/22)(h ¯ 2/2me )3/2 1/2 d, where dn () is the number of electrons in a unit volume, the energy of which lie in the small interval from to d. Then dn() C1/2d, where C is a constant. We can find the averaged value 具典 for free electrons by dividing the total energy in an unit volume by their concentration, i.e., F
具典
∫ dn 0 F
∫ dn 0
F
C ∫ 3 / 2 d 0 F
C ∫ 1 / 2 d 0
(2 5) F
5 2
(2 3) F
3 2
3 F 5
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Substituting this result into the expression for free electron gas pressure p (2/3) n (3/5) F we obtain p (2/5)1.32 1028 3.23 1019 1.71 109 Pa 1.68 104 atm. (4) The averaged potential energy of the electrostatic interaction of two point charges (refer to eq. (4.1.21)) 具U典 (e2)/(40具l典) where 具l典 is an averaged distance between two adjacent electrons of the degenerated electron gas (refer to Example E3.1) 具l典 n(1/3). Then 具U典 (e2n1/3) / (40) . We already know the 具典 (see above) therefore
5e2 n1 3 具U 典 e2 n1 3 5 . 40 3 F 12 0 F 具典 Substituting the numerical values into this expression we arrive at 具U 典 (1.32 1028 )1 3 5(1.6)2 2.81 具典 12 8.85 1012 3.23 1019
9.4
CRYSTAL DEFECTS
A crystal represents a complex quantum mechanical system with an enormous amount of particles with a strong interaction between them. If all the particles are located in space strictly ordered with the formation of an ideal three-dimensional crystal structure, then such a system possesses minimal free energy. In a real crystal, however, the ideal periodicity is often broken due to inevitable thermal fluctuations: some atoms break periodic array, abandon their ideal position and produce a defect. The frequency and amount of fluctuations are defined by the Boltzmann factor, i.e., they depend on temperature and binding strength or, in other words, on the depth of the potential well corresponding to the regular position of atoms. The role of defects in crystal properties is very high. On the other hand, new and improved old methods of defect crystal structure studies have appeared. This leads to the fact that, in recent decades, there has been extensive investigation into the physics of solids from the point of view of their deflection from perfection. Natural and new synthetic materials are being investigated. Let us distinguish point (zero-dimensional) and extended defects (dislocations). 9.4.1
Point defects
Point defects can appear in a small area of the crystal, not exceeding several internuclear distances. Due to thermal fluctuations, a single atom can abandon its ideal position and occupy a position in the so-called interstitials (a position, though having a potential well, is of small depth). The regular position in the lattice remains empty, and is called a vacancy. Such pair defects are referred to as Frenkel’s defects, they contain a vacancy in a regular array and interstitial atoms (Figure 9.23).
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+ Sh
–
+
–
+
–
+
–
–
+
–
+
+
–
+
–
F
+
–
–
+
–
+
Figure 9.23 The Frenkel (F) and Schotky (Sh) types of point defect in a crystal.
If an atom comes out onto a crystal surface and leaves a vacancy in its regular position, then such a single defect is referred to as a Schotky defect. The probability of a Schotky defect formation is proportional to the Boltzmann factor P⬃ exp(Esh/T), where Esh is the energy required for an atom’s removal from its regular position to the crystal surface. If N is the total amount of atoms and n is the vacancy number then P n/(Nn) exp (Esh/T ) or, at n N, n ⎛ Esh ⎞ ⬇ exp ⎜ . ⎝ T ⎟⎠ N
(9.4.1)
Estimations show that at T ⬇ 1000 K and vacancy activation energy E ⬇ 1 eV, the relative vacancy concentration is (n/N) ⬇ e12 ⬇ 105. Note that the relative vacancy concentration is proportional to exp(1/T). The number of Frenkel pair defects n depends on the number of regular positions in the crystal N and the number of interstitials N ⎛ EF ⎞ n ⬃ ( N N) exp ⎜ , ⎝ T ⎟⎠ where EF is the energy required for an atom’s disposition from the regular site into the interstitial. The vacancies in ionic crystals are mostly Frenkel defects. Crystal defects are not static. They wander around the crystal: a neighboring atom occupies the vacant position; the interstitial atom can occupy a regular site, etc. Only their relative number is determined by the Boltzmann factor.
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Along with such elementary defects, more complex ones can also appear; for example, divacancy, the formation of which is energetically more favorable in comparison with two individual vacancies. Besides, the mobility of a divacancy is higher than that of individual vacancies. The congestion of many vacancies leads to the formation of the so-called clusters. Vacancies strongly influence diffusion and related processes. Elastic displacements in the area surrounding a point defect decrease proportionally to 1/r3, where r is the distance from the defect. This shows that distortions in the neighborhood of a defect can be rather significant, but quickly falls down at distance. Possessing mobility, the point defects can interact with each other and with other defects. On meeting, the vacancy and interstitial atom can annihilate each other. An important role in vacancy formation is played by impurity atoms. They can replace regular atoms, and also be in interstitials. In particular, in the crystal structure KCl the replacement of K by Ca atoms leads to the occurrence of vacancies. Small amounts of impurities considerably change the electric properties of semiconductor crystals. 9.4.2
Dislocations
Apart from point defects, there also exist extended, in particular, one-dimensional edge, dislocations: a failure of a crystal lattice having in one direction a long (even macroscopic) dimension but small sizes in the other directions (one or several internuclear distances). Such defects play a very important role in the formation of many properties of crystals, in particular, mechanical properties. Quantitatively, the extended failure of the crystal lattice is described by the so-called Burgers vector. Let it be described in a simple cubic crystal. In Figure 9.24 a plane (001) of such a crystal is presented. In an ideal nondeformed lattice we shall choose a closed contour and number the atoms included in this contour (Figure 9.24a). Such a contour is called Burger’s contour. Travelling toward the east over the contour for two periods, south for two periods, then
0 0
1
1
2
0
0
2 7
7
3
1
2
b 7
w
6
5
3
3 4 6
5 (a)
4
4 6
5 (b)
(c)
Figure 9.24 Burger's vector in (a) an ideal, (b) an elastically deformed and (c) a crystal with edge dislocation.
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west and, finally north, we will arrive at the initial atom. In a numerical expression, the travel over the contour is 0–1–2–3–4–5–6–7–0. If a crystal is elastically deformed the atoms, being close to their ideal positions, are displaced by small distances (Figure 9.24b): the contour is deformed but still remains closed. Now create a defect in the crystal as a half-plane (i.e., an infinite plane in two directions limited by a line in the third side) and introduce it into the crystal parallel to similar crystallographic planes (Figure 9.24c). Now, traveling over Burger’s contour we will not come to the initial atom but to another point. To close the contour, it is necessary to draw a closing vector b, which is referred to as Burger’s vector. Note that Burger’s vector is perpendicular to the plane introduced. Dislocation is a linear crystal defect for which Burger’s vector is not zero. This definition is common for rather large numbers of a different kind of linear defect. The considered dislocation is referred to as an edge disposition; the extreme line (edge) of the induced plane is referred to as the line of dislocation (in Figure 9.24c this line in a point w is perpendicular to the plane of drawing). For such a dislocation, Burger’s vector is perpendicular to the dislocation line. A model of edge dislocation made of balls is presented in Figure 9.25. In Figure 9.26 a three-dimensional image of the edge disposition with Burger’s vector perpendicular to the plane is shown. The crystal above the dislocation edge is stretched and in its lower part is compressed. Now consider another kind of disposition, namely, a screw disposition. Let us imagine a pile of sheets of paper. Using a sharp knife, cut it from the center of the pile to its edge. Shift the free paper edges perpendicular to the sheets by the thickness of one sheet. Sticking the edges of the new planes to the edges of the old ones will result in a spiral. This spiral gives a graphic representation of screw dislocation. Its crystallographic image is given in Figure 9.27. Dislocations can move in crystals as do point defects; the edge dislocation is presented in Figure 9.28. Under the action of an external pressure the shift of the upper part of the
Figure 9.25 A ball model of the edge dislocation.
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Figure 9.26 A three-dimensional image of the edge dislocation.
Figure 9.27 A model of a screw dislocation.
crystal has forced the dislocation to be switched to the next atomic plane and, finally, to appear on the crystal surface. Similar movements can make screw dislocation too. There exist bigger crystal defects called small-angle boundaries. The scheme of such a boundary is depicted in Figure 9.29. It can be imaged by a set of number of edge dislocations. The presence of imperfections essentially influences the mechanical properties of crystals. Experience shows that theoretically calculated strength properties appear, as a rule, to be appreciably higher than experimental ones; this is caused by neglecting crystal imperfections. Table 9.3 lists the physical methods that allow the investigation of the real structure of crystals.
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Figure 9.28 A scheme of an edge dislocation movement.
b b
Figure 9.29 A low-angle edge. Table 9.3 Methods of investigation of crystal defects Method
Object thickness
Width of image
Maximal defects density (cm2)
Electron microscopy X-rays’ transmission X-rays’ reflection Metal sputtering onto surfacea Etching pitsb
100 nm 0.11.0 mm 2/50 m ⬇10 m
⬇10 nm 5 m 0.5 m 0.5 m
10111012 104105 106107 2 ·107
No limits
0.5 m
4 108
a
Sputtering of thin layers onto a surface and observation of dislocation by different methods. Surface etching and observation of etching pits by different methods.
b
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TRANSPORT PHENOMENA IN LIQUIDS AND SOLIDS
The laws of transport phenomena in states other than gases differ among themselves in many respects. This distinction is defined mainly by the difference in their atomic structure. This will be explained using the example of a monoatomic crystal with a cubic structure (refer to Table 9.1). The function of radial distribution G(r) has great value in this respect. To explain the physical sense of this function, we shall define one more concept: the coordination sphere, perhaps known to readers from chemistry. For this purpose we choose in the crystal any atom and superpose it with the coordinate system origin. Allocate around it a spherical layer of radius r and thickness r (r r); the layer volume is V 4r2r. The corresponding scheme is shown in Figure 9.30; for simplicity a two-dimensional section of a cubic lattice is used. If r r1 no atom falls onto the sphere. When r is less than the shortest interatomic distance a no atom falls into the spherical layer either. At r a all atoms which meet at this distance fall into the spherical layer (in a threedimensional cubic crystal there are six such atoms). With a further increase of r in an interval a r 兹苶2a , again no atoms fall in the layer, but at r 兹苶2a atoms do fall again in the layer; their number in the three-dimensional array will be 12. At r 兹苶3a in the layer eight atoms will fall. This procedure can be continued further. These spheres are referred to as coordination spheres and the numbers of atoms falling onto them as coordination numbers. Count the number of atoms N enclosed in the spherical layer 4r2r. Write this number as: N 4r 2 r n0 G(r ). Here n0 is the average over the whole crystal volume atomic concentration, and G(r) is the radial distribution function. Wherefrom: G(r )
N N n(r ) , 2 Vn0 4 r r n0 n0
(9.5.1)
where n(r) is the local atomic concentration in the spherical layer with radius r.
1 3
2
4
Figure 9.30 Coordination spheres in a crystal. A two-dimensional section of a cubic crystal is presented. Numbers denote the first, second, third and fourth coordination spheres.
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It can be seen from the last expression that when the atomic concentration in the allocated spherical layer is equal to average concentration n0, G (r) 1. The function of radial distribution is distinct from unity only in the case when the concentration of atoms in the allocated layer is different from n0. Hence G(r) characterizes the concentration deviation in a specified layer from the average value. The functions G(r) for gases, liquids and crystals are presented in Figure 9.31. In crystals atoms are arranged in the ordered lattice; they are at certain distances from each other. The G(r) function is different from zero when r is equal to one of the possible interatomic distances, i.e., when r touches the coordination sphere. In the resultant radial distribution function G(r) in a crystal is represented by a system of discrete peaks (Figure 9.31a). The regular arrangement of peaks in crystalline solids specifies the presence in a crystal of the long-range order. If atoms are at rest in their ideal positions, function G(r) is represented by a system of discrete, narrow spectral lines. Because of the thermal vibrations of atoms around their regular positions the lines are widened. In gases G(r) ⬅ 1; this means that there is a total disorder in gases. However, because gas atoms cannot approach each other closer than r d Эф in an area r def function G(r) falls to zero (Figure 9.31c). All this corresponds to an absence of even short-range order in gases; the full disorder in an arrangement of atoms. It has been experimentally established that function G(r) in liquids looks like that presented in Figure 9.31b: there are dim maxima at r (13) a, and further G(r) aspires to 1. This result indicates some order in an arrangement of atoms in the liquids at specified distances and the absence of order at large r. Such a situation corresponds to the shortrange order. It can be imagined as a presence in liquids of very small crystallites, completely disordered to each other. Due to the thermal motion of atoms these crystallites
G (c) 1
(b) 1
(a)
a
2a
r
Figure 9.31 A function of the atomic density radial distribution in (a) crystal, (b) liquids and (c) gas.
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continuously collapse and new ones are created instead. Each atom in a liquid spends some time (referred to as time of settled life) in a regular position and makes fluctuations around it. At this time the atoms of a liquid behave like atoms in a crystal. Later, the atom abandons its regular position and jumps to a new position remote from the first by a distance approximately equal to the interatomic distance in the liquid. During the jumping process, the atom of a liquid assimilates to a gas atom. If the temperature decreases, the time of settled life is increased; near the crystallization point this time increases more and the properties of the liquid come nearer to the properties of solids. On the contrary, if the temperature rises, coming nearer to the boiling point, the time of settled life decreases; and the liquid’s properties become similar to those of gas. On melting there is an increase in the specific volume of a substance, on an average by 3%. It is reasonable to assume that the increase in volume is caused by an increase in interatomic distances in the liquid in comparison with that in crystal. However, the compressibility dependence on pressure does not confirm this supposition. Since there is a short-range order in a liquid the local potential curve U(x) in its nearest surroundings resembles a periodic character (Figure 9.32). Atoms settle down in points of potential energy minimum. In a settled state an atom oscillates with an amplitude significantly smaller than the interatomic distance . It was found experimentally that the frequency of these fluctuations have the same order as in solids, namely v0 ⬇ 1013 sec1, and the oscillation period T0 (1/v0) ⬇ 10 13 sec. In order to leave the pseudo-equilibrium position an atom should overtake the potential barrier u. The probability of this process is proportional to the Boltzmann factor e(u / T ). In 1 sec, an atom makes v0 fluctuations; hence the frequency of jump over to the new position is v v0 exp((u/T )). Thus, the time of settled life can be obtained by (1 / v ) 0 exp(u / T ) F
u
2a
Figure 9.32 Atomic potential curve in a liquid.
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When a particle leaves a temporary equilibrium position and occupies a new one, it covers a distance . The particle’s average speed of wandering (chaotic motion) in liquids is then ⎛ u ⎞ ⬇ exp ⎜ ⎟ . ⎝ T ⎠ 0
Suppose that there is a concentration gradient in the liquid. In that case the chaotic wandering creates an excessive atom flow in the direction opposite to the concentration gradient; i.e., diffusion flow will take place. The diffusion coefficient for ideal gases is described by the formula D (1/3) , where is the molecular free path length, and is the average speed of thermal motion (refer to Section 3.3.8). In a liquid the role of free path length is played by the displacement and the role of average speed is (/). Therefore the diffusion coefficient in the liquid can be evaluated as
D
1 2 ⎛ u ⎞ exp ⎜ ⎟ . ⎝ T ⎠ 3 0
(9.5.2)
It can be seen from this expression that diffusion in a liquid sharply increases with temperature according to the exponential law. In gases the diffusion coefficient also rises with temperature, but not as fast: under the power D.T1/2 if heating is isochoric, and as D. T3/2 if heating is isobaric (see Section 3.7). Although the expressions obtained are approximate, they usually correctly estimate the order of diffusion coefficient provided that ⬇ 1010 m, v0 ⬇ 1013 sec1 and u (the activation diffusion energy) is approximately equal to the latent heat of melting. The data on diffusion parameters in liquids is given in Table 3.3. Theoretical treatment of a viscosity phenomenon of liquids can also be carried out proceeding from the same representations. Leaving aside the treatment itself, we write down the final result concerning the temperature dependence of the viscosity as Ae(u T ) ,
(9.5.3)
where A is a pre-exponential factor dependent on the parameters of the liquid and weakly dependent (in comparison with the exponent) on temperature. It can be seen that liquid viscosity falls very quickly when temperature increases. The result is not unexpected; such behavior is well known from our everyday experience. It essentially differs from the conclusion that was derived for ideal gas. The discussion on liquid properties can be used when considering solids. Diffusion in crystals is also defined by the speed of the atom’s wanderings, though it is greatly determined by a crystal’s imperfections. However, time of settled life in crystals is higher by some order of values than in a liquid because of the fact that practically all crystallographic positions are occupied (see Section 9.4). Atomic hopping (i.e., diffusion) is carried out on those points that are most preferable for this purpose: on vacancies, interstitials, along dislocations and other
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places in which the atomic order is destroyed. As a result, the intensity of diffusion in solids is much less than that in liquids. At the same time, the diffusion coefficient also increases exponentially with temperature and in some cases diffusion plays an appreciable role. Viscosity of solids is extremely high (see Table 3.3). For solids of different nature it differs essentially. It also depends on the structure of substances: degrees of crystallinity, the nature of chemical bonding, etc. Heat conductivity of solids also changes over a wide range. Similar to gases, the heat conductivity of liquids and crystals depends on their structure, especially of those particles that play the role of energy carriers. In crystals it can be phonons (refer to Section 9.3, in all solids) and electrons (electron gas in metals). As a result, the heat conductivity of solids changes over a wide range (see Table 3.3). We hope that readers can apply the representations given here to amorphous materials; it should be borne in mind that they have much in common with liquids, but their time characteristics are closer to solids. Unfortunately, there is no general theory allowing calculation of transport properties of condensed systems. This means that when deciding technical problems based on the use of Fick, Fourier and Newton laws, it is also necessary to use empirical laws and factors. However, it should be remembered that the essence of the phenomena and their physical sense remain the same, as for the description of the elementary model of an ideal gas.
9.6
SOME TECHNICALLY IMPORTANT ELECTRIC PROPERTIES OF SUBSTANCES
Ionic polarization The mechanisms of polarization are caused by the structure of molecules (previously considered in Chapter 4). There is a type of crystal polarization that has much in common with atomic polarization (refer to Section 4.2.4). In crystals, atoms are in an ordered array. As a result of interaction between neighboring atoms (i.e., chemical bond formation) redistribution of electron density occurs and atoms acquire an effective charge, positive and/or negative. The presence of the two kinds of atomic charge allows one to consider a crystal lattice consisting of two sublattices inserted into one another. Consider for example a crystal structure of cesium chloride (CsCl) (Figure 9.33). Ions Cs and Cl form two simple cubic sublattices shifted from each other along a spatial diagonal by a distance equal to half of its length. The total crystal electric dipole moment is zero; the crystal is not polarized. When an electric field is imposed, each sublattice is displaced from the other. The crystal then gets an uncompensated electric dipole moment, i.e., polarization of the whole crystal takes place. Such polarization is referred to as ionic polarization. Elasticity forces will be opposite to the sublattices’ displacement; these forces will compensate the action of the electric field. Thus, ionic polarization arises at elastic displacement due to the action of an electric field on positive and negative ions, shifting them from their equilibrium positions. By analogy with atomic polarizability, one can introduce an ionic polarizability with coefficient ion being attributed to each pair of oppositely charged ions. Under the order of value ion coincides with the value of at, i.e., ion1030 m3.
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ⴚ
ⴚ ⴚ
ⴚ ⴙ
ⴙ ⴙ
ⴙ
ⴚ
ⴚ ⴚ
ⴚ ⴙ
ⴙ ⴙ
ⴙ
Figure 9.33 Crystal structure of CsCl.
As a result of sublattice displacement, the polarization of an ionic crystal is accompanied by its deformation—the crystal is extended in the direction of the field. This crystal lengthening does not depend on the direction of the field: if the field direction is changed to the opposite direction, it will still lead to the same lengthening of the crystal. Such a change of crystal sizes by the action of the electric field is referred to as electrostriction. The relative crystal deformation l/l depends quadratic on the electric field strength (l/l ⬃ E2). The value of the sample length change is very small; for example, for SiO2 in a field strength E ⬃ 104 V/m the relative lengthening of a sample is 109. Currently, normal electrostriction has no practical application. However, there are crystals (e.g., Rochelle salt) in which the positive and negative sublattices are found to be asymmetrically disposed in the crystal; therefore these crystals are polarized even in the absence of an external field. Such polarization is referred to as spontaneous (see below). Charges are formed on the edges of spontaneously polarized crystals, however to measure it one should have a fresh chip. In due course ions of an opposite sign from the surrounding atmosphere neutralize the surface charges and prevent charges being found. When the crystal temperature changes, crystal deformation takes place. Consider an edge of a crystal on which the bounded positive charge exists; they are compensated by adsorbed negative ions. With an increase in temperature, crystal polarization will also increase leading to an increase in the density of the bounded charges, and adsorbed negative ions will no longer compensate them. Hence, a positive charge will be spread on this surface. Similarly, the negative charge will be concentrated on the opposite edge. Thus, the potential difference appears at heating between crystal sides. The phenomenon of the occurrence of electric charges on the edges of a crystal with a change in its temperature is referred to as a pyroelectric effect. A piezoelectric effect This effect exhibits the presence of spontaneous polarization of a crystal. G. Curie and P. Curie found that at mechanical compression or stretching of some crystals, the electric charges appeared on their edges. This phenomenon is referred to as a direct piezoelectric
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effect. Later, the opposite effect was also found experimentally, i.e., a reversed piezoelectric effect that consists of crystal deformation by applying an external electric field. At reversed piezoelectric effect, unlike electrostriction, the relative deformation of a crystal l/l depends linearly on the intensity of an electric field (l/l ⬃ E); consequently, if the crystal is extended in some direction of the electric field, the crystal will be compressed in the opposite field direction. In this case, the relative deformation of the crystal is by some order of values larger than at electrostriction. The piezoelectric effect is observed in noncentosymmetric crystals and is especially great in quartz, tourmaline, Rochelle salt, barium titanate, sugar, blende, and in some others. The most important at present is quartz, which has found wide application in practice, for example, in the widely known lighters. Consider the nature of the piezoelectric properties of quartz, SiO2. In this crystal the silicon atom bears a positive charge and the oxygen atom a negative charge. A freely grown quartz crystal represents a hexahedron prism topped with many-sided prisms (Figure 9.34). The axis of the prism Oz is an optical crystal axis. Directions Ox1, Ox2 and Ox3 are electric axes of the crystal; apparently all of them are equivalent. If one cuts out a quartz plate perpendicularly to the optical axis and compresses it along one of the electric axes, bounded charges will appear on the surface of the plate. On stretching, the signs of the charges change to its opposite. On compression or stretching along the optical axis Oz, a piezoelectric effect usually does not appear. Let us find the mechanism by which charges occur on crystal surfaces on deformation. The hexagonal atomic crystal structure is shown in Figure 9.35a where silicon atoms are z
x3
x1 x2
Figure 9.34 Crystal forms of SiO2 single crystal.
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x1 + 6
x1 1
+ −
+
+ x3
6
2
5
4− (a)
3
1
−
−
+
2
x2
x3
+ 5
4−
3
x2
(b)
Figure 9.35 A scheme of disposition of positive (Si) and negative (O) atoms in: (a) a free crystal; (b) a forced crystal.
designated by a “” sign and oxygen atoms are designated by a “” sign. If the crystal is compressed in the direction of one of the electric axes (Figure 9.35b) the Si1 ion will be wedged between ions O2 and O6 and the O4 ion will be wedged between Si3 and Si5 ions. Since in the absence of electric voltage, all charges compensate each other, the introduction of a positive charge of the Si1 ion creates an excessive negative charge on the edge of the crystal, and the displacement of the O4 ion creates an excessive positive charge. Hence, a negative-bounded charge appears on the top surface of the crystal and a positive charge appears on the bottom. On stretching, the signs of the charges will be opposite. The reverse piezoelectric effect can be similarly explained. On bringing a quartz plate into an electric field, it will be deformed due to displacement of the charges; the sign on deformation changes to the opposite when the direction of the field is changed. Direct and reverse piezoelectric effects have found very wide application in practice: for measuring pressure in rapidly proceeding processes, for transformation of electric vibrations in mechanical and in acoustoelectronics, etc. Ferroelectricity There are substances among dielectrics whose dielectric permeability in a narrow temperature interval is extremely high (⬃103 104) and depends nonlinearly on the electric field strength E. Such crystals are referred to as ferroelectrics; the Rochelle salt NaKC4H4O6.4H2O was the first to be discovered in the series by I.V. Kurchatov in the last century. A condenser with such dielectrics between plates has various capacitances, depending on the potential difference applied. The dependence of the polarization ≠ from the value of the electric field strength is given in Figure 9.36. Before application of an electric field, the ferroelectric is not polarized, i.e., ≠ 0 (point O). During the process of increasing the electric strength, the polarization changes according to curve ABC. If we now begin to reduce the strength E, the polarization decreases not along the previous curve CBOA, but along curve CBE. When the field strength falls to zero, the polarization appears to be ≠r.This polarization value is referred to as residual ferroelectric polarization. If the field is changed to the opposite direction and increased, it reduces polarization and at E EF the ferroelectric polarization will vanish to zero. The value of the residual field that removes ferroelectric polarization EF is referred to as a coercive force. With a further
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ᑬ B
ᑬHac ᑬr
C
E A
EF
D F
O
E
G
H
Figure 9.36 A hysteresis loop in ferroelectrics.
field increase the polarization changes along the curve FGH. If after point H, the field strength E again starts increasing in a positive direction to axis E, the polarization will change along the curve HDC. Thus, Figure 9.36 shows that the ferroelectric polarization change detains from the change of the polarizing field. This phenomenon is referred to as hysteresis and the whole closed curve is a hysteresis loop (similar to ferromagnetics). In general polarization, decreases with temperature, and ferroelectricity absolutely disappears at a temperature called the Curie point TC; the crystal transforms to a paraelectric state. The phenomena of magnetization/polarization in ferromagnetics and ferroelectrics have much in common. In the Russian literature, however, another definition is used, i.e., segnetoelectrics, in accordance with the name of the first-discovered compound of that type. Rochelle salt in Russian is called segnetova salt. The phenomenon was discovered by I.V. Kurchatov in 1932. There are many ferroelectric crystals with the common formula ABO3, natural and synthesized, on the basis of the first-discovered perovskite mineral CaTiO3 (Figure 9.37). All constituent atoms can be isomorphic substituted, both in hetero- or homovalence manner, the resulting compound possessing ferroelectric properties. In the paraelectric phase of these crystals, ions of small size (type B, black balls) are in the centers of ideal oxygen octahedrons. At temperatures lower than TC, due to specificity of interatomic interaction and features of atomic thermal vibration, some ions are displaced from their ideal positions, the structure becoming noncentrosymmetric. The ion of type B (or A) can displace along one of three symmetry axes—the second, third and fourth orders. Any of these displacements leads to the occurrence of a spontaneous dipole moment. This takes place simultaneously in a certain area of the crystal. Such areas with a certain macroscopic volume of spontaneous polarization are called domains. If the ferroelectric is not polarized as a whole, the distribution of the domains is random. When an external electric field is imposed the domains turn gradually along the field. Saturation is reached at large fields: all dipole moments of all domains are turned in one direction. Indeed, the picture is quite similar to that described in Section 5.3, which is devoted to ferromagnetism.
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Figure 9.37 A crystal type of perovskite CaTiO3.
All ferroelectrics are also piezoelectrics; therefore they have found wide application in sound and ultrasonic generators, microphones, ceramic condensers, etc. Electrets There are some dielectrics which, in the absence of an external electric field, can preserve the polarized state for a long time. These are referred to as electrets. An example of such a compound is wax, which consists of long molecules with a permanent electric dipole moment. If a small amount of wax is melted and, while it is not yet hardened, is placed in a strong electric field, the wax molecules will partly be oriented by the guidance given along the field; the wax remains in polarized state after it has hardened. Organic dielectrics possess the same property as well (naphthalene, paraffin, ebonite, mica, nylon, etc) and inorganic dielectrics (sulfur, boric glass, titanates of alkali earth materials of the perovskite type, etc.). Electrets are analogs to permanent magnets. They can be produced both as described above (thermoelectrets), or by the action of light on a photoconducting dielectric by the action of a strong electric field (photoelectrets). All electrets have a stable surface charge ,⬃ 100 C/m2. The time of an electret’s life, i.e., the time for which the surface charge will decrease in e time, lasts from several days to many years. Electrets are used as sources of a permanent electric field. They are also used in radio engineering (microphones and phones), in dose metering (electret dosimeters), in measuring techniques (electrostatic voltmeters), in computer facilities (memory elements), etc. Liquid crystals In modern techniques the so-called liquid crystals are gaining increasing importance. The liquid crystal state is characterized as being an intermediate between isotropic liquids and anisotropic solids: in some temperature interval they preserve some properties of a liquid (e.g., fluidity) and some properties of a crystal (e.g., anisotropy). The liquid crystal state is realized more often in substances whose molecules consist of a long flat atomic structure (rods) with included benzene rings (e.g., 4-met-oksi-benziliden-4-butil-aniline). Properties of liquid crystals are defined by different features of molecular packing in
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domains: they are built usually by the long axis along a single direction; whereas azimuthal orientation in different substances can behave differently. Depending on the molecular properties (e.g., the presence of dipole moments and orientation), the existence of the short- or long-range ordering in a molecule arrangement (refer to Section 9.5) and specificities of intermolecular interaction liquid crystals possess the whole scale of macroscopic properties. Practically all liquid crystals are dielectrics with strong anisotropy of electric properties, i.e., their dielectric permeability depends on direction (there are at least two value of : ⊥and 储). In an external electric field the liquid crystal molecules are oriented in such a way that the direction of the maximum dielectric permeability coincides with the external field direction. Almost all liquid crystals are diamagnetic. An exception is made with those substances whose molecules incorporate free radicals with permanent magnetic moments. Anisotropy of electric and optical properties of the liquid crystals, closely connected with anisotropy of their molecular structure, causes variety of their electrooptical properties. The external electric field can render a very strong influence on optical properties of liquid crystals, changing, for instance, their transparency and color. Therefore they have found very wide application as electronic indicator panel in microelectronics (calculators, watches and other devices). Such types of devices require very low voltage (parts of volts) and power ( W). In some cases optical properties of liquid crystals depend very strongly on temperature. Substances possessing such properties find application, for example, in industry and in medicine (for diagnostics of deviation from normal of the thermal mode of body parts).
PROBLEMS/TASKS 9.1. A crystal lattice is depicted in Figure T9.1. Find the Miller indexes of the crystallographic direction given in bold. z
z'
y
y' x x'
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9.2. A crystal lattice is depicted in Figure 9.5e. Find its Miller indexes. 9.3. Using the free electron model, calculate the maximum electron’s velocity in a potassium metal at T 0 K assuming the Fermi energy F 2.02 eV (see Example E9.5) and corresponding electron’s momentum pF. 9.4. Two nanoparticles with linear particle dimensions L1 10a and L2 20a (a is the lattice period) are at the same temperature T D/100, where D is the Debye temperature. Calculate the ratio of the internal energy U2/U1 for the case of the same amount of atoms (the zero energy can be neglected). 9.5. Two nanoparticles with linear particle dimensions L 10a and 20a (a is the lattice period) are at the same temperature T D/100, where D is the Debye temperature. Calculate the ratio of heat capacities C2(T)/C1(T) for the case of the same amount of atoms (the zero energy can be neglected). 9.6. A nanoparticle with linear particle dimensions L 10 a (a is the lattice period) is heated from temperature T D/100 up to T2 D/50, where D is the Debye temperature. Calculate the ratio of the internal energies U2(T2)/U1(T1) and compare the ratio with that of bulk crystal Uⴥ(T2)/Uⴥ(T1). (The zero energy can be neglected.) 9.7. Nanoparticles with linear dimensions L 10a and 20a (a is the lattice period) are heated (in average) from temperature T1 D/100 to T2 D/50 (D is the Debye temperature). Determine how much its heat capacity will change C(T2)/C(T1). Compare the result obtained with the analogous ratio Cⴥ(T2)/Cⴥ(T1) for bulk crystal. 9.8. Determine how much a nanoparticle’s heat capacity at temperature T D/100 (D is the Debye temperature) with linear dimensions L 10a (a is the lattice period) differs from that of Cⴥ of the same bulk crystal after converting to an equal number of atoms. 9.9. There are nanoparticles with linear dimensions L1 10a (a is the lattice period) and L2 20a of the same composition and temperature (T D/100). Estimate the ratio of heat transfer coefficients 1/2 of nanomaterials ( (1/3)C, CV being the heat capacity and is an averaged sound speed; assume that the free path in the nanoparticles is limited by particle dimensions L. 9.10. How many times larger is the ratio of free electrons to one metal atom at T 0 K in aluminum than in copper. The Fermi energy of these two metals is F,Al 11.7 eV and F,Cu 7.0 eV. 9.11. A nanoparticle with linear dimensions L 10a (a is the lattice period) is heated from a temperature T1 D/100 up to T2 D / 50, where D is the Debye temperature. Calculate the internal energies ratio U2(T2) / U1(T1) and compare the ratio with that of bulk crystal (the zero energy can be neglected).
ANSWERS 9.1. Indexes are [1,0,1]. 9.2. Indexes are (1,2,1). 9.3. max 兹2苶苶 F苶/苶 m 8.42 105 m/sec2 and pF 9.11 10318.42105 7.67 25 10 kg m/sec.
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9.4. (U2/U1) 26.2. 9.5. (C2/C1) 16.2. 9.6. [U2N(T2)/U1N(T1)] 405 and [Uⴥ(T2)/Uⴥ(T1)] 8. 9.7. [C(T2)/C(T1)] 130, [Cⴥ(T2)/Cⴥ(T1)] 8.
/
9.8. (Cⴥ CN) 41.5. 9.9. (1/2) 0.03. 9.10. Three times. 9.11. [U2(T2)/U1(T1)]N 405, [U2(T2)/U1(T1)]ⴥ 16.
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Appendix 1 System of Units Used in the Book
Units of physical properties measurements, based on the International System of units (SI) are described below. According to this system the basic units in mechanics are: meter (m), second (sec), kilogram (kg); an additional unit is the radian—a unit of measurement of a flat angle (rad) and that of a solid angle steradian (sr). All other units of the SI system are derived and can be obtained with the help of corresponding transformations: Area unit—square meter (area of a square with a 1 m side) [S] 1 m2 . –Volume unit—cubic meter (volume of a cube with a 1 m side) [V ] 1 m 3 . –Velocity unit—meter per second (velocity of uniform straight line motion; 1 m per 1 sec) [ y ] 1 msec. –Acceleration unit—meter per square second (acceleration when the uniform straight line motion velocity change is 1 m/sec) [ a ] 1 msec 2 . –Frequency unit—second to the power of minus one (at revolution) and hertz (Hz) (unit of oscillation frequency; the frequency when a single periodic process is accomplished in 1 sec) [] 1 sec1 . 581
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System of Units Used in the Book
–Angular velocity unit—1 radian per second (angular velocity when a uniform rotating body turns by 1 rad in 1 sec) [] 1 radsec. –Angular acceleration unit—1 radian per 1 sec in the second power (angular acceleration when angular velocity changes by 1 rad/sec in 1 sec) [] 1 radsec 2 . –Force unit—newton (N) (the amount of force required to give a 1 kg mass body an acceleration of 1 m/sec2) [ F ] 1 N 1 kg msec 2 . –Density unit—kilogram per cubic meter (the density of a uniform substance whose mass per 1 m3 is equal to 1 kg) [] 1 kgm 3 . –Pressure unit—pascal (Pa) (pressure produced by a force of 1 N acting on an area of 1 m2) [ P ] 1 Pa 1 Nm 2 . –Momentum unit—kg m/sec (a body of mass 1 kg moving translational with a velocity of 1 m/sec) [ p] 1 kg msec. –Force impulse—newton second (a force impulse produced by a force of 1 N for 1 sec) [ F t ] 1 N sec. –Work (energy) unit—joule (J) (the amount of work done when an applied force of 1 N moves in the direction of the force through a distance of 1 m) [ A] 1 J 1 N m. –Power unit—watt (W) (a watt is used to measure power or the rate of doing work; 1 W is a power of 1 J per second). [W ] 1 W(Jsec).
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583
–Torque unit—newton meter (moment of force produced by a force of 1 N relative to a point which is at distance of 1 m away from the force action line) [ M ] 1 N m. –Moment of inertia unit—kilogram square meter (moment of inertia of a material point 1 kg in mass relative to a rotation axis 1 m away) [ I ] 1 kg m 2 . –Angular momentum unit—kilogram square meter per second (angular momentum of a body with a moment of inertia in 1 kg m2 rotated with angular velocity 1 rad/sec) [ L ] 1 kg m 2 sec. In addition to the units presented above in molecular physics following units are also used. Unit of heat energy or heat—calorie; 1 cal 4.1868 J [Q] 1 cal. Heat capacity unit—calorie per kelvin (amount of heat to warm a body by 1 K) [C ] 1 calK. –Specific heat capacity unit—calorie per kilogram kelvin (amount of heat to warm 1 kg of a substance by 1 K) [Csp ] 1 calkg K. Mole heat capacity unit—calorie per mole kelvin (amount of heat to warm one mole of substance by 1 K) [Cmole ] 1 calmole K. In addition to the mechanical units of measurements, in the sections describing electricity and magnetism, one basic unit—the ampere (A) and a number of derivative units are used. In the SI system 1 A is defined as the force of that current which produces a specific force between two parallel infinitely long conductors which are 1 m apart, in 2 107 H/m.
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System of Units Used in the Book
Charge unit—coulomb (C) (the charge that runs through a cross-section of a conductor when a current of 1 ampere is flowing) [q ] 1 A 1 sec (Asec) (then an electric current I in 1 A corresponds to 1 coulomb transfer in 1 sec) [I ]
1C 1 sec
(A).
–Current density unit—ampere per square meter (electric current of 1 A per 1 m2 of crosssection of a conductor) [ j]
1A 1 m2
(Am 2 ).
–Electric field strength unit—volt/meter (electric field strength, acting on a point charge 1 C with a force 1 N) [E]
1N 1V 1C 1m
(kg mA sec 3 ).
–Electric displacement (induction) unit—coulomb/meter2 (electric field strength multiplied by 0) [ D] 0 E (A secm 2 ). –Electric field potential unit—volt (V) (electric field potential in which the charge of 1 C possesses potential energy 1 J) [] 1 J1 C (kg m 2 A sec 3 ). –Electric dipole moment unit—coulomb meter (dipole electric moment of a pair of opposite charges equal in value and being 1 m apart) [ p] 1 C 1 m(A sec m). –Electric quadrupole moment—coulomb meter2 (quadrupole electric moment of a system of two pairs of opposite charge equal in value of 1 C displaced alternately in square corners at side length 1 m) [Q ]1 C1 m 2 (A sec m 2 ).
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System of Units Used in the Book
585
Eelectric linear density unit—coulomb/meter (charge of 1 C uniformly distributed along a line of 1 m) [ ] 1 C1 m (A secm). Electric surface charge density unit—coulomb/meter2 (charge of 1 C, uniformly distributed over an area of 1 m2) [ ]
1C 1 m2
(A secm 2 ).
Electric volume charge density unit—coulomb/meter3 (charge, uniformly distributed in a volume of 1 m3) [ ]
1C 1 m3
(A secm 3 ).
–Dielectric polarization unit—coulomb/meter2 (a dielectric’s volumetric dipole moment) [ ¬]
1C 1 m2
(A sec m 2 ).
Dielectric susceptibility unit—dimensionless (polarization of isotropic dielectric in a unit field strength divided by 0) [ ]. Dielectric permeability unit—dimensionless (a value indicating by how much an averaged macroscopic field in a dielectric is less than an external field) [ ]
E0 . E
Polarization of a molecule unit—meter3 (a molecular dipole moment in a field of a unit strength divided by 0) [ ]
p 0 E
(m 3 ).
Electronic capacitance unit—farad (F) (capacitance of conductor, which is charged to potential 1 V receiving a charge of 1 C) [C ]
1C 1V
(A 2 sec 4 (kg m 2 ));
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System of Units Used in the Book
–Magnetic moment unit—ampere meter2 (electric current of 1 A flowing around an area of 1 m2) [M] 1 A 1 m 2 (A m 2 ). –Off-system unit—Bohr magneton (1 = 0.927 10–23 A m2). Magnetic field induction unit—tesla (T) (maximal magnetic force moment, acting on a unit magnetic moment) [ B]
1N 1 A m2
(kg (A sec 2 )).
Strength of magnetic field unit—ampere/meter (magnetic field induction, divided by 0) [H]
1A m
(Am).
Magnetization unit—amper/meter (moment of an unit volume moment) [ (]
1A (Am). 1m
Magnetic susceptibility unit—dimensionless (magnetization of an unit volume of a magnetic in an unit strength field) [ ]. –Specific magnetic susceptibility unit—meter3/kilogram (magnetization of a unit mass of a magnetic in a field of unit strength) [ sp ]
(m 3 kg).
Mole magnetic susceptibility unit—meter3/mole (magnetization of one mole of magnetics in a field of unit strength) [ M ]
M
(m 3 mole).
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587
Magnetic permeability unit—dimensionless (shows how many times greater is the magnetic field than an external magnetic field) []
B . B0
Magnetic flux unit—weber (Wb) (a magnetic field induction flux in 1 T through the surface of unit area) [ ] 1 T 1 m 2 1 Wb; (kg m 2 (A sec 2 )). Inductance unit—henry (H) (inductance of a conductor in which at a current of 1 A appears a total magnetic leakage of 1 Wb) [ L ] 1 Wb1 A; (kg m 2 (A 2 sec 2 )). Some important physical constants: Acceleration of free falling Gravitational constant Avogadro constant Mole gas constant Molar volume at normal conditions Boltzmann constant Elementar charge Electron mass Specific electron charge Light velocity in vacuum Stefan–Boltzmann constant Win shift constant Planck’s constant Rydberg constant First Bohr orbit radius Compton wavelength Bohr magneton Ionization energy of hydrogen atom Atomic unit of mass Nuclear magneton Electric constant Magnetic constant
g 9.81 m/sec2 G 6.67 1011 m3/(kg/sec2) NA 6.02 1023 mol1 R 8.31 J/(K mol) Vm 22.4 103 m3/mol
1.38 1023 J/K ⱍeⱍ 1.60 1019 C me 9.11 1031 kg (e/m) 1.76 1011 C/kg c 3.00 108 m/sec 5.67 108 W/m2 K4 C 2.90 103 m K h 6.63 1034 J sec; h/2 1.051034 J sec R 1.097 107 m1 a 5.29 1011 m C 2.43 1012 m B 9.27 1024 J/T Ei 2.16 1018; J 13.56 eV 1 a.u.m 1.66 1027 kg N 5.05 1027 J/T 0 0.885 1011 F/m 0 1.26 106 H/m
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Appendix 2 Gyroscope Precession in a Gravity Field
A symmetric body with a single motionless point and able to rotate at high angular velocity around an axis z⬘ passing through this motionless point is called a gyroscope. There are two types: balanced (the motionless point coincides with the center of inertia) and unbalanced (where this condition is not fulfilled). A child’s spinning top is a primitive example of an unbalanced gyroscope. Figure A2.1 shows an unbalanced gyroscope acquiring a rotation (precession) in a gravitational field. The pivot point O is a unique motionless point and the axis of rotation z⬘ passes through it. The gravity force is directed vertically downwards along axis z. The angle between axes z and z⬘ is denoted by and is assumed to be small. In the figure, for simplicity, the angular momentum vector L terminates in the center of mass C, the distance OC being lC. According to the basic equation of rotational motion dynamics, we can write dL ⫽ Mdt (refer to (1.3.57)). The force momentum (torque) of the gravitational force relative to point C is M = mglC sin. z z' d
g =
L
d dt
C
dL
mg O
The rotation of the gyroscope’s axis z⬘ relative to the vertical axis z is referred to as gyroscope precession. Under the action of the gravity force momentum the vector of the angular momentum L of the unbalanced gyroscope obtains an increment dL directed along 589
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Gyroscope Precession in a Gravity Field
the vector M (see (1.3.57) and Figure 1.19). Since dL is perpendicular to L the modulus of L is constant and only precession takes place. The gyroscope precession angular velocity g , as seen in Figure A 2.1, can be found as g ⫽
d mgᐉ C sin mgᐉ C ⫽ ⫽ . dt I z⬘ sin I z⬘
(A2.1)
Note that the angle is small and I and regarding both axes, z and z⬘, are approximately same. It can be seen that the precession angular velocity g is higher, the lower the gyroscope’s moment of inertia I and angular velocity z⬘; g does not depend on the angle between axis z and z⬘. Anyone can conduct an experiment using a child’s spinning top.
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Appendix 3 An Electrostatic Field of an Arbitrary Distributed Charge
Among the real problems the chemist can come across in practice, a simple situation with a discrete set of point charges is rarely seen. Any molecule consists of positively charged nuclei encircled by negative electrons, each particle being vibrated around positions of equilibrium. Therefore, the overall charge distribution is described in this case by the distribution function (r): dq ⫽ (r )dV .
(A3.1)
The charge density distribution (r) is of great importance because it permits the calculation of a wide number of molecular and crystal properties and enable us to follow the paths of chemical reactions. Consider a field created by the electric charge system described by the function (r) (refer to Figure A3.1). Our task is to calculate the electrostatic field created by this system in a certain point A. Direct an axis z of the Cartesian coordinate system in such a way that
z A
r
R
dV θ r' y x
591
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An Electrostatic Field of an Arbitrary Distributed Charge
it crosses point A. The electrostatic potential in point A is the superposition of contributions from all elements dq. (r⬘)dV ⬘ (r⬘)dV ⬘ ⫽∫ , R 冷 r ⫺ r⬘ 冨 V V
(r ) ⫽ ∫
(A3.2)
where r is the z component of the radius-vector of point A, r⬘ is the argument of the function (r⬘), ⱍRⱍ = ⱍr–r⬘ⱍ is the distance from the element dV to point A. The integration is over the coordinate r⬘ over the whole charge containing space. Denoting the angle between vectors r and r⬘ as and using the cosine theorem; we obtain R ⫽ (r2 ⫹ r⬘2 ⫺ 2rr⬘ cos)1/2. Then the integral can be rewritten as A ⫽ 冕(r⬘)dV⬘(r2⫹r⬘2 ⫺2rr⬘ cos)⫺1/2. If we calculate the field far from the origin (i.e., r⬘^r) the expression ⫺ 21
2 ⎞⎤ r⬘ 1 1 1 ⎡ ⎛ r⬘ ⫽ 2 ⫽ ⫹ ⫺ 2 cos ⎟ ⎥ 1 ⎢ ⎜ 2 1Ⲑ 2 2 R (r ⫹r⬘ ⫺2rr⬘ cos ) r ⎣⎢ ⎝ r r ⎠ ⎥⎦
can be decomposed into a series and can be expanded over the r⬘ orders
1 3 (1⫹ )⫺1Ⲑ 2 ⫽ 1⫺ ⫹ 2 ⫹ , 2 8 where ⎡ ⎛ r⬘ ⎞ 2 r⬘ ⎤ ⫽ ⎢⎜ ⎟ ⫺2 cos ⎥ . ⎝ ⎠ r ⎢⎣ r ⎥⎦ Summing up all the terms with the same order of r⬘/r and neglecting the terms of higher orders than quadratic, we obtain the expression 2 ⎤ 1 1 ⎡ r⬘ 1 ⫽ ⎢1⫹ cos ⫹ ⎛⎜ r⬘⎞⎟ ⫻ (3 cos2 ⫺1) ⎥ . ⎠ ⎝ R r⎣ r 2 r ⎦
Introducing this expression into eq. (A3.2) and taking into account that integration is accomplished over r⬘, we can obtain for A the sum 1 1 (r⬘)dV ⬘⫹ 2 ∫ r⬘ cos(r⬘)dV ⬘ ∫ r r 1 1 2 ⫹ 3 ∫ r⬘ ⫻ (3 cos2 ⫺1)dV ⬘. 2 r
A ⫽
(A3.3)
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An Electrostatic Field of an Arbitrary Distributed Charge
593
The magnitude of each of these integrals depends only on the properties of the electron density function (r⬘). Being calculated for the given system they can be expressed as numbers k0, k1 and k2 correspondingly. The dependence of A on ⱍrⱍ will be expressed by the sum
A =
k0 k1 k2 ⫹ ⫹ . r r2 r3
(A3.4)
The kn are referred to as the electric moments of the system. Let us analyze each term of this sum. The k0 value is expressed by an integral
k0 ⫽ ∫ (r⬘)dV ⬘.
(A3.5)
and is the total charge of the system concentrated in origin. It is referred to as a monopole moment or simply a monopole. For a neutral system k0 ⫽ 0. The coefficients k1 and k2, unlike k0, depend on charge distribution. The coefficient k1 characterizes an electric dipole moment
k1 ⫽ ∫ r⬘ cos (r⬘)dV ⬘.
(A3.6)
Since the value r⬘cos is z-coordinate of element dV⬘, this term characterizes the relative displacements of the positive and negative charges (r⬘)dV⬘ along this axis. Indeed, if one imagines a system consisting of two dissimilar charges q in points (0,0,z) and (0,0,⫺z) with z ⫽ 1/(2l), then a value r⬘cos⫽⫾(1\2)ᐉ can be factorized from the integral. The resultant expression 冕 (r⬘)dV⬘ will be equal to q and the whole coefficient k1, which is now equal to lq ⫽ p, composes the electric dipole moment oriented along the z-axis (see Section 4.1.5 and eq. (4.1.29)). The coefficient k2 1 2
k2 ⫽ ∫ r⬘2 (r⬘) (3 cos2 ⫺1)dV ⬘.
(A3.7)
is a so-called quadrupole moment. Try to understand what electron density distribution is described by such a factor. For spherically symmetric electron distribution k2 ⫽ 0. It follows from a specific type of k2 factor: keeping in mind that r⬘2 ⫽ x2 ⫹ y2 ⫹ z2 for the specified symmetry all three coordinates are equivalent, therefore x2 ⫹ y2 ⫹ z2 ⫽ 3z2 and, consequently, 3z2 ⫺ r⬘2 ⫽ 0. A flattened out electron density model is the charge q rotating around an z-axis at a level z ⫽ 0 at a distance r0 from the axis. Then ⫽ /2 and the expression in brackets becomes negative. Since r ⫽ const., then k2 is equal to ⫺r 02 q for a positive charge and ⫹r 02 q for a negative charge. It is reasonable to assume that as for
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An Electrostatic Field of an Arbitrary Distributed Charge
every “flattened out” distribution the quadrupole moment has such signs. It is easy to show that for the “extended” model distribution the signs will be the opposite. Expression (A3.4) shows that in the electrostatic field created by a particular system, the electric potential falls differently with distance (refer to Table 8.1) the higher the order of the moment, the sharper the potential falling down. Even neutral systems (atoms, molecules) create an electric field by means of which these systems interact with each other. Accordingly, the higher the order of the moment, the lower the energy of interaction; for example, dipole–dipole interaction is appreciably weaker than the interaction of monopoles (Coulomb interaction). All this information is useful in Chapter 8.
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Appendix 4 Langevin Theorem
Consider a system consisting of identical molecules weakly interacting with each other, characterized by a magnetic dipole moment M. An external magnetic field B acts on the system trying to orient all magnetic moments along z axis contrary to chaotic thermal motion. Allocate in this system a spherical volume V of a radius R. Suppose that there are a large enough number of molecules in this volume with all possible orientations of magnetic moments. Among all N molecules in the volume V we shall denote as dN() such molecules whose magnetic moments form an angle from up to d with a direction of vector B (Figure A4.1).
d R sinθ B z Rd
These dN() molecules account for a physically infinitesimal elementary volume dV(). Then the concentration of such molecules is n
dN . dV
(A4.1)
An equilibrium distribution of noninteracting particles in an internal force field is described by the Boltzmann formula n C exp(U/T ), where C is a normalizing coefficient. Since the potential energy of the dipoles in the external field is determined by a ratio (eq. (5.1.31)) (U MBcos), the equilibrium distribution of dipoles upon potential energies (i.e., on various orientations of the magnetic moments) in field B can be written as: n C exp
冸 T cos 冹 . MB
595
(A4.2)
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Langevin Theorem
From the last two equations we can derive ⎛
⎞
dN () C exp ⎜⎝ M B cos ⎟⎠ dV (). T
(A4.3)
Express an elementary volume dV(), occupied by a molecule the magnetic moment of which is directed at an angle , through the angles and d. Since dV() (V/4)d, where d is an elementary solid angle obtained by two coaxial cones with a common vertex at point 0 and openings of 2 and 2 d, then from the solid angle definition (4.1.10) and Figure A4.1, it follows that
d
dS 2( R sin )( R d) 2 sin d. R2 R2
Then dV() (1/2) V sind and (A4.3) could be rewritten as 1 ⎞ ⎛ MB dN CV exp ⎜ cos ⎟ sin d. ⎠ ⎝ T 2
(A4.4)
The constant C can be determined from the normalizing procedure: the integral from dN() over all possible orientation of angles (from 0 to ) must equal the total number of molecules N in volume V, that is:
∫ dN () N . 0
(A4.5)
To simplify further calculations mark (MB/T) as a and cos as x; then sin d dx. This change brings about the change in limits in (A4.5): instead of the inferior limit there will be 1 (cos 1), and the superior limit will be 1 (cos 1). After such transformation and change of a sign before an integral one arrives at 1
1 CV ∫ e ax dx N . 2 1 After these transformations we obtain
1 e a ea CV N, 2 a
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597
hence
C
2 aN (e a ea )V
2a n, (e a ea )
or C
where n N/V is the total molecules concentration. Then dN() can be rewritten as
dN()
a nVe a cos sin d. e ea a
(A4.6)
These dN() dipoles make a contribution of d(() to the general magnetization (. Taking into account the magnetization definition (5.2.3) d(() we can obtain
d (()
M cos dN ()
V
.
Changing in this expression dN() according to (A4.4), we can obtain
d (()
anM a cos e cos sin d. e ea a
The total magnetization can be found by integration 1
(
anM xe ax a e ea ∫1
dx.
Integration by parts (u x, dv eax dx) can give 1
e ax x e dx x ∫ a 1
+1
1
ax
-1
1 e ax dx. a ∫1
After this integration and substituting the limits: 1
∫
1
xeax dx
e a ea e a ea . a a2
(A4.7)
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Langevin Theorem
Magnetization ( can now be presented as
(
anM ⎛ e a ea e a ea ⎞ ⎟ e ea ⎜⎝ a a2 ⎠ a
⎛ e a ea 1 ⎞ or ( nM⎜ a . ⎝ e ea a ⎟⎠
The expression in brackets is referred to as the Langevin function L(a). Thus L
e a ea 1 . e a ea a
(A4.8)
Using the function L(a) we can finally write the magnetization ( as ( nM L (a )
(A4.9)
At limiting values of a (MB/T ), the L(a) function can be decomposed into the MacLoren series. At small a values we can have 1 1 2 5 L (a) a a 3 a . 3 45 945 This series is alternating-signed and therefore it diminishes rather quickly. If we limit ourselves to the first term, then L(a) (1/3)a. The expression already given can be obtained
(
nM2 B. 3T
The paramagnetic molar magnetic susceptibility at (MB/T ) ^ 1 is obtained M
0 N A M2 . 3T
(A4.10)
This expression coincides with eq. (5.2.23) derived from the very simple suppositions. The volumetric and specific susceptibilities can be calculated according to the formulas given above (refer to Section 5.2.2).
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Appendix 5 Maxwell Equations in Differential Form: Electromagnetic Wave Propagation in Vacuum
Vector algebra provides us with a good opportunity to write Maxwell equations in differential form, i.e., to characterize an electromagnetic field in a point. It allows us to see most clearly the physical sense of the equations and their importance for understanding the laws of electrodynamics. Let us start with the equation of a Gauss law, which in the integral form looks like: 养 DdS ∫ (r )dV. S
V
(A5.1)
Remember that in differential form, the divergence of a vector D in a point r is the limit to which the left-hand side of this equation tends under a contraction S (and V) to a point:
lim
V → 0
1 养 DdS div D(r ). V
(A5.2)
The symbol div means the sum of first particular derivatives. Therefore, ⎛ ⎞ div D(r ) ⎜ ⎟ D( xyz) D(r ). ⎝ x y z ⎠
Here an operator is introduced, well known in mathematics ⎛⎞ ⎛ ⎞ ⎛⎞ i ⎜ ⎟ j⎜ ⎟ k ⎜ ⎟ . ⎝ x⎠ ⎝ z ⎠ ⎝ y ⎠
Correspondingly, divD(r) is the scalar product of an operator and a vector D(r). In fact, the divergence is the flow of D vector “outflows” from a point r. Integrating the divergence 599
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Maxwell Equations in Differential Form
over the whole volume V, we arrive at the total power of the source, i.e., the flow through the closed surface S comprising the volume V.
∫ div D(r )dV 养S DdS.
V
This equation is referred in mathematics as the Ostrogradski theorem, from which an important expression originates: div D(r ) (r ).
(A5.3)
The divergence of a vector of an electric displacement in a point r (x, y, z) is equal to the density of an electric charge (that is the source power of the electrostatic field) in this point. This is the Causs theorem in differential form. It follows that the force lines of an electrostatic field proceed from a positive electric charge (a source), come to an end in a negative electric point charge (a drain) or go to infinity; at the drain divergency has a negative sign. Previous consideration (see Chapter 5) shows that there are no magnetic charges in nature, therefore one can write div B(r ) 0,
(A.5.4)
This expression is also a Maxwell equation. Hence, the magnetic field force lines are closed. The next Maxwellian equation is the law of electromagnetic induction. This equation describes the nature of producing the electric field E by variation of the magnetic field induction B ⎛ B ⎞
∫ E dl∫ ⎜⎝ t ⎟⎠ n dS. L
S
(A5.5)
Note that a rotor of vector E in a point r is the limit of the ratio of the electric field circulation E over the closed contour L, comprises an area S, to the area S while aspiring tightening contour L (and area S) to zero (see Figure 5.11), that is lim
S0
1 ∫ Edl div E(r). S L
(A5.6)
Integrating rot E upon surface S, we obtain circulation of vector E along a contour that comprises this area 养 Ed ᐉ ∫ rot EdS. L
S
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601
In mathematics this equation is referred to as Stokes’ theorem. Comparing this expression with (A5.5), rot E (r )
B(r ) B(r ), t
(A5.7)
that is, the rotor of the electric field strength in point r is equal to the time derivative from the magnetic induction in the same point. This implies that the induction electric field is a curling (vortical) field in contrast to the electrostatic potential field. Using the previous notions, we can write instead of rot E the vector product and E, that is (r). E] [ In the same way the next Maxwellian equation can be derived which connects the circulation of magnetic field strength and currents. It has the form rot H(r) D (r) jcond(r). There are no electric currents in vacuum ( jcond 0), therefore, the equation simplifies to rot H(r ) D (r )
(A5.8)
This means that the source of the magnetic field in the point r is the time changeable electric field (in the same point r). It is expedient to put all equations analyzed above together. div D(r ) (r ), div B(r ) 0
rot E(r ) B (r ) rot H(r ) D (r )
(A5.9)
To these equations it is expedient to add two, which connect the strength of both fields in vacuum and fields in an isotropic medium B 0 H,
D 0 E.
(A5.10)
The last equations are equitable only for isotropic media. In anisotropic media they have a tensor character. The last equation in the Maxwellian system is the relation between the strength of an electric field in a point r with the current density in the same point (Ohm’s law in differential form) j E.
(A5.11)
In these equations all electrodynamics is described! Try to estimate an electromagnetic wave’s propagation speed based on the Maxwellian treatment of electrodynamics. As usual, let us make the task simpler, i.e., we shall analyze a
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Maxwell Equations in Differential Form
certain physical model and look to see how far it corresponds with accepted representations. In this case our problem is the definition of the speed of propagation of an electromagnetic wave in vacuum. The Maxwell equation can be written in the form: ⎛ ⎞ 养 Eᐉ d ᐉ ⎜ B ⎟ , ⎝ t ⎠ L
(A5.12)
⎛ E ⎞ ⎛ D ⎞ ⎛ E ⎞ 0 ∫ ⎜ ⎟ dS 0 0 ∫ ⎜ ⎟ dS. ⎟ ⎝ ⎠ ⎝ t ⎠ n ⎠ t t S s
∫ Bᐉ dᐉ ⎜⎝ L
(A5.13)
Let us imaging an electromagnetic wave as successive steps of “constant” electric and magnetic fields (with vectors E and B fixed in their magnitudes, perpendicular to each other), running along an x-axis with planes of vectors oscillation E in x0y and B in x0z (see Figure A5.1(a)) with a “wave” front motion speed, which we should define. Consider that in some instances, the front of the “wave” reaches line 1–2. Allocate to planes x0y an imaginary rectangular contour defg and estimate the vector B flux through an area limited by the contour (Section 5.1.6, eq. (5.1.38)). In time dt the line 1–2 displaces to positions 4–3. The area A limited by the specified contour is A = c.t h, where h is the length of segment 1–2 and ct is the distance run by the front of the “wave” in time t. As the vector B everywhere is perpendicular to the plane x0y, d B ⎛ dA ⎞ Bhc dt Bhc B⎜ ⎟ ⎝ dt ⎠ dt dt 2 y
c
4
E
(A5.14)
c∆t e
d
2 E
∆A 1
c
h
x
d g
e
f
3
B
w g
x
(a) z
4
c∆t
(b)
B
1
3
f ∆A
In contrast, the circulation of the E vector along the contour 1–2–4–3, i.e., the left-hand side of eqs. (5.1.38, 5.4.6, 5.4.7 and A5.12), is equal to Eh. Therefore
养 Eᐉ d ᐉ Eh, L
(A5.15)
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since E = 0 on the segments 2–4, 4–3 and 3–1 (the wavefront has not yet reached them). From a comparison of eqs. (A5.14), (A5.15) and (A5.12), we can derive the ratio between E and B: (A5.16)
E cB.
Continuing to consider the same model, look what occurs in the plane x0z (Figure A5.1(b)). Again we shall allocate a rectangular contour (into planes x0z) and we shall make the same calculations as in the previous case. Using eq. (5.1.2), circulation of the vector B (A5.13) gives Bw (w is the length of segment 2–4); the vector E flux through area A (that is E) gives 00wct. Time derivation and successive cancellation on w gives B 0 0 Ec.
(A5.17)
If we substitute in this equation the E value from (A5.17) and canceling by B we obtain 00c2 = 1. Whence
c
1 00
.
(A5.18)
Three basic constants of electrodynamics appear connected with each other. Substituting values 0 and 0, we obtain for light speed in vacuum c 2.998 108 km/sec. The agreement of this value with that obtained experimentally was a triumph of Maxwell’ theory. A similar result can be obtained with a more exact model. It is also not so difficult to show that propagation of electromagnetic waves in a medium with dielectric susceptibility and magnetic susceptibility occurs with the speed
y
1 0 0
c
(A5.19)
Since the refraction index is the ratio of the light propagation in a medium to that in a vacuum n
1
(A5.20)
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Glossary of Symbols and Abbreviations
This glossary is intended to free the main text from multiple repetition of the explanation of the notation used. As a rule, an explanation is given on its first occurrence in the text and occasionally elsewhere. Since the vocabulary of physics is very broad, as well as the whole range of Greek letters, the same roman letters have been used several times, partly in different fonts. Vector values are given in bold. Roman letters: A a, a a, b, c B C Csp CM D D d, dhkl d苹 E e, ⏐e⏐ e e E EF F f(υ) f(ε) g g H h h,k,l I Iz i i i, j, k
force work acceleration lattice periods vector of magnetic field induction heat capacity of a body, system specific heat capacity molar heat capacity vector of electric field displacement (induction) diffusion coefficient crystallographic interplanar spacing molecule’s effective diameter vector of electric field strength absolute value of the electron charge (elementary charge) logarithmic natural base (exp) thermal efficiency energy, total mechanical energy Fermi energy vector of force Maxwell molecular velocity distribution molecule kinetic energy distribution free fall acceleration vector gyromagnetic ratio, Lande factor vector of magnetic field strength Planck’s constant Miller indexes electric current, moment of inertia (MI), intensity moment of inertia relative to z-axis number of degrees of freedom imaginary unit unit vectors (orts) of Cartesian coordinate 605
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j k £
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Glossary of Symbols and Abbreviations
vector of energy density wavevector (|k|⫽k⫽2/) kinetic energy, degrees of absolute thermodynamic temperature, performance factor Boltzmann constant L angular momentum’s vector Lz angular momentum relative to axis z l, ᐍ length, distance M molar mass M vector of force moment (torque) Mz force moment relative to axis z M vector of the magnetic dipole moment, electron atomic orbit or spin moment, nuclear magnetic moment m mass of a body, atom, molecule, total system’s mass n concentration n a unit vector of a normal NA Avogadro’s number a molar polarization p vector of the electric dipole moment p vector of momentum, pressure P power Q heat, activation energy Q, q electric charges R(r) a radial part of atomic wavefunction R molar refraction, heat emittance r(x, y, z) Cartesian radius vector r(r, , ) radius vector in a spherical coordinate system S area S entropy s wave polarization index T absolute thermodynamic temperature t, time T period U potential energy U internal energy of a molecular system V volume VM molar volume 〈〉 average speed of a particle rms ⫽ 兹冓苶2苶冔 root mean square of a particle’s speed prob most probable value of a particle’s speed velocity vector W thermodynamic probability Y(,)⫽ ()() angular part of the atomic wave function Z statistical sum
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Greek letters:
⫽/c
〈〉 0 〈osc〉 〈rot〉 o B N (x,y,z)
angle, molecular polarizability angle, a force constant the relative speed coefficient in special relativistic theory coefficient of anharmonicity; adiabatic index attenuation coefficient dielectric susceptibility, a micro-object energy average kinetic molecular energy angular acceleration vector, thermal process efficiency electric constant average energy of atomic oscillations in molecules average energy of molecule rotation coefficient of dynamical viscosity, coefficient of kinematic viscosity dielectric permeability, coefficient of molecular heat transfer wavelength, logarithmic attenuation decrement, mean free path length magnetic permeability, reduced molecular mass magnetic constant Bohr magneton nuclear magneton coefficient of kinematic viscosity, frequency ( ⫽ n speed of rotation) displacement from equilibrium position density of matter, electron density surface charge density unit vector of a tangent system’s time of life, relaxation time, linear charge density angular velocity, = 2 time-independent wave function magnetic flux linkage/time-dependant wave fuction flux angle, electric field potential magnetic susceptibility time-dependent wave function solid angle
Others fonts: ( ℜ ℘
magnetization polarizability (polarization vector) surface density of electrical current
Quantum numbers: n, ᐉ, mᐉ, s, ms quantum numbers of one-electron atom L, mL, J, mJ , S, ms quantum numbers of multielectron atom I, mI nuclear quantum numbers s, p, d ... one-electron states S, P, D ... many-electron atoms states
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Glossary of Symbols and Abbreviations
v j
a molecule vibrational quantum number a molecule rotational quantum number
Abbreviations: MP CM MI IRB IBB SF
material point center of mass moment of inertia ideal rigid body ideal black body superfine
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Index A A, see Ampere Absolute zero temperature 177 Absorption spectrum 492 Acceleration angular 12, 14, 40 average 4 center of mass 12 centripetal 25, 183 constant 9 due to gravity 20, 30, 587 in electric field in simple harmonic motion 108 instant 4 radial (normal) 7 tangential 5, 7, 14 Activation energy 186, 194, 446 Adiabatic process adiabatic index 199 with ideal gas 199 Alpha particle 325 Alternating current 305 Ampere (unit) definition 583 Ampere’s law 318 Ampere–Maxwell law 351 Amplitude 107, 112 Angle of incidence 363 of polarization 390 of reflection 363 of refraction 363 Angular acceleration 13, 14 displacement 12–14 force moment 51 torque 51, 52 frequency 107, 157 momentum 40, 41, 48–50 conservation of 71
orbital 332 spin 460, 498 vector form velocity 15 velocity 13, 14, 41 Antinode 158, 159 Approximations 76, 439 Archimedes’s buoyant force 127 Area, units of 581 Atmosphere (unit) 171 Atmosphere of earth 182, 406 Atom 332 Atomic mass unit 498 Atomic number 460 Avogadro constant 297, 339, 587 B B, see Magnetic field induction Barometric height distribution function 181–182 Beatings 117 Binding energy 512 Biot–Savart law 311 Birefringence 391 Bloch function 539 Bohr atom model 416–419 Bohr magneton 454, 586 Boltzmann constant 176, 219, 587 Boltzmann distribution at different temperatures 179, 181 Boltzmann factor 185–186, 551 Boson 542 Bragg’s law 386 Brewster’s law 388–389 C C, see coulomb Conservation law of charge 251 Capacitor displacement current in 254 parallel-plate 253, 254 609
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Carnot cycle 207–210 engine 211 microcycles 211 Celsius temperature 177 Center of mass acceleration of 12 of two particles 36 rigid body 47 symmetry and 37 Centripetal acceleration 35, 183, 185 Centripetal force 184, 185 Charge interaction force 75 unit 584 Chemical potential 542 Chemical shift 504–506 Chemical shift in ΓRS 514 Chemical shift in NMR 520 Circuit, electrical 307, 308 Critical angle total internal reflection 364 Clausius inequality entropy change 215 in nonequilibrium processes 214–216 Coherence 170 Collision center of mass 80 conservation of 79 conservation of energy in 81 elastic 80 head to head 80, 81 inelastic 88 momentum in 81 one-dimensional 81 Complex variable 110, 111 Conductivity electrical 280, 352 thermal 233, 236 Configurational space 541 Conservation energy law in thermodynamics 196 Conservation laws in collisions 79 of angular momentum 71 of energy 74 of linear momentum 69 of mechanical energy 67 Conservative force 60
Index
Constant-pressure process 198 Constant-volume process 198 Convection 181, 237 Cooper’s pairs 543 Coordinates, space-time 39 Coordination number 567 sphere 567 Coulomb 584 Coulomb’s law 252 and Gauss’ law 259–263 Crystal class 533 Crystal lattice 531–536 periods 556 Crystal structure 531–536 Crystallographic direction 535 Crystallographic plane 533–535 interplanar spacing 535 Curie point 345, 575 Curie temperature 346, 347, 575 Current alternating 305 charge carrier of 251, 280 displacement 350–354 Current density 306, 307 Current drift speed 306 Current loop as magnetic dipole 319 as magnetic moment 348 Cycles direct 205 reversed 205 Cyclic process nonreversible 214–216 reversible, as Carnot cycles 206, 207 D Damped harmonic motion energy 131–133 Damped oscillations 133–138 Debye unit 276 Deceleration 85, 473 Degree of freedom 106 Density electron density 285, 287, 288 linear 251, 263 surface 251, 262, 270, 283 volume, 251
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Index
Deuteron 86, 527 Diamagnetics 331, 337 Dielectric hysteresis loop 575 Dielectrics 280–282 Diffraction by grating 381–383 by single slit 379–381 Fraunhofer 379, 380 Huygens–Fresnel principle 378–379 of electrons 424–426 of neutrons 424–426 types of 379 X-ray 385–386 Diffraction grating 381–383 Diffusion coefficient D 235 Fick’s law 235 Dipole, electric, see Electric dipole Dipole, magnetic, see Magnetic dipole Dipole–dipole interaction 280 Dispersion of grating 383, 384 of light 395–398 Dispersion curves 545–550 acoustic branch 546, 548 optic branch 546, 548 Displacement of simple harmonic motion 108 Displacement current in capacitor 351 Distribution function calculation average value 172–174 Domains, ferroelectric 575 Domains, ferromagnetic 347–349 Doppler effect acoustic 154–156 for light 154 Drift speed of currier 306 E E, see Electric field Earth mass 62 Earth–moon system 99 Efficiency of engine 207, 209, 210 Elastic collision 80, 412 Electrets 576 Electric charge 251, 273, 305
611
Electric current 305–309 Electric charge in dielectrics 281 Electric dipole moment 276 potential energy 278–279 torque on 278 Electric displacement 284, 584 Electric field equipotential surfaces 275 flux of 261 induced 353 induced magnetic field 353 lines of force 253, 254 of electric quadrupole 501 of electromagnetic wave 33 of finite rod 256–257 of infinite cylinder 270 of infinite line 256–257 of infinite plate 263–264 of point charge 253, 254 of ring 258–259 of semi-infinite rod 300–301 of spherically symmetric charge 252, 267, 275 potential 29 Electric field potential capacitor 276 dipole 276–288 distribution of charge 251 of system of charges 274 point charge 274 ring 259 Electric potential superposition principle 274 Electric quadrupole moment 501–502 Electric strength flux 261 Electromagnetic force 29, 31 Electromagnetic induction 328–331 Electromagnetic radiation 353 Electromagnetic spectrum 354 Electromagnetic waves amplitudes of fields 353 energy density in 353 from dipole 353 Poynting vector 353 scale 354 sources of 351 speed of 353
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612
Index
wave equations 352–353 See also Light Electromotive force 308 Electron charge to mass ratio 325 magnetic dipole moment 332 orbital dipole moment 332 spin angular momentum 332 Electrons as quasi-particles 540 Electron paramagnetic resonance (EPR) 526–529 Electrons in crystals 537–545 Electrostatics 251–252 Electrostriction 572 Elevator, weight in 30 EMF, see Electromotive force Energy activation 186, 446, 447 binding 512 conservation 196 quantization of 457 quantized level 437–438 relationship with mass 97 rest 96 SI units of 582 simple harmonic motion 105 Energy density of electromagnetic radiation 353 Engine efficiency of 207 Entropy changes in 213 irreversible process 214–216, 220 reversible process 211, 212, 216, 220 statistical definition 212 Equation of state, ideal gas 175 Equilibrium in thermodynamics 169, 170, 230 Equipotential surface 275 Equipartitioning of energy on degree of freedom 194–195 Extraneous force 308
Fermi level 541 Fermion 541 Ferroelectrics (segnetoelectrics) 574, 575 Ferromagnetism 344–347 Fiber light guides 364 Fine interaction 467–468 Fine structure constant 468 First law of thermodynamics 194–205 in various processes 196, 197 Flux of electric field strength 261 magnetic field induction 328, 329 vector field 259, 328 Force as vector 20, 40 centripetal 185, 321 conservative 60 derived from potential energy 61, 63 dissipative 60 external 28, 52, 53 internal 53 non-conservative 60 unit of 582 Force constant 545, 546 Force fields 58–61 Forced oscillations 138–145 Fraunhofer diffraction 379, 380 Free expansion 196 Free fall acceleration 18, 20, 587 Free path length 235 Frequency and period 107 angular 107 of string vibration 160–161 resonant 140 simple harmonic motion 108 units of 581 Friction non-conservative nature 32, 238 Fresnel diffraction 378–379 Fuel combustion 26–28, 206
F
G
F, see Farad Fahrenheit temperature 177 Farad 585 Faraday 328–330, 587 Faraday’s law, see Electromagnetic induction
Galilean transformation 18–20 Gamma-resonance (Mössbauer effect) 510–513 Gas, ideal 174 Gas constant R, universal 181, 587
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Index
Gauss’ law applied to charge distributions 259–263 Gauss surface 261, 262 General theory of relativity 90–91 Grating, diffraction dispersion of 383 principal maxima of 382 resolving power 383, 384 secondary maxima 382 transmission 385 Gravitational constant 30, 587 Gravitational field gravitational mass 30 gravitational potential energy 61–62, 66 Gravity, acceleration of free fall 18, 20, 587 Gravity of earth 65, 66, 182 moon 182 Ground state 362 Group speed 398 H H, see Henry Harmonic motion damped 133–138 forced simple 138–145 See also Simple harmonic motion Harmonic oscillator 76, 129–131 Heat engine 206–207 pump 210–211 Heat capacity of crystals Debye model 552 Debye frequency 553, 554 Debye function 553, 554 Debye temperature 55, 554 Dulong–Petit law 551 Einstein model 551 Heat capacity of ideal gas in dissociation 202 versus experiment 204–205 Heat transport coefficient 237 Henry 587 Hertz 581 Hooke’s law 31 and potential energy 31, 62 elastic force work 52
613
Huygens’ principle diffraction 378–379 Hydrogen atom 416–419 Hysteresis loop in ferroelectrics 575 in ferromagnetics 348, 349 Hz, see Hertz I Ideal gas adiabatic process 199 and temperature 197 at constant pressure 213 at constant volume 213 average energy 176 general equation 175 heat capacities 197–204 in force field 178–180 internal energy 195 isotherms 200, 207, 208, 211 model 174 pressure 175 Impulse of force 21 Incidence, plane of 388 Index of refraction 297, 361, 363–365, 388, 392, 395, 396 total internal reflection 364, 393 Induced electric field, see Electromagnetic induction Induced EMF 329, 330 Induced magnetic fields 309–313 Inductance mutual 330 of solenoid 331 self 330 units 587 Induction, Faraday’s law of, see Electromagnetic induction Inelastic collision 88–89 Inertia, law of 17 Inertia, rotational 40 Inertial mass 30 Initial conditions, in harmonic motion 107, 108 Intensity in single-slit diffraction 380, 381 wave amplitude 406 Interference, from thin films 370–374
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Internal energy 88, 89, 195, 198 of ideal gas 186, 195, 196 of van der Waals’s gas 226 Internal forces 35, 38, 53, 61, 63, 70 International System of Units, see SI Invariance 68 Invariant quantities 18, 19 Ionic polarization 571–572 Irreversible process entropy change of 214–216, 220 Isotherm 200, 207, 208, 211, 222, 223 Isotope 183, 220, 244, 498 Isotropic material 147, 284, 286, 291, 295, 361, 392, 576 J Joule–Lenz law 328 Joule–Thomson effect 227–229
Index
Laue diffraction 385 Length, relativity 91 Lenz’s law 328 Light energy quanta 408, 409 momentum 412 polarization 386–395 polarized 386 quantization and emission of 387–388 sensitivity of eye to 361 speed of 392 unpolarized 362 visible 361 Linear density 251 Linear motion, with constant acceleration 7 Linear oscillator 129–131 Lines of force, electric field 275 Liquid crystals 576–577 Lorentz transformation 90
K M Kinematics 1–16 Kinetic energy center of mass 55 in simple harmonic motion 131 in transverse wave 152 of rotation 47, 55 relativistic 96 Kinetic theory pressure, ideal gas 175 temperature, ideal gas 177 Kirchhoff’s law 401 Knudsen flow 244 L Lattice defects clusters 563 dislocation 563–566 Burgers vector 564 edge 564, 565 linear 564 screw 564, 565 small angle boundaries 565 interstitials 561, 563 point 561–563 Frenkel 561, 562 Schotky 562 vacancies 561, 563
m, see meter Magnetic dipole 327–328 Magnetic splitting 347 Magnetic dipole moment of iron atom 347 of iron ions 347 Magnetic domain 347–349 Magnetic field circulating charges 305 electric current 305–309 electromagnetic wave 353, 354 flux 336 induction 309–313 induced electric fields 330 strength 307, 308, 310, 311, 320, 334, 342, 347, 348 Magnetic flux 328, 330, 331 Magnetic force between parallel wires 313 on moving charge 320–327 on wire with current 313 Magnetic monopole 310, 351 Magnetically ordered state 344–350 Magnetization 333, 336–344 Magneton Bohr 346, 453, 586, 587 nuclear 499, 516, 587
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Index
Magnets 348 Magnetic field induction 309–318, 586 Mass and energy 96, 97 and weight 30 atomic unit of 587 center of 36–38 equivalence principle 96, 97 gravitational 30 inertial 30 in relativity 91 Mass number 323, 498 Mass spectrometry mass spectrometer 322, 323 Maxwell’s equations 599–603 Maxwell energy distribution function 193–194 Maxwell velocity distribution function average 188, 189 most probable 188, 189 on molecular energy 189 root square mean 188, 189 Maxwell’s law 186–190, 195, 599–603 Mean free path 242 Mechanical energy kinetic 54 potential 61 potential energy curves 74–79 Miller indexes 534 Molar heat capacity at constant pressure 198 at constant volume 198 of ideal gases 198, 213 Molecular speeds 187, 188 Molecular mass 88, 131 Moment of inertia 40 –43 Momentum and Newton’s second law 20–29 angular, see Angular momentum conservation of 69, 71 kinetic energy 68 relativistic 91 velocity of center of mass 38 Moon 182 Mössbauer effect 510–513 Mössbauer spectroscopy 508–516 Mount Everest, potential energy 66 Mutual inductance units of 87 Mutual induction 330
615
N N, see Newton Nanoparticles 555–557 Natural frequency 140, 141, 396 Natural width of spectral line 510 Negative charge 253, 288 Neutral matter 251, 280, 342, 426, 497, 499 Neutron 425, 489, 497, 499 Newton 582 Newton rings 374 Newton’s first law 16–18 Newton’s law of gravity 30 Newton’s second law angular form 21 for particle 21 for system 22 in relativity 21, 95 momentum form 21 simple harmonic motion 119 Newton’s third law 29 Node 158, 159 Nonconservative forces 60 Nuclear forces 497, 499, 500 Nuclear magnetic resonance (NMR) 516–525 proton magnetic resonance (PMR) 518, 521 Nuclear magnetism 516–525 Nuclear physics nucleon model of nucleus 497–499 Nuclear quadrupole resonance (NQR) 525 Nucleon energy levels 499–500 form 500, 501 magnetic moment 499 mass 498 nuclear gyromagnetic ratio 498 nuclear magneton 499 quadrupole moment 501–502 size charge distribution radius 500–501 mass distribution radius 500–501 symmetry 500 Nucleus 332, 497–499 O Ohm’s law 307, 352 Orbital magnetic dipole moment 333, 342, 453
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Order long range 568 short range 568 Oscillation, center of 122 Oscillations damped 133–138 forced 138–148 simple harmonic 106–113 P Pa, see Pascal Parallel axis theorem 43–44 Paramagnetism 332, 340, 343 Pascal 582 Path independence and conservative forces 60 Pendulum physical 121–122 simple, mathematical 119–121 spring 118–119 Performance, thermal coefficient of 210 Period and frequency 107 of linear oscillator 119, 120, 122 of simple harmonic motion 132 Permanent magnetism 348 Permeability constant 309–310 Permittivity constant 293 Perovskite type crystal 575, 576 Perpendicular axis theorem 45 Phase of simple harmonic motion 107 Phase changes, on reflection 160, 368 Phase difference, in interference 370 Phase speed 151, 398, 546 Photo-effect external Einstein photo-effect law 407, 408 Physical kinetics collision cross-section 232 effective diameter at collisions 231 relaxation process, time of 230–231 Physical model 35 Physical pendulum 121–122 Physics, quantum 33, 479, 490, 510, 541 Piezoelectric effect 572–574 Piezoelectrics 572–574 Planck constant 404, 587 Planes, mirror 35 Plane motion 2 Plane of incidence 388
Index
Plane polarization 390 Polar material 280, 284, 291, 297 Polarization dielectric 286–292 reflection 388–389 Polarization of electromagnetic waves 362, 386–394 Polarized light plane 387, 389 refection 388–389 Polarizer 387–388 Polarizing angle 390 Postulate relativity 90 speed of light 90 Potential, electric, see Electric field potential Potential energy and work 61 electric 75 electric dipole 278 force 63 gravitational 62, 74 magnetic dipole 327 simple harmonic oscillator 131 Potential–energy curve 74–79, 186, 482 Potential difference 276 Power average 53 instantaneous 54 units of 582 Poynting vector 353 Precession 51, 337, 338, 477, 589–590 Pressure, kinetic theory of 175, 176 Prism 364, 365, 384, 391 Process adiabatic 199 cyclic 205 constant volume 198 irreversible 206 isobaric 198 reversible 206 Projectile motion 9 Proton 497–499 Pyroelectric effect 572 Q Quadrupole moment 501, 584, 593 Quadrupole resonance 525 Quadrupole splitting 506
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Quantization at emission of light 512 electron energy in atom 457–460 of angular momentum 455, 456 Quantum mechanics 423–496 Quartz crystal oscillator 137 R R (universal gas constant) 181, 487 Radial distribution function 567 Radian 581 Radiation, blackbody 398–402 Ratio of specific heats for gases 199 Ray 147 Rayleigh’s criterion 384 Real gas approximation corresponding states law 225 critical point 223 internal energy of 226 Joule–Thomson effect 227–229 inversion curve 229 inversion point 229 phase diagram of state 223, 224 van der Waals gas 221–226 real isotherms of 223 van der Waals equation 221, 222 in reduced parameters 225 Recoil 511, 512 Reduced amount of heat: entropy change in isobaric process 213 in isochoric process 213 Reference frames 21, 97 Reference system (frame) 1 Inertial 16 Noninertial 33 Reflection law of 363 phase change 368 polarization 388–389 thin film 371, 372 total internal 364 Refraction Huygens’ principle 378–379 index of 361, 363, 365 law of 363 Refrigerator 210–211 Relativity motion at high speed 90 speed summation law 90
617
Shortening of length 91 dilation of time 94 simultaneity 94 Resolving power of grating 384 Resonance 140, 497–529 Resonance absorption 508–510 Rest mass 426 Reversible process, entropy change in 213 Rigid body angular momentum and velocity 41 Rocket, propulsion 26, 27 Rolling motion, energy of 57 Root-mean-square molecular speed 188, 189 Rotation angular momentum 41 analogy with translation motion 50, 51 with constant acceleration 15 Rotational motion analogies to translational motion 50 kinematics 12 dynamics 16 Rotation of plane of 389–391 Rotator, rigid 47, 450 S Saturation ferromagnetic 349 paramagnetic 341 Second law of motion, see Newton’s second law Second law of thermodynamics 205–221 entropy 211–214 Selection rules 479, 480 Self-induction 331, see also Inductance Semiconductors acceptor type 544 donor type 544 intrinsic type 544 n-type 544, 545 p-type 544, 545 SI 581 Simple harmonic motion acceleration 108 amplitude 107 angular frequency 107 displacement 107
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equation of motion for 107 kinetic energy of 131 period of 106 phase 107 Simple harmonic oscillator period of 131 Single-slit diffraction 379–381 Snell’s law 363 Solenoid, magnetic field of 319, 320 Solid-state dynamics Born–Karman chain 545 Special relativistic theory 90–97 Specific heat capacity 197, 198, 583 Speed angular 57, 73 average 3, 8, 307, 570 molecular average 188 distribution of 186–190 most probable 188 root-mean-square 188 radial acceleration 14 speed and acceleration 95 Speed of light 603 Spherical symmetry 252, 268, 269, 270, 500 Spin, quantum number 460, 469, 480, 498, 517, 518, 527, 541, 542 Spring scale 30 Spring, force law 52 Spring, potential energy 62 Standing waves 157–160 Statistical thermodynamics 219–220 Statistics Bose–Einstein 542 classic 541 distribution function 541 Fermi–Dirac 541 quantum 540–543 Steradian 261, 581 String, waves on 160–161 Superconductivity 543 Superfine dipole–dipole interaction 523 Superfine interaction 515–516 Superposition principle 156, 157, 274, 277 Surface charge 576 Surface current 335 Symmetry operation 531 Syngony 533
Index
T T, see Tesla Tangent, unit vector of 2 Tangential acceleration 14 Temperature in kinetic theory 177–178 Temperature scale Celsius 177 Fahrenheit 177 Kelvin 177 Réaumer 177 Tesla 586 Thermal conductivity 233, 236 Thermal energy 197, 198, 206, 406 Thermal equilibrium 169, 407 Thermodynamics first law of 194–205 second law of 205–221 zeroth law of 170 Thermodynamic process 170, 206 Third law of motion 29 Thrust, of rocket 27, 28 Time of settled life 568 Time dilation 94 Torque and angular acceleration 48 and angular momentum 49 on electric dipole 278 on magnetic dipole 327 Total internal reflection 364 Trajectory of projectile 72 Transformation Galilean 18–20 Lorentz 90 space time 90 velocity 91 Translation 531, 532 Translational motion and rotational motion 50, 51 Transport phenomena in ideal gases 233–235 flow of G property 234 macroscopic representations 233–235 U Unit cell body-centered cell (BCC) 535
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face-centered cell (FCC) 535 primitive unit cell 535 Unit vector 2 V V, see Volt Vacuum heat transfer in vacuum 243–244 Velocity angular 13, 14 average 3 in simple harmonic motion 107 linear 14, 41 transformation 18, 19 Velocity space 187 Viscosity (internal friction) dynamical coefficient 239 kinematical coefficient 240 Volt 584 Voltage, see Electromotive force Volume, units of 581 W Water molecule 3, 45, 46, 170 Waves electromagnetic, see Electromagnetic waves interference 369–377 longitudinal 146 mechanical 145 sinusoidal 159 standing 157–160 string 160–163 transverse 146 transverse simple harmonic, energy of 151–154 transverse standing equation of 158 traveling, equation of 147–151
619
Wave equations 148, 150 Wave number 149, 492 Wave speed 155 Wavelength 148 Wavelength of light 354 Wavetrain 362 Weber 587 Weight 30 Wire magnetic field of 313 magnetic force on 316 Work–energy theorem 63 Work of gas in adiabatic process 199 in elementary processes 226 in isobaric process 198 in isochoric process 198 in isoprocesses 197–201 in isothermic process 200, 207 X X-rays bremschtralung 410 characteristic 410 X-ray diffraction 385–386 Z Zeeman effect 477–480 anomalous 477, 478 normal 477, 478, 480 nuclear 478 Zeroth law of thermodynamics 170 Zone theory conductivity band 539 energy bands 539 forbidden energy gap 539 hybrid band 539 valence energy band 539
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