ONS MANUAL
INCLUDES:
--'.I-Step Solutions to 25% of the text's
""'I"I""'''-Chapter Problems 'r
I!lMIi tions in the same two-column format as the worked examples in the text and the Study Guide •
Carefully rendered art to help you visualize each ...r()DI�em
STUDENT SOLUTIONS MANUAL TO ACCOMPANY
PHYSICS FOR SCIENTISTS AND ENGINEERS FIFTH EDITION
This solutions manual doesn't just give you the answers. It shows you how to work your way through the problems. The Solutions Manual includes: • •
•
Detailed step-by-step Solutions to 25% of the text's end-of-chapter Problems Solutions in the same two-column format as the worked examples in the text and the Problems and Solutions in the Study Guide Carefully rendered art to help you visualize the Problem and Solution
Other great resources to help you with your course work: Study Guide Gene Mosca, United States Naval Academy Todd Ruskell, Colorado School of Mines Vol. 1,0-7167-8332-0,Vol. 2,0-7167-8331-2 •
Begins with review of Key Ideas and Equations for each chapter
•
Tests your knowledge of the chapter's material with True and False Exercises and Short Questions and Answers
•
Provides additional Problems and Solutions to help you master your understanding of the chapter content
Be sure to visit the Tipler/Mosca Student Companion Web site at: www.whfreeman.com/tipler5e Accessible FREE OF CHARGE, the site offers: •
Online Quizzing
•
IIMaster the Concept" Worked Examples
•
Concept Tester Interactive Simulations
•
Concept Tester Quick Questions
•
Solution Builders
•
Applied Physics Video Clips
•
Demonstration Physics Video Clips
Student Solutions Manual for Tipler and Mosca's
Physics for Scientists and Engineers Fifth Edition Volume 2
DAVID MILLS Professor Emeritus College of the Redwoods with
CHARLES L. ADLER Saint Mary's College of Maryland EDWARD A. WHITTAKER Professor of Physics Stevens Institute of Technology GEORGE ZOBER Yough Senior High School PATRICIA ZOBER Ringgold High School
W. H. Freeman and Company
New York
Copyright
© 2004 by W. H. Freeman and Company
All rights reserved. Printed in the United States of America ISBN: 0-7167-8334-7 First printing 2003 W. H. Freeman and Company
41 Madison Avenue
New York, NY 10010 Houndrnills, Basingstoke RG21 6XS, England
Contents To the Student, v Acknowledgments, vii About the Authors, ix Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter
21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41
The Electric Field I: Discrete Charge Distributions,
1
The Electric Field II: Continuous Charge Distributions, Electric Potential,
51
Electrostatic Energy and Capacitance,
71
Electric Current and Direct-Current Circuits, The Magnetic Field,
135 189
29
97
159 219
Sources of the Magnetic Field, Magnetic Induction,
Alternating-Current Circuits,
Maxwell's Equations and Electromagnetic Waves,
275 291
Properties of Light, Optical Images,
Interference and Diffraction,
315
253
337 353
Wave-Particle Duality and Quantum Physics, Applications of the Schrodinger Equation, Atoms,
363 379 407
Molecules,
Solids and the Theory of Conduction, Relativity,
Nuclear Physics,
429
393
Elementary Particles and the Beginning
of the U ni verse,
447
To the Student This solution manual accompanies Physics
for Scientists and Engineers, 5e, by
Paul Tipler and Gene Mosca. Following the structure of the solutions to the Worked Examples in the text, we begin the solutions to the back-of-the-chapter numerical problems with a brief discussion of the physics of the problem, represent the problem pictorially whenever appropriate, express the physics of the solution in the form of a mathematical model, fill in any intermediate steps as needed, make the appropriate substitutions and algebraic simplifications, and complete the solution with the substitution of numerical values (including their units) and the evaluation of whatever physical quantity is called for in the problem. This is the problem-solving strategy used by experienced learners of physics, and it is our hope that you will see the value in such an approach to problem solving and learn to use it consistently. Believing that it will maximize your learning of physics, we encourage you to create your own solution before referring to the solutions in this manual. You may find that, by following this approach, you will find different, but equally
valid, solutions to some of the problems. In any event, studying the solutions
contained herein without having first attempted the problems will do little to help you learn physics. You'll find that nearly all problems with numerical answers have their answers given to three significant figures. Most of the exceptions to this rule are in the solutions to the problems on Significant Figures and Order of Magnitude and the problems dealing with nuclear physics. When the nature of the problem makes it desirable to do so, we keep more than three significant figures in the answers to intermediate steps and then round to three significant figures for the final answer. Some of the Estimation and Approximation Problems have answers to fewer than three significant figures.
Physics for Scientists and Engineers, 5e includes numerous spreadsheet problems. Most of them call for the plotting of one or more graphs. The solutions to these problems were generated using Microsoft Excel and its "paste special" feature, so that you can easily make changes to the graphical parts of the solutions.
v
Acknowledgments Charles L. Adler (Saint Mary's College of Maryland), Ed Whittaker (Stevens Institute of Technology, George Zober (Yough Senior High School) and Patricia Zober (Ringgold High School) are the authors of the new problems appearing in the Fifth Edition. Chuck, Ed, George, and Patricia saved me (dm) many hours of work by providing rough-draft solutions to these new problems, and I thank them for their help. Gene Mosca (United StatesNaval Academy and the co-author of the Fifth Edition) helped me tremendously by reviewing my work, helping me clarify many of my solutions, and providing solutions when I was unsure how best to proceed. It was a pleasure to collaborate with Gene in the creation of this solutions manual. All of us who were involved in the creation of this solutions manual hope that you will find the solutions useful in learning physics. We want to thank LayNam Chang (Virginia Polytechnic Institute), Brent A. Corbin (UCLA), Alan Cresswell (Shippensburg University), Ricardo S. Decca (Indiana University-Purdue University), Michael Dubson (The University of Colorado at Boulder), David Faust (Mount Hood Community College), Philip Fraundorf (The University of Missouri-Saint Louis), Clint Harper (Moorpark College), Kristi R. G. Hendrickson (University of Puget Sound), Michael Hildreth (The University ofNotre Dame), David Ingram (Ohio University), James J. Kolata (The University ofNotre Dame), Eric Lane (The University of Tennessee Chattanooga), Jerome Licini (Lehigh University), Laura McCullough (The University of Wisconsin-Stout), Carl Mungan (United StatesNaval Academy), Jeffrey S. Olafsen (University of Kansas), Robert Pompi (The State University of New York at Binghamton), R. J. Rollefson (Wesleyan University), Andrew Scherbakov (Georgia Institute of Technology), Bruce A. Schumm (University of Chicago), Dan Styer (Oberlin College), Daniel Marlow (Princeton University), Jeffrey Sundquist (palm Beach Community College-South), Cyrus Taylor (Case Western Reserve University), and Fulin Zuo (University of Miami), for their reviews of the problems and their solutions. Jerome Licini (Lehigh University), Michael Crivello (San Diego Mesa College), Paul Quinn (University of Kansas), and Daniel Lucas (University of Wisconsin-Madison) error-checked the solutions. Without their thorough and critical work, many errors would have remained to be discovered by the users of this solutions manual. Their assistance is greatly appreciated. In spite of their best efforts, there may still be errors in some of the solutions, and for those I (dm) assume full responsibility. Should you find errors or think of alternative solutions that you would like to call to my attention, I would appreciate it if you would communicate them to me by sending them to
[email protected].
Vll
It was a pleasure to work with Brian Donnellan, Media and Supplements Editor for Physics, who guided us through the creation of this solution manual.
Our thanks to Amanda McCorquodale and Eileen McGinnis for organizing the
reviewing and error-checking process. September
2003
David Mills
Professor Emeritus College of the Redwoods Charles
L. Adler
Saint Mary's College of Maryland Edward A. Whittaker
Professor of Physics Stevens Institute of Technology George Zober
Yough Senior High School Patricia Zober
Ringgold High School
\ YIll
About the Authors David Mills,
Professor Emeritus, College of the Redwoods, retired in May of
2000 after a teaching career of 42 years. He earned his bachelor's degree at
Humboldt State College, his master's degree at California State University-Hayward, and his doctoral degree at the University of Northern Colorado. His teaching career included experience with the Physical Science Study Committee materials, the Harvard Project curriculum, the Personalized System of Instruction, Microcomputer-Based Laboratory instruction, and the interactive-engagement movement in physics education. A 1996 NSF.lLI grant allowed him to transform instruction in physics at the College of the Redwoods from a traditional lecture-laboratory delivery system to one that was microcomputer based, eliminate the distinction between lecture and laboratory, and utilize interactive-engagement teaching and learning strategies. He authored the Test Bank to accompany Physics for Scientists and Engineers, 3e and 4e. He now lives in Henderson, NY and is an Adjunct Professor at the Community College of Southern Nevada.
Charles L. Adler is a professor of physics at St. Mary's College of Maryland. He received his undergraduate, masters, and doctoral degrees in physics from Brown University before doing his postdoctoral work at the Naval Research Laboratory in Washington, D.C. His research covers a wide variety of fields, including nonlinear optics, electrooptics, acoustics, cavity quantum electrodynamics, and pure mathematics. His current interests concern problems in light scattering, inverse scattering, and atmospheric optics. Dr. Adler is the author of over 30 publications. Edward A Whittaker
has been a professor of physics at Stevens Institute of Technology since 1984. His research interests include laser spectroscopy, quantum optics, and optical communications. In 2003 he was named an American Institute of Physics State Department Science Fellow.
George Zober
is a teacher of Advanced Placement Physics at Yough High School in western Pennsylvania. He serves as a Physics Consultant with the College Board and teaches Advanced Placement Physics Workshops at Wilkes University and Manhattan College during his summers. Patricia J. Zober
teaches Advanced Placement Physics at Ringgold High School in Monongahela, Pennsylvania and is a Physics Consultant for the College Board. Patricia presents Advanced Placement Workshops in Physics and during her summers is a faculty physics project leader with the Governor's School for the Sciences at Carnegie Mellon University.
IX
Chapter 21 The Electric Field 1 : Discrete Charge Distributions Conceptual Problems *1
••
Discuss the similarities and differences in the properties of electric charge
and gravitational mass. Similarities:
Differences:
The force between charges and
There are positive and negative charges but
masses varies as lIr2.
only positive masses.
The force is directly proportional to
Like charges repel; like masses attract.
the product of the charges or masses. The gravitational constant G is many orders of magnitude smaller than the Coulomb constant k. *5
••
Two uncharged conducting spheres with their conducting surfaces in contact
are supported on a large wooden table by insulated stands. A positively charged rod is brought up close to the surface of one of the spheres on the side opposite its point of contact with the other sphere.
(a) Describe the induced charges on the two conducting
spheres, and sketch the charge distributions on them. (b) The two spheres are separated far apart and the charged rod is removed. Sketch the charge distributions on the separated spheres. Determine the Concept Because the spheres are conductors, there are free electrons on
them that will reposition themselves when the positively charged rod is brought nearby.
(a ) On the sphere near the positively charged rod, the induced charge is negative and near the rod. On the other sphere, the net charge is positive and on the side far from the rod. This is shown in the diagram.
+
(b) When the spheres are separated and far
+
+
+
+
o
apart and the rod has been removed, the induced charges are distributed uniformly over each sphere. The charge distributions are shown in the
1
)
o
Chapter 21
2
diagram. *7
A positive charge that is free to move but is at rest in an electric field E will
(a) accelerate in the direction perpendicular to E . (b) remain at rest. (c) accelerate in the direction opposite to E. Cd) accelerate in the same direction as E. ( e) do none of the above. Determine the Concept The acceleration of the positive charge is given by ii
=
F m
=
E. Because qo and
!l.!!.. m
direction as the electric field. *8
•
m
are both positive, the acceleration is in the same
I (d) is correct. I
If four charges are placed at the corners of a square as shown in Figure 21-
33, the field E is zero at
(a) all points along the sides of the square midway between two charges. (b) the midpoint of the square. (c) midway between the top two charges and midway between the bottom two charges. (d) none of the above. -q
Figure 21-33 Problem 8
,0
}------{:+ +q
+q t;'-}: ------{, -
-q
Determine the Concept E is zero wherever the net force acting on a test charge is zero.
At the center of the square the two positive charges alone would produce a net electric field of zero, and the two negative charges alone would also produce a net electric field of zero. Thus, the net force acting on a test charge at the midpoint of the square will be zero.
I (b) is correct. I
*11
•
Two charges +q and -3q are separated by a small distance. Draw the electric
field lines for this system.
The Electric Field 1: Discrete Charge Distributions
3
Determine the Concept We can use the
rules for drawing electric field lines to draw the electric field lines for this system.
-3q
In the field-line sketch to the right we've
+q
assigned 2 field lines to each charge q.
*12
•
Three equal positive point charges are situated at the comers of an equilateral
triangle. Sketch the electric field lines in the plane of the triangle. Determine the Concept We can use the
rules for drawing electric field lines to draw the electric field lines for this system. In the field-line sketch to the right we've assigned 7 field lines to each charge q.
*14
·
The electric field lines around an electrical dipole are best represented by
which, if any, of the diagrams in Figure 2 1-34?
(c)
(d)
Figure 21-34 Problem 1 4 Determine the Concept Electric field lines around an electric dipole originate at the
positive charge and terminate at the negative charge. Only the lines shown in (d) satisfy this requirement. *15
··
I (d) is correct. I
A molecule with electric dipole moment
p
is oriented so that
p
makes an
angle e with a uniform electric field E that is in the direction of increasing x. The dipole is free to move in response to the force from the field. Describe the motion of the dipole. Suppose the electric field is nonuniform and is larger in the x direction. How will the motion be changed?
4
Chapter 21
Determine the Concept Because 8* 0, a dipole in a unifonn electric field will
experience a restoring torque whose magnitude is pEx sin e . Hence it will oscillate about its equilibrium orientation, 8= o. If 8« 1 , sin8� e, and the motion will be simple hannonic motion. Because the field is nonunifonn and is larger in the x direction, the
force acting on the positive charge of the dipole (in the direction of increasing x) will be greater than the force acting on the negative charge of the dipole ( in the direction of decreasing x) and thus there will be a net electric force on the dipole in the direction of increasing x. Hence, the dipole will accelerate in the x direction as it oscillates about 8 =0. *18
••
A metal ball is positively charged. Is it possible for it to attract another
positively charged ball? Explain. Determine the Concept Yes. A positively charged ball will induce a dipole on the metal
ball, and if the two are in close proximity, the net *19
force can be attractive.
··
A simple demonstration of electrostatic attraction can be done simply by tying a small ball of tinfoil on a hanging string, and bringing a charged wand near it. Initially, the ball will be attracted to the wand, but once they touch, the ball will be repelled violently from it. Explain this behavior. Determine the Concept Assume that the wand has a negative charge. When the charged
wand is brought near the tinfoil, the side nearer the wand becomes positively charged by induction, and so it swings toward the wand. When it touches the wand, some of the negative charge is transferred to the foil, which, as a result, acquires a net negative charge and is now repelled by the wand. Estimation and Approximation *23
··
A popular classroom demonstration consists of rubbing a "magic wand" made of plastic with fur to charge it, and then placing it near an empty soda can on its side ( Figure 2 1 -35). The can will roll toward the wand, as it acquires a charge on the side nearest the wand by induction. Typically, if the wand is held about 1 0 cm away from the can, the can will have an initial acceleration of about 1 rnIs2• If the mass of the can is 0.0 1 8 kg, estimate the charge on the rod.
Soda can
Figure 21-35 Problem 23
The Electric Field 1: Discrete Charge Distributions
5
Picture the Problem We can use Coulomb's law to express the charge on the rod in terms of the force exerted on it by the soda can and its distance from the can. We can apply Newton's 2nd law in rotational form to the can to relate its acceleration to the electric force exerted on it by the rod. Combining these equations will yield an expression for Q as a function of the mass of the can, its distance from the rod, and its acceleration.
Use Coulomb's law to relate the force on the rod to its charge Q and distance r from the soda can: Solve for Q to obtain: ( 1)
L reenter
Apply can:
of mass
=fa to the
Because the can rolls without slipping, we mow that its linear acceleration a and angular acceleration a are related according to: Because the empty can is a hollow cylinder: Substitute for I and Fto obtain:
a and
FR=fa a
a=R where R is the radius of the soda can.
f=MR2
where M is the mass of the can.
solve for
Q� r7Q
Substitute for Fin equation ( 1 ):
Substitute numerical values and evaluate Q:
Q=
= I 14l nC I
Electric Charge *27
·
How many coulombs of positive charge are there in 1 kg of carbon? Twelve
grams of carbon contain Avogadro's number of atoms, with each atom having six protons and six electrons. Picture the Problem We can find the number of coulombs of positive charge there are in 1 kg of carbon from
Q = 6nce,
where
nc is the number of atoms in
1 kg of carbon and the factor of 6 is present to account for the presence of 6 protons in
6
Chapter 21
each atom. We can find the number of atoms in lkg of carbon by setting up a proportion relating Avogadro's number, the mass of carbon, and the molecular mass of carbon to nco Express the positive charge in terms of the electronic charge, the number of protons per atom, and the number of atoms in 1 kg of carbon: Using a proportion, relate the number of atoms in 1 kg of carbon
nc, to Avogadro's number and the
nc
-
N
A
m
-�-n ---.' c
M
-
-
NA mC M
molecular mass M of carbon: Substitute to obtain:
Substitute numerical values and evaluate Q:
Q
=
6(6.02 X 1023 atoms/mol)(1 kg)(1.6 x 10-19 C) = 1 4.82 X 107 C 1 0.012 kg/mol
Coulomb's Law *32
••
A point charge of -2.5 jiC is located at the origin. A second point charge of 6
jiC is at x = 1 m, Y = 0.5 m. Find the x and y coordinates of the position at which an electron would be in equilibrium. Picture the Problem The positions of the
charges are shown in the diagram. It is
Y,m
apparent that the electron must be located along the line joining the two charges. Moreover, because it is negatively charged, it must be closer to the -2.5 jiC than to the 6.0 jiC charge, as is indicated in the figure. We can find the x and y coordinates of the electron's position by equating the two electrostatic forces acting on it and solving for its distance from the origin. We can use similar triangles to express this radial distance in terms of the x and y coordinates of the electron.
The Electric Field 1: Discrete Charge Distributions
7
Express the condition that must be satisfied if the elech-on is to be in equilibrium: Express the magnitude of the force that ql exerts on the electron: Express the magnitude of the force that q2 exerts on the electron: ql
Substitute and simplify to obtain:
_
l q2 1
�+JL25mY -7 Substitute for ql and q2 and simplify: Solve for r to obtain:
(-L4m-2) r2 + (22361m-1) r +125m = 0 r=2_036m and
r=-OA386m Because r < 0 is unphysical, we'll consider only the positive root Use the similar triangles in the diagram to establish the proportion involving the y coordinate of the
2_036m O_5m L12m
�
electron: Solve for
Ye:
Use the similar triangles in the diagram to establish the proportion involving the x coordinate of the
Ye =O_909m Xe 2_036m 1m L12m _
electron: Solve for
Xe:
The coordinates of the electron's position are:
Xe =L82m (xe,yJ=1 (-L82m,-O.909m) I
8
Chapter 2 1
*33
··
A charge o f -1.0 ;...tC is located at the origin; a second charge o f 2.0 ;...tC is
located atx = O,y
=
0. 1 m; and a third charge of 4.0;...tC is located atx = 0.2 m, y = O.
Find the forces that act on each of the three charges. Picture the Problem Let ql represent the
charge at the origin,
Y,m qz
q2 the charge at (0, 0. 1
m), and q3 the charge at ( 0.2 m, 0). The diagram shows the forces
F1.2
acting on each of the charges. Note the action-and-reaction pairs. We can apply
=
"
2 p.C
"
"
FZ•1
"
Coulomb's law and the principle of
"
"
"
superposition of forces to find the net force acting on each of the charges. Express the net force acting on q1 : Express the force that q2 exerts on ql:
Substitute numerical values and evaluate
F",
�
F2,1 :
(8.99xI0' N ·m'/C' )(2pC) tO.ll�! m (-O.lm)] (1.80N)] �
i:
Express the force that q3 exerts on ql:
Substitute numerical values and evaluate
F",
�
3,1
=
kq3ql r r3,13 3,1
F3 1 :
(8.99x10' N ·m'/C' )(4pC) tO.2IpCm J (-0.2m)i (0.899N)i F; : i; =\ (O.899N)i (1.80N)) I �
Substitute to find
Express the net force acting on q2:
+
-
-
-
F2 = F3,2 + F;,2
The Electric Field
1:
Discrete Charge Distributions
9
because F.,,2 and F2,I are action-and-reaction forces. Express the force that
q3 exerts on q2:
F32, =kq33q2 r3,2 r32, =kq�q2 r32,
[(-O.2m)i (O.lm)J]
Substitute numerical values and evaluate
F",
+
F3,2 :
(8,99 10' N, )(4 jiC) (0.(222jiC4m))' [(- 0,2m)i (0, lm)j1 =(-1.28N)i (0.640N)J
�
m 'IC '
X
+
+
Find the net force acting on
q2:
-(1.80N)J =(-1.28N)i (0.640N)J -(1.80N)J (-1.28N)i-(1.16N)J 1
F2 =F32, =1
+
F3, I are an action-and-reaction pair, as are F23, and F3,2 ' express the net force acting on q3:
Noting that
-
F., ,
3
and
-(0.899N)i -l(-1.28N)i (0.640N)JJ
F3 =FI3, F23, =-F3,1 - F32, = 1 =1 +
(0.381N)i -(0.640N)J
+
The Electric Field
*37
·
A charge of 4.0 J-LC is at the origin. What is the magnitude and direction of
the electric field on thex axis at (a) x = 6 m, and (b)x = - 10 m?
(c) Sketch the function Ex versusx for both positive and negative values ofx. (Remember that Ex is negative when E points in the negativex direction.) Picture the Problem Let
q
represent the charge at the origin and use Coulomb's law for
E due to a point charge to find the electric field atx = 6 m and -10 m. (a) Express the electric field at a point P located a distancex from a charge
q:
-() kq =-2 rpo
E
X
A
x
'
10 h C apter 21 09N .m2/C2 )(4,LlC ) �
E-(6m)= (8.99xl
Evaluate this expression for x=6 m:
=
(6mY
I
I (999N/C)i I
(b) Evaluate Eatx=-lO m:
(c) The following graph was plotted using a spreadsheet program:
500
250
� «5'
a
-250
-500
*38
·
-2
-1
x
a
(m)
2
Two charges, each +4 pC, are on thex axis, one at the origin and the other at
x = 8 m. Find the electric field on thex axis at (a)x=-2 m, (b) x = 2 m, (c) x = 6 m, and
(d) x = 10 m. x.
(e) At what point on thex axis is the electric field zero? if) Sketch Ex versus
Picture the Problem Let q represent the charges of +4 fI2 and use Coulomb's law for
Edue to a point charge and the principle of superposition for fields to find the electric field at the locations specified. Noting that ql =q2, use Coulomb's law and the principle of superposition to express the electric field due to the given charges at a point P a distancex from the origin:
The Electric Field 1 : Discrete Charge Distributions
)
()
(1
kql kq, 1 r + kql E(x) =Eq, (x +Eq, x = 2 rq"p+ q � r r "p = fJ " p 8m x 2 x x2 ( 8m-xY -
( ) 2)(71 rq"p + ( 1 x rq"p = (36kN'm 8m -Y J -
-
�
IC
�
�
�
fJ"P
�
(a) Apply this equation to the point atx = -2 m:
ii,(-
2m) (36kN �
l[ (2�)' (-i)+ (1O�)' (-il] I (-936kN/C)i I
m' IC
�
(b) Evaluate Eatx = 2 m:
E(2m)� (36kN.m'/cl[ (2�)' i)+ ( (6�)' (-I)H (S.oOkN/c)i I (c) Evaluate E atx
=
6 m:
E(6m)� (36kN ·m'/Ct6�)' �)+ (2�)' (-I)�] I (-S.oOkN/c)i I (d) Evaluate E atx
=
1 0 m:
E(10m)� (36kN m'/ctl 0�)' (I) + (2�)J)] I (935kN/C)i I �
(e) From symmetry considerations:
E(4m)= @]
(f) The following graph was plotted using a spreadsheet program:
11
J
12
Chapter 21
·4
*42
••
x
4
8
(m)
12
A point charge of +5.0 j.1C is located at x = -3.0 cm, and a second point
charge of -8.0 j.1C is located at x = +4.0 cm. Where should a third charge of +6.0 j.1C be
placed so that the electric field at x = 0 is zero?
Picture the Problem If the electric field at x = 0 is zero, both its x and y components must
be zero. The only way this condition can be satisfied with the point charges of +5.0 j.1C and -8.0 j.1C are on the x axis is if the point charge of +6.0 j.1C is also on the x axis. Let the subscripts 5, -8, and 6 identify the point charges and their fields. We can use Coulomb's law for Edue to a point charge and the principle of superposition for fields to determine where the +6.0 j.1C charge should be located so that the electric field at x = 0 is zero. Express the electric field at x = 0 in terms of the fields due to the charges
E()O = Es,£ E_8,£ E6,£ +
+
=0
of +5.0 j.1C, -8.0 j.1C, and +6.0 j.1C : Substitute for each of the fields to obtain: or
kqs kq6 ( ) kq_8 ( r_8
-
2 �
l� +
-
2 �
-l� +
� )-0
- -l 2
-
Divide out the unit vector i to obtain: Substitute numerical values to obtain:
5
6
_-_8_ =0
(3cmY (4cmY r6 =12.38cm I
T he Electric Field 1: Discreteh C arg e Distributions *45
··
A 5-p.C point charge is located at x = 1 m, Y = 3 m; and a
-
4 ,uC point charge -
is located at x = 2 m, Y = -2 m. (a) Find the magnitude and direction of the electric field
at x = -3 m, Y = 1 m. (b) Find the magnitude and direction of the force on a proton at
x = -3 m, Y = I m.
Picture the Problem The diagram shows the electric field vectors at the point of interest
P due to the two charges. We can use Coulomb's law for E due to point charges and the superposition principle for electric fields to find p . We can apply F to find the
E
E q=
force on a proton at ( -3 m, 1 m). Y,m 3
./
/q2 = SI-'C
2
--�---+��r----r---+-x,m 2 -3 -2 �'-.. '-..
-1
'-.
'-..
-2
'-..
'-..
'--. ,/, = -4 f.LC
(a) Express the electric field at
l and q
( -3 m, 1 m) due to the charges
q2 :
Evaluate
EI :
C) qk rl p = (8.99X109N .m 2C/ 2)(-4,u EI =l , 2 2 (S) m + (3) m p
]
[
m m +(3)) (-S)1 �(SY m +(3Y m 1'j� 5 7 + 0.514 = (-1.06kN/ C)(-0.81 ))= (0.908kN/ c)1 ( 0 544 kNC)) / ?
�
+
E 2:
Evaluate
-
13
.
14 h C apter 21 Substitute and simplify t o find
Ep:
Ep =(0.908kN/ C)i + (-0.544kN/ C)} (-2.0lkN/ C)i (-1.0lkN/ C)) C)) =(-1.1OkN/ C)i (-1.55kN/ +
+
+
The magnitude of
Ep is:
The direction of Ep is:
Y C + (1.55kN/ Ep = �(1.10kN/ Y C C ! = ! 1.90kN/ e
E
(
)
C =! 2350 ! =tan- I -1.55kN/ -1.10kN/ C
(
)
Note that the angle returned by your calculator for
C . tan_1 -1.55kN/ -1.10kN/ C
IS
the
reference angle and must be increased by
1800 to yield BE.
(b) Express and evaluate the force on a proton at point P:
C))J c)1+ (-1.55kN/ F q=Ep =(1.6x10-19C )l(-1.10kN/ =(-1.76x10-'6i N) + (-2.48X10-16} N) The magnitude of F is:
The direction of F is:
eF
]
(
! =tan- I -2.48X10-16N =! 2350 -1.76x10-16N '
where, as noted above, the angle returned
(
by your calculator for
tan- I -2 ·48X10-16N -1.76x10-16N
)
is the reference
angle and must be increased by yield BE.
1800 to
The Electric Field 1: Discrete Charge Distributions *48
•••
15
Two positive point charges + q are on the y axis at y = +a and y = -a as in
Problem 44. A bead of mass m can-ying a negative charge -q slides without friction along a thread that runs along the x axis. (a) Show that for small displacements of x « a, the bead experiences a restoring force that is proportional to x and therefore undergoes simple harmonic motion. (b) Find the period of the motion. Picture the Problem In Problem 44 it is shown that the electric field on the x axis, due to
equal positive charges located at ( 0, a) and (O,-a), is given by
Ex
=
2kqx(x2 + a2 t/2. We can use T = 2Jr�m/ k' to express the period of the motion
in terms of the restoring constant Ii. (a) Express the force acting on the on the bead when its displacement from the origin is x: Factor a2 from the denominator to obtain:
Fx =
For x « a: i.e., the bead experiences a linear restoring force.
(b) Express the period of a simple harmonic oscillator:
T
=
2Jr 1m
fk7
Obtain Ii from our result in part (a):
Substitute to obtain:
Motion of Point Charges in Electric Fields *50
·
(a) Compute elm for a proton, and find its acceleration in a uniform electric
field with a magnitude of 100 N/C. (b) Find the time it takes for a proton initially at rest in such a field to reach a speed of O.Ole ( where e is the speed of light).
16 C a er h pt
21
Picture the Problem We can use Newton's 2nd law of motion to find the acceleration of
the proton in the uniform electric field and constant-acceleration equations to find the time required for it to reach a speed of O.Ole and the distance it travels while acquiring this speed.
1.6 10-19 C 1.67x10-27 kg = 19.58x107 Clkg 1 Fnet eE a=--=e=
(a) Use data found at the back of your text to compute elm for an electron:
X
------
mp
Apply Newton's 2nd law to relate the
mp
acceleration of the electron to the electric field:
mp
10-19 C){l00N/C) a= (1.6x1.67 x10-27 kg =19.58 109 m1s2 1 The direction of the acceleration of a proton is in the direction of the electric field.
Substitute numerical values and evaluate a:
X
(b)
Using the definition of a
acceleration, relate the time required for an electron to reach O.Ole to its
a
acceleration: Substitute numerical values and evaluate f..t: *54
••
A particle leaves the origin with a speed of 3 x 10 6 m/s at 35 ° to the x axis. It -
moves in a constant electric field E =
Eyj. Find Ey such that the particle will cross the �
x
axis atx = 1.5 em if the particle is (a) an electron, and (b) a proton.
Picture the Problem We can use constant-acceleration equations to express the x and y coordinates of the particle in terms of the parameter t and Newton's 2nd law to express
the constant acceleration in terms of the electric field. Eliminating the parameter will yield an equation for y as a function of x, q, and m that we can solve for Ey. Express thex and y coordinates of the particle as functions of time:
x {vcosB)t =
h T e Electric Field 1: Discreteharg C e Distributions and y
Apply Newton's 2nd law to relate the acceleration of the particle to the net
a
={vsinB)t
-
-t ai2
Ey Fnet,y q =--=- y m m
force acting on it: Substitute in the y-coordinate equation to obtain: Eliminate the parameter t between the two equations to obtain: Set y = 0 and solve for
Y
q t =(vsin) B t E ---y 2 2m
y
E x2 = ( tan)x B - q 2mv2 cos2 B Y
mv2 sin2B = Ey x q
Ey :
Substitute the non-particle specific data to obtain:
sin 7 0° E y - m(3qx106(O.mist OIS) m _
=(S.64x1014m /s2) m q
(a) Substitute for the mass and charge of an electron and evaluate
Ey :
(b) Substitute for the mass and charge of a proton and evaluate
*58
·
Ey:
kg s ) 9.11x10-31 E y =(S.64x1014m 12 1.6x10-1C9 =13.21kN/C I - 7 kg 1.67 x102 ml 2 14 s ) = (S 64x 10 . Ey 1.6x10-1C9 =IS.89MN/C I
A dipole of moment 0.5 e·nm is placed in a uniform electric field with a
magnitude of
4.0x104N/C. What is the magnitude of the torque on the dipole when (a)
the dipole is parallel to the electric field, (b) the dipole is perpendicular to the electric field, and (c) the dipole makes an angle of 30° with the electric field? (d) Find the potential energy of the dipole in the electric field for each case. Picture the Problem The torque on an electric dipole in an electric field is given by
. i = PxE and the potential energy of the dipole by U -pE. =
Using its definition, express the torque
i=pxE
17
18
Chapter
21
on a dipole moment in a uniform
and
electric field:
T
E p = sin8
where () is the angle between the electric dipole moment and the electric field. (a) Evaluate ,for () = 0°:
T
(b) Evaluate ,for ()
90°:
T
30°:
T
(c) Evaluate dor ()
=
=
(d) Using its definition, express the potential energy of a dipole in an
E p = si n 0° =� =(O.Se · n)( m 4.0 x10 4N/ )sin C 90° =13.20x10-24N'm 1 = (0.Se . n)( m 4.0xl0 4N/ ) C sin30° = 11.60 xl 0-24N .m 1
U
=-p·i =p E - cos8
electric field: 0°:
U
90°:
U
Evaluate U for () = 30°:
U
Evaluate U for ()
Evaluate U for ()
*59
··
=
=
=-(O.Se . n)( m 4.0x104N/ )cosoo C =1-3.20xl0-24J 1 =-(O.Se . n)( m 4.0x104N/ )cos90° C =0 )cos30° =-(O.Se · n)( m 4.0x104N/ C =1-2.77x10-24 J I
For a dipole oriented along the x axis, the electric field falls off as
lIx3 in the
lIi in the y direction. Use dimensional analysis to prove that, in any direction, the field far from the dipole falls off as llr3•
x direction and
Picture the Problem We can combine the dimension of an electric field with the
dimension of an electric dipole moment to prove that, in any direction, the dimension of
[ L 1/ 3] and, hence, the electric field far from the dipole falls
the far field is proportional to off as
lIr3.
Express the dimension of an electric field:
] [ E ] = [ kQ 2 [ L]
The Electric Field
1: Discrete Charge Distributions
Express the dimension an electric dipole moment: Write the dimension of charge in terms of the dimension of an electric dipole moment: Substitute to obtain: This shows that the field E due to a dipole 3 p falls off as lIr . General Problems
*63 · ( a) What mass would a proton have if its gravitational attraction to another proton exactly balanced out the electrostatic repulsion between them? ( b) What is the true ratio of these two forces? Picture the Problem We can equate the gravitational force and the electric force acting on a proton to find the mass of the proton under the given condition.
( a) Express the condition that must be satisfied if the net force on the proton is zero: Use Newton's law of gravity and Coulomb's law to substitute for Fg
and Fe:
Solve for m to obtain:
m= e
H;
Substitute numerical values and evaluate m:
N·m 2Cf 2 = i 1.86xlO-9kg i. m= (1.6xlO-19C ) 6.68.99xl09 7 x 10-11 N·m 2fkg2 . ( b) Express the ratio of Fe and Fg:
ke2
Gm� � r
2
ke2
Gm�
19
20 h C apter 21 Substitute numerical values to obtain:
(8.99xl09N .m 2C/ 2)(1.6xlO-19C) =1 1.24xl036 g )(1.67 x 10-27 k) g (6.67 x 10-11N · m 2 /k2 *66
••
In copper, about one electron per atom is free to move about. A copper penny
has a mass of 3 g. (a) What percentage of the free charge would have to be removed to give the penny a charge of 15 fLC? (b) What would be the force of repulsion between two pennies carrying this charge if they were 25 cm apart? Assume that the pennies are point charges. Picture the Problem We can find the percentage of the free charge that would have to
be removed by finding the ratio of the number of free electrons ne to be removed to give the penny a charge of 15 fLC to the number of free electrons in the penny. Because we're assuming the pennies to be point charges, we can use Coulomb's law to find the force of repulsion between them. (a) Express the fractionJof the free charge to be removed as the quotient
f =N ne
of the number of electrons to be removed and the number of free electrons: Relate Nto Avogadro's number, the mass of the copper penny, and the molecular mass of copper:
N -=-:::: ::>N=NA M NA M m
m
Relate ne to the free charge Q to be removed from the penny:
Q
f= -e = N� A M
Substitute numerical values and evaluate!
(-15,uC )(63.5g /m ) ol =3.29xlO-9 =1 3.29x 10- 7 0/0 I f =- ()3( g 1.6xl 0-1C9 )(6 .02 x 1023 m orl)
h T e Electric Field 1: Discreteh C arg e Distributions
21
(b) Use Coulomb's law to express the force of repulsion between the two pennIes: Substitute numerical values and evaluate F:
F = (8.99xl0 *69
••
9
N·m C/ 2
)(9.38x 1013 Y(1.6xlO-1C9 Y 1 = 32. 4 N I (0 25 m 2)
2
.
A positive charge Q is to be divided into two positive charges ql and q2.
Show that, for a given separation D, the force exerted by one charge on the other is greatest ifql q2
= =1Q·
Picture the Problem We can use Coulomb's law to express the force exerted on one
charge by the other and then set the derivative of this expression equal to zero to find the distribution of the charge that maximizes this force. Using Coulomb's law, express the force that either charge exerts on the other: Express q2 in terms of Q and ql: Substitute to obtain:
Differentiate F with respect to ql and set this derivative equal to zero for extreme values:
-dF = k -d[ql (Q-ql )] d d l D2 ql q = 2 [ql (- l) + Q - q]l =0
; for extrem a
Solve for ql to obtain: To determine whether a maximum or a minimum exists at ql
=1Q ,
differentiate F a second time and evaluate this derivative at ql
= 1Q:
d -2 l d2F k - [ 2 2 d D qd l Q q] ql = k2 (-2) D 0 _
R 2 points (a) away from the center of the shell in both regions.
29
30
Chapter 22
( b) toward the center of the shell in both regions. (c) toward the center of the shell for r < R, and is zero for r > R2 . (d) away from the center of the shell for r < R , and is zero for r > R 2 . Determine the Concept We can apply Gauss's law to determine the electric field for r R , and
r
> R 2 • We also know that the direction of an electric field at any point is
r=
r
Integrate this expression from R/12 to = R to obtain:
59
-
a = 12R
= r' r'= r2 . x( ) = -21lk(Jo f 2r 2 dr -Jx +r ' �� 2nka{ N +R' � �x' +� J
x( ) -21lk(Jo !...- dr where -Jx 2 + dV_
V_
Substitute and simplify to obtain:
R
R/.fi
{� �' + 2nka{�x, +R' � �x' +�' J � 2nk R the potential on the axis of a disk charge approaches kQlx, where Q = o'1rR2 is the total charge on the disk. [Hint: Write (x2 + R2)1/2 = x(l + R21x2)112 and use the binomial expansion.]
*58
··
The potential on the axis of a disk charge of radius R and charge density 0" is given by V = 2nko- l(x2 + R2 r - x1-
Picture the Problem
Express the potential on the axis of the disk charge: Factor x from the radical and use the binomial expansion to obtain:
Substitute for the radical term to obtain:
provided x > > R. Find the greatest surface charge density O"max that can exist on a conductor before dielectric breakdown of the air occurs. *63
·
Electric Potent ial
61
We can solve the equation giving the electric field at the surface of a conductor for the greatest surface charge density that can exist before dielectric breakdown of the air occurs. Picture the Problem
Relate the electric field at the surface of a conductor to the surface charge density: Solve for CTunder dielectric breakdown of the air conditions:
(J"max
Substitute numerical values and evaluate CTmax:
(J"max
=Eo Ebreaddown = (8.85 10-12C2/N . m2)(3 MV /m ) = 1 26.6 j1C/m2 1 X
*66 Calculate the potential relative to infinity at the center of a uniformly charged sphere of radius R and charge Q. Picture the Problem We can find the potential relative to infinity at the center ofthe sphere by integrating the electric field for 0 to We can apply Gauss's law to find the electric field both inside and outside the spherical shell. •••
00.
The potential relative to infinity the center of the spherical shell is: Apply Gauss's law to a spherical surface of radius r < R to obtain:
'" R (1) JErRd r o R d = Er 0) and back-biasing ( V < 0). Show that if you plot In Ivs V for forward biasing (using V> 0.3 V or so), you get nearly a straight line, What is the slope of the line?
106
Chapter 25
Picture the Problem A spreadsheet program to plot 1 as a function of V is shown below. The formulas used to calculate the quantities in the columns are as follows:
Cell B1 AS A6 B5
Algebraic Form 10 V (mV)
ContentlF onnula 2 -200 AS +25 $B$l * (EXP(A5/25) - 1 )
V (mV) -200.0 - 1 7S.0
V+t..V
Jo(eVI25I1lV -1)
I (nA) -2.00 -2.00
7S.0 1 00.0 The following graph was plotted using the data in spreadsheet table shown above. 110 90
�
5
....
70 50 30 10 -10
-200
-150
-100
-50
V (mY)
0
50
100
A spreadsheet program to plot In(1) as a function of V for V> 0.3 V follows. The formulas used to calculate the quantities in the columns are as follows: Cell Bl AS A6 BS
ContentIFormula 2 300 AS + 1 0 LN($B$l *(EXP(AS/2S) - 1 ))
Algebraic Form 2 nA V
-l)J InlJo(eVl25mv V + t..V
'Electric Current and D irect-Current Ci r cuits -
A I 0=
"
;-
Y (mY) 300 310 320 330 340 350 ,
'
.
970 980 990 1000
,
B
107
C nA
2 In(I) 12,69 13.09 1 3 .49 1 3 .89 14.29 14.69 -,;-
-,il:
39.49 39.89 40.29 40.69
Ie' -,,'
'.;:',�'!
A graph of ln(J) as a function of V follows. Microsoft Excel' s Trendline feature was used to obtain the equation of the line. 45 40 35 30 25 � ..E 20 �
15 10 5 0
300
400
500
For V» 25 mY:
600
700
900
1000
eV 125 mY -1 eV 125mY >:::
and
1 Take the natural logarithm of both sides of the equation to obtain:
800
V(mV)
� �
1oeVI2
5mY
In (J ) = In (Ioev/25mY ) 1 V = In (Jo ) + 25rnV
which is of the form y = mx + b, where m =
1
25 rnV
=
I 0.04(rnVt I
.L--� ---'----'-----'-
in agreement with our graphical result.
108 Chapter *58
25
The space between two concentric spherical-shell conductors is filled with a
material that has a resistivity of 1 0 Q·m. If the inner shell has a radius of l.5 cm and the outer shell has a radius of 5 cm, what is the resistance between the conductors? (Hint: Find the resistance of a spherical-shell element of the material of area 4m,2 and length dr,
9
and integrate to find the total resistance of the set of shells in series.) The diagram shows a cross-sectional view of the concentric Picture the Problem
spheres of radii a and b as well as a spherical-shell element of radius r. Using the Hint we can express the resistance dR of the spherical-shell element and then integrate over the volume filled with the material whose resistivity p is given to find the resistance between the conductors. Note that the elements of resistance are in senes.
-
_P � dR -- P dr 4nr2 A
Express the element of resistance dR: Integrate dR from r
=
a to r = b to
obtain:
R
Substitute numerical values and evaluate R:
_ 109 n . m 4Jr
=
( 1.5cm1 _5_cml_J
I 3 .7 1 109 n I x
Temperature Dependence of Resistance *60
·
A
tungsten rod is 50 cm long and has a square cross-sectional area with sides
of l .0 mm. (a) What is its resistance at 20°C? (b) What is its resistance at 40°C? Picture the Problem We
can use R = pLIA to find the resistance of the rod at 20°e.
Ignoring the effects of thermal expansion, we can we apply the equation defining the temperature coefficient of resistivity,
a,
to relate the resistance at 40°C to the resistance at
20°e. (a) Express the resistance of the rod at 20°C as a function of its
R20 = P20L
A
' El ectric Current and Direct-Current C i rcuits
109
resistivity, length, and cross sectional area:
)
evaluate R20:
0'S R20 =(S.SXIO-SQ .m) Imm 2 ( = / 27.SmQ /
(b) Express the resistance of the rod at 40°C as a function of its
R40 =P40 AL
Substitute numerical values and
resistance at 20°C and the temperature coefficient of resistivity
a:
=P2o[1+ a(tc - 20Co)].:f. A =P20 AL +P20 AL a(tc -20Co) =R20 [1+a(tc -20Co)]
Substitute numerical values (see Table 25- 1 for the temperature coefficient of resistivity of tungsten) and evaluate R40:
*65
•••
A wire of cross-sectional area A, length
L" resistivity Ph and temperature
al is connected end to end to a second wire of the same cross-sectional area, length L2, resistivity 0., and temperature coefficient a2, so that the wires carry the same current. (a) Show that if PILIal+0.L2a2= 0, the total resistance R is independent of coefficient
temperature for small temperature changes. (b) If one wire is made of carbon and the other is copper, find the ratio of their lengths for which R is approximately independent of temperature. Picture the Problem
Expressing the total resistance of the two current-carrying (and
hence warming) wires connected in series in terms of their resistivities, temperature coefficients of resistivity, lengths and temperature change will lead us to an expression in
PILlal+0.L2a2= 0, the total resistance is temperature independent. In part (b) we can apply the condition that PILIal + 0.L2a2 = ° to find the ratio of the lengths of the
which, if
carbon and copper wires.
(a) Express the total resistance of these two wires connected in series:
R =RI +R2 LI (1+alL\T)+P2 L2 ( 1+a 2L\T)+.l[PILI(1+alL\T)+P2 L2(1+a 2L\T)] = PI A A A
110
Chapter
25
Expand and simplify this expression to obtain:
the te mperature. (b) Apply the condition for temperature independence obtained in (a) to the carbon and copper WIres:
Solve for the ratio of Leu to L e:
Leu Le
_
Peae Peuaeu
Substitute numerical values (see Table 25-1 for the temperature coefficient of resistivity of carbon and copper) and evaluate the ratio of Leu to Le:
- 0.5 10-3 K-1 = I 64 I --8-Q�3�-1 . -7-x-10- . m .9- -x-10---3 -K---' 0.6 V, the differential resistance of a diode (Problem 54) is
*147
=
•••
approximately Rd
=
(25 mY)!I, and for V< 0, Rd increases exponentially with
this result to justify the approximation given in Problem 55.
IVI.
Use
We can use the definition of differential resistance and the expression for the diode current given in problem 54 to express R d and establish the required results.
Picture the Problem
The differential resistance Rd is given by: From Problem 54, the current in the diode is given by:
1 = 10 (e
V/25m Y
- 1)
Substitute for Ito obtain:
= For V> 0.6 V, equation (1) becomes:
1
� �
25 mV e-V/25mV
1o e
10
V/25mV
(1)
1 34
Chapter 25 I -V/25mV � 10 e V/25 mV � - => e
Solve for the exponential factor to obtain:
10
Substitute in equation (2) to obtain:
Rd �
Examination of equation (2) shows that, for V This result, together with that for V
>
Problem 5 5 .
25 mV
0 ) a nd two portions of the current sheet labeled I, and 12 ,
(a) What is
the direction of the magnetic field B at P due to the two portions of the current shown?
(b) What is the direction of the magnetic field
B at point P due to the entire sheet?
What is the direction of B at a point below the sheet (y < O)? the rectangular curve shown in Fi gure -
above the sheet is given by B =
-
(c)
(d) Apply Ampere's law to
27-52b to show that the magnetic field at any point �
1 f.1oAi .
(a) i.- w --oj
(b)
Figure
)(
)(
x
x
)(
x
x
27-52 Problem 72
Picture the Problem In
parts (a),
(b), and (c) we can use a right-hand rule to determine
the direction of the magnetic field at points above and below the infinite sheet of current.
In part
(d) we can evaluate {B . if around the specified path and equate it to We and
solve for B.
(a )
its in the j direction JIsince At P the magnetic field points to the right li.e., ' \' vertical components cancel.
(b)
Because the sheet is infinite, the same argument used in (a) applies; B is in the i direction.
l
-
A
-
(c) Below the sheet the magnetic field points to the left, i.e., in the i direction. The vertical components cancel. (d) Express iii
0
if , in the
counterclockwise direction, for the given path:
{ii o de = 2 fii o de + 2 fii o de parallel
..l.
Sources of the Magnetic Field
fB . cif = o
For the paths perpe ndicular to the -
-
sheet, B and
l 75
d.e are perpendicular
.1
to each other and:
fB . d£ = Bw parallel
For the paths parallel to the sheet, -
B and
d.e are in the same direction -
and: Substitute to obtain:
Solve for
paral el B = .l rIlo A
B:
2
and
Babove = 1 - t floAf I Magnetization and Magnetic Susceptibility
*79
••
A cyli nder of magnetic material is placed in a long solenoid of n turns per
unit length and current I. The values for magnetic field given below. Use these values to plot B versus
B withi n the material versus nI are
Bapp and Km versus nI.
nI, AIm
0
50
100
150
200
500
1000
10,000
B, T
0
0.04 0.67
1.00
1.2
1.4
1.6
1.7
Picture the Problem
We can use the data in the table and
Bapp. We can fi nd Km using B = Km Bapp . We can find the applied field
Bapp = flo nI to plot B versus
Bapp
for a long solenoid using: Km can be found from
Bapp and B
usmg:
B Km = -Bapp
The following graph was plotted using a spreadsheet program. The abscissa values for the graph were obtained by multiplying nl by J.1o.
B initially rises rapidly, and then becomes
nearly flat. This is characteristic of a ferromagnetic material.
1 76
Chapter
27 2.0 ����----'-�--���'---��--�-r-----�
E
'l:l
1 .6
1
12
16
0.8
II! '
6· ""'_ .
..
d
I !
.
'·. .';,,"-" 1
I
" r"
!-
iI •
- I
L
'. Y .
0.4 0. 0
.'
0.000
'''1'
0.002
,
1
0.004
'I
0.006
I
0.008
B app (T)
· r
0.0 1 0
0.01 2
0.014
The graph of Km versus nI shown below was also plotted using a spreadsheet program. Note that Km becomes quite large for small values of nI but then diminishes. A more revealing graph would be to plot BI(n!), which would be quite large for small values of nI and then drop to nearly zero at nI = 10,000 Aim, corresponding to saturation of the magnetization.
6000 5000 4000 :..:5: 3000 2000 1 000 o
r" "' . o
2000
4000 III
6000
(AIm)
8000
"'?'T
1 0000
Atomic Magnetic Moments
*82
••
Nickel has a density of
8.7 g/cm3 and a molecular mass of 58.7 glmol.
Nickel ' s saturation magnetization is given by !JoMs = 0.61 T. Calculate the magnetic moment of a nickel atom in Bohr magnetons.
Sources of the Magnetic Field
1 77
Picture the Problem We can find the magnetic momen t of a nickel atom J-l from its relationship the saturation ma gnetization Ms using Ms = n I' where 11 is the number of
molecules.
11,
in tum, can be found from Avogadro ' s number, the density of nickel, and
its molar mass using n
= NAMP .
- nr Ms il
Express the saturation magnetic field in terms of the number of molecules
or
Ms I' = _ n
per unit volume and the magnetic moment of each molecule: Express the number of molecules per unit volume in terms of Avo gadro ' s number NA, the molecular mass M, and the density
p:
Substitute and simpli fy to obtain:
Substitute numerical values and evaluate
J-l:
-- --'-- ----: :...� ..- '_rr_ --____:_'i T)--'---58.7 x 10,--3 kg/mol ----,,...,-!:-: ---== I' -r-4;r x 10-7--=-N/(0.61 A 2 6.02 1023 atoms/mol =
Express the value of 1 Bohr magneton: Divide
J-l by J-lB to obtain:
X
I'B = 9.27x lO-24 A · m 2 I' 5.44xlO-24 A · m 2 = 0.587 I'B 9.27 xlO-24 A · m 2 =
or
I' = I 0.587 1'B I
Paramagnetism
*86
··
Assume that the magnetic moment of an aluminum atom is 1 Bohr magneton. The density of aluminum is 2.7 g/cm3 , and its molecular mass is 27 g/mo1. (a) Calculate Ms and JioMs for aluminum. (b) Use the results of Problem
84 to calculate Xm at T
=
300
K. (c) Explain why the result for Part (b) is larger than the value listed in Table 27- 1 .
1 78
Chapter
27
Picture t h e Problem
In (a) we can express the saturation magnetic field in terms o f t he
number of molecules per unit volume and the magnetic moment of each molecule and use n = NA P 1M to express the number of molecules per unit volume in terms of Avogadro's
number NA, the molecular mass M, and the density p. We can use XI1l from Problem 84 to calculate Xm .
=
J.-loJ.-LMs/3kT
(a ) Express the saturation magnetic field in terms of the number of molecules per unit volume and the magnetic moment of each molecule: Express the number of molecules per unit volume in terms of Avogadro' s number NA, the molecular mass M, and the density
p:
Ms
Substitute to obtain:
_
-
NA P
M
J...lB
Substitute numerical values and evaluate Ms : Ms =
_
-
(6.02 x 1 023 atoms/mol)(2 .7 x 103 kg/m3 )(9 .27 x 10-24 A · m2 )
! S.S8x 105 AIm !
27g/mol
and
(b) From Problem 84 we have:
Xm
=
J...loJ.-lMs 3kT
Substitute numerical values and evaluate Xm: Xm
(c)
=
1 0-7 N/A 2 9.27 x10-""'--24IKA--'r;105 AIm--L ! S .23 x 10-4 ! ·(m-2LiS.S8x -"4/'Z"x --.,. -------,3-.1.3L--'-8-1-x1 0-23 J 3 00 K ) =
! In calculating Xm in (b)we neglected any diamagnetic effects. !
179
Sources of the Magnetic Field Ferromagnetism For annealed iron, the relative permeability Km has its maximum value of approximately 5 500 at Bapp 1 .57x 10-4 T. Find M and B when Km is maximum.
*90
·
=
Picture the Problem
We can use
B Km Bapp =
to find B and M
=
(Kill - l)Bapp/ Jl.o
to
find M. Express B in tenns of M and Km: Substitute numerical values and evaluate B:
B
= =
(5500)(1.57 x l 0-4 T ) 1 0.864T 1
Relate M to Km and Bapp :
Substitute numerical values and evaluate M:
*96
··
M
(5500)(1.57 x 10-4 T ) 47r x 10-7 N/A 2 = 16.87 x 105 Aim 1
=
Two long straight wires 4-cm apart are embedded in a unifonn insulator that
has a relative penneability of Km = 1 20. The wires carry 40 A in opposite directions. (a) What is the magnetic field at the midpoint of the plane of the wires?
(b) What is the force
per unit length on the wires? Picture the Problem
Because the wires carry equal currents in opposite directions, the
magnetic field midway between them will be twice that due to either current alone and
will be greater, by a factor of Km, than it would be in the absence of the insulator. We can
use Ampere ' s law to find the field, due to either current, at the midpoint of the plane of
dF -
the wires and
=
-
-
Id.e x B to find the force per unit length on either wire.
(a) Relate the magnetic field in the insulator to the magnetic field in its absence : Apply Ampere' s law t o a closed circular path a distance
r
from a
current-carrying wire to obtain:
180
Chapter 27
Solve for Bapl' to obtain:
Because there are two current carrying wires, with their currents in opposite directions, the fields are additive and: Substitute numerical values and evaluate B:
_
B-
=
(b) Express the force per unit length experienced by either wire due to the current in the other: Apply Ampere' s law to obtain:
F f!
120(4Jr 10-7 N/A 2 )(40 A) Jr(0.02 m) I 96.0mT I
=
X
B1
1.8 . ie
=
B (2 nr)
=
Ji0 1c
=
Ji0 1
where r is the separation of the wires. Solve for B:
Substitute to obtain:
Substitute numerical values and evaluate
F f!
F _ 120(4Jr x 10-7 N/A 2 )(40 A Y £ 2Jr(0.04m) =
I 0.960 N/m I
General Problems
*101
•
In Figure 27 -5 5 , find the magnetic field at point P, which is at the common
center of the two semicircular arcs.
Sources of the Magnetic Field
Figure
27-55 Problem
Picture the Problem
181
1 01
Let out of the page be the positive x direction. Because point P is
on the line connecting the straight segments of the conductor, these segments do not
contribute to the magnetic field at P. Hence, the resultant magnetic field at P will be the sum of the magnetic fields due to the current in the two semicircles, and we can use the expression for the magnetic field at the center of a current loop to find
B
p .
Express the resultant magnetic field at P: Express the magnetic field at the center of a current loop:
B
=
f..Lo1 2R
where R is the radius of the loop.
Express the magnetic field at the center of half a current loop: -
Express
-
B\ and B 2 : and
jj
2
=
_
f..LoI i 4R 2
Substitute to obtain:
* 1 04
••
A power cable carrying 50 A is 2 m below the earth' s surface, but the cable's
direction and precise position are unknown. Show how you could locate the cable using a compass. Assume that you are at the equator, where the earth's magnetic field is north. Picture the Problem
0.7 G
Depending on the direction of the wire, the magnetic field due to
its current (provided this field is a large enough fraction of the earth' s magnetic field)
1 82
Chapter 27
will either add to or subtract from the earth' s field and moving the compass over the ground in the vicinity of the wire will indicate the di rection of the current. Apply Ampere ' s law to a circle of radius r and concentric with the center of the wire:
BlVire = f.10
Solve for B to obtain:
1
2w
Substitute numerical values and evaluate BlVire :
Express the ratio of Bwi re to Bearth:
(47r x lO-7 N/A 2 )(50A) 2 7r(2 ) = 0.0500G _
Bwire -
Bwire = 0.05G 7 Bearth 0 . G
m
::::>
7%
Thus, the field of the current-carrying wire should be detectable with a good compass. If the cable runs east-west, its magnetic field is in the north-south direction and thus either adds to or subtracts from the earth's field, depending on the current direction and location
of the compass. Moving the compass over the region one should be able to
detect the change. If the cable runs north-south, its magnetic field is perpendicular to that of the earth, and moving the compass about one should observe a change in the direction of the compass needle. A very long straight wire carries a current of 20 A. An electron 1 cm from the center of the wire is moving with a speed of 5 .Ox 1 06 mls. Find the force on the
* 108
••
electron when it moves (a) directly away from the wire, (b) parallel to the wire in the
direction of the current, and (c) perpendicular to the wire and tangent to a circle around the wire . Picture the Problem
Chose the coordinate system shown t o the right. Then the current i s
in the positive z direction. Assume that the electron i s a t ( 1 cm, 0, 0). W e can use
F-
=
- -
-
-
q v x B to relate the magnetic force on the electron to v and B and B
=
1 21
f. 0 - j to _
47r
r
express the magnetic field at the location of the electron. We'll need to express v for each of the three situations described in the problem in order to evaluate F = q v
�
xB.
Sources of the Magnetic Field '\
,.
-
-
-
\
I Y' \
/
cm
B
\
q
-
/
183
/
(1,0,0) -
I
/
-
-
x, cm
F = qv x B
Express the magnetic force acting on the electron: Express the magnetic field due to the current in the wire as a function of distance from the wire:
/-Lo 2 1 ': = 2q/-LoI (v- X ] ) F- = qv x--] 4Jr 4nr
Substitute to obtain:
,:
-
r
(a) Express the velocity of the electron when it moves directly away from the wire:
( X ] ) = 2q/-LoIv k� F- = 2q/-LoI 4nr 4nr ; VI
Substitute to obtain:
Substitute numerical values and evaluate
,:
F:
2(4Jr X 10-7 N/A2 )(- 1.6 x 10-19 C)(5 X 106 rnIs) (20 A)k 4Jr(0.01m) = 1 (- 3.20 x 10-16 N)k 1 _
F-
(b) Express
v when the electron is
traveling parallel to the wire in the direction of the current: Substitute in equation (1) to obtain:
( k� X ] ) = - 2q/-LoIv ; F- = 2q/-LoI 4nr 4nr V
,:
I
(1)
184
Chapter
27
Substitute numerical values and evaluate F :
F=
_
2 (47r 10-7 N/A 2 )(-1. 6 10-19 C )(5 106 m/s)(20A)i = I (3 . 20 10-1 6 N)i I 47r(0.01m) X
x
(c) Express v when the electron is
traveling perpendicular to the wire and
X
v
x
= vj
tangent to a circle around the wire: Substitute in equation
*1 1 1
••
Figure
(1)
27-60
to obtain:
I (vJ� x J�) = fQl F- = 2q4f-lo � nr
shows a bar magnet suspended by a thin wire that provides a
0.8
restoring torque -K8. The magnet is 1 6 em long, has a mass of kg, a dipole moment of f1 = 0. 1 2 Am2 , and it is located in a region where a uniform magnetic field B can be
established. When the external magnetic field is
0.2
T and the magnet is given a small
angular displacement /18, the bar magnet oscillates about its equilibrium position with a period of 0.500 s. Determine the constant K and the period of this torsional pendulum when B = O .
B
Figure
27-60 Problem
Picture the Problem
111
We can apply Newton' s 2nd law for rotational motion to obtain the
differential equation of motion of the bar magnet. While this equation is not linear, we can use a small-angle approximation to render it linear and obtain an expression for the square of the angular frequency that we can solve for K when there is an external field and for the period T in the absence of an external field. Apply
IT = I
a
to the bar magnet
when B *- 0 to obtain the differential equation of motion for the magnet:
-
K() -
d 2 () j.1Bsin() = 1 dt 2
where I is the moment of inertia of the magnet about an axis through its point of suspenSIOn.
Sources of the Magnetic Field For small displacements from equilibrium (8
«
1 ):
Rewrite the differential equation as:
d 2e + ( K+ j.iB )e 0 I dt2
or
=
Because the coefficient of the linear
(1/
term is the square of the angular
=
K + j.iB I
(1)
frequency, we have: Express the moment of inertia (see
L2 1 = ...L 12 m
Table 9-1) of the bar magnet about an axis through its center: Substitute to obtain:
())
2 K+ j.iB L2
= --'--- I
Ti m
Solve for K to obtain:
Substitute numerical values and evaluate K :
Substitute B = 0 and equation
0) =
(I ) to obtain:
2mT in
Solve for T:
Substitute numerical values and evaluate T:
T
=
=
7r(0.16 m) 3(0.2460.8kg N . mlrad) I 0. 523s I
1 85
1 86 Chapter 27 An iron bar of length 1 .4 m has a diameter of 2 em and a uniform magnetization of 1 .72x 1 06 Aim directed along the bar ' s length. The bar is stationary in *] 1 6
••
space and is suddenly demagnetized so that its magnetization disappears. What is the rotational angular velocity of the bar if its angular momentum is conserved? (Assume that Equation 27-27 holds where m is the mass of an electron and q = -e.)
Picture the Problem
We can use the definition of angular momentum and Equation 27-
27, together with the definition of the magnetization M of the iron bar, to derive an expression for the rotational angular velocity of the bar j ust after it has been demagnetized.
Assuming its angular momentum to
be conserved, use the definition of L
L
= 10)
to express the angular momentum of the iron bar j ust after it has been demagnetized: Solve for the angular velocity
cu:
Assuming that Equation 27-27 holds yields:
L
0) = 1 L
= 2m J1 = 2me MV 2me Mm- 2 ,e q
=
e
e
where r is the radius of the bar and ,e its length.
Modeling the bar as a cylinder, express its moment of inertia with respect to its axis: Substitute to obtain:
0) = Substitute numerical values (see Table 1 3 - 1 for the density of iron) and evaluate
* 1 18
••
cu:
A relatively inexpensive ammeter, called a tangent galvanometer, can be
made using the earth ' s field. A plane circular coil of N tums and radius R is oriented such that the field Be it produces in the center of the coil is either east or west. A compass is
Sources of the Magnetic Field
1 87
placed at the center of the coil. When there is no cw-rent in the coil, the compass needle
points north. When there is a current J, the compass needle points in the direction of the
resultant magnetic field B at an angle e to the north. Show that the current J is related to e and to the horizontal component of the earth's field Be by
Picture the Problem
Note that
Be and Be are perpendicular to each other and that the
resultant magnetic field i s at an angle e with north. We can use trigonometry to relate Be and Be and express Be in terms of the geometry of the coil and the current flowing in it.
Express Be in terms of Be:
Be = Be tanB
where e i s the angle of the resultant field from north.
\
Express the field Be due to the current in the d-oil: \
Solve for J:
•••
2R
where N is the number of turns.
Substitute t6- obtain:
*125
N I Be = Jio
NJi I 0_ = Be tanB 2R
_ _
1=
2RBe
__
Jio N
tanB
A disk of radius R carries a fixed charge density
a
and rotates with
angular velocityw. (a) Consider a circular strip of radius r and width dr with charge dq. Show that the current produced by this strip dI (oi2 7r) dq = war dr. (b) Use your result from Part (a) to show that the magnetic field at the center of the disk is B = t Jio CJ"OJR . ( c) =
Use your result from Part (a) to find the magnetic field at a point on the axis of the disk a distance x from the center.
188 Chapter 2 7 Picture the Problem
The diagram shows
the rotating disk and the circular strip of radius r and width dr with charge dq. We can use the definition of surface charge density to express dq in terms of r and dr and the definition of current to show that dI = OJCJr
dr. We can then use this current and
expression for the magnetic field on the axis of a current loop to obtain the results caned for in (b ) and (c). (a) Express the total charge dq that
dq = CJ"dA = 2 1CCJrdr
passes a given point on the circular strip once each period: Using its definition, express the current in the element of width dr:
(c) Express the magnetic field dBx at a distance x along the axis of the disk due to the current loop of radius r and width dr:
Integrate from r = 0 to r = R to
obtain:
(b) Evaluate Bx for x = 0 :
d 1C dr dI = q = 2 CJr = I (j)ardr I 2 1C dt (j)
Chapter 28 Magnetic Induction Conceptual Problems
*1
•
A conducting loop lies in the plane of this page and carries a clockwise
induced current. Which of the following statements could be true? (a) A constant magnetic field is directed into the page. (b) A constant magnetic field is directed out of
the page. (c) An increasing magnetic field is directed into the page. (d) A decreasing
magnetic field is directed into the page. (e) A decreasing magnetic field is directed out of
the page. Determine the Concept
We lmow that the magnetic flux (in this case the magnetic field
because the area of the conducting loop is constant and its orientation is fixed) must be changing so the only issues are whether the field is increasing or decreasing and in which direction. Because the direction of the magnetic field associated with the clockwise current is into the page, the changing field that is responsible for it must be either increasing out of the page (not included in the list of possible answers) or a decreasing field directed into the page.
*6
·
I (d) is correct . I
If the current through an inductor were doubled, the energy stored in the
inductor would be (a) the same. (b) doubled. (c) quadrupled. (d) halved.
(e) quartered.
Determine the Concept
Um
=
The magnetic energy stored in an inductor is given by
-t LI 2 . Doubling I quadruples Um . I (c) is correct. I
A pendulum is fabricated from a thin, flat piece of metal. At the bottom of its *1O · arc, it passes between the poles of a strong permanent magnet. In Figure 28-42 a, the metal sheet is continuous, whereas in Figure 28-42b, there are slots in it. The pendulum with slots swings back and forth many times, but the pendulum without slots comes to a stop in no more than one complete oscillation. Explain why.
N
(b)
(a) Figure
28-42 Problem 1 0
1 89
190
Chapter 28
Determine the Concept
In the configuration shown in (a), energy i s dissipated by eddy
currents from the emf induced by the pendulum movement. In the configuration shown
in (b), the slits inhibit the eddy currents and the braking effect is greatly reduced. Estimation and Approximation
* 1 3 ·· A physics teacher attempts the following emf demonstration. She has two of her students hold a long wire connected to a voltmeter. The wire is held slack, so that there is a large arc in it. When she says "start", the students being rotating the wire in a large vertical arc, as if they were playing jump-rope. The students stand 3 .0 m apart, and the sag in the wire is about 1 . 5 m. (You may idealize the shape of wire as a semi-circular arc of diameter d = 1 . 5 m.) The induced emf from the jump rope is then measured on the voltmeter. (a) Estimate a reasonable value for the maximum angular velocity which the students can rotate the wire. (b) From this, estimate the maximum emf induced in the wire. The magnitude of the earth' s magnetic field is approximately 0.7 G. (c) Can the students rotate the jump-rope fast enough to generate an emf of 1 V? (d) Suggest modifications to the demonstration that would allow higher emfs to be generated. Picture the Problem We
can use Faraday' s law to relate the induced emf to the angular velocity with which the students turn the jump rope.
(a) It seems unlikely that the students could tum the "jump rope" wire faster than 5 revolutions per second. This corresponds to a maximum angular velocity of: (b) The magnetic flux ¢Jm through the rotating circular loop of wire varies sinusoidally with time according to: Because the average value of the cosine function, over one revolution, is Yz, the average rate at which the flux changes through the circular loop is: From Faraday' s law, the magnitude of the induced emf in the loop is: Substitute numerical values and evaluate E:
OJ = 5
rev s
x 2 n rad = I 3 1 .4 rad/s. I rev
¢Jm = BAsin OJt d¢m = BAOJcosOJt dt d¢m l = l.BAOJ = l.nr 2 BOJ 2 dt 2
and
av
Magnetic Induction 191 (c)
No. To generate an emf of 1 V, t he studenls would have to o ta te thejump rope about 500 times fast er.
(d)
The use of multiple strands of lighterwire(so that the compositewire could be rotated at the same angular speed) looped several times around wouldincrease the induced emf.
r
Magnetic Flux
*17 ·
A circular coil has 25 turns and a radius of 5 cm. It is at the equator, where
the earth' s magnetic field is 0.7 G north. Find the magnetic flux through the coil when its
plane is (a) horizontal, (b) vertical with its axis pointing north,
(c) vertical with its axis pointing east, and (d) vertical with its axis making an angle of 3 0° with north. Picture the Problem Because the coil defines a plane with area A and B is constant in
magnitude and direction over the surface and makes an angle 8 with the unit normal vector, we can use NBA to find the magnetic flux through the coil.
rpm =
cos e
Substitute for N, B, and A to obtain:
(
)
n(5 10-2 mY cose rpm = NBn r2 cos e = 25 0.7 G· + 10 G = (1 .37 lO-s W b)cos e X
X
(a) When the plane of the coil is horizontal, 8= 90° :
(b) When the plane of the coil is vertical with its axis pointing north, ()= 0° :
(c) When the plane of the coil is vertical with its axis pointing east, 8= 90°:
(d) When the plane of the coil is vertical with its axis making an angle of 30° with north, 8= 3 0° :
rpm = (l.3 7 10-s W b )cos90° =0 X
rpm = (1 .3 7 10-s W b)cos 00 = 11 .37 x lO-s W b I X
rpm = (1 .37 x lO-s W b)cos90° =0 rpm = (1 .3 7 x lO-s W b)cos30° = 11 . 19x lO-s W b I
192 Chapter 28 *24 ·· A long straight wire carries a current 1. A rectangular loop with two sides parallel to the straight wire has sides a and b, with its near side a distance d from the straight wire, as shown in Figure 28-45 . (a) Compute the magnetic flux through the rectangular loop. (Hint: Calculate the flux through a strip of area dA b dx and integrate from x = d to x = d + a.) (b) Evaluate your answer for a 5 cm, b = 1 0 cm, d = 2 cm, and 1= 20 A. =
=
n
b
Figure 28-45 Problem 24 Picture the Problem We can use the hint to set up the element of area dA and express the flux d
¢m through it and then carry out the details of the integration to express rAn.
(a) Express the flux through the strip of area dA: Express B at a distance x from a long, straight wire:
dCPm = BdA
where dA = bdx.
B
Substitute to obtain:
Integrate from x = d to x = d + a:
(b) Substitute numerical values and evaluate
¢m :
flo 21
flo 1
4Jr
2Jr X
_ ------
X
Magnetic Induction 193 Induced EMF and Faraday's Law •
A uniform magnetic field B is established perpendicular to the p1ane of a loop of radius 5 cm, resistance 0.4 n, and negligible self-inductance. The magnitude of *27
B is increasing at a rate of 40 mT/s. Find (a) the induced emf [; in the loop, (b) the
induced current in the loop, and (c) the rate of joule heating in the loop.
Picture the Problem We can find the induced emfby applying Faraday' s law to the loop. The application of Ohm ' s law will yield the induced current in the loop and we can find the rate of joule heating using P (a) Apply Faraday' s law to express the induced emf in the loop in tenns
= 12 R .
d AB)= AdB = ;rrR2 dB ( lEI= dq)mt = !£ dt dt dt
of the rate of change of the magnetic field: Substitute numerical values and evaluate
lEI:
(b) Using Ohm's law, relate the induced current to the induced
lEI= Jr(0.OSmY(40mT/s)= 1 0.3 14 mV 1
1= ER = 0.3 14mV = 1 0.78S mA 1 O.4n
voltage and the resistance of the loop and evaluate J:
(c) Express the rate at which power is dissipated in a conductor in terms of the induced current and the resistance of the loop and evaluate
p
= 12 R = (0.78S mAY(0.4n) = 1 0.247 flW 1
P: *31
••
A 1 00-tum circular coil has a diameter of 2 cm and resistance of 50 n. The
plane of the coil is perpendicular to a uniform magnetic field of magnitude
1 T. The direction of the field is suddenly reversed. (a) Find the total charge that passes
through the coil. If the reversal takes 0. 1 s, find (b) the average current in the coil and (c) the average emf in the coil. Picture the Problem We can use the definition of average current to express the total charge passing through the coil as a function of Jay. Because the induced current is
proportional to the induced emf and the induced emf, in tum, is given by Faraday' s law,
we can express �Q as a function of the number-of turns of the coil, the magnetic field, the resistance of the coil, and the area of the coil. Knowing the reversal time, we can find the average current from its definition and the average emf in the coil from Ohm' s law.
194
Chapter 28
(a) Express the total charge
that passes through the coil in terms of the induced current: Relate the induced current to the induced emf:
£ 1 = 1av = R
Using Faraday's law, express the induced emf in terms of rPnl: Substitute and simplify to obtain:
l1¢m
I1Q = £ I1t = M I1t = 2 ¢m R R R _
=
=
2 NBA
- R NBtrd2 2R
=
-
2 NB( : d2
)
-'----rr/dt and compare your
answer with that obtained in Part (a). Picture the Problem We can use the expression for a motional emf and Ampere' s law to express the net emf induced in the moving loop. We can also use express the magnetic flux through the loop and apply Faraday' s law to obtain the same result. (a) Express the motional emf induced in the segments parallel to
c
= B(x)vb
the current-carrying wire: Using Ampere' s law, express
B(d + vt) and B(d + a + vt):
B(d+vt) = and
JL o I 2rc(d+vt )
B(d + a+ vt) = Substitute to express wire and
&1 for the near
&2 for the far wire:
JL o I 2rc(d+ a + vt)
JLo lvb = 1 2rc(d+vt)
c
and c
1
o lvb
JL - 2rc(d+ a+vt)
Magnetic Induction c
Noting that the emfs both point
=
C, - c2
199
f.1ol-.,."- vb� = -:... 2n(d+vt) 2n(d+a+vt)
upward and hence oppose one another, express the net emf induced in the loop:
The motion of the segments perpendicular to the long wire does not change the flux through the rectangular loop. Consequently, these segments do not contribute to the the induced emf.
= _ drpmdt drpm = B(x)dA= B(x)bdx B(x)= f.12m01 lb f d rpm = fB(x)dx=� 2n x = f.12nolb In[d+d+a+vt vt]
(b) From Faraday' s law we have:
c
Express the magnetic flux in an area of length b and width vdt:
where, from Ampere' s law,
Substitute and integrate from
x= d+vt to d+a + vt :
d+a+vt
d+a+vt
d+vt
d+vt
�
Differentiate with respect to time and simplify to obtain: c
= _ �dt [f.12nolb In d+d+a+vt vt] _ f.12nolb�dt [In d+d+a+vt vt] )[(d+vt)v-(d+2a+vt)v]J =_f.12:rr0lb[( d+d+vt a+vt (d+vt) f.lolbv [(d+vt)- (d+a+vt)]= f.1olbv [ 1 1] =-� (d+vt)(d+a+vt) -� d+a+vt-d+vt f.1olbv [1 1] =� �-d+a+vt =
Inductance *54
1=10
·
A coil with self-inductance
L carries a current I, given by
sin 2;ift. Find and graph the flux rpm and the self-induced emf as functions of time.
200
Chapter 28
Picture the Problem We can apply
rpm
=
LI to find
rPm and Faraday' s law to find the
self-induced emf as functions of time.
rpm
Use the definition of self-inductance to express rPm:
=
LI
=
I LIo
sin
2ift I
The graph of the flux rPm as a function of time shown below was plotted using a spreadsheet program. The maximum value of the flux is LIo and we have chosen 21if = 1 rad/s.
1.0 "..,.....".,..,.,...�
-1.0
::; .p.:.J:.!�::.:;..:�4����E:;L 2
o
3 t
Apply Faraday' s law to relate and dI/dt:
&,
L,
4
6
5
(s)
dI dt
d [ . d.] I 8m2'':!t dt
& = -L- = -L-
0
The graph of the emf & as a function of time shown below was plotted using a
spreadsheet program. The maximum value of the induced emf is 21ljLIo and we
have chosen 21if= I rad/s.
Magnetic Induction
201
-1.0 -f""'''''--'-�--+-''--' �-+' '' � �"'--I'��="4=-' ---' =--1 ''=---''----' 'F'''''' o
2
t
*57
••
4
3
5
6
(s)
A long insulated wire with a resistance of
18 Wm is to be used to construct a
resistor. First, the wire is bent in half, and then the doubled wire is wound in a cylindrical form as shown in Figure 28-53. The diameter of the cylindrical form is 2 cm, its length is
25 cm, and the total length of wire is 9 m. Find the resistance and inductance of this wire wound resistor.
Figure 28-53 Problem 57 Picture the Problem Note that the current in the two parts of the wire is in opposite directions. Consequently, the total flux in the coil is zero. We can find the resistance of the wire-wound resistor from the length of wire used and the resistance per unit length. Because the total flux in the coil is zero: Express the total resistance of the WIre:
L=0 R = (18 �)L= (18 �}9m)= 1 162Q 1
202 Chapter 28 Magnetic Energy
*61
··
In a plane electromagnetic wave, such as a light wave, the magnitudes of the
electric fields and magnetic fields are related by E
=
cB, where
c =
1/�Co 110 is the speed
of light. Show that in this case the electric energy and the magnetic energy densities are equal. Picture the Problem We can examine the ratio of c
=
Urn to
UE with E
= cB and
1/�Co 110 to show that the electric and magnetic energy densities are equal. B2
Express the ratio of the energy
urn
density in the magnetic field to the energy density in the electric field:
Substitute E
=
uE
cB:
_
2 110 .1 G o E2 2 C'
B2 ""'0 "0 E2 /I C'
1
Substitute for c:
*64 ·· You are given a length d of wire which has radius a, and told to wind it into an inductor in the shape of a cylinder with a circular cross-section of radius r. The windings are to be as close together as possible without overlapping. Show that the selfm . ductance 0f th·IS m . ductor IS .
L
(rdj
= 110 4a).
Picture the Problem The wire of length d and radius a is shown in the diagram, as is the inductor constructed with this wire and whose inductance L is to be found. We can use the equation for the self-inductance of a cylindrical inductor to derive an expression for
L.
The self-inductance of an inductor with length .e, cross-sectional area A,
(1)
Magnetic Induction and number of turns per unit length
203
n IS:
The number of turns N is given by:
N-2a 2a1 ;rr N(2;rr ) (�)2;rr 2a a
N= �
The number of tums per unit length
n=
n IS:
Assuming that a « r, the length of the wire d is related to nand r:
d
=
e
=-
r =
r=
r
e
Solve for .e to obtain:
Substitute for e, A, and n in equation ( 1 ) to obtain:
RL Circuits *69
••
In the circuit of Figure 28-29, let &0 = 1 2 V, R O. At time t
=
3 n, and L = 0.6 H. The
0.5 s, find (a) the rate at which the battery supplies power, (b) the rate of joule heating, and (c) the rate at which energy is being switch is closed at time t
=
=
stored in the inductor. Picture the Problem We can find the current using
I = If (1 - e -tlr 1
where
If = &oIR ,and ,= LIR , and its rate of change by differentiating this expression with respect to time.
Express the dependence of the current on If and Evaluate Ir and
r:
r:
and 'f
Substitute to obtain:
=
R 3Q O.2s L 0.6H = =
204
C hapte r
Express
28
dl/dt:
J(O.S s) = (4 A)(l - e-s(o.s
(a) Find the current at t = 0.5 s :
s)s-J
= 3.67 A
)
p(O.Ss) = J(0.5s) & = (3.67 A) (12 V) =144. 0 W I
The rate a t which the battery supplies power at t = 0.5 s is:
�(O.Ss) = [J(O.Ss)Y R = (3.67 Ay(30) = 1 40.4W I
(b) The rate ofjoule heating is:
dUL = � [1.LJ2 ]= LJ dJ dt dt dt 2
(c) Using the expression for the magnetic energy stored in an inductor, express the rate at which energy is being stored:
dUL = � [1.LJ2 ]= LJ dJ dt dt 2 dt
Substitute for L , 1, and dJ/dt to obtain:
Substitute numerical values and evaluate
Evaluate this expression for
t = 0.5 s :
dU dt
L :
dUL = ( 8W)(l- e-s(O.5s)s-J )e-S(o.ss)s-J dt 4 = (48W)(l - e-2.S )e-2.S =13.62W I
Remarks: Note that, to a good approximation, dUJdt = P - PJ•
*75 ••• Given the circuit shown in Figure 28-56, assume that the switch S has been closed for a long time so that steady currents exist in the inductor, and that the inductor L has negligible resistance. (a) Find the battery current, the current in the 100 n resistor
Magnetic Induction
205
and the current tlu'ough the inductor. (b) Find the initial voltage across the inductor when the switch S is opened. (c) Using a spreadsheet program, make graphs of the CUlTent and voltage across the inductor as a function of time.
10V
+
100 Q
2H
Figure 28-56 Problem 75 Picture the Problem The self-induced emf in the inductor is proportional to the rate at which the current through it is changing. Under steady-state conditions, dlldt = 0 and so
the self-induced emf in the inductor is zero. We can use Kirchhoff s loop rule to obtain the current through and the voltage across the inductor as a function of time. (a) Because, under steady-state
10V
conditions, the self-induced emf in
and
the inductor is zero and because the inductor has negligible resistance,
I
=
-
(IOQ)I
=
0
10 V I 1. 00 A I . =
IOQ ·
we can apply Kirchhoff s loop rule to the loop that includes the source, the 10-0 resistor, and the inductor to find the current drawn from the battery and flowing through the inductor and the 10-0 resistor: By applying Kirchhoff s j unction rule at the junction between the
II oo-n resistor = Ibattery Iinductor = @] -
resistors, we can conclude that: (b) When the switch is closed, the current cannot immediately go to zero in the circuit because of the inductor. For a time, a current will circulate in the circuit loop between the inductor and the 100-0 resistor. Because the current flowing through this circuit is initially 1 A, the voltage drop across the 100-0 resistor is initially
1100 V . I Conservation of energy (Kirchhoff s loop rule) requires that the voltage drop across the inductor is also 1100 V. I
206
Chapter 28
(c) Apply Kirchhoffs loop rule to the RL circuit to obtain: R
1
1(t)- 1 e-[.' - 1 e--;: L 2H
The solution to this differential
-
equation is:
where
-
0
T =
-
R
=
0
--
lOOn
=
0.02s
A spreadsheet program to generate the data for graphs of the current and the voltage across the inductor as functions of time is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell B1 B2 B3 A6 B6
),Jl:,:r, L"
2
,3'"
[�4/
56 .7 ", 8" 9' 10 Iv 1 1 12 32 33 34 :35 36
Formula/Content 2 1 00 1 0 $B$3 *EXP((-$B$2/$B$ 1)* A6) A �I"" {." B L= 2 R= 1 00 1 0= 1 t 0.000 0.005 0.0 1 0 0.0 1 5 0.020 0.025 0.030 ""ill,
0. 1 3 0 0. 1 35 0. 1 40 0. 1 45 0. 1 50
H
C
Algebraic Form L R 10
to
1 0e
R --I
L
,,_
ohms A
let) 1 .00E+00 7 .79E-01 6.07E-01 4.72E-01 3 . 68E-01 2. 87E-01 2 .23E-01
Vet) 1 00.00 77.88 60.65 47.24 3 6.79 28.65 22.3 1
1 .5 0E-03 1 . 1 7E-03 9 . 1 2E-04 7 . 1 0E-04 5 .53E-04
0. 1 5 0. 1 2 0.09 0.07 0.06
,., '�
The following graph of the current in the inductor as a function of time was plotted using the data in columns A and B of the spreadsheet program.
Magnetic Induction
207
'\-��+--'=������ -':'=�-=--1
0.0
0.00
0.03
0.06
t
0.09
0.12
0.15
(5)
The following graph of the voltage across the inductor as a function of time was plotted using the data in columns A and C of the spreadsheet program. 100
80
� ;".
60
40
20
0 0.00
0.03
0.06
t
0.09
0.12
0.15
(5)
For the circuit of Example 2 8- 1 3 , find the time at which the power *80 •• dissipation in the resistor equals the rate at which magnetic energy is stored in the inductor. Picture the Problem If the current is initially zero in an LR circuit, its value at some later time t is given by
I = If (1 e-t/r 1 where If = eo!R and T = LIR is the time constant -
for the circuit. We can find the time at which the power dissipation in the resistor equals the rate at which magnetic energy is stored in the inductor by equating expressions for these rates and using the expression for 1 and its rate of change. Express the rate at which magnetic energy is stored in the inductor: Express the rate at which power is
dUL = � [.lLI 2 ]= LI dI dt dt 2 dt
208
Chapter 28
dissipated in the resistor: Equate these expressions to obtain :
Simplify to obtain:
(1)
1= Ir(l- e-t/r )
Express the current and its rate of change:
and
dI = dt =
(1 - e-t/r )= -I e-t/r ( _!) If !!..dt f I e-t/r f r
r
Substitute in equation (1) to obtain: or
1- e-t/r = e-t/r 1 = 2 e-t/r t - -dnl. -
Solve for t: Using
T=
=>
3 3 3 f.1s from Example 28-
1 1, evaluate t to obtain:
2
t = -(333 ,us)ln1 = I 23 1f.1s I
General Problems
*85
••
Figure 28-58 shows an ac generator. It consists of a rectangular loop of
dimensions a and b with N turns connected to slip rings. The loop rotates with an angular velocity OJ in a uniform magnetic field B. (a) Show that the potential difference between the two slip rings is
E=
NBabOJ sin
at what angular frequency value is 110 V?
OJ must
CtJt.
(b) If a = 1 cm, b = 2 cm, N = 1000, and B = 2 T,
the coil rotate to generate an emf whose maximum
Magnetic Induction 209 I�·---b (JJ
1
�I
x-x B
x
x
X
x
x
X
N turns
Figure 28-58 Problems 85 and 86 Picture the Problem We can apply Faraday's law and the definition of magnetic flux to derive an expression for the induced emf in the coil (potential difference between the slip rings). In part (b) we can solve this equation for wunder the given conditions. (a) Use Faraday' s law to express the induced emf: Using the definition of magnetic flux, relate the magnetic flux
C =_ drpm dt rpm ( t)= NBAcos (f)t
through the loop to its angular velocity: Substitute to obtain:
c
sin{f)t = 1
(b) Express the condition under which & = &max: Solve for and evaluate condition:
w under
= - !!:...-[NBAcos{f)]t dt = - NBabm(- sinmt) = I NBab{f)sinmt I
this
{f)= Cmax NBab
--
1 l 0V - (1 000)(2T)(0.01m)(0.02 m) = I 275 radls I
*88 ·· Show that the effective inductance for two inductors LJ and L 2 connected in parallel, so that none of the flux from either passes through the other, is given by
21 0 C hapter 28
Picture the Problem We can use the common potential difference across the parallel combination of inductors and the fact that the current into the parallel combination is the sum of the currents through each inductor to find an expression of the equivalent inductance. Define L eff by:
�
=
Leff
dI/ dt
or
dI dt Relate the common potential difference across the inductors to
£
1
=
their inductances and the rate at
and
which the current is changing in
£
each:
2
1
(1)
dI1 -'-1 dt
(2)
=
=
£ __ Leff
T
L
2
dI2 dt
=
Because the current divides at the
I I) +12
parallel junction:
and
dI dt
dI1 dt
-=-
Solve equations (2) and (3) for dI/dt and dI21dt and substitute to obtain: Express the relationship between an emf & applied across the parallel
combination of inductors and the emfs &1 and
&2 across the individual
inductors:
Substitute to obtain:
Substitute in equation ( 1 ) and solve for II Leff:
(3)
dI2 dt
+-
Magnetic Induction
211
*89 ·· Figure 28-59a shows an experiment designed to measure the acceleration of gravity. A large plastic tube is encircled by a wire, which is arranged in single loops separated by a distance of 1 0 cm. A strong magnet is dropped through the top of the loop. As the magnet falls through each loop, the voltage rises, then rapidly falls through 0 to a large negative value as the magnet passes through the loop, and the returns to O. The shape of the voltage signal is shown in Figure 28-59(b). (a) Explain how this experiment works. (b) Explain why the tube cannot be made of a conductive material. (c) Qualitatively explain the shape of the voltage signal in Figure 28-59b. (d) The times at which the voltage crosses 0 as the magnet falls through each loop in succession are given in the table in the next column. Use these data to calculate a value for g. (a)
_�...-.Magnet
Tube
Oscilloscope
(b)
ovr-------�r_--�
TIme
Figure 28-59 Problem 89 Loop Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Zero crossing time (s) 0.01 1 1 89 0.063 1 33 0. 1 0874 0. 1 4703 0. 1 805 2 0.2 1 025 0.23 85 1 0.26363 0.28853 0.3 1 1 44 0.33494 0.3 5476 0.37592 0.39 1 07
212
Chapter 28
Picture the Problem
(a)
As the magnet passes through the coil, it induces an emf because of the changing flux through the coil. This allows the coil to" sense" when the magnet is passing through it.
(b)
One cannot use a cylinder made of conductive material because eddy currents induced in it by a falling magnet would slow the magnet.
( c)
As the magnet approaches the loop, the flux increases, resulting in the increasing voltage signal. When the magnet is passing the coil, the flux goes from increasing to decreasing, so the induced emf becomes zero and then negative. The time at which the induced emf is zero is the time at which the magnet is at the center of the coil.
(d) Each time represents a point when the distance has increased by 1 0 cm. The following graph of distance versus time was plotted using a spreadsheet program. The regression curve, obtained using Excel ' s "Add Trendline" feature, is shown as a dashed line. 1.4
1.2 1.0 ...-- 0.8
S �
0.6
0.4 0.2 0.0 0.1
0
0.2 t
0.4
(5)
The coefficient of the second-degree term is
g
0.3
t g.
= 2(4.9257m1s2 ) = 1 9.85m1s2
Consequently,
I
Magnetic I *91
n du
ct ion 213
Suppose the coil of Problem 90 is rotated about its vertical centerline at
constant angular velocity of2 radls. Find the induced current as a function oftime. Picture the Problem We can apply F araday ' s law and the definition of magnetic flux to
derive an expression for the induced emf in the coil. We can then apply Ohm's law to find the induced current as a function of time. Note that only halfofthe loop is in the magnetic field.
I( t) C:R( t)
Apply Ohm ' s law to relate the
=
induced current to the induced emf:
(1)
Use Faraday' s law to express the induced emf:
rpm ( t)=
Using the definition of magnetic
NBAcos wt
flux, relate the magnetic flux through the loop to its angular velocity:
C:( t)= - � [NBAcos wt] dt = - NBAw(- sinwt) = NBAw sin wt
Substitute to obtain:
. Wt I( t) NBAw Slll R
Substitute in equation ( 1 ) to obtain:
Substitute numerical values and evaluate
=
I(t):
. (2 radJ)s t I( t)= ( SO)(1. 4T)( O . 252m )(Q O.15 m)(2radls)SIll 4 I (0.350A )sin(2rad/s)t I =
*96 ••• Figure 2 8-62 shows a rectangular loop of wire, 0.30 m wide and 1 .50 m long, in the vertical plane and perpendicular to a uniform magnetic field B = 0.40 T, directed inward as shown. The portion of the loop not in the magnetic field is 0. 1 0 m long. The resistance of the loop is 0.20 n and its mass is 0.05 kg. The loop is released from rest at
t
=
O. (a) What is the magnitude and direction of the induced current
when the loop has a downward velocity v? (b) What is the force that acts on the loop as a result of this current? (c) What is the net force acting on the loop? (d) Write the equation of motion of the loop. (e) Obtain an expression for the velocity of the loop as a function
214 Chapt er 28 o f time. ( f) Integrate the expression obtained in Part (e) t o find the displacementy a s a function of time. (g) Using a spreadsheet program, make a graph of the positiony of the loop as a function of time for values ofy between 0 m and 1.4 m (i.e., when the loop
leaves the magnetic field). At what time t doesy = 1.4 m? Compare this to the time it
would have taken if B = O .
fx----x----x- - - x-: ;x
x
x
;x
x
x
:x
x
x
x;
x
x;
,
:x
x
:x
x
:x
x
,
x
x;
x:
Figure 28-62 Problems 96 and 97 Picture the Problem We can use 1= c/R and & = Bv f to find the current induced in the loop and Lenz' s law to determine its direction. We can apply the equation for the force
on a current-carrying wire to find the net magnetic force acting on the loop and then sum the forces to find the net force on the loop. Separating the variables in the differential equation and integrating will lead us to an expression for v(t) and a second integration to an expression foryet). We can solve the latter equation fory =
1.40 m to find the time it
takes the loop to exit the magnetic field and our expression for v(t) to find its exit speed.
Finally, we can use a constant-acceleration equation to find its exit speed in the absence of the magnetic field. (a) Relate the magnitude of the induced current to the induced emf
1=
£
R
and the resistance of the loop: Relate the induced emf to the
£= Bvl!
motion of the loop: Substitute for & to obtain:
1=
I �I! I v
As the loop falls, the flux into the page decreases. The direction of the induced current is such that its magnetic field opposes this decrease, i.e., clockwise. (b) Express the velocity-dependent force that acts on the loop in terms
Magnetic Induction
215
of the current in the loop: Substitute for I to obtain:
Apply dF = Idf!. x B to the horizontal portion of the loop that is in the magnetic field to conclude that the net magnetic force is upward. Note that the magnetic force on the left side of the loop is to the left and the magnetic force on the right side of the loop is to the right. (c) The net force acting on the loop is the difference between the downward gravitational force and the upward magnetic force: (d) Apply Newton' s 2nd law of motion to the loop to obtain its equation of motion:
or
Factor g to obtain an alternate form of the equation of motion:
(e) Separate the variables to obtain:
dv
----
gor
B2f2
--
mR
= dt
v
dv = dt a-bv
--
where
Integrate v ' from 0 to v and t ' from o to t:
Vf-dv' S I ' = dt
o
B2f!.2
a= g and b = - mR
a -bv,
0
=>
1( )
a-bV --In -- = t b a
Chapter 28
216
Transform from logarithmic to exponential fom1 and solve for v to
ob tain : .
Notmg that
vt
V{t)= I (1- e-'/r ) I
a
= - , we have: b
vt
where
if) Write
v
as
r
v
v
a
g
=--.t... =--.t....
dyldt and separate
variables to obtain: Integrate
y' from 0 to y and t' from 0 to t:
fo dy'= Vt f(l- e-"/r }it'
Y
I
0
y{t)= I VI [t - (1 - e-l/r )] I
and
r
y
(g) A spreadsheet program to generate the data for graphs of position as a function of time is shown below. The formulas used to calculate the quantities in the columns are as follows:
t
Cell Bl B2 B3 B4 BS
F ormulalContent O.OS 0.2 0.4 0.3 $B$ 1 *$B$7*$B$2/($B$}I'2*$B$4"2)
Algebraic Form
B6 B7 AlO B I0 CI0
$B$SI$B$7 9.8 1 0.00 $B$S*(AI0-$B $6*( 1 -EXP(-AI01$B$6))) 0.S*$B$7*AI0"2
r
R= B= L= vt=
0.2 0.4 0.3 6.8 1 3 0.694 9.8 1
m mls s mls"2
m
R B L
VI t y
g I
t2
-zg
Magnetic Induction r
9 ..10
,11
;, 1 2 ,,1 3
. 't, 14 15
"? , ;1 6-,;"
'':''
'
"
"
','
""
,
17
18 \;1 9 ',,"'10 21" 22 1'"/ ;;,' 2 3 OJ, 24 25 26
;'27-
.
,
28 29
3'0
' ''' ,'f
,.;
t 0.00 0.05 0.10 0. 1 5 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1 .00
Y
Y (no B)
0.000 0.0 1 2 0.047 0 . 1 03 0.179 0.273 0.384 0.51 1 0.654 0.809 0.978 1 . 1 59 1 .3 51 1 .553 1 .764 1 .985 2 .2 1 4 2 .451 2 .695 2.946 3 .202
0.000 0.0 1 2 0.049 0.1 10 0. 1 96 0.307 0.44 1 0.60 1 0.785 0.993 1 .226 1 .484 1 .766 2.072 2.403 2.759 3 . 1 39 3 .544 3 .973 4.427 4.905
Examining the table, we see that y = 1 .4 m when t :::::
217
I 0.60 I s.
The following graph shows y as a function of t for B 7:- 0 (solid curve) and B = 0 (dashed curve).
0.0
0.2
0.4
I
0.6
(5)
0.8
1.0
Chapter 29 Alternating-Current Circuits Conceptual Problems *1
•
As the frequency in the simple ac circuit in Figure 29-27 increases, the rms
current through the resistor (a) increases. (b) does not change. (c) may increase or decrease depending on the magnitude of the original frequency. (d) may increase or decrease depending on the magnitude of the resistance.
(e) decreases.
R
Figure 29-27 Problem 1 Determine the Concept Because the rms current through the resistor is given by
ImlS *5
=
cmlS/ R
•
and both
&rms
and R are independent of frequency,
I (b) is correct. I
If the frequency in the circuit in Figure 29-29 is doubled, the capacitive
reactance of the circuit will (a) increase by a factor of 2 . (b) not change. (c) decrease by a factor of2. (d) increase by a factor of 4. (e) decrease by a factor of 4.
-
c
Figure 29-29 Problem 5 Determine the Concept The capacitive reactance of an capacitor varies with the frequency according to Xc
*9
••
=
1/ OJC.
Hence, doubling
(i)
will halve Xc.
I (c) is correct. I
Making LC circuits with oscillation frequencies of thousands of hertz or
more is easy, but making LC circuits that have small frequencies is difficult. Why?
219
220
Chapter
29
Determine the Concept To make an LC circuit with a small resonance frequency
requires a large inductance and large capacitance. Neither is easy to construct. *1 2
·
Are there any disadvantages to having a radio tuning circuit with an
extremely large Q factor? Determine the Concept Yes; the bandwidth must be wide enough to accommodate the modulation frequency. Estimation and Approximation *18
··
The impedances of motors, transformers, and electromagnets have inductive
reactance. Suppose that the phase angle of the total impedance of a large industrial plant is 25° when the plant is under full operation and using 2.3 MW of power. The power is supplied to the plant from a substation 4.5 km from the plant; the 60 Hz rms line voltage
at the plant is 40,000 V. The resistance of the transmission line from the substation to the plant is 5.2 Q. The cost per kilowatt-hour is 0.07 dollars. The plant pays only for the
actual energy used. (a) What are the resistance and inductive reactance of the plant' s total
load? (b) What is the current in the power lines and what must be the rms voltage at the substation to maintain the voltage at the plant at 40,000 V? (c) How much power is lost in transmission? (d) Suppose that the phase angle of the plant' s impedance were reduced to 1 8° by adding a bank of capacitors in series with the load. How much money would be saved by the electric utility during one month of operation, assuming the plant operates at full capacity for 1 6 h each day? (e) What must be the capacitance of this bank of capacitors? Picture the Problem We can find the resistance and inductive reactance of the plant's total load from the impedance of the load and the phase constant. The current in the power lines can be found from the total impedance of the load the potential difference across it and the rms voltage at the substation by applying Kirchhoff's loop rule to the substation-transmission wires-load circuit. The power lost in transmission can be found from Ptrans = I!lSRtrans. We can find the cost savings by finding the difference in the power lost in transmission when the phase angle is reduced to 1 8°. Finally, we can find the capacitance that is required to reduce the phase angle to 1 8° by first finding the
capacitive reactance using the definition of tan 8 and then applying the definition of
capacitive reactance to find C.
Rtr•n,
=
5.2.Q
Z = 1� + iXt.
(irms
=
40 kV
0=25°
Alternating-Current Circuits (a) Relate the resistance and inductive reactance of the pla nt's total load to
Z and 5:
Express Z in terms of the current I in the power lines and voltage &""s at
the plant: Express the power delivered to the plant in terms of CmlS' 1m1s and J and ,
solve for 1ffi1S:
R
=
Zcosb
and
XL = Zsinb Z EI
mlS
=
P.v = E
mlS
ImlS
cos b
and
Pay I m1S = ----
(1)
E cos b mlS
Substitute to obtain:
Substitute numerical values and evaluate Z:
Substitute numerical values and evaluate R and XL:
Z= Z
=
&�
cosb
P.v
( 40kVY cos 25° 630n 2.3 MW =
R (630n)cos25° =
=
and
XL (630n)sin25° =
(b) Use equation ( 1 ) to find the current in the power lines:
I
=
Ztot
=
ffi1S
1 571n 1
=
2.3 MW ( 40kV)cos250
1 266n I
=
I 63: 4A I
Apply Kirchhoff s loop rule to the circuit: Solve for Csub: Evaluate
Ztot:
Substitute numerical values and evaluate Csub:
Esub
=
= =
�R2 + X�
�(57 1nY +(266ny 630n =
(63 .4AX5.2n + 630n) 1 40.3kV I
22 1
222
Chapter
29
(c) The power lost in transmission IS:
2 R trans = (63 . 4 A)2 (S . .Ptrans = 1 mlS =
1 20.9 kW I
') Q)
(d) Express the cost savings ;j.C in
terms of the difference in energy consumption per-unit cost
(P25° - PI 8o);j.t and the
U
of the energy:
Express the power list in transmission when 0 = 1 8°: Find the current in the transmission lines when 0 = 1 8° : Evaluate
11 80 =
2.3 MW = 60.S A (40 kV )cos1 8°
� 80 :
Substitute numerical values and evaluate ;j.C:
!1C = (20.9kW - 1 9.0kW)(16h1d)(30dlrnonth)($0.07 / kW . h) = 1 $63.84 1 Relate the new phase angle 0 to the
inductive reactance XL, the reactance
tan 0 = xL
- XC
R
due to the added capacitance Xc, and the resistance of the load R: Solve for and evaluate Xc:
Xc
=
XL
- R tan 0
= 266Q - (S71 Q)tan1 8° = 80.S Q
Substitute numerical values and evaluate C: Alternating Current Generators *21
·
A 2-cm by 1 .S-cm rectangular coil has 300 turns and rotates in a magnetic
field of 4000 G. (a) What is the maximum emf generated when the coil rotates at 60 Hz? (b) What must its frequency be to generate a maximum e mf of 1 1 0 V? Picture the Problem We can use the relationship
Emax
= 21iNBAf to relate the
maximum emf generated to the area of the coil, the number of turns of the coil, the
Alternating-Current Circuits magnetic field in whic h the coil is rotating, and the frequency at which i t rotates.
emax =
(a) Relate the induced emf to the
=
NBA (1) 2TiNBAf
(1)
magnetic field in which the coil is rotating: Substitute numerical values and evaluate
Cmax :
(b) Solve equation ( 1 ) forf
f-
emax
2TiNBA
Substitute numerical values and evaluate f
1 l0V - 2 JZ" (300)(0. 4 T)(2 10-2 )(1 .5 10-2 )
f-
x
m
m
x
-
-
I 486Hz I
Alternating Current in a Resistor * 23
•
A l OO-W light bulb is plugged into a standard 1 20-V (rms) outlet. Find (a)
Inns, (b) Imax, and (c) the maximum power. Picture the Problem We can use and
Pmax = 1max cmax to fmd Pmax.
P.v = emlJmlS to find Inns, Imax = .J21nns to find Imax,
(a) Relate the average power delivered by the source to the rms voltage across the bulb and the rms current through it: Solve for and evaluate Inns:
Inns
=
P.v
emlS
=
lOO W 1 20V
=
I 0.833A I
(b) Express Imax in terms of Inns: Substitute for Inns and evaluate Imax :
( c) Express the maximum power in terms of the maximum voltage and
1max
=
.J2(0 . 8 33A )= I l.I8A I
223
224 Chapter 29 maximum current: Substitute numerical values and evaluate
Pmax =
(1 . 18A)J2(1 20V) 1 200 W I =
Alternating Current in Inductors and Capacitors *29
•
An emf of 1 0 V maximum and frequency 20 Hz is applied to a 20-J.LF
capacitor. Find (a) Imax and (b) ImlS •
Picture the Problem We can use lmax = &maxlXC and Xc = 1 1 we to express Imax as a function of &max,j, and C. Once we've evaluate Imax, we can use
Im1S = ImaJ J2 to find Im1S.
Express Imax in terms of &max and Xc:
max Imax = & XC
Express the capacitive reactance:
Xc
= _1_ = _1_ OJC 2JifC
Substitute to obtain: (a) Substitute numerical values and evaluate
lmax:
(b) Express Im1S in terms ofImax :
= 2Jr(20s -I )(20,uF)(lOV) = I 25. 1 mA I lmax = 25 . 1 mA = I i7 . 8 mA I IJlllS =
Imax
J2
J2
.
.
LC and RLC Circuits without a Generator *32
•
Show from the definitions of the henry and the farad that
unit s-I . Picture the Problem We can use XL = (f)L and Xc = 1 1 we to show the
1/.JLC 1/.JLC
has the
has the
unit S- I . Alternatively, we can use the dimensions of C and L to establish this result. Substitute the units for L and C in the expression
1/.JLC
to obtain:
Alternating-Current Circuits
[e] = [Q] [V ]
A l ternatively, use the defining
equation (C
=
225
QI V) for capacitance
to obtain the dimension of C: Solve the defining equation = dJ/ dt ) for inductance to
(V L
obtain the dimension of L :
Le :
Express the dimension of 1/..J
1 [V ][T]2 [Q] [Q] [V ]
Le has units of �
Because the SI unit of time is the second, we've shown that 1/..J
*37 ·· An inductor and a capacitor are connected, as shown in Figure 29-30. With the switch open, the left plate of the capacitor has charge Qo. The switch is closed and the charge and current vary sinusoidally with time. (a) Plot both Q versus t and J versus t and explain how to interpret these two plots to illustrate that the current leads the charge by 90°. (b) Using a trig identity, show the expression for the current (Equation 29-3 8) leads the expression for the charge (Equation 29-39) by 90° . That is, show that
J = -Jpeaksin OJ{ = Jpeakcos (OJt + 1C ).
2
Figure 29-30 Problem 3 7 Picture the Problem Let Q represent the instantaneous charge on the capacitor and apply Kirchhoff s loop rule to obtain the differential equation for the circuit. We can then solve this equation to obtain an expression for the charge on the capacitor as a function of time and, by differentiating this expression with respect to time, an expression for the current as a function of time. We' ll use a spreadsheet program to plot the graphs.
226
Chapter
29
Apply Kirchhoff s loop rule to a clockwise loop just after the switch is closed: Because 1 =
dQ/dt : Q(t) = Qo c os (cut - 8)
The solution to this equation is:
where
cu =
JLC1
Because Q(O) = Qo, 8= 0 and:
Q(t) = Qo cos cut
The current in the circuit is the derivative of Q with respect to t:
1
=
dQ = !!.- [Qo cos cut] = -cuQo sin cut dt dt
(a) A spreadsheet program was used to plot the following graph showing both the charge on the capacitor and the current in the circuit as functions of time. L, C, and Qo were all arbitrarily set equal to one to obtain these graphs. Note that the current leads the charge by one-fourth of a cycle or 90°.
o
-1 o
2
4
6 t
8
10
(s)
(b) The equation for the current is:
1 = -cuQo sin cut
The sine and cosine functions are related through the identity:
- sin B = cos B + ;
Use this identity to rewrite equation (1):
1
( )
=
(1)
( )
-cuQo sin cut = cuQo cos cut + %
showing that the current leads the charge by 90°.
Alternating-Current Circuits
227
RL Circui ts w ith a Generatol" *44 ·· A resistor and an inductor are connected in parallel across an emf C = Cmax as shown in Figure 29-32 . Show that (a) the current in the resistor is lR = (cmax /R) cos 0Jl, (b) the current in the inductor is h ( Cmax /XL) cos (wt - 90°), and (c) 1 = lR + h = Imax cos (wt - 5), where tan 8= R/XL and Imax = &max/Z with =
Z-2 = R-2 + XZ2 .
R
L
Figure 29-32 Problem 44 Picture the Problem We can apply Kirchhoff s loop rule to obtain expressions for lR and
lL and then use trigonometric identities to show that 1 = lR + h
tan 8 = R/XL and Imax = cma.jZ with
Z-2 = R-2 + XZ2 .
= Imax cos (wt - 5), where
(a) Apply Kirchhoff s loop rule to a clockwise loop that includes the source and the resistor: Solve for h
E IR - �coS - R OJt
(b) Apply Kirchhoff s loop rule to a
Emax
clockwise loop that includes the
because the current lags the potential
source and the inductor:
difference across the inductor by 90°.
Solve for h:
IL =
cos(OJt - 900)- IL XL = 0
�laXL cos(cot -90°)
(c) Express the current drawn from the source in terms of lmax and the phase
constant 8:
Use a trigonometric identity to expand cos(wt - 5):
1 = Imax (cos OJt cos 6' + sin OJtsin 8) = Imax cos OJt cos 8 + Inlax sin OJt sin 6'
228
Chapter 29
From our results in (a):
1
=
1R + 1L
= £max cos cut
£max XL
cos(cut 9 0°)
+
=£
� COS
R
A useful trigonometric identity is:
R
-
cut + £max sm. cut XL
A cos cut + B sin cut = .JA 2 + B 2 cos(cut -8)
where
8 = tan -1 B A
Apply this identity to obtain:
{ ) (�' ) co,("" -o)
I � . &;'
'
'+
(1)
and
8= Simplify equation ( 1 ) and rewrite equation (2) to obtain:
£max XL tan-I I
£max R
=
tan
-l
( J R
XL
(2)
{ ) (�) co,("" - o) G) +UJ co,("" - o) �G), co,("" - o)
I � . &;'
'
� &="
'
+
'
� &=,
=1
£max
where
Z
tano �
cos(cut -8)
I l R XL
and
_1_2 = _1_2 + _1_2 Z
R
XL
Filters and Rectifiers *48 ·· The circuit shown in Figure 29-3 5 is called an RC high-pass filter because it transmits signals with a high-input frequency with greater amplitude than low-frequency
Alternating-Current Circuits signals. If the input voltage is �n
VOUI ( t )= VH cos (OJt - 5) where Vpeok VH = ----;======= 1+
229
( t )= VpCOk cos OJt, show that the output voltage is
(_)2 1
OJRC
C
----f l------r-Figure 29-35 Problem 48 Picture the Problem We can use Kirchhoff s loop rule to obtain a differential equation relating the input, capacitor, and resistor voltages. We'll then assume a solution to this equation that is a linear combination of sine and cosine terms with coefficients that we can find by substitution in the differential equation. Repeating this process for the output side ofthe filter will yield the desired equation. Apply Kirchhoff s loop rule to the input side of the filter to obtain:
Substitute for Yin and I to obtain:
Because Q
=
C V:
Substitute for dQldt to obtain:
V;n - V - IR = 0
where V is the potential difference across the capacitor.
dQ Vpeak co s OJt - V - R -= 0 dt dQ � dV = [CV]= C dt dt dt dV Vpeak cOS OJt - V - RC-= 0 dt the differential equation describing the potential difference across the capacitor.
Assume a solution of the form:
V= � cos OJt + � sin OJt
Substitution of this assumed solution and its first derivative in the differential equations, followed by equating the coefficients of the sine and cosine terms, yields two coupled linear equations:
Vc + OJR C� = Vpeak and
� - OJRC� = 0
230 Chapter 29 Solve these equations simultaneously to obtain:
vc
1
=
Va l + (cuRCY pe k
and
cuRC Vpeak 1 + (cuRCY
Vs =
Note that the output voltage is the voltage across the resistor and that it is phase shifted relative to the input voltage:
VH COS{cut - 5) where VH is the amplitude of the signal.
Assume that VH is of the form:
VH (t) = Vc cos cut + Vs sin cut
The input, output, and capacitor voltages are related according to:
=
VOllt
VH (t ) = �n (t ) - V(t )
Substitute for VH (t ) , Vpeak (t ) , and V {t ) and use the previously established values for Vc and Vs to obtain:
Vpeak - Vc
Vc
=
Vs
= -
and
�
Substitute for Vc and Vs to obtain:
vc - (cuRCY Vpeak 1 + (wRCY and cuRC vs = Vpeak 1 + (cuRCY
VH, vc, and Vs are related according to the Pythagorean relationship:
VH = Jv� + v�
_
Substitute for Vc and Vs to obtain:
cuRC V Jl + (wRCY peak Vpeak =
VH =
_rC"�c)'
Show that the average power dissipated in the resistor of the high-pass filter of Problem 48 is given by
*53
··
(
Vp�ak {wRCY P,ve = 2R 1 + {wRCY Picture the Problem
J
We can express the instantaneous power dissipated in the resistor
Alternating-Current Circuits
23 1
and then use the fact that the average value of the square of the cosine function over one cycle is 1'2 to establish the given result. The instantaneous power pet) dissipated in the resistor is:
PCt) =
VO�,t R
The output voltage Vout is: From Problem 4 8 :
Substitute i n the expression for pet) to obtain:
V.H
=
(_1)2 Vpea k
---;==��=
1+
v. 2 R [
OJRC
PCt) = --1:L COS2 (OJt
-
0)
( )2
V,;.'
1 R 1+ _ OJRC
rS' (wt - O)
Because the average value of the square of the cosine function over one cycle is 1'2:
Simplify this expression to obtain:
p
ave
- 2R ( 1 +(wRCY (OJRCY J -
Vp�ak
*57 ·· Using a spreadsheet program, make a graph of VL versus/ rd27r and 5 versus/for the low-pass filter of Problem 55. Use R = 1 0 kQ and C = 5 nF. =
Picture the Problem We can use the expressions for VL and 5 derived in Problem 56 to plot the graphs of VL versus/and 0 versus/for the low-pass filter of Problem 55. We'll simplify the spreadsheet program by expressing both VL and 5 as functions ofh dB . From Problem 56 we have:
L = �1 + (OJRCY
v
Vpeak
and
0 = tan-l (OJRC)
Rewrite each of these expressions in terms ofh dB to obtain:
2 ( L J 1+ h
dB
232 Chapter 29
( ]
and 8
= tan- J (2;ifR C) tan- J L =
J3dB
A spreadsheet program to generate the data for graphs of VL versus/and 5 versus/for
the low-pass filter is shown below. Note that Vpeak has been arbitrarily set equal to 1 V. The formulas used to calculate the quantities in the columns are as follows: Cell Bl B2 B3 B4 B8
F ormula/Content 2 .00E+03 5 .00E-09 1 (2 *PIO * $B$ l * $B$2)",- 1 $B$3/SQRT( 1 +((A8/$B$4)",2))
C8
ATAN(A8/$B$4)
D8
C8 * 1 80IPIO
A graph of Vout as a function of / follows:
Algebraic Form R C Voeak
h dB
Vpeak
( ) tan-' ( J
�
1+
J hdB
L hdB
'
5 in degrees
Alternating-Current Circuits
23 3
1 .0
0.8
�
-:::.g..
0.6
0.4
0.2
0.0 0
10
20
30
40
50
30
40
50
f (kHz)
A graph of 6 as a function of f follows:
';;) '" f! OJ) '" 2� ., '0
60
40
20
o
10
20
f (kHz)
*59
•••
frequency
Show that the trap filter, shown in Figure 29-3 7, acts to reject signals at a
0) 1/..JL C =
.
How does the width of the frequency band rejected depend on
the resistance R?
\(in
R
Figure 29-37 Problem 5 9 Picture the Problem We can apply Kirchhoff s loop rule to both the input side and output side of the trap filter to obtain an expression for the impedance of the trap. Requiring that the impedance of the trap be zero will yield the frequency at which the circuit rejects signals . Defining the bandwidth as
/),.0) 10) =
-
{j)lrap
1 and requiring that
234 Chapter 29
IZlrap 1 = R will yield an expression for the bandwidth and reveal its dependence on R. Apply Kirchhoff s loop rule to the output of the trap circuit to obtain:
VOtlt - JXL - JXc = 0
Solve for VOtlt:
= J (XL + Xc ) = JZtrap where Ztrap = XL + Xc Vout
(1)
Apply Kirchhoffs loop rule to the input of the trap circuit to obtain:
�n �n = -----""--J = ------"-'--R + XL + Xc R + Ztrap
Solve for I:
Substitute for I in equation ( 1 ) to obtain:
Because
Vout
= VIn
Ztrap R + Ztrap
XL = i wL and
-i
Xc = - : wC Note that Ztrap = 0 and Vout = 0 provided: Let the bandwidth /j. OJ be: Let the frequency bandwidth to be defined by the frequency at which
IZtrap I = R . Then:
Because
wtrap =
Solve for
2 2
m - mtrap :
OJ ;::: OJtrap, m - trap 2 mtrap :
Because W
.1 .JLC
�
(2)
1 mL - - = R mC or
o/LC- 1 = mRC
(�J2 mtrap
- 1 = mRC
Alternating-Current Circuits Substitute in equation (2) to obtain :
tJ.
W W = I W W.mp 1 = R C ,�,"P
2 I 2L I
_
=
R
23 5
LC Circuits with a G enerator *64
•• •
L C circuit
One method for measuring the compressibility of a dielectric material uses an with a parallel-plate capacitor. The dielectric is inserted between the plates and
the change in 'resonance frequency is determined as the capacitor plates are subjected to a compressive stress. In such an arrangement, the resonance frequency is 1 20 MHz when a dielectric of thickness 0 . 1 em and dielectric constant
K =
6.8 is placed between the
capacitor plates. Under a compressive stress of 800 atm, the resonance frequency decreases to 1 1 6 MHz. Find Young's modulus of the dielectric material. Picture the Problem We can use the definition of the capacitance of a dielectric-filled capacitor and the expression for the resonance frequency of an LC circuit to derive an expression for the fractional change in the thickness of the dielectric in terms of the resonance frequency and the frequency of the circuit when the dielectric is under compression. We can then use this expression for /).t/t to calculate the value of Young' s modulus for the dielectric material. Use its definition to express Young' s modulus of the dielectric material: Letting t be the initial thickness of the dielectric, express the initial capacitance of the capacitor: Express the capacitance of the capacitor when it is under compreSSIOn: Express the resonance frequency of the capacitor before the dielectric is
Y
-
_
tJ.P
(1)
_ - --
strain
tJ.t/t
Co KEo A =
t
Cc KEo A =
W
0
compressed: When the dielectric is compressed:
e
s tr s s --
-
t - tJ..t
)CoL - �KE� AL 1
1
--- - --= = = =
c - )CcL - �KEo AL
W
1
1
= - -- - --= = =
t - tJ..t
236 Chapter 29 Express the ratio of We to
eva
and
we Wo
simplify to obtain:
�KEOt �1- l:J.t �K Eo AL
_
-
_
AL
-
t
t - l:J.t
Expand the radical binomially to
We Wo
obtain:
=
( 1 )1/2 _
l:J.t t
(
I1t 2 1 - We W t =
1
_
l:J.t 2t
t.
provided I1t « Solve for I1tlt:
�
o
J
Substitute in equation ( 1 ) to obtain:
y
Substitute numerical values and evaluate Y:
-
_
=
(800 atm) (1 0 1 . 325 kPaJatm) 2 1 1 16 MHZ 120 MHz ! 1.22 x 109 N/m 2 !
(
J
RL C Circuits with a Generator *69
Show that the formula �v
··
=
R&�, j
Z2
gives the correct result for a circuit
containing only a generator and (a) a resistor, (b) a capacitor, and (c) an inductor. Picture the Problem The impedance of an ac circuit is given by
Z
=
�R 2
+
(XL - Xc Y
. We can evaluate the given expression for Pay first for
XL = XC = ° and then for R
(a) For X= O, Z = R and:
(b), (c) If R
=
=
0.
0, then:
Remarks: Recall that there is no energy dissipation in an ideal inductor or capacitor.
Alternating-Current Circuits
237
*74 " FM radio stations have carrier frequencies that are separated by 0.20 MHz. When the radio is tuned to a station, such as 1 00 . 1 MHz, the resonance width of the receiver circuit should be much smaller than 0.2 MHz, so that adjacent stations are not received. Iffo = 1 00. 1 MHz and 11/= 0.05 MHz, what is the Q factor for the circuit? Picture the Problem We can use its definition, circuit.
Q = fo /l1f to find the Q factor for the
Express the Q factor for the circuit:
Substitute numerical values and evaluate Q: *79
"
Q=
1 00. l MHz 0.05 MHz
�
=�
In the circuit shown in Figure 29-42, the ac generator produces an rms
voltage of 1 1 5 V when operated at 60 Hz. What is the rms voltage across points (a) AB,
(b) BC, (c) CD, (d) AC, and (e) BD?
137 mH
SO Q
Figure 29-42 Problem 79 Picture the Problem We can find the rms current in the circuit and then use it to find the potential differences across each of the circuit elements. We can use phasor diagrams and our knowledge the phase shifts between the voltages across the three circuit elements to find the voltage differences across their combinations.
(a) Express the potential difference between points A and B in terms of
VAB = IrmsXL
(1)
Irms and XL: Express Irms in terms of c and Z:
£
ImlS = Z
£
�R 2 + (XL - XC Y
238 Chapter 29 Eva luate XL and Xc to
obtain:
XL
and
XC
= 27ifL = 2Jr(60s-1 )(1 3 7 mH) = 5 1 .6 0 1 1 _ -_ - 27ifC 2Jr(60s-1 )(25 JLF) = 1 06. 1 0 _
_
. 1 15V �(500Y + (5 1 .6 0 - 1 06. 1 0Y = 1 .55 A
Substitute numerical values and evaluate Irms:
I =
Substitute numerical values in equation (1) and evaluate VAB :
VAB (1 .55 A)(S 1 . 6 0) 1 80.0V I
mlS
=
=
(b)
VBC = IrrnsR = (1 .55 A)(500)
(c) Express the potential difference
VCD = Irms Xc = (1 .55 A)(1 06. 1 0)
Express the potential difference between points B and C in terms of Irms and R: between points C and D in terms of Irrns and Xc: (d) The voltage across the inductor lags the voltage across the resistor as shown in the phasor diagram to the right: Use the Pythagorean theorem to find
VAC :
(e) The voltage
across the inductor lags the voltage across the resistor as shown in the phasor diagram to the right:
= 1 77.5V I = 1 1 64V I V;IC
VAC = �V1B + V�c
= �(80.0 VY + (77.5 VY = l l l 1 V I
Altemating-CUlTent Circuits
239
Use the Pythagorean theorem to find VBD:
*84
••
Show that Equation 29-48 can be written as
Picture the Problem We can substitute for XL and Xc in Equation 29-48 and simplify the resulting equation to obtain the given equation for Imax. Equation 29-48 is:
Imax
=
.
�R 2 + (Cmax ) XL - X 2 C
Substitute for XL and Xc to obtain:
Simplify algebraically to obtain:
Imax
cmax
(
)
= ---;========== = ====
1 R 2 + o i L2 1 - / o Le
2
Cmax
Cmax
*88
••
A method for measuring inductance is to connect the inductor in series with a
known capacitance, a known resistance, an ac ammeter, and a variable-frequency signal generator. The frequency of the signal generator is varied and the emf is kept constant until the current is maximum. (a) If C = 1 0 fiF, cmax = 1 0 V, R = 1 00 fl, and I is maximum at 0)= 5 000 rad/s, what is L? (b) What is lmax ? Picture the Problem We can use the fact that when the current is a maximum, XL = Xc, to find the inductance of the circuit. In (b), we can find Imax from �ax and the impedance of the circuit at resonance.
240 Chapter 29 (a) Relate XL and Xc at resonance:
Solve for L to obtain:
Substitute numerical values and evaluate L :
(b) Noting that, at resonance,
L=
1
(S O O Os-I )(lOflF)
Imax
0, express Imax in terms of the applied emf and the impedance of
X=
=
&max Z
=
10 V l oo n
=
=
I ·
4 00 mH
I
1 O.lO O A 1
the circuit at resonance: *90
••
In the circuit shown in Figure 29-44, R
= 1 0 Q, RL
=
30 Q, L =
1 50 mH, and C = 8 flF; the frequency of the ac source is 1 0 Hz and its amplitude is 1 00 V. (a) Using phasor diagrams, determine the impedance of the circuit when switch S is closed. (b) Determine the impedance of the circuit when switch S is open. (c) What are the voltages across the load resistor RL when switch S is closed and when it is open? (d) Repeat Parts (a), (b), and (c) with the frequency of the source changed to 1 000 Hz. (e) Which arrangement is a better low-pass filter, S open or S closed?
a
R
b
Figure 29-44 Problem 90 Picture the Problem Because we'll need to use it repeatedly in solving this problem, we 'll begin by using complex numbers to derive an expression for the impedance Zp of the parallel combination of C with L and RL in series. The total impedance of the circuit is then Z = R + Zp. We can apply Kirchhoff s loop rule to obtain expressions for the voltages across the load resistor with S either open or closed.
Use complex numbers to relate Zp to
RL, XL, Xc: and
Alternating-Current Circuits 241 1 ---1 1 = --+ Zp -iXc RL+ iXL L+i(XL- Xc) _ RXCXL - iRLXC -iRLXC Zp = RXCXL L+i(XL- Xc) -iRLXC RL-i(XL-Xc) Zp = RXCXL L+i(XL- XC) RL-i(XL-Xc)
or
Multiple the numerator and denominator o f this fraction by the complex conjugate o f
RL i(XL- Xc): +
Simplify to obtain: (1)
(a)
S is closed.
Because L is
shorted: Evaluate
Xc:
Substitute numerical values in
equation ( 1 ) and evaluat e Zp, and 0:
Z, IZI,
1 _ _- ( 1 Xc _- _ 2ife 27r lOs-1)(8,uP) = 1.99kO Zp = 300-i(0.4520), Z = 400-i(0.4520) , IZI = �(400/ +(0.4520/ = 140.00 1
and
In Problem 29-77 we showed that for a parallel combination o f a resistor and capacitor, the phase angle 0 is given by:
Substitut e numerical values and
evaluate 0:
5 = tan -1 ( - 0.400 4 520 ) = 1-89.40 1 No phasor diagram is shown because it is
242
Chapter
29 impossible to represent it to scale.
( b)
S is open; i.e., the inductor is
in the circuit.
Find XL :
Substitute numerical values in
equation (1) and evaluat e Zp, Z, and 5:
IZI,
XL = = 27ifL = 2Jr(1 0s-1 )(0.15H) = 9.42£1 zp = 30. 3 £1+ i (9. 0 1£1) , = 40.3 £1+i(9.0H2), IZI = �(40.3 £1)2 + (9.01£1)2 = 1 41. 3 £1 1 9.0 1£1J () = tan-I ( X) = tan-I ( 40.3£1 = 1 12.60 1 wL
z
and
R
The phasor diagram for this case is shown to the right .
( c)
S is closed.
Apply Kirchhoff s
loop rule to a loop including the
£-IR -VR L
=0
source, R, and RL:
VR = 1= Z L
Express the current I in the circuit :
£-IR
£
Substitut e and simplify to obt ain:
From (a) we have:
Zp = 30£1 -i(0.452£1) , = 40£1 -i(0.452£1) , IZI = 40. 0 £1, {) = tan-I (-0��52 ) = _0.647o::::0: 0 Z
and
i c s 243 =(1-�}lOOV)COS[(20S-' )m] 1 (7 5V)cos[(20s-l)m] 1 -IR - IXL- RL = 0 Alternatin g-Current C
Substitute numer ical values to obtain :
r u it
V"
=
S
is open.
Apply Kirchhoffs loop
rule to a loop including the source, R, L, and Solve for
RL
&
V
when S is open:
VR : I.
Express the current I in the circuit:
1= & Z
Substitute to obtain :
Substitute numerical values and
evaluate Zp and Z:
=
=
and s: u =
Substitute numerical values and evaluate
VRL :
30. 30 +i(9 . 0 10), 40.30 +i(9 . 0 10), 41. 30, tan -I ( XL ) = tan _ 1 ( 9.420 ) R+R L 40. 30
Zp Z IZI=
. 420 ) = (1 _ 100+9 41. 30 (100 V) cos[(20 S-I )m - 13.2°] =1 (53.0V)cos[(20s-l)m-13.2°] I XL 2Jr(1 000s-1 )(0.15H) 9420 Xc = 2 Jr(1000s1 -I )(8,LlF) = 19 . 90
VRL
X
(d) Find XL and Xc when f= 1 000 Hz:
S
is closed. XL
to:
=
and
= 0, and Zp simplifies
=
244 Chapter 29 Substitute numerical values in
equ ati on 1 ( ) and evaluate Zp, and 8:
Z, IZI,
=9. 170 -i (13. 8 0) , = 19.170 - i (13.80) , = �(19.17 oy + (13.80Y =1 23.6 0 1 o = tan-, ( -19.113.780° ] = 1 _ 3s.70 1
Zp Z \ZI
and
A phasor diagram for this circuit is shown to the right,
S is o pe n.
Substitute numerical
values in equation ( 1 ) and evaluate
Zp, Z, IZI, and 8:
=0.01400-i (20.30) , = 10.00 -i(20.30) , Izl = �(1O· 00r + (20. 30r = 1 22.6 0 1 Zp
Z
and Find the total impedance, its magnitu de, and phase angle f or the circuit:
= 10. 0 0 -i (20AO) , �(1O.00r + (20.40Y = 1 22. 7 0 I 8�tan-' ( - ���(} 1 - 63. 9 0 I
Z Z
=
and
The phasor diagram is shown to the right.
(e)
\
The load voltage at the higher frequency is much more attenuated with S open, while opening S does not reduce the low frequency load voltage significantly. Therefore, S open is the better arrangement for a low - pass filter.
*97
•••
(a) Show that Equation 29-47 can be written as
tan 5
= Q(o/(j)(j)-o (j)2 ) 0
(b) Show that near resonance
Alternating-Current Circuits
245
())
(c) Sketch a plot of 6'v ersus x, where x = wi %, for a circuit with high Q and for one with
low Q .
Picture the Problem
We can manipulate Equation 29-47 into a form that has the ratio of
L to R in it and then use the definition of Q to eliminate L and R. In ( b) we can approximate
u/ - ()); , near resonance, as 2OJofj,0J and substitute in the result from ( a) to
obtain the desired result. ( a) From Equation 29-47:
tan 8
OJL -ljOJC = OJ2 L R OJR 2 ))2 = L(OJ -1/ L ) = L(( - OJ; ) OJR OJR
-ljC
=
C
Express Q in terms of
%,
L and R :
Solv e for LlR to obtain:
Substitute to obtain:
( b) Near resonance:
Q = OJoRL L Q R OJo
-=
tan 8 =
Q(OJ2 - OJ;) ())())o
( 1)
OJ2 - OJ; = (OJ+OJo)(())-OJo) � 2())0fj,())
Substitute in equation ( 1) to obtain:
(c) A following graph of 6' as a function of x = wi % w as plotted using a spreadsheet program. The solid curve is for a high-Q circuit and the dashed curve is for a low-Q circuit.
246 Chapter 29 1.5 1.0
.g �
0.0
-1.0 -1.5
0.0
*102
0.5
1.0 x
2.0
1.5
One method for measuring the magnetic susceptibility of a sample uses an
•••
LC c ircuit consisting of an air-core solenoid and a capacitor. The resonant frequency of the c ircuit w ithout the sample is determined and then measured again w ith the sample inserted in the solenoid. Suppose the solenoid is 4 cm long, 0 . 3 cm in diameter, and has 400 turn s of fine wire. Assume that the sample that is inserted in the solenoid is also long and fills the air space. Neglect end effects. (In practice, a test sample oflrnown
4
cm
susceptibility of the same shape as the unknown is used to calibrate the instrument.) ( a) What is the inductance of the empty solenoid?
(b)
What should be the capacitance of the
capacitor so that the resonance frequency of the c ircuit without a sample is 6.0000 MHz?
(c) When a sample is inserted in the solenoid, the resonance frequency drops to 5.9989
MHz. Determine the sample 's susceptibility. Picture the Problem
We can use L
=
JLon2 Af to determine the inductance of the empty
solenoid and the resonance condition to find the capacitance of the sample-free c ircuit when the resonance frequency of the circuit is 6.0000 MHz. By expressing L as a function offo and then evaluating dfoldL and approximating the derivative with !1foiM ,
w e can evaluate X from its definition. ( a) Express the inductance of an air core solenoid:
Substitute numerical v alues and evaluate L: L
=
2 1Z' (0. 003m)2(0.04m) I 35.5,uH I (41Z' 10-7 N/A2)( 0.400 ) 04m 4 X
=
247
Alt ernating-Current Circuit s (b) Express the condition for resonance in the LC circuit :
or
1_ 2..riL0 = _ 2lifoC
( 1)
/1:/
Solve for C to obt ain:
Substitut e numerical values and evaluate C:
(c) Express the sample ' s
susceptibility i n terms of L and M: Solve equation ( 1) forfo:
Differentiat e 10 with respect to L:
1 C - 47r2(6 MHz)(3S. S,uH) --/ 119 =M X
;if
/
( 2)
L
1 27r-JLC dlo 1 d 1 dL 27r.,Jc dL 47r.,Jc 1 = _ 10 47rL-JLC 2L 10 =
r-J/2 =
=
_
L-3/2
=
Approximate dfoldL by f:.folM :
M -=-- or-=--
!110 M
Substitute i n equation ( 2) t o obt ain:
x=
Substitute numerical values and
x=
evaluate x:
10
2L
!110 10
2L
-2 !11100 _2( S.99 896.M0HZ000-6.MHz0000o MHZ ) = 1 3. 67 xlO-4 1
The Transformer *105
•
An ac voltage of 24 V is required for a device whose impedance is 1 2 Q.
( a) What should the turn ratio of a transformer be, so the device can be operated from a 1 20-V line?
(b)
Suppose the transformer is accidentally connected reversed ( i.e., with the
secondary winding across the 1 20-V line and the 1 2-0 load across the primary). How much current will then flow in the primary winding?
Chapter 29
248
Picture the Problem Let the subscript I denote the primary and the subscript 2 the
secondary. We can use
V2N1
=
V;N2 and N/l NJ2 to find the tum ratio and the =
primary current when the transformer connections are reversed. (1)
(a) Relate the number of primary and secondary turns to the primary and secondary voltages: Solve for and evaluate the ratio
N2INI:
(b)
Relate the current in the primary
to the current in the secondary and
N2 NI I1
_
=
24 V -I! 1 V; 120V 5
V2
_
N2
I N 2 1
to the turns ratio: Express the current in the primary winding in terms of the voltage across it and its impedance:
I
2
=
V2 2
Z
Substitute to obtain: Substitute numerical values and evaluate II:
*110
••
An audio oscillator (ac source) with an internal resistance of2000 Q and an
open-circuit rms output voltage of 12 V is to be used to drive a loudspeaker with a resistance of 8 Q. What should be the ratio of primary to secondary turns of a
transformer, so that maximum power is transferred to the speaker? Suppose a second identical speaker is connected in parallel with the first speaker. How much power is then supplied to the two speakers combined? Picture the Problem Note:
In a simple circuit maximum power transfer from source to
load requires that the load resistance equals the internal resistance of the source. We can use Ohm's law and the relationship between the primary and secondary currents and the primary and secondary voltages and the turns ratio of the transformer to derive an expression for the turns ratio as a function of the effective resistance of the circuit and the resistance of the speaker(s).
Alternating-Current Circuits 249 Express the effective loudspeaker resistance at the primary of the
R
efT
=
�
1
I
transformer:
Express
I] in terms of h N], and N2:
1] 12 N 2 N] =
Substitute to obtain:
(1 )
(2)
Express the power delivered to the two speakers connected in parallel: Find the equivalent resistance Rsp of the two 8-0 speakers in parallel:
1 1 1 2 1 Rsp 80 80 80 40 Rsp = 40
-=-
+ -=- = --
and
Solve equation (1) for RefT to obtain:
Substitute numerical values and evaluate RefT:
Find the current drawn from the source:
V = 12V = 4.00mA I] =Rtot 20000 9990 +
250
Chapter 29 �p
Substitute numerical values in equation (2) and evaluate the power
=
(4mAy(999Q)= 1 16.0mW I
delivered to the parallel speakers: General Problems *115
Figure29-47 shows the voltage Vversus time t for a square-wave voltage.
••
If V = 12 V, (a) what is the rms voltage of this waveform? (b) If this alternating
o
waveform is rectified by eliminating the negative voltages, so that only the positive voltages remain, what now is the rms voltage of the rectified waveform?
v
Va
t
Figure 29-47
Problem 115
Picture the Problem
The average of any quantity over a time interval f:...T is the integral
of the quantity over the interval divided by f:...T. We can use this definition to find both the
average of the voltage squared,
(V 2 t and then use the definition of the rms voltage.
(a) From the definition of Vrn1s we have: Noting that
- Vo2 V0 2 , evaluate =
(b) Noting that the voltage during the second half of each cycle is now zero, express the voltage during the first half cycle of the time interval
+f:...T:
Express the square of the voltage during this half cycle:
Alternating-Current Circuits 251 Calculate
(V2 t by integrating V2
from t = 0 to t =
by 6.T:
t t1T
and dividing
(V2)
av
=
V02
t1T
r,L\7 dt 1
=
V02 [t]1t.T
t1T
0
=
� 2
V2 0
Substitute to obtain:
*120
••
Repeat Problem 1 1 9 if the resistor R is replaced by a 2-j.iF capacitor.
Picture the Problem
We can apply Kirchhoff's loop rule to obtain an expression for
charge on the capacitor as a function of time. Differentiating this expression with respect to time will give us the current in the circuit. We can then find Imax and Imin by
considering the conditions under which the time-dependent factor in I will be a maximum
or a minimum. We can use the maximum value of the current to find IrnlS• Apply Kirchhoff's loop rule to obtain: Substitute numerical values and solve for q(t):
Differentiate this expression with respect to t to obtain the current as a function of time:
Express the condition that must be
E1 + E2
q(t) (2 ,uF(2,uF)(20)(18V )V)cos(1 131 ) = (40flC)cos(1 131s-l)t 361lC I = ddtq = �[ dt (40flC)cos(1 131s-l)t 36 flC] =1-(45. 2 mA)sin(1 131s-l)t 1 sin(1 131 )t = 1 1- 45. 2 m A 1 sin(1 131s-l)t = - 1 = 1 45. 2 m A 1 =
mlmmum:
Imin
and
maXImum:
Imax
average value of the sine function over a period is zero:
Iav
=
-I
+
=
satisfied if the current is to be a
capacitor as an open circuit and the
+
-I S
and
Because the dc source sees the
S
+
satisfied if the current is to be a
Express the condition that must be
- qC(t) = 0
@]
t
252 Chapter 29 Because the peak CUlTent is 45 .2 rnA:
Inns
=
Ji 45 �A I 32. 0 mA I =
=
Chapter 30 Maxwell's Equations and Electromagnetic Waves Conceptual Problems *1
•
True or false:
(a) Maxwell's equations apply only to fields that are constant over time. (b) The wave equation can be derived from Maxwell's equations.
(c) Electromagnetic waves are transverse waves.
(d) In an electromagnetic wave in free space, the electric and magnetic fields are in phase.
(e) In an electromagnetic wave in free space, the electric and magnetic field vectors E and B are equal in magnitude.
(j) In an electromagnetic wave in free space, the electric and magnetic energy densities are equal. (a) False. Maxwell's equations apply to both time-independent and time-dependent fields. (b) True
(c) True (d) True
(e) False. The magnitudes of the electric and magnetic field vectors are related according toE= cB. (j) True *4
•
Are the frequencies of ultraviolet radiation greater or less than those of
infrared radiation? Determi ne the Concept
The frequencies of ultraviolet radiation are greater than those of
infrared radiation (see Table 30-1). *8
• A helium-neon laser has a red beam. It is shone in tum on a red plastic filter (of the kind used for theater lighting) and a green plastic filter. (A red theater-lighting filter transmits only red light.) On which filter will the laser exert a larger force?
A red plastic filter absorbs all the light incident on it except for the red light and a green plastic filter absorbs all the light incident on it except for the green light. If the red beam is incident on a red filter it will pass through, whereas, if it is
Determine the Concept
253
254
Chapt er
30 absorbs more a greater force on the green filter.
incident on the green filter it will be absorbed. Because the green filter energy than
does
the
red filter, the laser beam will
exert
Estimation and Approximation *12 ·· Estimate the radiat ion pressure force exerted on the earth by the sun, and compare the radiation pressure force to the gravit at ional attract ion of the sun. At the earth's orbit the intensity of sunlight is 1 .37 kW/m2 •
We can find the radiation pressure force from the definition of pressure and the relationship between the radiation pressure and the intensity of the radiation from the sun. We can use Newton' s law of gravit ation to find the gravitational force the sun exerts on the earth. Picture the Problem
The radiation pressure exerted on the earth is given by:
� =�
Fr
=>
= �A
where A is the cross- sectional area of the earth. Express the radiation pressure in terms ofthe intensity of the radiation I from the sun: Substituting for Pr and A yields:
= In R2 7 m2 )(6370kmY Fr = n(13 0W/ 3 xl08 mls = ! 5. 82x108 N ! Fr
Substitute numerical values and evaluate
Fr:
The gravitational force exerted on the earth by the sun is given by:
Gmsun me arth r2 where r is the radius of the earth's orbit .
F
Substitute numerical values and evaluate F
=
c
=
F:
10-11 . m2 I kg2 )(1.99 1030 kg)(5.98 1024 kg) = 3. 5 3 x 1022 N (1. 5 1011 m) Fr 5. 82 108 N = 1. 6 5 10-14 Fr F 3 . 5 3 1022 N F: The gravitational force is greater by a factor of approximately 1 014.
(6. 6 7
X
N
Express the ratio of the force due radiation pressure t o the gravitational force
X
X
X
X
X
X
Maxwell's Equations and Electromagnetic Waves 255 'k13 Repeat Problem 12 for the planet Mars. Which planet has the larger ratio of radiation pressure to gravitational attraction? Why? Picture the Problem We can find the radiation pressure force from the definition of pressure and the relationship between the radiation pressure and the intensity of the radiation from the sun. We can use Newton's law of gravitation to find the gravitational force the sun exerts on Mars.
The radiation pressure exerted on Mars is given by:
� =�
�
Fr
= �A
where A is the cross-sectional area of Mars. Express the radiation pressure on Mars in terms of the intensity of the radiation IMars from the sun: Substituting for Pr and A yields:
Express the ratio of the solar constant at the earth Iearth to the solar constant IMars at Mars:
I = ( J2 I = I I Mars
rearth
earth
rMars
�
Mars
earth
( J2 rearth
rMars
Substitute for IMars to obtain:
Substitute numerical values and evaluate Fr: Fr
1 I m J 2 = 1 7.09 x1 0 7 N I = 7r(1370W/3x m1028)ml(33s95km)2 ( 2.1. 520X10 9x 1011 m (0 ) =
The gravitational force exerted on Mars by the sun is given by:
F
Gmsun mMars
=
GmSW1
.1 1 mearth
r2 r2 where r is the radius of Mars' orbit.
Substitute numerical values and evaluate F: F
= (6.67 X10-1 N· m2 Ikg2(2.)(1.2999X101x 1031 m0 kg) )(0.11)(5. 9 8 x 1024 kg) = 1. 66 x 102 1 N due 9 X10 7 N = 4. 27 X10-1 4 _ 1.7 .606x1021 N
Express the ratio of the force radiation pressure Fr to the gravitational force F:
Fr F
256
Chapter Because the ratio of these forces is for the earth and 4.2 7 for Mars, Mars has the larger ratio. The reason that the ratio is higher for Mars is that the dependence of the radiation pressure on the distance from the Sun is the same for both forces whereas the dependence on the radii of the planets is different. Radiation pressure varies as , whereas the gravitational force varies as (assuming that the two pla nets have the same density, an assumpti on that is nearly true). Consequently, the ratio of the forces goes as Because Mars is smaller than earth, the ratio is larger. 30
1.65 x l 0-14
x 1 0-14
(r - 2 ) ,
R2
R3
1 R 2 I R 3 = R- •
*14 ·· In the new field of laser cooling and trapping, the forces associated with radiation pressure are used to slow down atoms from thermal speeds of hundreds of meters per second at room temperature to speeds of just a few meters per second or slower. An isolated atom will absorb radiation only at specific resonant frequencies. If the frequency of the laser-beam radiation is one of the resonant frequencies of the target atom, then the radiation is absorbed via a process called resonance absorption. The effective cross-sectional area of the atom for resonant absorption is approximately equal to A?, where A is the wavelength of the laser beam. (a) Estimate the acceleration of a rubidium atom (atomic mass 85 glmol) in a laser beam whose wavelength is 780 nm and intensity is 10 W/m2 • (b) About how long would it take such a light beam to slow a rubidium atom in a gas at room temperature (300 K) down to near-zero velocity? We can use Newton' s 2nd law to express the acceleration of an atom in terms of the net force acting on the atom and the relationship between radiation pressure and the intensity of the beam to find the net force. Once we know the acceleration of an atom, we c an use the definition of acc eleration to find the stopping time for a rubidium atom at room temperature. Picture the Problem
(a) Apply obtain:
I F=ma to the atom to
F=ma
where beam.
F is the force exerted by the laser
The radiation pressure Pr and intensity of the beam I are related acc ording t o: Solve for
F to obtain: F
Substitute for in the expression of Newton' s 2nd law to obtain: Solve for a:
F=
fA = e
f}}
-=ma e
f},}
a= me
IA} e
Maxwell's Equations and Electromagnetic Waves Substitute numerical values and evaluate a: a=
(b)
257
lOW/m2 ) (780nm)2 = 1 1.44xI Osm/s2 1 1 mol (85� mol x 6. 02 xl023 particles J (3X 108 m/s) = (
U sing the definition of
acceleration, express the stopping
Vfinal
I1t
a
time I':..t of the atom: Because Vfinal � 0:
I1t
Using the rms speed as the initial
�
speed of an atom, relat e Vini!ial to the t emperature of the gas:
Vini!ial
Substitut e in the expression for the
I1t
st opping time to obt ain:
- vini!ial
- v ini!ial a
= = �3:T VmlS
= _� �3kT a
In
Substitut e numeric al values and evaluate I':.. t:
I1t
= - 1.44x 1105 m /s2 31. 38x10-23 lmol(300K) = 1 2. 06ms I (85� molX 6. 02 x 1023 particles J JIK
___ ____
Maxwell's Displacement Current
*19 ··
Current of lOA fl ows into a capacitor having plates with areas of 2 0.5 m . (a) What is the displacement current between the plates? What is dEldt
(b) B· de
between the plates for this current? (c) What is the line integral of
around a circle
of radius 10 cm that lies within the plates and parallel to the plates? Picture the Problem
We can use the conservation of charge to find fct, the definitions of
the displacement current and electric flux to find dEldt, and Ampere' s law to evaluate
B· de
around the given path.
(a) From conservation of charge we know that :
Id
= = I IO.OA I I
258
Chapter
30
( b) Express the displa cemen t cutTent
h
Substitute for
e A elE d =E 0 leltrpe =E0 �[EA]=E 0 elt elt
1
elE elt
dEldt:
Substitute numerical values and evaluate dEldt:
�
Eo A
dtelE = (8. 8 5 X10-1 2 C2lOAI N · m2 )(0. 5 m2 ) = 1 2.26X101 2 � I {B if = !lol enclosed .
(c) Apply Ampere' s law to a circular path of radius r between the plates and parallel to their surfaces to obtain: Assuming that the displacement current is uniformly distributed:
Ienclosed Id Tr
r2 = A
=>
Ienclosed _
7r:
r2 Id
A
where R is the radius of the circular plates. Substitute for
lenclosed to obtain:
Substitute numerical values and evaluate
*23
{B. if :
•••
Show that the generalized form of Ampere ' s law (Equation 30-4) and the Biot Savart law give the same result in a situation in which they both can be used. Figure 30-13 shows two charges +Q and -Q on the x axis at x = -a and x = +a, with a current
1
=
- dQldt along the line between them. Point P is on the y axis at y = R. (a) U se the
Biot-Savart law to show that the magnitude of B at point P is B
(b)
= !l27r:R.ola .JR2 1 a2 +
Consider a circular strip of radius r and width dr in the yz plane with its center at the
origin. Show that the flux of the electric field through this strip is
Maxwell's Equations and Electromagnetic Waves 259 ExdA = il a r2 Eo
(
+
a2
t/2 rdr
(C) Use your result from Part ( b) to find the total flux CPe through a circular area of radius
R. Show that
( d) Find the displacement current h and show that
(e) Finally, show that Equation 30-4 gives the same result for B as the result found in Part ( a) .
Figure 30-13
y
p
Problem 23
Picture the Problem
We can follow the step-by-step instructions in the problem
statement to show that Equation 30-4 gives the same result for B as that given in Part ( a). ( a) Express the magnetic field at P using the expression for B due to a straight wire segment:
where
. eI sm
Substitute for sinel and sin� to obtain:
B
=
Ji
s
. e2 m
=
I
.,j
a
R2 +a2
p - 4JZ'o R .,jR2 +a2 _
Jiola
2a
2JZ'R .,j
1
R 2 + a2
260
Chapter
30
(b) Express the electric fl ux through
the circular strip of radius rand width dr in the yz plane:
The electric field due to the dipole IS:
Substitute for Ex t o obtain:
d¢e = ExdA =
=
2kQa
(2
2 W2 (2rcrdr )
r +a J
2Qa
(2
2 W2 (2rcrdr)
4rc Eo r +a J
=
(c) Multiply both sides of the expression for ¢e by Eo: Integrate r from 0 to R to obt ain: , Eo
- Rf (2
¢ e - Qa
o
r
rdr
-
(d) The displacement current is defined to be:
The t ot al current is the sum of I and
h
(e) Apply Equation 30-4 (the generalized form of Ampere's law) to obtain:
(
-1
2 - Qa -JR 2 +a 2 +a 2 )1
1
+a
J
-
Q
(1 - -J 2
a
R +a
2
J
Maxwell's Equations and Electromagnetic Waves 261 B=�(I+I) 2Jr R B-- � [ I a 2Jr R .JR2+a2 J
Solve [or B:
d
Substitut e for I + ld from (d) to obt ain:
= f.1oIa 1 W.JR2+a2
Maxwell's Equations and the Electromagnetic Spectrum
*25
·
Find the wavelength for (a) a typical AM radio wave with a frequency of
1000 kHz and (b) a typical FM radio wave with a frequency of 100 MHz. Picture the Problem
We can use c =fA to find the wavelengths corresponding to the
given frequencies. Solve c =fA for A :
(a) Forf= 1000 kHz :
A=
3 X 108m1s = 300 mI 1000 x103 S-I I
(b) Forf= 100 M Hz :
*26
•
What is the frequency of a 3-cm microwave?
Picture the Problem
wavelength. Solve
c =fA
We can use c =fA t o find the frequency corresponding to the given
forf
Substitute numerical values and evaluatef
f= !:...A 3x m1s =10 I oHz = 1 10.0 H f= 3x108 . G zI 102_ m .
262 Chapter
30
Elect.-ic Dipole Radiation *32
At a distance of 30 Ian from a radio station broadcasting at a frequency of 0.8 3 MHz, the intensity of the electromagnetic wave is 2 x 10-1 W/m2. The transmitting •••
antenna is a vertical dipole. What is the total power radiated by the station? Picture the Problem
The intensity of radiation from an electric dipole is given by
lo( sin2 B)/r2, where B is the angle between the electric dipole moment and the position
vector r. We can integrate the intensity to express the total power radiated by the antenna and use this result to evaluate 10. Knowing 10 we can find the total power radiated by the
station. From the definition of intensity we
dP = IdA
have:
and
Ptot = ffI(r, e)dA
where, in polar coordinates,
sinededrp 21"[ = �ot f f I(r, e) r2 sin ededrp o0
dA = r2
Substitute for dA to obtain:
Express the intensity of the signal as
(1)
a function of rand B: Substitute for lr( , 8):
�ot = 10
2""
f fSin3 eded¢
o0
From integral tables we find that:
Substitute and integrate with respect to ¢ to obtain:
From equation ( 1) we have:
Substitute to obtain:
2" 4 8n 2" =-1 Io f drp=-Io[] rpo �ot=0 3 0 4
Io =
3
I(r,e)r2 e
sin2
8n I(r,e)r2 tot = 3 e
p
sin2
or, because B= 90°,
3
Maxwell's Equations and Electromagnetic Waves 263 ( ?"Ol =�I ) 2 ?"ot = ¥ (2X10- 13 W/m 2 )(30km )2 = / 1. 5 1mW / 8Jr .:>
r r
Substitute numerical values and
evaluate Ptot:
Energy and Momentum in an Electromagnetic Wave
*38
·
The rms value of the magnitude of the magnetic field in an electromagnetic
wave is Brms
=
0.245 ,uT . Find (a) EmlS, ( b) the average energy density, and (c) the
intensity. Picture the Problem
Given BmlS, we can find Erms using Erms = cBrms. The average energy
density of the wave is given by Uav = E mlSBrmslf.1oC and the intensity of the wave by 1= UaVC
(a) Express Enns in terms of BmlS: Substitute numerical values and
Erms
evaluate Erms:
= ( x 108 mls)(0.245,uT) = 1 Vim 1 3
73.5
(b) The average energy density Uav is given by: Substitute numerical values and evaluate Uav:
Uav
= (4Jr(x10-Vi7 N/mA) (20.)(3245x10,uT8 )mls) =1 1 73.5
47.8nllm3
(c) Express the intensity as the product of the average energy density and the speed of light in a vacuum: Substitute numerical values and
evaluate 1:
)( x 10 8 m/s ) 1=( = 1 14. 3 W /m2 I 47.8nllm3 3
264 Chapter 30 *41
··
An AM radio station radiates an isotropi c sinusoidal wave with an average
power of 50 kW. What are the amplitudes of Emax and Bmax at a distance of (a) 500 m, ( b) 5 km, and (c) 50 km?
Picture the Proble m
of
Pav
We can use 1=
Pavl4trr2
and the distance r from the station .
and 1 = ErmsBnn/f.1o to express Errns in terms
Express the intensity I of the
radiation as a function of its average power and the distance r from the station: The intensity is also given by:
I
=
EmlSBrms E� E�13X =
flo
_ 4rc r 2 E
�
Equate these expressions to obtain:
Solve for Emax:
=
cflo
2cflo
�x IIo 2cr
�(!)r
Emax �� =
(a) Substitute numerical values and evaluate Emax for r = 500 m: Emax
(SOOm)
=
U se Bmax = Emaxlc to evaluate Bmax:
4rc 10 -7 N/A2 (SOkW) ( 1 !3.46V/m I 500m J 3.46V/m 1 11. Sn T I B 3X 108 rnIs x
=
2rc
max
=
=
(b) Substitute numerical values and evaluate Emax for r = 5 km:
Use Bmax = Emaxlc to evaluate Bmax:
Bmax
=
0.346 VIm /l.lSn T / 3X 108 rnIs
(c) Substitute numerical values and evaluate Emax for r = 50 km:
=
Maxwell's Equations and Electromagnetic Waves 265 Use Bmax
*43
=
EmaJc to evaluate Bmax:
••
Bmax
. 0 346 Vim = I 0.115nT I = 03xl0 8 mls
Inst ead of sending power by a 750-kV, 1 000-A transmission line, one desires
to beam this energy via an electromagnetic wave. The beam has a uniform intensity within a cross-sectional area of 50 m2• What are the rms values of the electric and the magnetic fields? Picture the Problem
We can use 1
of 1. We can then use Brms
=
=
ErmsBmlSlJ.1o and Bm1s = EmlS/c to express Erms in t erms
EmlS/c t o find BmlS .
Express the intensity 1 of the
radiation as a function of its average
EmlS BmlS
I
=
I
= =
_
E!lS Cllo
f-Lo
power and the distance r from the station:
U se the definition of intensity t o relat e the intensity o f the electromagnetic wave to the power
p A
Itrans.line V A
in the beam: Substitute for 1 to obtain:
Cf-Lo1trans.line V
EmlS
=
Brms
7S. 2kV/m = 1 0.2S1mT I = 3xl0 8 m 1s
A
Substitut e numerical values and evaluate EmlS:
E rms
=
266 Chapter 30 *45
The electric fi eld of an electromagnetic wave oscillates in the y direction and
••
the Poynting vector is given by
where x is in meters and t is in seconds. ( a) What is the direction of propagation of the wave?
(b)
Find the wavelength and the frequency. ( c) Find the electric and magnetic
fields. Picture the Problem We
can determine the direction of propagation of the wave, its
wavelength, and its frequency by examining the argument of the cosine function. We can find E from
lSi
= E2
/f.1oc and B from B
=
Elc. Finally, we can use the definition of the
Poynting vector and the given expression for S to find E and B . -
-
-
Because the argument of the cosine function is of the [onn kx the wave propagates in the positive direction. 2lr 10m-1 (b)
( a)
-
illt,
x
Examining the argument of the
cosine function, we note that the
k=
wave number k of the wave is: Solve for and evaluate A:
A
=
A
=
2�_1 I 0.628m I 10 =
Examining the argument of the cosine function, we note that the angular frequency
OJ of the wave
is:
Solve for and evaluatef to obtain:
f = 3xl02lr9s-1 1477 MHz I =
(c) Express the magnitude of Sin terms of E: Solve for E: Substitute numerical values and evaluate E:
E=
�f.1oCISI
Maxwell's Equations and Electromagnetic Waves Because
S(x,t) =(100W/m2)cos2[lOx-(3x109)t]i and S = _1- Ex B: Jlo
U se B = Elc to evaluat e B: -
Because S with
B
267
=
1
-
Jlo
B=
194 Vim = 0.6 7 Jl 4 3x108m/S
T
E x B , the direction of B must be such that the cross product of E
is in the positive x direction:
The Wave Equation for Electromagnetic Waves
*52
•••
( a) Using argument s similar to those given in the text, show that for a plane
wave, in which E and B are independent ofy and z,
aEz
__
ax
aBy
= --
at
and
aBy
--=Jl 0 E 0 aEz ax
--
at
(b) Show that Ez and By also satisfy the wave equation. Picture the Problem
We can use Figures 30-10 and 30- 1 1and a derivation similar to that
in the text to obtain the given results. In Figure 30- 1 1, replace Bz by Ez• For i'll" small:
Eva luate the line integral of
Earound the rectangular area i'll"fu:
Express the magnetic flux through the same area :
- 1JE · de;:::
aE z ax
&!:J.z
__
( 1)
268
Chapter 30
Apply Faraday's law to obtain:
B = o(B && ) at y at
a E ·d.e�--
1-
-
i
n
dA
--
Substitute in equation ( 1 ) to obtain:
aEaxz _ aBaty
or
In Figure 3 0- 1 0, replace
Ey By by
and evaluate the line integral of
B
around the rectangular area &�:
Evaluate these integrals to obtain:
(b)
provided there are no conduction curre nts.
aBaxy _ aEatz
-- - flo Eo --
Using the first result obtained in
Ez
(a), find the second partial derivative of
with respect to x:
Use the second result obtained in (a) to obtain:
2 1 / c , aax2E2 c21 aat2E2z aax (aBaxy)_- 0 0 axa (aEz) -at
or, bec ause f..1oEo = z
__
Using the sec ond result obtained in
By
(a), find the second partial derivative of
with respect to x:
Use the second result obtained in (a) to obtain:
=
---
___
II E-
Maxwell's Equations and Electromagnetic Waves 269 ?
or, because Po Eo = l/c-, (iBy
--
8x2
1
82 By
c2 8t2
---
General Problems
*57 ··
A circul ar l oop of wire can be used to detect electromagnetic waves. Suppose
a 1 00-MHz FM radio station radiates 50 kW uniformly in all directions. What is the maximum rms voltage induced in a l oop of radius 30 cm at a distance of 1 05 m from the station? Picture the Problem
The maximum rms voltage induced in the loop is given by
Crms = AwBo / J2, where A is the area of the loop, Bo is the amplitude of the magnetic
field, and
OJ
is the angular frequency of the wave . We can use the definition of density
and the expression for the intensity of an electromagnetic wave to derive an expression
for Bo.
The maximum induced rms emf occurs when the plane of the l oop is perpendicular to
B:
From the definition of intensity we have:
Crms =
AwBo
J2
JiR2wBo
(1)
where R i s the radius of loop of wire .
I =� 47r r2 where r is the distance from the transmitter.
The intensity is al so given by:
Substitute to obtain:
Solve f or Bo:
Substitute in equation ( 1 ) to obtain:
1=
EoBo B�c = 2J.10 2J.10
270
30
Chapter
Substitute numerical values and evaluate E
*61
-
_
nns
··
given by E)
6fms:
(0.3m-=Y( 100MHz) "':" -J22 (1Os- -m"""T")
-'--
---"-
The electric fields of two harmonic waves of angular frequency =
E)O, cos(k)x - w/)J
and
w)
and
OF.!
are
E2 E2,0 COS(k2X - w2t + 5)J . Find (a) the =
instantaneous Poynting vector for the resultant wave motion and (b) the time-average Poynting vector. If the direction of propagation of the second wave is reversed so E2 =
E2 0 COS(k2X + w2t + 5)J , find (c) the instantaneous Poynting vector for the
resultant wave motion and (d) the time-average Poynting vector. Picture the Problem
We can use the definition of the Poynting vector and the -
relationship between Band
-
E to find the instantaneous Poynting vectors for each of the
resultant wave motions and the fact that the time average of the cross product term is zero for
lV) :;t OF.!,
vectors.
and 12 for the square of cosine function to find the time-averaged Poynting -
(a) Because E) and the
-
E2 propagate in
x direction:
-
-
"
E x B = fLoSi
=>
-
,..
B = Bk
Expre ss B in terms of E) and E2: Substitute for E) and E2to obtain:
Express the instantaneous Poynting vector for the resultant wave motion:
X =
=
�(E),o cos(k)x - w)t)+ E2,0 COS(k2X - w2t + 8))k
2 _1_ (E)'0 cos(k)x - w)t)+ E2 0 COS(k2X - w2t + 5)\ (J x k) ,
flo c
_1_ �c
J
[E)�o cos2(k)x - w)t)+ 2E) OE20 cos(k)x - w/) '
X
,
2 COS(k2X - w2t + 5)+ E�,o cos (k2X - w2t + 5)] i
Maxwel l's Equations and Electroma gnetic Wave s
(b)
The time average of the cross product term is zero f or lVj :;r 0>.1, and the time aver age of the square of the c osine ter ms is 1'2:
1 Sav = - --
2 /-ioC
[2 2]:E].o + E2.0 I
(c) In this case B2 = Bk bec ause the wave with k = k2 prop agates in the The magnetic field is then: -
2 71
�
-
�
-
i direction.
Express the instantaneous Poynting vector for the resultant wave motion:
s=
1 (E]0 cos(k]x - OJ]t)+ E20 COS(k2X - OJ2t + ))J x �(E1,0 cos(k]x - OJ]t)- E2,0 COS(k2X + OJ2t + o))ic
_ _
f-Lo
'
0
,
(d) The time average of the square
Sav
of the cosine terms is 1'2:
1 [E 20 - E220]:-I 1 ,
=
--
2/-ioC
'
*62
·· At the surface of the earth, there is an approximate average solar flux of 0 .75 2 kW/m . A family wishe s to c onstruct a solar energy conversion system to p ower their home . If the conversion system is 30 percent efficient and the family needs a maximum of 25 kW, what effecti ve surface area is needed for perfectly absorbing collectors? Picture the Problem We can use
the definitions of power and intensity to express the
area of the surface as a function of P, 1, and the efficiency
Use the definition of power to relate
E & IA& = t
the requ ired surface area to the
p=
intensity of the solar radiation:
where
Solve for A to obtain:
G
A =� 1&
Substitute numerical values and evaluate A:
G.
is the efficiency of the system.
272 Chapter *65
30
A long cylindrical conductor of length L, radius
•••
and resistivity p carries a
a,
steady CUlTent I that is uniformly distributed over its cross-sectional area. ( a) Use Ohm's law to relate the elect ric field E in the conductor to I, p, and a.
(b)
Find the magnetic field
B just outside the conductor. (c) Use the results for Part (a) and Part (b) to compute the
(i jj);f.1o at (the edge of the conductor). In what direction is S ? (d) Find the flux 1S n dA through the surface of the conductor into the conductor,
Poynting vector
S=
x
r= a
and show that the rate of energy flow into the conductor equals iR, where R is the
resistance of the cylinder. (Here Sn is the inward component of surface of the conductor.) Picture the Problem
-
S perpendicular to the
We can use Ohm's law to relate the electric field E in the conductor
to I, p, and a and Ampere's law to find the magnetic field B j ust outside the conductor.
Knowing
- E and B we can find S and, using its normal component, show that the rate of
energy flow into the conductor equals iR, where R is the resistance . (a) Apply Ohm ' s law to the cylindrical conductor to obtain:
v = IR = IpL = IpL2 = EL A TCa
Solve for E:
(b)
Apply Ampere 's law to a
circular path of radius a at the
fji . ie
=
B{2TCa) = f.1oIenclosed = f.101
surface of the cylindrical conductor: Solve for B to obtain:
(c) The electric field at the surface of the conductor is in the direction of the current and the magnetic field at the surface is tangent to the surface. Use the results of (a) and
(b)
and the right-hand rule to
evaluate
S:
- 1 1(
- S=-ExB f.1o Ip f.101 "tangent = - -- "parallel 2TCa f.1o TCa 2
J
�
X(
J
�
where f is a unit vector directed radially outward from the cylindrical conductor.
(d) The flux through the surface of the conductor into the conductor is:
273
Maxwell's Equations and Electromagnetic Waves Substitute for 5n, the in ward
component of S and simplify to obtain : ,
pL pL . Smce R = = : A T[{1 2 -
*67
•••
--
fSll dA = �
Small particles might be blown out of solar systems by the radiation pressure
of sunlight. Assume that the particles are spherical with a radius r and a density of 1 g/cm3 and that they absorb all the radiation in a cross-sectional area of ;rr2 . The particles are a distance R from the sun, which has a power output of 3 .83 x l 0 26 W. What is the radius r for which the radiation force of repulsion just balances the gravitational force of attraction to the sun? Picture the Problem
We can use a condition for translational equilibrium to obtain an
expression relating the forces due to gravity and radiation pressure that act on the particles. We can express the force due to radiation pressure in terms of the radiation pressure and the effective cross sectional area of the particles and the radiation pressure in terms of the intensity of the solar radiation. We can solve the resulting equation for r. Apply the condition for translational equilibrium to the particle: ( 1) The radiation pressure Pr depends on the intensity ofthe radiation I: The intensity of the solar radiation at a distance R is:
Substitute to obtain:
Substitute for Pr, A, and m in equation ( 1) : Solve for R t o obtain:
Substitute numerical values and evaluate r:
r = 16rc p3Pc GM
-----
s
274 Chapter
30
= I 0. 5 74 *69
3 108 m/s 6.67 x 10-1 1 N · m2 2 I kg
X
JLll1
I
1 .99
x 1030
kg
An intense point source of light radiates 1 MW isotropically. The source is
•••
located 1 m above an infinite, perfectly reflecting plane. Detenrune the force that acts on the plane. Picture the P roblem
ring of radius
r
Let the point source be a distance a above the plane. Consider a
and thickness
dr
in the plane and centered at the point directly below the
light source. Express the force of force on this ring and integrate the resulting expression to obtain
F.
The intensity anywhere along this infinitesimal ring is PI4Jr (r2 + a 2) and the element of force
on this ring of area 2Jr
dF
rdr
is given by:
F
d
P rdr r 2 +a 2
_
-
(
c
a 2 r +a 2
) .J
Pardr 2 r 2 +a 2
= ---;---� -,-
c
)1
{
where we have taken into account that only the normal component of the incident radiation contributes to the force on the plane, and that the plane is a perfectly reflecting plane.
Integrate
dF r from
=
r
0 to =
00 :
1
From integral tables:
a
Substitute to obtain:
F = (!) F = 3x110M8W = 1 3 . 3 3mN I Pa c
Substitute numerical values and evaluate
F:
a
=
P
mls
e
Chapter 3 1 Properties of Light Conceptual Problems
*5
••
The density of the abnosphere decreases with height, as does the index of
refraction. Explain how one can see the sun after it has set. Why does the setting sun appear flattened? Determine the Co ncept
The change in abnospheric density results in refraction of the
light from the sun, bending it toward the earth. Consequently, the sun can be seen even after it is just below the horizon. Also, the light from the lower portion of the sun is refracted more than that from the upper portion, so the lower part appears to be slightly higher in the sky. The effect is an apparent flattening of the disk into an ellipse.
*10
••
We learned in Chapter 30, Section 30-3 , that an oscillating electric dipole
produces electromagnetic radiation (see Figure 3 0-8). Assuming that the light reflected off and refracted into the surface of a piece of transparent material is caused by such dipoles, show that the condition for Brewster's angle (Equation 3 1 -2 1 ) is exactly the same as saying that the refracted ray is perpendicular to the axis of the radiating dipoles for light polarized in the plane of incidence. Determine the Co ncept
The diagram shows that the radiated intensity for a dipole is
zero in the direction of the dipole moment. Because the dipole axis is in the same direction as the polarization, for light polarized parallel to plane of incidence, the dipole axis will point in the same direction as the reflected wave, i.e., in the direction described by Brewster's law. As the diagram indicates, there is zero field in the direction of the refracted ray. On the other hand, if the incoming wave is polarized perpendicular to the plane of incidence, the dipole axis will never point along the direction of propagation for the reflected or refracted wave. Unpolarized incident ray Polarized reflected
roy
Slightly polarized refracted ray
27 5
276
Chapter 3 1
*1 4
It is a common experience that on a calm, sunny day one can hear voices of
persons in a boat over great distances. Explain this phenomenon, keeping in mind that sound is reflected from the surface of the water and that the temperature of the air just
above the water' s surface is usually less than that at a height of 1 0 m or 20 m above the water. Picture the Problem
The sound is reflected specularly from the surface of the water (we
assume it is calm). It is then refracted back toward the water in the region above the water because the speed of sound depends on the temperature of the air and
is greater at the
higher temperature. The pattern of the sound wave is shown schematically below.
Sources of Light *24
··
Singly ionized helium is a hydrogen-like atom with a nuclear charge of 2e. Its energy levels are given by Ell = -4Eofn2 , where Eo = 1 3 .6 eV. If a beam of visible white light is sent through a gas of singly ionized helium, at what wavelengths will dark lines be found in the spectrum of the transmitted radiation? Determine the Concept
The energy difference between the ground state and the first
excited state is 3Eo = 40. 8 eV, corresponding to a wavelength of 30.4 nm. This is in the far ultraviolet, well outside the visible range of wavelengths. There will be no dark lines in the transmitted radiation. The Speed of Light
*29 ··
In Galileo's attempt to determine the speed of hght, he and his assistant were
located on hilltops about 3 km apart. Galileo flashed a light and received a return flash
from his assistant. (a) If his assistant had an instant reaction, what time difference would
Galileo need to be able to measure for this method to be successful?
(b)
How does this
time compare with human reaction time, which is about 0.2 s? Picture the Problem
We can use the distance, rate, and time relationship to find the time
difference Galileo would need to be able to measure the speed of light successfully.
(a) Relate the distance separating Gahleo and his assistant to the speed of light and the time required for it travel to the assistant and back to Galileo:
D
=
cf:.t
Propeliies Solve for t1t:
Substitute numerical values and
evaluate t1t:
(b) Express the ratio of the human
reaction time to the transit time for ,
the light:
!:'t
=
!:' t
=
277
D
-
c
2 3 m ( k ) x 3 l 08 mls
!:, t rcaction
or
of Light
!:'t
!:'treaction
=
=
0.2 s
20 ,us
11
=
=
04 !:'t
I 20.0 ,us I 1 04
I
Reflection and Refraction *31
••
A ray of light is incident on one of a pair of mirrors set at right angles to each
other. The plane of incidence is perpendicular to both mirrors. Show that after reflecting off each mirror the ray will emerge in the opposite direction, regardless of the angle of incidence.
Picture the Problem The diagram shows
ray 1 incident on the vertical surface at an
angle 81> reflected as ray 2, and incident on
the horizontal surface at an angle of
incidence � . We'll prove that rays 1 and 3
are parallel by showing that 81 = 84 , i.e., by showing that they make equal angles with the horizontal . Note that the law of
reflection has been used in identifying
equal angles of incidence and reflection.
We know that the angles of the right
triangle formed by ray 2 and the two
mirror surfaces add up to 1 80°: The sum of fh. and � is 90°: Because 81
=
82 :
The sum of 84 and � is 90°: Substitute for � to obtain:
3
278
Chapter
*37
31
Light is incident nonnally on a slab of glass with an index of refraction n =
1 .S . Reflection occurs at both surfaces of the slab. Approximately what percentage of the
incident light energy is transmitted by the slab? Picture the Problem Let the subscript
1
refer to the medium to the left (air) of the
rt l
first interface, the subscript 2 to glass, and
=
1
1"13
= 1
II,l
the subscript 3 to the medium (air) to the
-
right of the second interface. Apply the
equation relating the intensity of reflected
---
II
light at nonnal incidence to the intensity of
the incident light and the indices of
13
refraction of the media on either side of the interface to both interfaces. We'll neglect
multiple reflections at glass-air interfaces. Express the intensity of the
transmitted light in the second medium:
Express the intensity of the
transmitted light in the third
medium:
Substitute for 12 to obtain :
Solve for the ratio VII :
Substitute numerical values and evaluate 1311) :
13 I)
5 )2 (1. 5 -1)2 -[1 (1-1. ][ ] 1 1+1.5 1. 5 +1
=
0.922 I 92.2% I =
Properties of Light 279 *40
Figure 3 1 -5 6 shows a beam of light incident on a glass plate of thickness d
and index of refraction
(a ) Find the angle of incidence so that the perpendicular
11.
separation between the ray reflected from the top surface and the ray reflected from the bottom surface and exiting the top surface is a maximum. (b) What is this angle of incidence if the index of refraction of the glass is 1 .60? What is the separation of the two beams if the thiclmess of the glass plate is 4.0 cm?
Figure 31 -56
Air Problem 40
Picture the Problem
l __ e � 1 I I I I I I
Let x be the
perpendicular separation between the two rays and let .e be the separation between the points of emergence of the two rays on
AiI
the glass surface. We can use the geometry of the refracted and reflected rays to express x as a function of .e,
d, er,
d
Glass
and �. Setting the derivative of the resulting equation equal to zero will yield Air
the value of � that maximizes x. (a) Express .e in terms of angle of refraction er:
d
and the
Express x as a function of .e, en and �:
d,
Differentiate x with respect to �:
f.
=
2dtanBr
Chapter 31 dx = 2e1 --d (tan Br cos Bj ) 2d( - tan Br sm. Bj sec 2 Br cos Bj dBr d8 J d8 d8 n J sin Bj = n2 sin Br nJ n2 n, sinBj = nsinBr
280
=
-
I
I
+
Apply Snell's law to the air-glass interface:
or, since
Differentiate implicitly with respect to� to obtain:
__
I
= 1 and
(1)
(2)
=
or
dBr = dBj
- --
Substitute in equation ( 1 ) to obtain:
1 - sin 2 Bj 1 . B sm. Br cos 2 B -sm n
Substitute j
for
and
j
for
to
obtain:
cos 2 Br /cos 2 Br dx = 2d( 1 - sin 2 Bj sin 2 Bj cos 2 Br = 2d (1 - sm. 2 B. - sm. 2 B. cos 2 Br ) dBj ncos3 Br ncos3 Br J ncos3 Br 1 - sin 2 Br cos 2 Br : dx dB. 1 n Bj sin Br -si n 2d [1 -sm. 2 B. - sm. 2 B. (1 - -sm 1 . 2 B. )] = dBj ncos3 Br n2 lIn2,
Multiply the second term in parentheses by
and simplify to obtain:
-
--
I
Substitute
I
for
I
Substitute
for
dx
--
Factor out
to obtain:
I
I
simplify, and set equal to zero to obtain:
I
Properties of Light 281
dx 2d3 [si e - 2n2 sin2 e; +n2 ] = O for extrema de; n 3 cos er n4
;
If dxld81 = 0, then it must be true
that:
Solve this quartic equation for (), to
obtain:
(b) Evaluate Bt for
n
=
1 .60:
B; sin -(1.6 1 48 . 5 ° I 2d taner cose; =
=
In (a) we showed that: Solve equation (2) for
x =
8r:
Substitute numerical values and
evaluate
8r:
Substitute numerical values and
evaluate x:
er sin-I ( _1.I-6 sin48 . 5 0 ) 27. 9° 2(4cm)tan2 7.9°cos48. 5 ° 1 2. 8 1cm I =
=
x =
=
Total Internal Reflection
*46
••
An optical fiber allows rays of light to propagate long distances through total
internal reflection. As shown in Figure 3 1 -57, the fiber consists of a core material with index of refraction n2 and radius b, surrounded by a cladding material of index
n3
< nz.
The numerical aperture of the fiber is defined as sin 8J , where 81 is the angle of incidence
of a ray of light impinging the end of the fiber that reflects off the core-cladding interface at the critical angle. Using the figure as a guide, show that the numerical aperture is
given by
��
n - n� assuming the ray is incident from air. (Hint: Use of the Pythagorean
theorem may be required. )
282
Chapter 3 1 Incident ray f13 --c;:::"""'"".,....,.,-:"7;:-"'7;-:-�-=:=:��
r-=-���""""-""
11,
Figure
31-57 Problems 46, 47, and 48.
Picture the Problem We can use the geometry of the figure, the law of refraction at the
air-n ) interface, and the condition for total internal reflection at the n)-n2 interface to show that the numerical aperture is given by
Referring to the figure, note that:
�n� - n� .
n3
.
SIn ec
n2
and
the right triangle to obtain:
or
a2
=-
c
b2
-+- =
c
b
2
c
1
a
!:
Substitute for !!:.- and � to obtain: c c
sin B
c
2
�l- : �l-
Solve for - : c
�
c
b
SIn . e2 Apply the Pythagorean theorem to
a
= -=
c
2
�
n
;
n 22
Use the law of refraction to relate 8) and �:
Substitute for sin � and let n) = 1
(air) to obtain:
Dispersion
*51
••
A beam of light strikes the plane surface of silicate flint glass at an angle of
incidence of 45°. The index of refraction of the glass varies with wavelength, as shown in
the graph in Figure 3 1 -26. How much smaller is the angle of refraction for violet light of wavelength 400 nm than that for red light of wavelength 700 nm?
Properties of
Light
283
Picture the Problem We can apply Snell's law of refraction to express the angles of
refraction for red and violet light in silicate flint glass. \
Express the difference between the angle of refraction for violet light
11
=
(1)
B Br red - Br violet ,
,
and for red light:
Apply Snell ' s law of refraction to the interface to obtain:
sin 45 0 n sin Br =
B
SIn. -, ( J21 n )
AB
= SIn. , ( J21nred J - SIn-, ( J2n1violet J
r =
Substitute in equation ( 1 ):
Substitute numerical values and
evaluate 11 8 :
Ll
MJ
.
-
Sin -' ( )2(: 60)) -Sin-' ( )2(:.66)) = 26.23 0 - 25. 2 1 0 1 1. 020 1 =
=
Polarization
*60 ··
A stack of N + 1 ideal polarizing sheets is arranged with each sheet rotated
by an angle of Jr/(2N) rad with respect to the preceding sheet. A plane, linearly polarized light wave of intensity
10
is incident normally on the stack. The incident light is polarized
along the transmission axis of the first sheet and is therefore perpendicular to the
transmission axis of the last sheet in the stack. (a) Show that the transmitted intensity through the stack is given by the expression
10 COS2 N(�).
(b) Using a spreadsheet or
graphing program, plot the transmitted intensity as a function of N for values of N from 2
to 1 00. (c) What is the direction of polarization of the transmitted beam in each case? Picture the Problem Let In be the intensity after the nth polarizing sheet and use
1 = 10 cos2 B
to find the ratio of 1n+' to In.
(a) Find the ratio of
1,,+1
to In:
I
-.!!±!.. =
In
COS2 2N 1r
284 Chapter
31
1
N+I
Because there are N such reductions
11
of intensity:
=
1N+I 10
=
) COS2 N(� 2N
and
IN+I
10
=
cos 2N (2N) n
IN+/lo
as a function of N is shown below. The (b) A spreadsheet program to graph formulas used to calculate the quantities in the columns are as follows:
�.
I,
•
1 2 3 4 5,
co '
"'�§
96 97 98 99
WO
A graph of
1110
Algebraic Form N N+ 1
ContentIForrnula 2 A2 + 1 (cos(PIO/(2* A2)Y'(2* A2)
Cell A2 A3 B2
COS2N ( 2:V) B
A N 2 3 4 5
0.250 0.422 0.5 3 1 0.605
95 96 97 98 99 1 00
0.974 0 .975 0.975 0.975 0.975 0.976
1110
."
as a function of N follows.
t lf
c
1 .0
".,-,=-r .,.....,.,...,
0.6
-· H!.o---f��=:!.�����f;::;:';,....;:;,p�-+-�-H'-"""4
0.5
H'---1I--'-':!";!-'''''::':-*
o
10
20
30
40
50 N
60
70
80
90
1 00
(c)
*62
Properties of Light In each case, the pol arization of the transmitted beam is perpendicular to that of the incident beam. ••
285
Show that a linearly polarized wave can be thought of as a superposition of a
right and a left circularly polarized wave.
Picture the Problem A circularly polarized wave is said to be
right circularly polarized
if the electric and magnetic fields rotate clockwise when viewed along the direction of
propagation and left circularly polarized if the fields rotate counterclockwise.
Ex = Eo cos OJt and Ey = Eo sin OJt or Ey = -Eo sin OJt
For a circularly polarized wave, the x
and y components of the electric
field are given by:
for left and right circular polarization,
respectively. For a wave polarized along the x
Eright + EJeft -
aXIS:
*64
x
••
-
= Eo cos OJt i +Eo cos OJt i = I 2Eo cos OJt i I ,..
"
Show that the electric field of a circularly polarized wave propagating in the
direction can be expressed by
i Eo sin(kx mt)J + Eo cos(kx -OJt)k =
-
Picture the Problem We can use the components of E to show that E is constant in -
-
time and rotates with angular frequency
w.
Express the magnitude of E in terms of its components:
Substitute for Ex and Ey to obtain:
E = �[Eo sin(kx - OJt)Y + [Eo cos(kx -OJt)] 2 = �E� [sin 2 (kx -OJt) + cos 2 (kx -OJt)] = Eo and the i vector rotates in the yz plane with angular frequency w.
Chapter
2 86
31
General P roblems *68
••
Figure 3 1 -5 8 shows two plane minors that make an angle e with each other.
Show that the angle between the incident and retlected rays is 2 e.
Figure
31-58 Problem 68
Picture the Problem
Angle ADE is the
angle between the direction of the incoming ray and that reflected by the two minor surfaces. Note that triangle ABC is
A
isosceles and that angles CAB and ABC are equal and their sum equals e. Also from the law of reflection, angles CAD and CBD equal angle ABC. Because angle BAD is twice BAC and angle DBA is twice CBA, angle ADE is twice the angle e.
*71
··
A swimmer at the bottom of a pool 3 m deep looks up and sees a circle of
light. If the index of refraction of the water in the pool is 1 .3 3 , find the radius of the circle. Picture the Problem
We can apply Snell' s law to the water-air interface to express the
critical angle ee in terms of the indices of refraction of water (nl) and air (n2) and then relate the radius of the circle to the depth Air,
n2
Water,
=
1
n,
=
d
1 .33
of the swimmer and ee.
r--
1
r
--.. 1
Properties of Light Relate the radius of the circle to the depth
d
of the point source and the
r =
287
d tan Be
critical angle ee:
Apply Snell's law to the water-air interface to obtain: Solve for ee:
Substitute for ee to obtain:
Substitute numerical values and evaluate r:
*76 ·· A Brewster window is used in lasers to preferentially transmit light of one polarization, as shown in Figure 3 1 -59. Show that if is the polarizing angle for the
/
/
Bp I nl n2 interface, then Bp 2 is the polarizing angle for the n2 n I interface.
Figure
31-59 Problem 76
Picture the Problem
Let the angle of refraction at the first interface by el and the angle
of refraction at the second interface be fh.. We can apply Snell's law at each interface and eliminate el and
n2 to show that fh. = Bt>2.
Apply Snell' s Brewster' s law at the
nl-n2 interface:
tan B
PI
-
_
2
n nl
_
288
Chapter 3 1
Draw a reference triangle consistent with Brewster's law:
Apply Snell 's law at the n l -n2 interface: Solve for B, to obtain:
Referring to the reference triangle we note that:
8I
e,
.
(
sin -,
(-:-: -�r=n=�n=:==n=; J
nl . 8 = sm _I �sln �
PI
J
i.e., B, is the complement of �" Apply Snell's law at the n2-n, interface: Solve for (h. to obtain:
Refer to the reference triangle again to obtain:
Equate these expressions for 8, to obtain:
n2 sin
*79
••
(a) For a light ray inside a transparent medium that has a planar interface
with a vacuum, show that the polarizing angle and the critical angle for internal reflection
satisfy tan Bp
=
sin Be. (b) Which angle is larger?
Properties of Light 289 P i c t u r e the Problem
angle to show that tan
We can apply Snell's law at the critical angle and the polarizing
Bp Be. = sin
(a) Apply Snell's law at the medium-vacuum interface:
sin Be sin 90° 1 tan BP tan Bp 1 n
For
B1
= Bp, n 1
=
n, and n2 = 1 :
=
=
n
= _ 2 =
n
1
n
=>
n
=
1
Because both expressions equal one: (b) For any value of
*85
••
B:
Suppose rain falls vertically from a stationary cloud 1 0,000 m above a
confused marathoner running in a circle with constant speed of 4 rnIs. The rain has a terminal speed of 9 rnIs. (a) What is the angle that the rain appears to make with the vertical to the marathoner? (b) What is the apparent motion of the cloud as observed by the marathoner? (c) A star on the axis of the earth's orbit appears to have a circular orbit of angular diameter of 4 1 .2 seconds of arc. How is this angle related to the earth's speed in its orbit and the velocity of photons received from this distant star? (d) What is the speed of light as determined from the data in Part (c)? Picture the Problem
The angle that the rain appears to make with the vertical, according
to the marathoner, is the angle whose tangent is the ratio of Vrunner to Vrain . The circular motion of the star is analogous to the circular motion of the cloud with Vrutmer = Vearth and Vrain = c.
(a) The angle that the rain appears to make with the vertical to the
B tan-I [ =
marathoner is given by: Substitute numerical values and evaluate
B:
(b) The cloud moves in a circle whose radius is given by:
B
=
R
=
J tan -I [ 4 mJS J Vrwmer vram
9 mJs
HtanB
=
1 24 . 0 ° 1
290
Ch er ap t
31
!
Substitute numerical values and
R = (1 0 km)tan 24° = 4.45 km
evaluate R : (c) Here Vrunner = Vearth and
(1)
Vrain = c :
where e = 1 (angular (d) From equation (1):
c
Convert 20.6" to degrees:
= vearth
tan e
=
evaluate c:
c:
diameter)
2 nRearth-sun T.arth
20.6" = 20.6" x
Substitute numerical values and
Substitute numerical values and evaluate
I
tan
� 60"
e
x � = 5 .722 x 1 0-3 0 60'
Chapter 32 Optical Images Conceptual Problems· *4
Under what condition will a concave mirror produce (a) an upright image, (b) a virtual image, (c) an image smaller than the object, and (d) an image larger than the object? ••
Determine the Concept Let s be the object distance andfthe focal length of the mirror.
(a) If
s
< f, the image is virtual, upright, and larger than the object.
(b) If
s
< f, the image is virtual, upright, and larger than the object.
(c) If s > 2f, the image is real, inverted, and smaller than the object. (d) Iff< s < 2f, the image is real, inverted, and larger than the object. *9
•
Under what conditions will the focal length of a thin lens be (a) positive and
(b) negative? Consider both the case where the index of refraction of the lens is greater than and less than the surrounding medium. Determine the Concept
(a) The lens will be positive if its index of refraction is greater than that of the surrounding medium and the lens is thicker in the middle than at the edges. Conversely, if the index of refraction of the lens is less than that of the surrounding medium, the lens will be positive if it is thinner at its center than at the edges. (b) The lens will be negative if its index of refraction is greater than that of the surrounding medium and the lens is thinner at the center than at the edges. Conversely, if the index of refraction of the lens is less than that of the surrounding medium, the lens will be negative if it is thicker at the center than at the edges. *14
•
If an object is placed 2 5 cm from the eye of a farsighted person who does not
wear corrective lenses, a sharp image is formed (a) behind the retina, and the corrective lens should be convex. (b) behind the retina, and the corrective lens should be concave. (c) in front of the retina, and the corrective lens should be convex. (d) in front of the retina, and the corrective lens should be concave.
291
292
Chapter 32
Determine the Concept The
eye muscles ora farsighted person lack the ability to shorten the focal length of the lens in the eye sufficiently to form an image on the retina of the eye. A convex lens (a lens that is thicker in the middle than at the circumference)
will bring the image forward on to the retina.
I (a) is correct. I
*17 · The image of a real object formed by a convex mirror (a) is always real and inverted. (b) is always virtual and enlarged. (c) may be real. (d) is always virtual and diminished.
Determine the Concept Referring to the ray diagram show below we note that the image
is always virtual and diminished.
I (d) is correct. I \
\
\ --
-)\--
-
I I
F
-_
c
Explain the following statement: A microscope is an object magnifier, but a telescope is an angle magnifier.
*21
·
Determine the Concept Microscopes ordinarily produce images (either the intermediate
one produced by the objective or the one viewed through the eyepiece) that are larger than the object being viewed. A telescope, on the other hand, ordinarily produces images that are much reduced compared to the object. The object is normally viewed from a great distance and the telescope magnifies the angle subtended by the object. Estim ation and Approximation
Estimate the maximum value that could be usefully obtained for the magnification of a simple magnifier, using Equation 32-20. (Hint: Think about the smallest f ocal length lens that could be made from glass and still be used as a magnifier.) *24
••
Picture the Problem Because the focal length of a spherical lens depends on its radii of
curvature and the magnification depends on the focal length, there is a practical upper limit to the magnification. Use equation 32-20 to relate the magnification M of a simple
Xnp M =-
f
Optical Images 293 magnifier to its focal lengthf Use the lens-maker's equation to relate the focal length of a lens to its radii of curvature and the index of refraction of the material from which it is constructed: For a plano-convex lens, Hence:
r2
1 ( - l)
-= n
f
(
I
1
--rl
r2
J
= 00.
Substitute in the expression for M and simplify to obtain:
1)
Note that the smallest reasonable value for rl will maximize M. reasonable smallest value for the radius of a magnifier is 1 cm. Use this value and n = 1.5 to estimate
A
( )( ) Mrnax 1.5-1lcrn25 cm = @}] =
Plane Mirrors
Two plane mirrors make an angle of 90°. The light from an object point that is arbitrarily positioned in front of the mirrors produces images at three locations. For each image location draw two rays from the object that, upon one or two reflections, appear to come from the image location. *27
·
Determine the Concept Draw rays of light
from the object that satisfy the law of reflection at the two mirror surfaces. Three virtual images are formed, as shown in the adjacent figure. The eye should be to the right and above the mirrors in order to see these images.
I ��� ;±;;;;;;;l I
J I
//
1/
if
/
/
/
/ /
/ I I I I I I I I I' I' •
Spherical Mirrors
concave spherical mirror has a radius of curvature of 24 cm. Draw ray diagrams to locate the image (if one is formed) for an object at a distance of (a) 55 cm, (b) 24 cm, (c) 12 cm, and Cd) 8 cm from the mirror. For each case, state whether the *30
··
A
294
Chapter 32
image is real or virtual; upright or inverted; and enlarged, reduced, or the same size as the object. The easiest rays to use in locating the image are 1) the ray parallel to the principal axis and passes through the focal point of the mirror, the ray that passes through the center of curvature of the spherical mirror and is reflected back on itself, and 2) the ray that passes through the focal point of the spherical mirror and is reflected parallel to the principal axis. We can use any two of these rays emanating from the top of the object to locate the image of the object. Picture the Problem
(a) The ray diagram is shown to the right. The image is real, inverted, and reduced.
(b) The ray diagram is shown to the
The image is real, inverted, and reduced.
right.
( c) The ray diagram is shown to the right. The object is at the focal point of the mirror.
The image is real, inverted, and the same size as the object.
The emerging rays are parallel and do not form an image.
Optical (d) The ray diagram is shown to the right.
Images 29S ./
-7+
J J J ---L
?
_
The image is virtual, erect, and enlarged . dentist wants a small mirror that will produce an upright image with a magnification of 5.5 when the mirror is located 2.1 cm from a tooth. (a) What should the radius of curvature of the mirror be? (b) Should the mirror be concave or convex? *35
.
A
Picture the Problem We can use the mirror equation and the definition of the lateral
magnification to find the radius of curvature of the mirror. (a) Express the mirror equation:
1 = -1 =-2 -s1 + s' r f
Solve for r:
r = 2ss'
The lateral magnification of the mirror is given by:
m = --s'
Solve for s':
s' = -ms
Substitute for s' in equation (1) to obtain:
ms r = -2 I-m
--
s'+s
Substitute numerical values and evaluate r: (b)
(1)
s
---
)(2.1 cm ) = l s .13cm r = -2(S.S I I-S.S
The mirror must be concave. A convex mirror always produces a diminished virtual image.
concave mirror has a radius of curvature 6 crn. Draw rays parallel to the axis at 0.5 cm, 1 cm, 2 cm, and 4 cm above the axis, and find the points at which the reflected rays cross the axis. (Use a compass to draw the mirror and a protractor to find the angle of reflection for each ray.) (a) What is the spread ox of the points where these *39
••
A
296
Chapter 32
rays cross the axis? (b) By what percentage could this spread be reduced if the edge of the milTor were blocked off so that parallel rays more than 2 cm from the axis could not strike the mirror? Picture the Problem
(a) The figure to the right shows the mirror and the four rays drawn to scale. Using a calibrated ruler, the spread of the crossing points is & � 1.0 cm. Note that the triangles formed by the center of curvature, the point of reflection on the mirror, and the point of intersection of the reflected ray and the mirror axis are isosceles triangles. Express the equal angles of the isosceles triangles:
,
,
,
,
,
, ,
,
, ,
,
,
,
,
/
.:{;��;;;==::::-=--_-__:�-- �_----rc -
C
Or
=
Sin-I(�)
where y is the distance of the incoming ray from the mirror axis andR is the radius of curvature of the mirror. Using the law of cosines, the distance between the point of intersection and the mirror is given by: Evaluate d for y/R = 2/3:
= 1 .975 cm Evaluate d for y/R = 1112:
= 2.990cm Express the spread &:
5x = 2.990cm-1 .975 cm = l l .Ol cm I
in good agreement with the result obtained above.
(b) Evaluate d for y/R
Optical Images 297
= 1/3: =
Express the new spread &':
Ox'
Express the ratio of &' to &:
&'
2.81 8cm 2.990cm-2.81 8cm 0 . 172 cm
=
=
=
Ox
0 . 172 cm = 17.0% 1 .01cm
By blocking off the edges of the mirror so that only paraxial rays within 2 cm of the mirror axis are reflected, the spread is reduced by 83.0%. Images Formed by Refraction *44
··
A very long glass rod of3.5-cm
diameter has one end ground to a convex
spherical surface of radius cm. Its index of refraction is 1.5. (a) A point object in air is on the axis of the rod35 cm from the surface. Find the image and state whether the image
7. 2
is real or virtual. Repeat (b) for an object 6.5 cm from the surface and (c) an object very far from the surface. Draw a ray diagram for each case. Picture the Problem We can use the equation for refraction at a single surface to find the
images corresponding to these three object positions. The signs of the image distances will tell us whether the images are real or virtual and the ray diagrams will confirm the correctness of our analytical solutions. Use the equation for refraction at a single surface to relate the image and object distances: Here we have n] = and n2 n 1 .5. Therefore:
1
=
=
Solve for s':
(a) Substitute numerical values (s =35 cm and r = 7. 2 cm) and evaluate s':
n]
-+
s
n2
-
s'
n
1
-+- =
s s'
s,
=
s' = =
n? - n 1
= --=--
...!..
_
r
(1)
n-1
--
r
nrs s(n-1 )-r
--,__.
(1.5 )(7.2 cm )(35cm ) (35 em)(1.5-1)-(7.2 em)
1 3 6 .7 em I
where the positive distance tells us that the
298 Chapter 32 image is 36.7 cm in back of the surface and is
(b) Substitute numerical values
(s 6.5 cm and r = 7.2 cm) and evaluate Sf: =
I real. I
(1 .5 )(7.2 em )(6.5 em) 6.5 em 1 . 5 - 1 - 7.2 em = 1 -17. 8 em I
:""" --' )""" s/ = -;-���-----!" )�(....,..--( )(
where the minus sign tells us that the image is 17.8 cm in front of the surface and is
I virtual. I
(c) When s = 00, equation (1) becomes: Solve for s/:
Substitute numerical values and evaluate Sf:
n n-1
s/
r
nr s/ =--
n-1 (1 . 5 ) (7 .2 em) = I· 21.6 em I. s/ = 1 .5-1
i.e., the image is at the focal point, is
I real, I and of zero size. •
F
Optical Images 299 A glass rod 96 cm long with an index of refraction of 1.6 has its ends ground *49 to convex spherical surfaces of radii 8 cm and 16 cm. A point object is in air on the axis of the rod 20 cm from the end with the 8-cm radius. (a) Find the image distance due to refraction at the first surface. (b) Find the final image due to refraction at both surfaces. (c) Is the final image real or virtual? ••
Picture the Problem We can use the equation for refraction at a single surface to find the
images due to refraction at the ends of the glass rod. The image formed by the refraction at the first surface will serve as the object for the second surface. The sign of the final image distance will tell us whether the image is real or virtual. (a) Use the equation for refraction at a single surface to relate the image and object distances at the first surface:
n2
nl
n2 - nl
-+ - = ----"-_...!... r s Sf
(1)
Solve for Sf:
Substitute numerical values and evaluate Sf:
(1. 6 )(Scm )(20em) s f -,----)--"--'-
Because s' 0 and y' = -2.00 em, the image is real, inverted, and diminished in agreement with the ray diagram. *59
··
Two converging lenses, each of focal length 10 cm, are separated by35 cm.
An object is 20 cm to the left of the first lens. (a) Find the position of the final image using both a ray diagram and the thin-lens equation. (b) Is the image real or virtual? Is the image upright or inverted? (c) What is the overall lateral magnification of the image?
302
Chapter
32
We can apply the thin-lens equation to find the image formed in the first lens and then use this image as the object for the second lens.
Picture the Problem
(a) The parallel, central, and focal rays were used to locate the image formed by the first lens and the parallel and central rays to locate the image formed by the second lens.
__
______
L
Apply the thin-lens equation to express the location of the image
(1)
formed by the first lens: Substitute numerical values and evaluate SI' : Find the lateral magnification of the first image:
S' = 1
{IO cm){20 em) = 20cm 20cm-IOcm
s/ 20cm ml= --= - --- =- I 20cm s
Because the lenses are separated by 35 cm, the object distance for the second lens is 35 cm 20 cm 1 5 cm. Equation -
=
( 1 ) applied to the second lens is: Substitute numerical values and evaluate S2':
/ (IOcm s =
){I5 cm) = 30cm
15cm-IOcm
and the final image is object. Find the lateral magnification of the second image:
sz'
! 85 . 0cm i from the
30cm
m2 = --= - --- = -2 1 5 cm s
Because S'2 0 and m = m1m2 = 2, the image is real, erect, and twice the size of the obj ect. >
Optical Images 303 The overall lateral magnification of the image is the product of the magnifications of each image: An object is 1 5 cm in front of a positive lens of focal length 15 cm. A second positive lens offocal length 1 5 cm is 20 cm from the first lens. Find the final image and draw a ray diagram. *64
••
We can apply the thin-lens equation to find the image formed in the first lens and then use this image as the object for the second lens.
Picture the Problem
Apply the thin-lens equation to express the location of the image formed by the first lens: Substitute numerical values and evaluate s/: With
SI
'
=
ro,
the thin-lens equation
applied to the second lens becomes:
(1)
s'
=
15cm-15 cm
1
� S2
(15 cm)(15 cm)
=
f
l
2
=>
s
z
'
=
=
<Xl
1 1 1 5.0 cm I 2 =
A ray diagram is shown below:
The final image is 50 cm from the object, real, inverted, and the same size as the object. Aberrations
Chromatic aberration is a common defect of (a) concave and convex lenses. (b) concave lenses only. (c) concave and convex mirrors. (d) all1enses and mirrors. *70
·
Determine the Concept
Chromatic aberrations are a consequence of the differential
refraction of light of differing wavelengths by lenses.
I (a) is correct. I
304
Chapter 32
The Eye
The Model Eye 1: A simple model for the eye is a lens with variable power P located a fixed distance d in front of a screen, with the space between the lens and the screen filled by air. Refer to Figure 32-60. The "eye" can focus for all values of s such that xnp � s � Xfp. This "eye" is said to be normal if it can focus on very distant objects. (a) Show that for a normal "eye," the minimum value of P is *73
••
Pmi
n
1
=d
(b) Show that the maximum value of P is Pmax
1 +1 d
=xnp
(c) The difference
=
Pmax - Pmin is
called the accommodation. Find the minimum power and accommodation for a model eye with d = 2.5 cm and Xnp = 25 cm.
A
p
Screen_
d-----+l Model eye ,
Figure 32-60 Problems 73, 74, and 75 Picture the Problem The thin-lens equation relates the image and object distances to the
power of a lens.
)
(a Use the thin-lens equation to
relate the image and object distances to the power of the lens: Because s' = d and, for a distance object, s = 00:
Prrlln
=
1 lL] ;r m =
O t
p ical Image s 305
(b) If xnp is the closest distance an object could be and still remain in clear focus on the screen, equation ( 1 ) becomes: (c) Use our result in (a) to obtain : Use the results of (a) and (b) to express the accommodation of the model eye: Substitute numerical values and evaluate A:
= OD 2.5cm 1 40 . I 1 1 --1 = -1 A = P -po =--+d d Xnp Xnp
Pmin
=
1
max
m1l1
1 1 4.00D I _= A =_ 25cm
If two point objects close together are to be seen as two distinct objects, the images must fall on the retina on two different cones that are not adjacent. That is, there must be an unactivated cone between them. The separation of the cones is about 1 JLID. Model the eye as a uniform 2.S-cm-diameter sphere with a refractive index of 1 .34. (a) What is the smallest angle the two points can subtend? (See Figure 32-6 1 .) (b) How close together can two points be if they are 20 m from the eye?
*79 ·
Figure 32-61
Problem 79
We can use the relationship between a distance measured along the arc of a circle and the angle subtended at its center to approximate the smallest angle the two points can subtend and the separation of the two points 20 m from the eye. Picture the Problem
(a) Relate Bmin to the diameter of the eye and the distance between the activated cones: Solve for
Bmin:
Bmin
= -2,liIll d-eye
Chapter 32
306
Substitute numerical values and evaluate �"ill:
2 JLI11 = I 80.0 Jifad I emil = 2.5cm
(b) Let D represent the separation of
D = Remin
the points R = 20 m from the eye to obtain:
= (20m )( 80 Jifad) = 1 1. 60mm 1
The Simple Magnifier
person with a near-point distance of 30 cm uses a simple magnifier of power 20 D. What is the magnification obtained if the final image is at infinity? *85
·
A
Picture the Problem We
(M =
can use the definitions of the magnifying power of a lens xnp /I ) and of the power of a lens (P = 1/1 ) to find the magnifying power of the
given lens.
The magnifying power of the lens is given by:
Xnp =Pxnp M=j where P is the power of the lens.
Substitute numerical values and evaluate M: *90
••
M = (20m- 1 )(0.3m) 1 6 .00 1 =
(a) Show that if the final image of a simple magnifier is to be at the near
point of the eye rather than at infinity, the angular magnification is given by
x M =�+ 1 I
(b) Find the magnification of a 20-D lens for a person with a near point of 30 cm if the final image is at the near point. Draw a ray diagram for this situation. Picture the Problem
We can use the definition of the angular magnification of a lens
x
and the thin-lens equation to show thatM =� +
I
(a ) Express the angular
magnification of the simple magnifier in terms of the angles subtended by the object and the image:
1. ( 1)
Optical Images 307 Solve the thin-lens equation for s: Because the image is virtual:
s = s'fs' -
--
f
S = -xnp ,
Substitute to obtain:
Express the angle subtended by the object: where y is the height of the object. Express the angle subtended by the Image: Substitute for s to obtain:
Substitute in equation ( 1) and simplify:
(b) In terms of the power of the magnifying lens: The magnification of a 20-D lens for a person with a near point of 30 cm and the final image at the near point IS:
A ray diagram for this situation is
shown to the right:
M = (0.3 m)(20m-1)+ 1 = �
308
Chapter 32
The Microscope
A microscope has an objective of focal length 8.5 mm and an eyepiece that *93 gives an angular magnification of 1 0 for a person whose near point is 25 cm. The tube length is 1 6 cm. (a) What is the lateral magnification of the objective? (b) What is the magnifying power of the microscope? ••
Picture the Problem
The lateral magnification of the objective is
magnifying power of the microscope is M = m M o e' (a) The lateral magnification of the
objective is given by: Substitute numerical values and evaluate mo:
(b) The magnifying power of the microscope is given by:
mo - L/ fa and the =
L m =-a fa m o = - 16c m = 1 -1 .881 8.5 mm
M = m oMe where Me is the angular magnification of the lens.
Substitute numerical values and evaluate M:
M = (-1.88)(10) = 1 - 18 .8 1
A compound microscope has an objective lens with a power of 45 D and an eyepiece with a power of 80 D. The lenses are separated by 28 cm. Assuming that the final image is formed 25 cm from the eye, what is the magnifying power? *95
••
Picture the Problem The
magnifying power of a compound microscope is the product of the magnifying powers of the objective and the eyepiece. (1)
Express the magnifying power of the microscope in terms of the magnifying powers of the objective and eyepiece: From Problem 82, the magnification of the eyepiece is given by: The magnification of the objective is given by:
L m o =-fo
Optical Images where 111
Substitute to obtain:
o
L
=
309
D- fo - fe
o =_ D - J+-J: fo
e
Substitute for me and mo in equation ( 1 ) to obtain: Substitute numerical values and evaluate M: M
]
(
=[ (80D )(0.2 5 m)+ 1 ] - 28c m-2.22c m-l .25c m = 1_ 2321 2 . 22c m
The Telescope
The 200-in (5 . 1 -m) mirror of the reflecting telescope at Mt. Palomar has a •• focal length of 1 .68 m. (a) By what factor is the light-gathering power increased over the 40-in (l.OI6-m) diameter refracting lens of the Yerkes Observatory telescope? (b) If the focal length of the eyepiece is 1 .25 cm, what is the magnifying power of this telescope?
*99
Because the light-gathering power of a mirror is proportional to its area, we can compare the light-gathering powers of these mirrors by finding the ratio of their areas. We can use the ratio of the focal lengths of the objective and eyepiece lenses to find the magnifying power of the Palomar telescope. Picture the Problem
(a) Express the ratio of the light gathering powers of the Palomar and Yerkes mirrors:
P.Palomar PYerkes
_
APalomar mirror AYerkes mirror
Jr
2
Jr
2
- 4 d Palomar mirror d "4 Yerkes mirror
_ d;alomar mirror d�erkes mirror Substitute numerical values and evaluate PPaloma/PYerkes:
Ppalomar PYerkes
or Ppalomar
(b) Express the magnifying power of the Palomar telescope:
= (200inr = 25.0 (40 in) = I (25.0)PYerkeS
310
Chapter 32
Substitute numerical values and evaluate M:
M=-
1 .68 m = 1 -134 1 1.25c m
General Problems *105
•
A camera uses a positive lens to focus light from an object onto film. Unlike
the eye, the camera lens has a fixed focal length, but the lens itself can be moved slightly to vary the image distance to the image on the film. A telephoto lens has a focal length of 200 mm. By how much must it move to change from focusing on an object at infinity to an object at a distance of 30 m? We can express the distance lis that the lens must move as the difference between the image distances when the object is at 30 m and when it is at infinity and then express these image distances using the thin-lens equation.
Picture the Problem
Express the distance lis that the lens must move to change from focusing on an object at infinity to one at a distance of 3 0 m: Solve the thin-lens equation for s': Substitute and simplify to obtain:
Js s , = -s- f jj,s = Js30 _
Substitute numerical values and evaluate lis:
*110
••
jj,s
S30
-f
S30
-f
Js30
30m -1] [30m-0 .2 m
=
(200mm)
=
1 1 .3 4mm 1
A scuba diver wears a diving mask with a face plate that bulges outward
with a radius of curvature of 0.5 m. There is thus a convex spherical surface between the water and the air in the mask. A fish is 2.5 m in front of the diving mask. (a) Where does the fish appear to be? (b) What is the magnification of the image of the fish?
Optical I mages 3 1 1 Picture the Problem We can use the equation for refraction at a single surface to locate the image of the fish and the expression for the magnification due to refraction at a spherical surface to find the magnification of the image.
(a) Use the equation describing refraction at a single surface to relate the image and object distances:
!2+� ' s
s
=
nz -nl r
Solve for s':
Substitute numerical values and evaluate s':
s
'
m) = (1 - 1 .33(1 ) (0.5 m)(2.5 ) (2.5 m) - (1 .33 ) (0.5 m) = 1 -0. 839m 1
Note that the fish appears to be much closer to the diver than it actually is.
(b) Express the magnification due to refraction at a spherical surface: Substitute numerical values and evaluate m:
n I s' m = - -n2 s m =
_
(1 .33 ) (- 0 . 83 9 m) 1 0.446 1 (1)(2.5 m) =
Note that the fish appears to be smaller than it actually is. (a) Find the focal length of a thick, double convex lens with an index of refraction of l.5, a thickness of 4 cm, and radii of +20 cm and -20 cm. (b) Find the focal length of this lens in water. *115
••
Here we must consider refraction at each surface separately. To find the focal length we imagine the object at s = 00, and find the image from the first refracting surface at s' I. That image serves as the object for the second refracting surface. We'll find that this is a virtual image for the second refracting surface, i .e., S2 is negative. Using the equation for refraction at a single surface a second time, we can locate the image formed by the second refracting surface by the virtual object at S2' The location of that image is then the focal point of the thick lens. We'll let the numeral 1 denote the first surface and the numeral 2 the second surface. In part (b) we can proceed as in part (a) (except that now nl 1.33 for the first refraction and nz = 1.33 for the second refraction) to determine the focal length in water, which we denote byfw. Picture the Problem
=
312
Chapter
32
(a) Use the equation for refraction
at a single surface to relate s, and
s,':
For s,
!i + � =
s, s,'
112
-
11,
1)
= 00:
Solve for
21] S/ = -11-"--'112 -11,
SI':
Substitute numerical values and evaluate
SI' :
The object distance second lens is:
S2 for the
m) = 60.0c m S ' = (1 .5)(20c 1 .5-1 1
S2 =-(s,' -4c m) = -(60c m-4c m) =-56c m
Solve the equation for refraction at a single surface for
S2'- -112r2s2 - (112 111 )S2 - 11lr2
Substitute numerical values and evaluate
5'
S2' :
52' :
_
= 5z' + 2c m = 1 9.3c m + 2c m = I 21 .3c m I
f
(b) Substitute numerical values in
m) = 17 6c m 5, = (1 .5)(20c 1.5- 1 .33
The object distance second lens is:
S2 for the
I
52 =-(5/ -4c m) =-(17 6c m-4c m) =-17 2c m
Substitute numerical values in equation (2) and evaluate
5z' =
S'2:
(1.33 ) (- 20c m) (- 17 2c m) =77 .2c m (1 .33 -1 .5)(-172c m)-(1 .5)(- 20c m)
Because fw is measured from the center of the lens:
(2)
(1 ) (- 20c m)(- 56c m) 2 - (1-1 .5)( - 56c m)- (1 .5)(- 20c m) = 19.3c m
Because f is measured from the center of the lens:
equation (1) and evaluate 5/ :
(1)
fw
= 5z' + 2c m =77.2c m + 2c m = 1 7 9.2c m I
Optical Images 313 Rem arks: Note that if we use the expression given in Problem 114 we obtain
j.v = 83.3 em, in only moderate agreement with the exact
result given above.
When a bright light source is placed 30 em in front of a lens, there is an upright image 7.5 e m from the lens. There is also a faint inverted image 6 em in front of the lens due to reflection from the front surface of the lens. When the lens is turned around, this weaker, inverted image is 1 0 cm in front of the lens. Find the index of refraction of the lens. *120
•••
The mirror surfaces must be concave to create inverted images on reflection. Therefore, the lens is a diverging lens. Let the numeral 1 denote the lens in its initial orientation and the numeral 2 the lens in its second orientation. We can use the mirror equation to find the magnitudes of the radii of the lens' surfaces, the thin-lens equation to find its focal length, and the lens. maker' s equation to find its index of refraction. Picture the Problem
Solve the mirror equation for
I'll:
Substitute numerical values and evaluate I'll: Solve the mirror equation for
2 1'1 1= s] S]S/ , +S] I'll 2(30c mX 6c m) = 1O.Oc m
6c m +30c m
=
hi:
Substitute numerical values and evaluate h�: Solve the thin-lens equation for!
Substitute numerical values and evaluate! Solve the lens-maker' s equation for n to obtain:
i mXIOc m) = 1 5. 0c m h 2(30c Oc =
I m +30c m
j= SS' s' + s j= (30cm ) (-7 .5cm ) =-lO.Oc m -7 .5c m + 30c m
314
Chapter 32
Because the lens is a diverging lens, rJ - 1 0 cm and r2 = 1 5 cm. Substitute numelical values and evaluate n:
=
n
=
(
1 1 1 _ _ l_ (- IOcm ) - 10cm IS cm
=�
J
+
The lateral magnification of a spherical mirror or a thin lens is given by m = -s' ls. Show that for objects of small horizontal extent, the longitudinal magnification is approximately _m2. (Hint: Show that ds'/ds = - s,2/i.) *125
•••
Picture the Problem We
due to a change in s.
examine the amount by which the image distance s' changes
Solve the thin-lens equation for s':
Differentiate s' with respect to s:
ds' � dsd[(1 -1J-I -; ] �- (� 1- 1 �-7�-m H I The image of an object of length will have a length m2/j,s. 1 ds
s'
f
/j,s
S
'2
2
-
Chapter 33 Interference and Diffraction Conceptual Problems *1
When destructive interference occurs, what happens to the energy in the light
•
waves? The energy is distributed nonuniforrnly in space; in some regions the energy is below average (destructive interference), in others it is higher than average (constructive interference). Determine the Concept
*6 • A loop of wire is dipped in soapy water and held so that the soap film is vertical. (a) Viewed by reflection with white light, the top of the film appears black. Explain why. (b) Below the black region are colored bands. Is the first band red or violet? ( c) Describe the appearance of the film when it inriewed by transmitted light.
(a) The phase change on reflection from the front surface of the film is 1 80°; the phase
change on reflection from the back surface of the film is 0°. As the film thins toward the top, the phase change associated with the film's thickness becomes negligible and the two reflected waves interfere destructively.
(b) The first constructive interference will arise when t = Al4. Therefore, the first band will be violet (shortest visible wavelength). (c) When viewed in transmitted light, the top of the film is white, since no light is reflected. The colors of the bands are those complementary to the colors seen in reflected light; i.e., the top band will be red.
A double-slit interference experiment is set up in a chamber that can be evacuated. Using monochromatic light, an interference pattern is observed when the chamber is open to air. As the chamber is evacuated, one will note that (a) the interference fringes remain fixed. (b) the interference fringes move closer together. (c) the interference fringes move farther apart. (d) the interference fringes disappear completely. *10
·
Determine the Concept
Ym
= m
J..L
d
,
The distance on the screen tomth bright fringe is given by
where L is the distance from the slits to the screen and d is the separation of
the slits. Because the index of refraction of air is slightly larger than the index of refraction of a vacuum, the introduction of air reduces A, to Aln and decreases Ym. Because the separation of the fringes is Y", - Ym-), the separation of the fringes decreases
315
3 16 Chapter 33 and I (b) is correct. I Estimation and Approximation *12 It is claimed that the Great Wall of China is the only human object that can be seen from space with the naked eye. Make an argument in support of this claim based on the resolving power of the human eye. Evaluate the validity of your argument for observers both in low-earth orbit (�400 km altitude) and on the moon. Picture the Problem We'll
assume that the diameter of the pupil of the eye is
5 mm and that the wavelength of light is 600 nm. Then we can use the expression for the
minimum angular separation of two objects than can be resolved by the eye and the relationship between this angle and the width of an object and the distance from which it is viewed to support the claim. Relate the width w of an object that can be seen at a height h to the critical angular separation ac:
tanac = -h
Solve for w:
W = htanac
The minimum angular separation a.: of two point objects that c
d
= -
wherem 19 =
Substitute numerical values to obtain: or
1 5 . 46 Jll1l < d < 5.7 5 Jll1l 1
A film of oil of index of refraction n = 1 .45 floats on water (n 1.33 ). When illuminated with white light at normal incidence, light of wavelengths 700 nm and 500 *26
••
=
nm is predominant in the reflected light. Determine the thickness of the oil film. Picture the Problem Because the index of refraction of air is less than that of the oil, there is a phase shift of trrad (11) in the light reflected at the air-oil interface. Because
the index of refraction of the oil is greater than that of the glass, there is no phase shift in the light reflected from the oil-glass interface. We can use the condition for constructive interference to determinem for It = 700 nm and then use this value in our equation describing constructive interference to find the thickness t of the oil film. Express the condition for constructive interference between the waves reflected from the air-oil interface and the oil-glass interface: Substitute for obtain:
l' and
solve for It to
Substitute the predominant wavelengths to obtain:
2t +11' = 1',21',31',... or
2t
- 1 1 I 3 1 I S '} I - 2"/\',2"/\' ,2" /\',
.
- (m + 2"/\' )
_ . .
1
1 I
(1 )
where It' is the wavelength of light in the oil andm = 0, 1, 2, . . .
1 =� m +l. 2
700nm
2nt 2nt and 500nm =-= -m +l. m +1. 2
2
320
Chapter 33
Di vide the first of these equations by the second to obtain:
700 nm 500nm m
Solve for m :
=
2nt m + -.L2
= __
_
2nt
m + l2
__
m + -.L 2
2 for A- 7 00 nm =
Solve equation (1) for t:
t - (2 + "21 ) 7 00nm) _- I 603 nm 1 2 (1 .45
Substitute numerical values and evaluate t: Newto n ' s Rings
A Newton's ring apparatus consists of a plano-convex glass lens with radius of curvatureR that rests on a flat glass plate, as shown in Figure 33-42. The thin film is air of variable thickness. The pattern is viewed by reflected light. (a) Show that for a thickness t the condition for a bright (constructive) interference ring is *27
··
t
=
A- m 0,1, 2,... (m+ t)-, 2
=
(b) Apply the Pythagorean Theorem to the triangle of sides r, R - t, and hypotenuseR to
show that for t < < R, the radius of a fringe is related to t by r
=
-J2tR
(c) How would the transmitted pattern look in comparison with the reflected one? (d) Use R = 1 0 m and a diameter of 4 cm. How many bright fringes would you see if the apparatus were illuminated by yellow sodium light (A :::::: 590 nm) and viewed by reflection? (e) What would be the diameter of the sixth bright fringe? if) If the glass used in the apparatus has an index of refraction n = 1 .5 and water (nw 1 .33) is placed =
between the two pieces of glass, what change will take place in the bright-fringe pattern?
R
Figure 33-42
Problem 27
I
I
Interference and Diffraction 321 This arrangement is essentially identical to a "thin fi lm" configuration, except that the "film" is air. A phase change of 1 800 ( 1-A ) occurs at the Picture the Problem
top of the flat glass plate. We can use the condition for constructive interference to derive the result given in (a) and use the geometry of the lens on the plate to obtain the result given in (b). We can then use these results in the remaining parts of the problem. (a) The condition for constructive interference is:
2t + 1-A = ,,1., 2,,1.,3,,1., ...
or
2t = 1- A, t A, -t A, .. . = (m + 1-),,1.
where A is the wavelength of light in air and m = 0, 1 , 2, . . .
Solve for t:
A = 0,1, 2, t = (m + 1- )-,m ... 2
(1)
(b) From Figure 33-39 we have: or
R2 = r 2 + R2 -2Rt + t 2
For t « R we can neglect the last term to obtain: Solve for r : (c)
r = I �2Rt I
(2)
The transmitted pattern is complementary to the reflected pattern.
(d) Square equation (2) and substitute for t from equation ( 1 ) to obtain: Solve for m :
r2 1 m = --RA 2
Substitute numerical values and evaluate m :
Y - ! = 67 m = (1 ° m(2cm )(S 90nm) 2
(e) The diameter of the mlh fringe is:
D = 2r
and so there will be � bright fringes. = 2�(m + 1-)RA
322
C
hapter 33
=
D = 2 �(5 + t)(lO m)(S 90nm) = l . 1 4cm
Noting thatm 5 for the sixth fringe, substitute numerical values and evaluate D:
I
1
The wavelength of the light in the film becomes Itair / 444 nm The separation between fringes is reduced and the number of fringes that will be seen is increased by the factor l .3 3 . n =
(j)
.
n=
Two-Slit Interference Pattern
Two narrow slits separated by 1 mm are illuminated by light of wavelength 600 nm, and the interference pattern is viewed on a screen 2 m away. Calculate the number of bright fringes per centimeter on the screen. *30
·
Picture the Problem The number of bright fringes per unit distance is the reciprocal of
the separation of the fringes. We can use the expression for the distance on the screen to themth fringe to find the separation of the fringes. Express the number N of bright fringes per centimeter in terms of the separation of the fringes:
I N=_
(1)
Lly
Express the distance on the screen to themth and (m + 1 )st bright fringe: Subtract the second of these equations from the first to obtain:
U ily =
Substitute in equation ( 1 ) to obtain:
N=�
Substitute numerical values and evaluate N:
N=
d
AL
X ) = 1 8 .33cm -1 I (600nm 2m I
White light falls at an angle of 30° to the normal of a plane containing a pair of slits separated by 2.5 J-lffi . What visible wavelengths give a bright interference maximum in the transmitted light in the direction normal to the plane? (See Problem 34.) *35
••
Picture the Problem Let the separation of the slits be d. We can find the total path
difference when the light is incident at an angle ¢ and set this result equal to an integer multiple of the wavelength of the light to relate the angle of incidence on the slits to the direction of the transmitted light and its wavelength.
Interference and Diffraction 323 !'::,. £ =
Express the total path difference:
d sin ¢ + d si n e
!'::,.£ = m A
The condition for constructive interference is:
where m is an integer.
d sin ¢ + d sin e = rnA
Substitute to obtain: Divide both sides of the equation by d to obtain:
. d.. • n Slll 'f' + Slll u
Set B= 0 and solve for A:
A
Substitute numerical values and simplify to obtain:
A =
=
=
rnA
d
-
d sin¢ m
(2.5 ,um)sin 30° 1 .25 ,um m
=
rn
Evaluate A for positive integral values of m :
From the table we can see that 625 the electromagnetic spectrum.
m
A (nm)
1 2 3 4
1250 625 417 3 13
run
and 4 1 7
run
are in the visible portion of
Diffraction Pattern of a Single Slit
*39 Measuring the distance to the moon (lunar ranging) is routinely done by firing short-pulse lasers and measuring the time it takes for the pulses to reflect back from the moon. A pulse is fired from the earth; to send it out, the pulse is expanded so that it fills the aperture of a 6-in-diameter telescope. (a) Assuming the only thing spreading the beam out to be diffraction, how large will the beam be when it reaches the moon, 382,000 km away? (b) The pulse is reflected off a retroreflecting mirror left by the Apollo 1 1 astronauts. If the diameter of the mirror is 20 in, how large will the beam be when it gets back to the earth? (c) What fraction of the power of the beam is reflected back to the earth? (d) If the beam is refocused on return by the same 6-in telescope, what fraction of the original beam energy is recaptured? Ignore any atmospheric losses. 00
Picture the Problem The diagram shows the beam expanding as it travels to the moon
and that portion of it that is reflected from the mirror on the moon expanding as it returns to earth. We can express the diameter of the beam at the moon as the product of the beam
324
Chapter
33
divergence angle and the distance to the moon and use the equation describing diffraction at a circular aperture to find the beam divergence angle. We can follow this same procedure to find the diameter of the beam when it gets back to the earth. In Parts (c) and (d) we can use the dependence of the power in a beam on its cross-sectional area to find the fraction of the power of the beam that is reflected back to earth and the fraction of the original beam energy that is recaptured upon return to earth.
I I
/ � --.. --..
\
11
v, 1\
dt,l,seo ,
\
--.. --..
�
\
\T \
/ -� \ " ----/- -
D >::: 8 L
(a) Relate the diameter D of the beam at the moon to the distance to
the moon L and the beam
divergence angle B :
The angle B subtended by the first diffraction minimum is related to the wavelength
/L sin 8 1 .22 --=
A of the light and the
dtelescope
diameter of the telescope openi�g
dtelescope by : Because B «
1 , sin B �
B and:
/L 8 >::: 1 .22 -dtelescope
Substitute for B in equation obtain:
( 1 ) to
D=
1.22L/L dtelescope
Substitute numerical values and evaluate D:
D = (3. 8 2 x 10 (b)
)-, _8 ) ___ 22--,-(S_0_0_ nm--,._ m 6 . 12.54em 1m 10 em
The portion of the beam reflected back to the earth will be that portion incident on the mirror, so the diffraction angle is:
mx
m
x
2
/L 8 � 1 .22-dmirror
=
I 1 . S 3km I
Interference and Diffraction 325
[
The beam will expand back to:
DI = L 1 .22 � dnllrror
]
Substitute numerical values and evaluate D' :
D' = (3.82 x l 08 m ) (c) Because the power of the beam is proportional to its cross-sectional area, the fraction of the power that is reflected back to the earth is the ratio of the area of the mirror to the area of the expanded beam at the moon:
1 .22(500 nm ) = 1 459m I 2.54em 1m :-20 m' x in x 2 10 em pi P
_
Jr
2
'4 dmirror
�llITOT
Jr
Abeam
4
D
Substitute for D to obtain: pi
dmirror 1 .22LA =
P
dtelescope
Substitute numerical values and evaluate P'IP:
2 =
(
( "; )
d,,=,d.,=op,
1 .22LA
(
)
' 2 5 cm ) ) ( (20 in 6 in � pi = 8 P 1 .22(3. 8 2 1 0 m)(500nm ) X
= 1 1.10x l 0-7 1 (d) The angular spread of the beam from reflection from the 20-in mirror is given by:
, ,
d
A 8 :::::; 1 .22 -dmirror
The diameter D' of the beam on return to earth will be:
D' :::::; 1 .22L -A
Letting pI! represent the power intercepted by the telescope, we have:
P /I
dmirror
pi
_
L1
-'telescope
f
Jr
_
2 dtelescope -4 _ __
---'-
J 2
(1)
32 6
Chapter
33
Substitute for D' and simplify:
p" = P'
[ dtelescopedmirror )2
(2)
1.22L?
Multiply equation (2) by equation ( 1 ) and simplify to obtain: p" P ' = p" = P' p P
2 [ dtelescopedrnirror ) [ dmirrordtelescope )2
1 .22L?
[ dmirrordtelescope )
1 .22L?
4
1 .22L?
4 ' ) 25 cm � (20in )(6 in) ( 8 1 .22(3. 82 10 m)(500nm )
Substitute numerical values and evaluate P"IP:
=
p" p
x
=
1 1 .21
X
1 10 - 4
Interference-Diffraction Pattern of Two Slits
* 43 •• Suppose that the central diffraction maximum for two slits contains 1 7 interference fringes for some wavelength of light. How many interference fringes would you expect in the first secondary diffraction maximum? Determine the ConceptPicture the Problem There are 8 interference fringes on each
side of the central maximum. The secondary diffraction maximum is half as wide as the central one. It follows that it will contain 8 interference maxima. Using Phasors to Add Harmonic Waves
* 46
•
Find the resultant of the two waves EI = 4 sin 0Jt and E
2
=
3 sin ( 0Jt + 600).
Picture the Problem Chose the coordinate system shown in the phasor diagram. We can
use the standard methods of vector addition to find the resultant of the two waves.
The resultant of the two waves is of the form:
E
=
R
sin (
mt +
6')
Interference and D i ffraction
-
Express the x component of
R:
R , = 4 + 3 cos 60° = 5 . 5 0
Express the y component of
R:
R = 0 + 3 si n 60° = y
Find the magnitude of
R:
( J R
2 . 60
I tan- ( 2. 60 ) = 25.3 ° 5.50
Find the phase angle 5between R and E1 :
0=
t
Substitute to obtain:
E=
1 6.08 sin(mt 25 .3 °) I
an -I
y
R x
=
327
+
Remarks: We could have used the law of cosines to find R and the law of sines to
find &
*52 ••• For single-slit diffraction, calculate the first three values of ¢ (the total phase difference between rays from each edge of the slit) that produce subsidiary maxima by (a) using the phasor model and (b) setting dJ/d¢ = 0, where l is given by Equation 33-19. Picture the Problem We can use the phasor diagram shown in Figure 3 3 -26 to determine
the first three values of ¢ that produce subsidiary maxima. Setting the derivative of Equation 33-19 equal to zero will yield a transcendental equation whose roots are the values of ¢ corresponding to the maxima in the diffraction pattern. (a) Referring to Figure 33-26 we see
that the first subsidiary maximum occurs when: A minimum occurs when: Another maximum occurs when: Thus, subsidiary maxima occur when:
(b) The intensity in the single-slit diffraction pattern is given by:
¢ = 5n ¢ = (2n + 1 )n; n = 1, 2, 3, ...
and the first three subsidiary maxima are at ¢ = 3 n; 5 n; and 71r.
1=1 o
( sin
1¢ 2 1¢
J
Set the derivative of this expression equal to zero for extrema:
3 28 Ch apter 33 d1 - = 210 d¢
( . "'J [ sm - If' I
2
i¢
I '" - If' cos 4
. ",]
- -) - sm -
( I
'" If'
i¢
I
2
2
I
2
If'
0 for relative maxima and minima
=
Simplify to obtain the transcendental equation: Solve this equation numerically (use the "Solver" function of your calculator) to obtain:
¢
=
I 2.86n; 4.92n; and 6.94n I
Remarks: Note that our results in (b) are smaller than the approximate values found in (a) by 4.80%, 1 .63 %, and 0.865% and that the agreement improves as n increases. Diffraction and Resolution
Two sources of light of wavelength 700 nm are separated by a horizontal *55 • distance x. They are 5 m from a vertical slit of width 0.5 mm. What is the least value of x for which the diffraction pattern of the sources can be resolved by Rayleigh's criterion? Picture the Problem We can use
Rayleigh's criterion for slits and the geometry of the diagram to express x in terms of 2, L, and the width a of the slit.
-, x L Referring to the diagram, relate a." L, and x: For slits, Rayleigh' s criterion is:
Equate these two expressions to obtain: Solve for x:
x i':j
a
C L
a
= a
C
x
L X =
A
A a
-AL a
Interference an d D i ffraction
Substitute numerical values and evaluate x :
3 29
(700 nm ) (5 m ) 1 7.00 mm 1 0.5 mm
x
The star Mizar in Ursa Major is a binary system of stars of nearly equal *60 magnitudes. The angular separation between the two stars is 14 seconds of arc. What is the minimum diameter of the pupil that allows resolution of the two stars using light of wavelength 550 nm? ••
We can use Rayleigh' s criterion for circular apertures and the geometry of the diagram to obtain an expression we can solve for the minimum diameter D of the pupil that allows resolution of the binary stars. Picture the Problem
1 T
.... .. t-. -
Your pupil
\
L -----.� I
A
(a) Rayleigh's criterion is satisfied provided:
a c = 1 .22 -
Solve for D:
D = I .22�
D
ac
550 nm 14"x 610 x 8 d 3 00" 1 00 I = 1 9 . 89 rnm :::d cm
Substitute numerical values and evaluate D:
D l .22 =
7r fa
---
--
Diffr action Gratings
*62 · With the diffraction grating used in Problem 6 1 , two other lines in the firstorder hydrogen spectrum are found at angles 81 x l O-2 rad and fh. = l .32 l O 1 rad. Find the wavelengths of these lines. =
9.72
x
-
We can solve rn A for A. with m = 1 to express the location of the first-order maximum as a function of the angles at which the first-order images are found.
Picture the Problem
d sin e
=
330
Chapter
33
The interference maxima in a diffraction pattern are at angles 8 given by: Solve for A:
d sin B
=
inA
where d is the separation of the slits and In = 0, 1 , 2, . . . A
= d sin B In
Relate the number of slits N per centimeter to the separation d of the slits:
N = !...d
Let m = and substitute for d to obtain:
d = sin B N � sin(9.72x 10-2 rad) 1 = I = 48 5nm I 2000 cm� sin(1 .32 X 10- l rad) 1 = = 65 8 nm I 2000cm-1
1
Substitute numerical values and evaluate Al for 81 = 9.72 rad:
X 10-2
Substitute numerical values and evaluate Al for 8 2 = 1 .32 rad:
xl 0-1
A
*67 ·· A diffraction grating with 4800 lines per centimeter is illuminated at normal incidence with white light (wavelength range of 400 nm to nm). For how many orders can one observe the complete spectrum in the transmitted light? Do any of these orders overlap? If so, describe the overlapping regions.
700
d sin B = rnA, rn = 1 , 2, 3, .. . to
Picture the Problem We can use the grating equation
express the order number in terms of the slit separation d, the wavelength of the light A, and the angle e. The interference maxima in the diffraction pattern are at angles 8 given by:
d sin B = rnA, rn = 1, 2, 3, . ..
Solve for m :
d sin B m = --A
If one is to see the complete spectrum:
sin B � 1
and
d m �A
Interference and Diffraction Evaluate 117 ma, :
Because m max
1
lnmax
=
4800 cm-1
Am3x
----
331
1
4800 cm-1 = 2.98 700 nm
----
2.98, one can see the complete spectrum only for m = 1 and 2.
Express the condition for overlap:
Because 700 run < 2 x 400 run, there is no overlap of the second - order spectrum into the first - order spectrum; however, there is overlap of long wavelengths in the second order with short wavelengths in the third - order spectrum. *71 ·· Mercury has several stable isotopes, among them 198Hg and 202Hg. The strong spectral line of mercury, at about 546.07 nm, is a composite of spectral lines from the various mercury isotopes. The wavelengths of this line for 1 98Hg and 202Hg are 546.07532 nm and 546.07355 nm, respectively. What must be the resolving power of a grating capable of resolving these two isotopic lines in the third-order spectrum? If the grating is illuminated over a 2-cm-wide region, what must be the number of lines per I centimeter of the grating? Picture the Problem We can use the expression for the resolving power of a grating to
find the resolving power of the grating capable of resolving these two isotopic lines in the third-order spectrum. Because the total number of the slits of the grating N is related to width w of the illuminated region and the number of lines per centimeter of the grating and the resolving power R of the grating, we can use this relationship to find the number of lines per centimeter of the grating The resolving power of a diffraction grating is given by:
R
=
Substitute numerical values and evaluate R:
R
= =
Express n, be the number of lines per centimeter of the grating, in terms of the total number of slits N of the grating and the width w of the
n =
A = mN IflAI
-
(1)
546.07532 15 46.07532 - 546.0735 51
1 3.09 x l 05 I N
w
332
Chapter 3 3
grating: From equation ( 1 ) we have :
N= R
m
Substitute to obtain:
Substitute numerical values and evaluate n :
n =
n
_
-
R
--
mw
3.09 1 05 (3 )(2 cm) x
_
-
I 5 15 .
x
1 0 4 em
-I 1
General Problems
*77 · A long, narrow horizontal slit lies 1 mm above a plane mirror, which is in the horizontal plane. The interference pattern produced by the slit and its image is viewed on a screen 1 m from the slit. The wavelength of the light is 600 nm. (a) Find the distance from the mirror to the first maximum. (b) How many dark bands per centimeter are seen on the screen? Picture the Problem We can apply the condition for constructive interference to find the
angular position of the first maximum on the screen. Note that, due to reflection, the wave from the image is 1 800 out of phase with that from the source. (a) Because Yo « L, the distance from the mirror to the first maximum is given by: Express the condition for constructive interference: Solve for e :
For the first maximum, m = 0 and:
Substitute in equation ( 1 ) to obtain:
(1)
d sin e = (m + t)/l" m
=
0, 1,2, . . .
Interference and Di ffraction
Because the image of the slit is as far behind the mirror's surface as the slit is in front of it, d 2 mm. Substitute numerical values and evaluate yo :
Yo
The number of dark bands per centimeter is the reciprocal of the fringe separation: Substitute numerical values and evaluate n:
_
) -) [(1.- ) 6002 mmnm ] .
sm
?
= I 0 . 150 mm I
=
(b) The separation of the fringes on the screen is given by:
(1 - m
333
}.L L'-.y = d n
d 1 ==-
n=
�y }.L
2 mm = I 3.33x10 m (600nm) (1m ) 3
)
I
*82 •• A thin layer of a transparent material with an index of refraction of 1 .30 is used as a nonreflective coating on the surface of glass with an index of refraction of 1 .50. What should the thickness of the material be for it to be nonreflecting for light of wavelength 600 nm? A
Picture the Problem Note that reflection
at both surfaces involves a phase shift of 1r rad. We can apply the condition for destructive interference to find the thickness t of the nonreflective coating.
=
600 nm
Air Coating Glass
The condition for destructive interference is:
2t = (m + t)Acoatina = (m + t) �
Solve for t:
t = (m + t)
Evaluate t for m = 0:
t = (2 ) 600nm = 1 1 15 nm I 2(1 .30)
o
ncoating
Aair 2ncoating
1.
*87 ·· The Impressionist painter Georges Seurat used a technique called point illism, in which his paintings are composed of small, closely spaced dots of pure color, each about 2 mm in diameter. The illusion of the colors blending together smoothly is
334
Chapter 3 3
produced in the eye o f the viewer by diffraction effects. Calculate the minimum viewing distance for this effect to work properly. Use the wavelength of visible light that requires the greatest distance so that you're sure the effect will work for all visible wavelengths. Assume the pupil of the eye has a diameter of 3 mm . ,
We can use the geometry of the dots and the pupil of the eye and Rayleigh's criterion to find the greatest viewing distance that ensures that the effect will work for all visible wavelengths. Picture the Problem
f.------
Referring to the diagram, express the angle subtended by the adjacent dots: Letting the diameter of the pupil of the eye be D, apply Rayleigh's criterion to obtain: Set e =
ac
to obtain:
Solve for L: Evaluate L for the shortest wavelength light in the visible portion of the spectrum:
L
----i�
d () � -
L
ac
d L
= 1 .22-A D
=
L=
1 .22
� D
Dd
1( .22A )( ) 2 mm = 1 12.3m I L = 3 mm(400nm ) 1 .22
*88 ••• A Jamin refractometer is a device for measuring or for comparing the indexes of refraction of gases. A beam of monochromatic light is split into two parts, each of which is directed along the axis of a separate cylindrical tube before being recombined into a single beam that is viewed through a telescope. Suppose that each tube is 0 .4 m long and that sodium light of wavelength 589 nm is used. Both tubes are initially evacuated, and constructive interference is observed in the center of the field of view. As air is slowly allowed to enter one of the tubes, the central field of view changes to dark and back to bright a total of 1 9 8 times. (a) What is the index of refraction of air? (b) If
Interference and D i ffraction
335
the fringes can be counted to ± 0 .25 fringe, where one fringe is equivalent to one complete cycle of intensity variation at the center of the field of view, to what accuracy can the index of refraction of air be determi ned by this experiment? It is given that with one tube evacuated and one full of air at l -atm pressure, there are 1 98 more wavelengths of light in the tube full of air than in the evacuated tube of the same length. We can use this condition to obtain an equation that expresses this difference in terms of L, A,,, and Ao. We can obtain a second equation Picture the Problem
relating An, n, and Ao ( IL"
IL0 =_ ) and solve the two equations simultaneously to find n
n.
= 1L0 n
(a) The wavelengths are related by:
IL"
The number of wavelengths in length L is the length L divided by the wavelength. Thus:
L L -= 1 98 IL" 1L0
Substitute for A,,:
Solve for An to obtain: Substitute numerical values and evaluate n :
n = 1 + 1 9 81L0 L 89 = 1 1 .0002916 1 n = 1 + 1 9 8 5OA m
( run]
(b) Replace 1 98 with 1 9 8 ± 0.25 and assume that the uncertainties in L and Ao are
negligible:
n = 1 + � (1 98 ± 0.25) = 1 1 .0002916 ± 0.0000004 1
Chapter 34 Wave-Particle Duality and Quantum Physics Conceptual Problems
*1
•
The quantized character of electromagnetic radiation is revealed by
(a) the Young double-slit experiment. (b) diffraction of light by a small aperture. (c) the
photoelectric effect. (d) the J.J. Thomson cathode-ray experiment. Determine the Concept The Young double-slit experiment, the diffraction of light by a
small aperture, and the J.J. Thomson cathode-ray experiment all demonstrated the wave nature of electromagnetic radiation. Only the photoelectric effect requires an explanation based on the quantization of electromagnetic radiation.
I (e )is correct. I
*5 The work function of a surface is ¢. The threshold wavelength for emission of photoelectrons from the surface is (a) he/¢. (b) ¢/hI (c) hjl¢. (d) none of the answers are correct. Determine the Concept The threshold wavelength for emission of photoelectrons is
related to the work function of a metal through ¢ = he/At . Hence At
I ( )is correct. I a
=
he/¢ and
*11 · Explain why the maximum kinetic energy of electrons emitted in the photoelectric effect does not depend on the intensity of the incident light, but the total number of electrons emitted does depend on the intensity of the incident light. Determine the Concept In the photoelectric effect, an electron absorbs the energy of a
single photon. Therefore, Kmax = hf- ¢, independently of the number of photons incident on the surface. However, the number of photons incident on the surface determines the number of electrons that are emitted. Estimation and Approximation
*16 ·· Students in an advanced physics lab use X rays to measure the Compton wavelength, A.c. The students obtain the following wavelength shifts � - II, as a function of scattering angle () () � - II,
45° 0.647 pm
135° 90° 1 80° 75° 1 .67 pm 2.45 pm 3 .98 pm 4.95 pm
337
338
Chapter 3 4
Use their data to estimate the value for the Compton wavelength. Compare this number with the accepted value. Picture the Problem From
ILz - /l,
=
the Compton-scattering equation we have Ac (1 - cos 8), where Ac hi mec is the Compton wavelength. Note that this =
equation is of the form y = mx + b provided we let y = � - AI and x = 1 - cos e. Thus, we can linearize the Compton equation by plotting �A = ,,1,2 - A, as a function of 1 - cos 8 . The slope of the resulting graph will yield an experimental value for the Compton wavelength. (a) The spreadsheet solution is shown below. The formulas used to calculate the
quantities in the columns are as follows: Cell A3 B3 C3
Formula/Content Algebraic Form 45 e (deg) 1 - cos e 1 - cos(A3 *PIOI1 80) 6.47E"-13 �A = ILz - A, e (deg) 45 75 90 135 1 80
1- cose
�-AI
0.293 0.741 1 .000 1 .707 2.000
6.47E- 1 3 1 .67E- 1 2 2.45E- 1 2 3 .98E- 1 2 4.95E-12
The following graph was plotted from the data shown in the above table. Excel' s "Add Trendline" was used to fit a linear function to the data. The regression equation is
�A
=
2.48 x 1 0-1 2 (1 - cos 8) - 1 .03 x 1 0-13 6.0E- 1 2 5.0E-12 � 25 E .."!
f'! 0;
-0
4.0E-12 3.0E- 1 2 2.0E-12 1 .0E- 1 2 O.OE+OO
0.0
0.5
1 .0
l -cos(theta)
1 .5
2.0
Wave-Parti cle Dual i ty and Quantum Physics
From the regression line we note that the experimental value for the Compton wavelength AC,exp is:
.1c,exp =
I 2.48
X
1 0- 1 2
m
339
I
The Compton wavelength is given by: Substitute numerical values and evaluate k:
.1 = 1 2 40 eV �nm = 2 . 43 x l O -1 2 C 5 . 1 1 x 1 0' eV
Express the percent difference between k and k,exp:
% diff =
m
AC,exp - Aexp AC,exp _1 = Aexp Aexp 12 2 .48 x 1 0 - 1 = 1 2. 0 6 % = 2.43 x 1 0- 1 2 m
m
1
*17 Baseball, tennis, golf, and soccer are sports that involve placing a ball in play with a certain speed. Estimate which of these sports has a ball with the smallest de Broglie wavelength when the ball is moving with the highest speed typically created by a professional athlete. ••
The de Broglie wavelength of an object is given by A = hlp, where p is the momentum of the obj ect. Picture the Problem
The de Broglie wavelength of an object, in tenus of its mass m and speed v, is: The values in the following table were obtained using the internet:
The de Broglie wavelength of a baseball, moving with its maximum speed, is:
A=� mv
Type of ball Baseball Tennis Golf Soccer
m
(g) 142 57 57 250
Vmax (rn/s )
44 54 42 31
6 . 6 3 x 1 0-34 J . S 1 . 0 6 x 1 0-34 A= (0 . 1 42 kg )(44 m1s ) =
m
340
Chapter
34
Proceed as above to obtain the values shown in the table:
Type of ball
In
(g) (mls) 142 44 54 57 57 42 250 3 1
Baseball Tennis Golf Soccer
A
Vmax
(m) 1 .06 x l O -34 2 . 1 5 x 1 0-34 2.77 x l O -34 0.85 5 x l O-34
Examination of the table indicates that the soccer ball has the shortest de Broglie wavelength.
The Particle Nature of Light : Photons
*20 · Find the photon energy for light of wavelength (a) 450 nm, (b) 550 nm, and (c) 650 nm. Picture the Problem We can use E = he/A to find the photon energy when we are given
the wavelength of the radiation.
(a) Express the photon energy as a
function of wavelength and evaluate E for A = 450 nm:
E=
he A
= l 2 40 eV · nm = 1 2.76 eV 450 nm
(b) For A = 5 5 0 nm:
. E = 1 240 ev nm = 1 2 . 2 5 ev 550 nm
I
(c) For A
l E = 2 40 eV · nm = 1 1 .91 eV 650 nm
I
=
650 nm:
I
*23 • Lasers used in the telecommunications network typically have a wavelength near 1 .5 5 pm. How many photons per second are being transmitted if such a laser has an output power of 2.5 mW? Picture the Problem The number of photons per unit volume is, in turn , the ratio of the
power of the laser to the energy of the photons and the volume occupied by the photons emitted in one second is the product of the cross-sectional area of the beam and the speed at which the photons travel; i.e., the speed oflight. Relate the number of photons emitted per second to the power of
N
=
P
E
=
PA he
Wave-Particle Duality and Quantum Physics
341
the laser and the energy of the photons: Substitute numerical values and evaluate N:
N-
(2.
)(
)
1.55 ,um 34 6.63x10- J. s x108m/ s 5mW
( )(3 = 1 1.95xlOI6 s-1 I _
)
The Photoelectric Effect *28
••
When a surface is illuminated with light of wavelength 780 nm, the
maximum kinetic energy of the emitted electrons is 0.37 eV. What is the maximum kinetic energy if the surface is illuminated with light of wavelength 410 nm? Picture the Problem
We can use Einstein's photoelectric equation to find the work
function of this surface and then apply it a second time to find the maximum kinetic energy of the photoelectrons when the surface is illuminated with light of wavelength 365 nm. Use Einstein's photoelectric equation to relate the maximum
Kmax
=
he T-¢
kinetic energy of the emitted electrons to their total energy and the work function of the surface: Using Einstein's photoelectric equation, find the work function of the surface:
¢ = E - Kmax =
=
he K T - max
1240eV · nm
-O.37eV
780nm
=1.22eV
Substitute for ¢ and 2 and evaluate
Kmax
=
1240eV ·nm -1 .22eV 410nm
1
= 1.80eV
I
Compton Scattering *32
•
Compton used photons of wavelength 0.0711 nm. (a) What is the energy of
these photons? (b) What is the wavelength of the photon scattered at e = 180°? (c) What is the energy of the photon scattered at this angle?
342
Chapter 34
Picture the Problem We can use the Einstein equation for photon energy to find the energy find
of both
the incident and scattered photon and
the Compton scattering equation to
the wavelength of the scattered photon.
(a) Use the Einstein equation for photon energy to obtain:
E= he = 1240eV·run = 1 17 .4 keV I 0.0711run A,
(b) Express the wavelength of the scattered photon in terms of its pre scattering wavelength and the shift in its wavelength during scattering: Substitute numerical values and evaluate Az:
-3 4J·s = ,,1,2 = 0.0711run (9 11x610. 63xlO 1 -3 kg )(3x 10 m/s )(1-COS180 ) I0 .0760run I . +
(c) Use the Einstein equation for photon energy to obtain:
8
0
E= he =1240eV·run = 11 6 . 3 keV I ,,1,2 0.0760run
Electrons and Matter Waves *39
··
An electron, a proton, and an alpha particle (the nucleus of a helium atom)
each have a kinetic energy of 150 keY. Find (a) their momenta and (b) their de Broglie wavelengths. Picture the Problem
The momenta of these particles can be found from their kinetic
energies and speeds. Their de Broglie wavelengths are given by A= hlp. (a) The momentum of a particle p, in terms of its kinetic energy K, is
p =.J2mK
given by: Substitute numerical values and evaluate Pe:
p,
(
= � = 2�.1lx10-" kg ) 150 keVx 1. 6x��-" C
= 12 .09xlO-22N .s I
)
Wave-Particle Duality and Quantum Physics
34 3
Substitute numerical values and evaluate Pp:
Pp
(
= �2m pK = 2(1.67 X10-27 kg ) 150 kevx 1.6X10-19 C eV = IS.95x10-2IN.s I
J
Substitute numerical values and evaluate Pa:
P.
�
�2m.K
�
(
J(
2 4 ux 1 . 66x �O- " kg 150 kevx 1.6x:�-" C
= 11.79 x10-20 N·s I
J
(b) The de Broglie wavelengths of the particles are given by: Substitute numerical values and evaluate
Ap:
Ap
=
� Pp
J· s = 6.63x10-34 = 8.95 x10- 21 N . s 17.41x10-14 m I Substitute numerical values and evaluate Ae:
Ae
=� Pe
= 6.63x10-34 J· s = 13.17x10-12 m I 2.09x10-22 N· s Substitute numerical values and evaluate Aa:
Aa
=
� Pa
= 6.63x10-3 4J·s =1 3.70x10-14 m 1 1.79x lO-20 N· s *42
•
A proton is moving at v = O.003c, where c is the speed of light. Find the
electron's de Broglie wavelength. Picture the Problem
this proton.
We can use its defmition to calculate the de Broglie wavelength of
344
Chapter 34
Use its definition to express the de Broglie wavelength of the proton:
Substitute numerical
values and evaluate Ap:
3x10-34J·s = A= ( .67x10-6.6 27 kg)[0.003(3x 108 m/s)] 10·441 pm 1 1
*47
•
An electron microscope uses electrons of energy 70 keV. Find the
wavelength of these electrons.
Picture the Problem
We can use
A
=
1.}t/
nm,
where K is in eV, to find the
wavelength of 70-ke V electrons. Relate the wavelength of the electrons to their kinetic energy: A
Substitute numerical values and
evaluate Il:
=
1.226 �70xl03 eV
nm =
! 4.63 pm I
A Particle in a Box
*52
··
Use a spreadsheet program or graphing calculator to plot IjI(x) and the
probability distribution
.
PIcture the Problem
1f/2(X) of a particle in a box for the states
The wave functlOn for state n ISlf/ll .
.
n =
1, 2, and 3.
{2 sm. nTlX ' The () �L x =
following graphs were plotted using a spreadsheet program.
L
Wave-Particle Duality and Quantum Physics
The graph of If!(x) for
n =
1 is shown below:
;:::-"'----::---r--""'--:;;: '"l : , >�
1.0
c:
.9 '"
=
2
is shown below:
h
.",,"
OC:,
.1
0.5
0.0
�
-0.5
-1.0
1'"
'I- - ', "'"'
0.0
0.2
0.4
0.6
0.8
1.0
0.6
0.8
1.0
xlL
The graph of
1f/2(X) for n = 2 is shown below: 1.0.,.,.."..-.."....,�
0.8
o ';;:j
t:: �
� �
.D
2
P-
and <X2 > as a function
Picture the Problem
of n .
From Problem 63 we have
x
=
L
-
and /
=
-
7r
W ave-P ar ticle Du ali ty and Qu antum P hysics 349
0.25 +-�+---'--'I=:"'--+----"'-+--'---:'-+--+-'--1--i 10
19
28
37
46
n
55
64
73
82
91
100
General Problems *67
•
A light beam of wavelength 400 nm has an intensity of 100 W/m2 .
(a) What is the energy of each photon in the beam? (b) How much energy strikes an area of 1 cm2 perpendicular to the beam in 1 s? (e)How many photons strike this area in 1 s? Picture the Problem
We can use the Einstein equation for photon energy to find the
energy of each photon in the beam. The intensity of the energy incident on the surface is the ratio of the power delivered by the beam to its delivery time. Hence, we can express the energy incident on the surface in terms of the intensity of the beam. (a) Use the Einstein equation for photon energy to express the energy
Epholon
he = hf = -;:
of each photon in the beam: Substitute numerical values and
evaluate Epholon:
(b) Relate the energy incident on a
surface of area A to the intensity of
the beam:
1240eV·ru n = 3.10e � I 400run I
Epholon =
E=IAM
350
Chapte r 34
Substitute numerical values and evaluate E:
E
(100W/m2 )(10-4 m2 )(ls) = O.OIJx leV 1.60 xl 0-19 J = 1 6.25 1016eV 1 =
X
(c) Express the number of photons striking this area in 1 s as the ratio of the total energy incident on the surface to the energy delivered by
N= E 6.25x1016ev Ephoton 3.10eV = 1 2.02x10161
each photon: *73
••
Suppose that a 100-W source radiates light of wavelength 600 nm uniformly
in all directions and that the eye can detect this light if only 20 photons per second enter a dark-adapted eye with a pupil 7 mm in diameter.How far from the source can the light be detected under these rather extreme conditions? Picture the Problem
We can relate the fraction of the photons entering the eye to ratio of
the area of the pupil to the area of a sphere of radius R. We can find the number of
photons emitted by the source from the rate at which it emits and the energy of each photon which we can find using the Einstein equation. Letting r be the radius of the pupil,
Nentering eyethe number of photons per second entering the eye, and Nemitted
Nenteringeye Aeye Nemitted 47rR2
the number of photons emitted by
the source per second, express the fraction of the light energy entering
the eye at a distance R from the source:
Solve for R to obtain:
Find the number of photons emitted by the source per second: Using the Einstein equation, express the energy of the photons:
R
= !... NNemitted 2 enteringeye
p Nemitted= � photon
Ephoton = -he ;:
( 1)
W ave-P ar ticle Du ali ty and Qu an tum P hysics Substitute numerical values and
= 1 240 eV·
EphOlon
evaluate Epholon: Substitute and evaluate Nemitted:
-(
and evaluate R:
2.0 7 eV
lOOW
)(
_
3 .5mm
3 .0 2xl020 s -'
2
20 s -'
=
)
2.0 7 eV 1.60xl0-J9 J/eV 2 3.0 2 X 10 0 S-J
= R
=
600run
Nemitted
Substitute for Nemittcd in equation (1)
nm
35 1
= 1 6.80 x103 km I *79
••
The Pauli exclusion principle states that no more than one electron may
occupy a particular quantum state at a time. Therefore, if we wish to model an atom as a collection of electrons trapped in a one-dimensional box, each electron in the box must have a unique value of the quantum number n. Calculate the energy that the most
energetic electron would have for the uranium atom with atomic number 92, assuming
the box has a width of 0.05 nm. How does this energy compare to the rest-mass energy of the electron? Picture the Problem
We can use the expression for the energy of a particle in a well to
find the energy of the most energetic electron in the uranium atom. Relate the energy of an electron in
2
(
h2
E" = n 8mL 2
the uranium atom to its quantum number n:
J
Substitute numerical values and evaluate E92:
E 92 = ( )
922
[
? - 4 6 .6 3xl0 3 J· s 8 9 .11xlO-3I kg 0 .0 5run
(
y
X
l eV
9 1.6xl0-' J
]
=
I 1·28MeV I
The rest energy of an electron is:
2 Express the ratio of E92 to mec :
E 92 m G2 e
=
1.28MeV 0 .512MeV
=
2.50
Chapter
352
34
T he energy of the mostenerge tic elec tron is approxi mately 2.5 ti mes the rest - m assenergy of an elec tron. *83
••
and state 11
(a) Show that for large n, the fractional difference in energy between state +
1 for a particle in a one-dimensional box is given approximately by
n
(b) What is the approximate percentage energy difference between the states 111 =
1000 and 112 = 1001? (c ) Comment on how this result is related to Bohr's
correspondence principle.
Picture the Problem
We can use the fact that the energy of the nth state is related to the
energy of the ground state according to E n
=n
2
E1 to express the fractional change in
energy in terms of n and then examine this ratio as n grows without bound. (a) Express the ratio
(Ell
+
1-
En)/EII:
for n» 1 (b) Evaluate E]OO] - E 000 . E1000
1.
E100] - E]OOO E]OOO
l':::
I
_2_ = 0.2% 1000
ally,the e ne rgy isco ntinuous. Fo rve ry l arge values of n, the (c) eClneassic rgy diffe re nce betwee n adj ace ntlevelsisi nfi nitesi mal.
I
Chapter 35 Applications of the Schrodinger Equation Conceptual Problems *4
•
The Schrodinger Equation could be applied equally well to baseballs as to
electrons yet we would never analyze the motion of a baseball with a wave function. Explain why this is the case by estimating the quantum mechanically predicted lowest energy level of a baseball trapped inside a locker. You can treat the locker as if it were a one-dimensional infinite potential well. What value of the quantum number n would you need for a ball rolling around in the locker, after you toss it in, so that the kinetic energy is approximately equal to the quantum mechanically calculated energy? Picture the Problem
Assume a mass of 1 50 g for the baseball, 30 cm for the width of the
locker, and 1 cmls for the speed of the ball, and equate the kinetic energy of the ball and the quantum-mechanical energy and solve for the quantum number n. The allowed energy states of a particle of mass m in a
I-dimensional infinite potential well
of width L are given by: The kinetic energy of the ball is:
1 2 K=-mv 2
For EI/ = K:
Solve for the quantum number n:
Substitute numerical values and evaluate n:
n=
2 mvL
h
--
(
)(
)(
)
m/s 2 0. 15 kg 0.0- 1...". 0.3 m ..,----"--'--"n = ---'---...=:..4 =
Evaluate equation (2) for
M = 0. 572 eV:
1 2 40 e V
.
run
0.572 eV
=
I 2 168 I run
The positions of these lines on a horizontal linear scale are shown below with the wavelengths and transitions indicated.
7�4 6�4 1
5�4
1
1
--- --------- --------------------------------------------- -----
2 168 nm 2627 nm
4052 nm
*29 ••• In the center-of-mass reference frame of a hydrogen atom, the electron and nucleus have equal and opposite momenta of magnitude p. (a) Show that the total kinetic
energy of the electron and nucleus can be writtenK = p
2 /(2 j.l) where
j.l = meM / (M +me) is called the reduced mass, me is the mass of the electron, and M is the mass of the nucleus. (b) For the equations for the Bohr model of the atom, the motion of the nucleus can be take into account by replacing the mass of the electron with the reduced mass. Use Equation 36-14 to calculate the Rydberg constant for a hydrogen atom with a nucleus of mass M = mp. Find the approximate value of the Rydberg constant by letting M go to infinity in the reduced mass formula. To how many figures does this approximate value agree with the actual value. (c) Find the percentage correction for the ground-state energy of the hydrogen atom by using the reduced mass in Equation 36-16. Remark: In general, the reduced mass for a two-body problem with masses mJ and m2 is given by j.l =
m1m2 ml +m2
Picture the Problem
We can express the total kinetic energy of the electron-nucleus
system as the sum of the kinetic energies of the electron and the nucleus. Rewriting these kinetic energies in terms of the momenta of the electron and nucleus will lead to K =
p2l2mr.
(a) Express the total kinetic energy of the electron-nucleus system: Express the kinetic energies of the electron and the nucleus in terms of their momenta:
2
2
K e =J!...-and K = L
2me
n 2M
Ato ms 369 Substitute to obtain:
�I :�, I
provided we define Ji = meAt/eM + m )
e.
(b) From Equation 36-14 we have: (1)
where
2 C= k e_4 4JZ"C1i3 __
Use the Table of Physical Constants at the end of the text to obtain:
C =1.204663 1037 m -I / kg X
ForH:
-C RHSubstitute numerical values and evaluate RH:
RH
kg = (1.204663x 1037 m -I / kg ) 9.11xlO-31 9.11xlO-3I kg 1+1.67 x 10-27 kg
Let M -+ CXJ in equation (1) to obtain
RH,approx
RH,approx:
=Cme
Substitute numerical values and evaluate RH,approx:
RH,approx
=
(1.204663x 1037 m-I / kg )(9.11x10-31 kg ) = \1.097448 x 107 m-I \
370 Chapter 36 R H and RH,ap Tox agree to three significant figures. P
(c) Express the ratio of the kinetic energy K of the electron in its orbit about a stationary nucleus to the kinetic energy of the reduced-mass systemK':
1
Substitute numerical values and
K
evaluate the ratio of the kinetic
K'
1
1+
energIes:
3 9.11x10- 1 kg 1.67 X 10-27 kg
= 0.999455
or K
=
0.999455K'
and the correction factor is the ratio of the
I
masses or 0.0545%
I
Remarks: The correct energy is slightly less than that calculated neglecting the motion of the nucleus.
*30 ·· The Pickering series of the spectrum ofHe+ (singly-ionized helium) consists of spectral lines due to transitions to the n = 4 state ofHe+. Experimentally, every other line of the Pickering series is very close to a spectral line in the Balmer series for hydrogen transitions to n = 2. (a) Show that this is true. (b) Calculate the wavelength of a transition from the n = 6 level to the n = 4 level ofHe+, and show that it corresponds to one of the Balmer lines.
We can use Equation 36-15 with Z = 2 to explain how it is that every other line of the Pickering series is very close to a line in the Balmer series. We can use the relationship between the energy difference between two quantum states and the wavelength of the photon emitted during a transition from the higher state to the lower state to find the wavelength of the photon corresponding to a transition from the n = 6 to the n = 4 level ofHe+. Picture the Problem
(a) From Equation 36-15, the energy levels of an atom are given by:
ForHe+, Z= 2 and:
Eo
En
= _Z2
En
=-4n2
n2 where Eo is the Rydberg constant (13.6 eV).
Eo
Atoms 371 Because of this, an energy level with even principal quantum number 11. in l-k!- will have the same energy as a level with quantum number 11.12 in H. Therefore, a transition between levels with principal quantum numbers 2m and 2p in He+ will have almost the same energy as a transition between level m and pin H. In particular, transitions from 2m to 2p = 4 in He+ will have the same energy as transitions from In to n = 2 in H (the Balmer series). (b) Transitions between these energy levels result in the emission or absorption of a photon whose wavelength is given by:
(1)
E6
=
and
E4 Substitute for E6 and E4 in equation (1) and evaluate A:
/L
=
4 (13 ·:2eV )
_
eV ) · 4 (13 : 2
_
=
=
-1.51eV
-3.40eV
1240eV· nm
=
(
)
=
I 656nm I
-1.51eV - - 3.40eV which is the same as the n = 3 to transition in H.
n =
2
Quantum Numbers in Spherical Coordinates *35
(a)
··
Find the minimum value of the angle e between L and the z axis for and (c) £ =50.
£ l ( b) £ 4 =
,
=
,
Picture the Problem
The minimum angle between the z axis and L is the angle between
the L vector for m = i! and the z axis. Express the angle e as a function of
Lz and L:
Relate the z component of L to me
and i! :
Express the angular momentum L: Substitute to obtain:
B
=
cos
-l(�i!(i! l)n ] -l(�T+i IT] Hz
+
=
cos
372 Chap ter 36
(a) Fore = 1:
(b) For f! = 4:
e=cos-I(�� (50)=�
(c) For f! = 50:
Quantum Theory of the Hydrogen Atom *39
·
(a) If spin is not included, how many different wave functions are there
corresponding to the first excited energy level
n =
2 for hydrogen? (b) List these
functions by giving the quantum numbers for each state. Picture the Problem
We can use the constraints on n, £, and m to detennine the number
of different wave functions, excluding spin, corresponding to the first excited energy state of hydrogen.
= ol O r
For n = 2:
£
(a) For £ = 0, me= ° and we have:
1 state
For £= 1, me = -1, 0, +1 and we
3 states
have: Hence, for n = 2 we have: (b) The four wave functions are summarized to the right.
s s
14 t te 1 a
n
£
me
(n, £ , me)
2
°
°
(2,0,0)
2
1
-1
(2,1,-1)
2
1
°
(2,1,0)
2
1
1
(2,1,1)
*44 ·· Show that the ground-state hydrogen wave function (Equation 36-33) is a solution to SchrMinger's equation (Equation 36-21) and the potential energy function (Equation 36-26).
Atoms 373
.
solutlOn to
=
We wish to show that '1',, 0,0
11. 2 2 -0 ( r2 -0 ) (r --2mr or or
Picture the Problem
If/
+U
L ff
=
); (:, f' .,z'I"" Ce,z'I"" (r ) - --. kZe2 r =
where U
E If/ ,
=
is a
Because the
ground state is spherically symmetric, we do not need to consider the angular partial derivatives in Equation 36-21. The normalized ground-state wave function is:
II' '1' 1 0 0
, ,
r
-- ( J e - Ce I "1/ 7r
I O
r2:
Multiply both sides of this equation by
r
ao
-Zr/ao _
-Zr/ao
o '-- - C -0 [e ] - -C -Z e ----'-' or or ao r2 0 or = -C -Z r2e If/ , ,D
Differentiate this expression with respect to to obtain:
3/2
_1_ �
_
-zr/ao _
-Zr/ao
-Zr/ao
If/I,O,O
ao
Differentiate this expression with respect to to obtain:
2 Z Z r 2z 0 ( ) 2 2 C -J - -- r e - [--- +r ( J ]ce 2 2 2 e k Z r C 11. ] C 2z 2 C e e --- ----+r e 2 J ( r 2mr [
-o (r 2 &
O lf/I ,D,O &
_
� &
-zr/ao _
�
-Zr/ao
�
Substitute in Schrodinger' s equation to obtain: Z
ao
Solve for E:
Because
ao =
ao
-Zr/ao
-Zr/ao _
E
-Zr/ao
11.--2 2 : mke
= Because this is the correct ground state energy, we have shown that Equation 36-33, is a solution to Schrodinger's Equation 36-21 with the potential energy function Equation 36-26.
374 Chapte r 36 The Spin-Orbit Effect and Fine Structure *50
·
The potential energy of a magnetic moment in an external magnetic field is given by U = jL . B . (a) Calculate the difference in energy between the two possible -
-
orientations of an electron in a magnetic field B
= l.SO Tk. (b) If these electrons are �
bombarded with photons of energy equal to this energy difference, "spin flip" transitions can be induced. Find the wavelength of the photons needed for such transitions. This phenomenon is called electron spin resonance . Picture the Problem
The energy difference between the two possible orientations of an
electron in a magnetic field is 2J1B and the wavelength of the photons required to induce
a spin-flip transition can be found from helM. The magnetic moment JiB associated with
the spin of an electron is 5.79x l O-5 eVrr. (a) Relate the difference in energy
M
between the two spin orientations in terms of the difference in the potential energies of the two states:
= 2J1l3 = 2( S.79x10- 5eV/ T )(O.6 T) = 1 6 .9 SxlO- 5eV I
(b) Relate the wavelength of the photon needed to induce such a transition to the energy required: Substitute numerical values and evaluate A.:
A
. 6.9SxlO-)eV = 1 1.78 cm I
= 124 0eV :un =1.78xl07nm
The Periodic Table *56
•
Write the electron configuration of (a) carbon and (b) oxygen.
Determine the Concept We
can use the atomic numbers of carbon and oxygen to
determine the sum of the exponents in their electronic configurations and then use the rules for the filling of the shells to find their electronic configurations. (a) The atomic number Z of carbon is 6. So we must fill the subshells of the electronic
configuration until we have placed its 6 electrons. This is accomplished by writing
1 1s 2 2s 22p2 I
(b) The atomic number Z of oxygen is 8. So we must fill the subshells of the electronic
Atoms 375 configuration until we have placed its 8 electrons. This is accomplished by writing l s 2 2s 2 2 p 4
\
\
Optical Spectra and X-Ray Spectra *61
·
(a) Calculate the next two longest wavelengths in the K series (after the Ka
line) of molybdenum. (b) What is the wavelength of the shortest wavelength in this series? Picture the Problem When
an electron from state n drops into a vacated state in the n
=
1 shell, a photon of energy M = En - E ) is emitted. We can find the wavelength of this photon using A he/ M . The second and third longest wavelengths in the K series
=
correspond to transitions from n
=
3 to
n =
1 and n = 4 to
wavelength to the transition from n = 00 to n = 1. Express the wavelength of the emitted photon in terms of the
A
n =
he En - E)
=
1 and the shortest
1240eV .run En - E)
energy transition within the atom: Express the energy of the nth energy state:
En
=
-( Z-lY nE�
where n = 1, 2, . . . Substitute to obtain:
1240eV .run - ( Z-lY :� - (- ( Z -lY �o) 1240eV .run
(a) Evaluate this expression with n =
3 and Z = 2 to obtain:
4
�
= =
1240eV .run (42-1Y(13 . 6eV{1- 3\ )
I O.0610run I
376 Chapte r 36 Use n
=
4 and Z = 42 to obtain:
A,
=
Of
1240eV . 1) (42 -1Y(13.6eV) 1- 4"2
= I 0 .0578 nm I
1240eV · nm '" (42-lY( 1 3.6eV)(1-0) I 0.0542nm I
A
(b) The shortest wavelength in the series corresponds to the largest energy difference between the initial
=
=
and final states. Repeat the calculation in part (a) with n = 00 to
obtain:
General Problems *68
··
We are often interested in finding the quantity ke21r in electron volts when r
is given in nanometers. Show that ke2 = Picture the Problem
1 .44 eY-nm.
We can show that ki =
ground state energy of an atom for ke2•
1.44 eV ·nm by solving the equation for the
Express the ground state energy of an atom as a function of k, e, and ao: Solve for ke2 : Substitute for Eo and ao to obtain:
*71
••
ke2
2 (1 3 .6eV)(0.0529 nm) = Il .44eV . nm I =
The combination of physical constants a = e2kllic, where k is the Coulomb
constant, is known as the fine-structure constant. It appears in numerous relations in
atomic physics. (a) Show that a is dimensionless. (b) Show that in the Bohr model of
hydrogen v" = ca in, where Vn is the speed of the electron in the stationary state of quantum number n.
Picture the Problem
We can show that a is dimensionless by showing that it has no
units. In part (b) we can use Bohr's 3Td postulate and the expression for the radii of the
Bohr orbits, together with the definition of a, to show that the speed of the electron in a stationary state of quantum number n is related to a according to
VI1 =
ca In.
Atoms 377
(a) Express the units of a:
Because a i s unitless, it is also dimensio nless. (b) Apply the quantization of angular momentum postulate to
v"
=
obtain: The radii of the Bohr orbits are given by:
nh m rn
-
r" = n 2
v"
mke2 nh
mn 2
h2
m ke2
�
v" =
ke2
--
nh
--
ke2 a
Solve for v,,:
h2
=
Divide this expression by the definition of a to obtain:
m kZe2
or, because Z = 1 for hydrogen,
r" = n 2
Substitute and simplify to obtain:
h2
nh e2k he
e
n
� n
*74 · A Rydberg atom is one in which an outer-shell electron is placed into a very high excited state (n � 40 or higher.) S uch atoms are useful for experiments which probe the transition from quantum mechanical behavior to classical. Furthermore, these excited states have extremely long lifetimes (i.e., the electron will stay in this high excited state for a very long time.) A hydrogen atom is in the n = 45 state. (a) What is the ionization energy of the atom when it is in this state? (b) What is the energy level separation (in e V) between this state and the n = 44 state? (c) What is the wavelength of a photon resonant with this transition? (d) How large is the atom in the n = 45 state?
Picture the Problem The ionization energy of the electron is the magnitude of the energy of the atom in the given state. We can use E = -EoIn2, where Eo is the ground-state energy, to find the energy levels in the 44th and 45th states and, hence, the energy level separation between the states. The wavelength of a photon resonant with this transition
378
Chapt e r
36
can be found from A helM. We'll approximate the size of the atom in the by finding the radius of the outer-shell electron. =
(a) The energy of the atom in its nth state is: The energy of the atom in the n = 45 state is: The ionization energy is the negative of the energy in the state n = 45 : (b) The energy level separation between the n = 45 and n = 44 state IS:
n =
45 state
Eo EI/ = -n2 13.6eV E45 = - (45Y = -6.72meV Eionizing = -E45 = 16.72meV I E45�44 -
(13(45.6eV)2
__
=
13.6 ev
_
(44)2
13.09x10-4 eV I
)
(c) The photon wavelength is:
Substitute numerical values and evaluate A:
A=
.
1240 eV 01x106 run ! 3.09x10-4 eV = !4.
(d) The radii of the Bohr orbits are given by:
r =n 2Z
Substitute numerical values and evaluate the radius of the 45th Bohr orbit:
r = (45Y
run
ao
0.05�9 run = !107nm
!
Chapter 37 Molecules Conceptual Problems *1
•
Would you expect the NaCI molecule to be polar or nonpolar?
Determine the Concept Yes. Because the center of charge of the positive Na ion does
not coincide with the center of charge for the negative CI ion, the NaC I molecule has a permanent dipole moment. Hence, it is a polar molecule . *5
••
The elements on the far right column of the periodic table are sometimes
called noble gases because they virtually never react with other atoms to form molecules. However, this behavior is sometimes modified if the resulting molecule is formed in an electronic excited state. An example is ArF. When it is formed in the excited state, it is written ArF* and is called an excimer (for excited dimer). Refer to Figure 37- 1 3 and discuss how this diagram would look for ArF in which the ArF ground state is unstable but the ArF* excited state is stable. Remark: Excimers are used in certain kinds of lasers. Determine the Concept The diagram would consist of a non-bonding ground state with
no vibrational or rotational states for ArF (similar to the upper curve in Figure 37-4) but for ArF* there should be a bonding excited state with a definite minimum with respect to inter-nuclear separation and several vibrational states as in the excited state curve of Figure 37-1 3 . Estimation and Approximation *14
Repeat Problem 1 3, finding the quantum number
••
v
and spacing between
adj acent energy levels for a S-kg mass attached to lS00-N/m spring vibrating with an amplitude of 2 cm. Hint: Pick
v
so that the quantum energy formula (Equation 37-18)
gives the correct energy for the given system. Then find the energy increase for the next highest energy level. Picture the Problem We can solve Equation 37- 1 8 for v and substitute for the frequency of the mass-and-spring oscillator to estimate the quantum number vand spacing between adjacent energy levels for this system.
The vibrational energy levels are given by Equation 37-1 8 : Solve for
v:
Ev
= (v+t)hl
where v= 0, 1 , 2, . . .
1 Ev v=--hi 2 379
380
Chapte r
37 or, becaus e Ev v>::;
The vibrational energy of the object attached to the spring is:
Ev
=
v=
l..2 kA2
kA2
2hf l /k f=_ 2;r �;;
The frequency of oscillation f of the mass-and-spring oscillato r is given by:
Substi tute numerical values and evaluate
Set
hf
_
where A is the amplitude of its motion.
Substitute for E v in the expression for v to obtain:
v=
v» 1,
--
v:
n(0.02_�Y �(Skg)(lS00N/m) = I 1 .64 x 1032 I 6 . 63 x 1 0 J·s
v= 0 in Equation 37- 1 8 to
2
adjacent energy levels: Substitute numerical values and evaluate
Eov:
Remarks: Note that our value for
/k 4n � ;;
EOv = 2. hif = �
express the spacing between
_6.63 x 10-34J . s lS00N/m S kg 4n = 1 9.14 x 10-34J I
EOv -
vjustifies our assumption that v »
1.
Molecular Bonding ·
*18
The equilibrium separation of the HF molecule is 0.09 1 7 nm, and i ts
measured electric dipole moment is 6.40x ionic?
OO( J.
10-30 em. What percentage of the bonding is
Picture the Problem The percentage of the bonding that is ionic is given by
1
Pmeas PIOO
Molecules 381 Express the percentage of the bonding that is ionic:
Pe rcent
Express the dipole moment for
P IOO
1 00 % ionic bonding: Substitute to obtain:
=
io ni c bo ndi ng
=
( )
I OO
er
Pe rcent ioni c bonding =
Pmeas P IOO
100( P;as )
Substitute numerical values and evaluate the percent ionic bonding:
Pe rcent ion ic bonding *24
•••
=
'C m · [(1.60x6.4100_�/C0);0.0917nm )]
I OO
=
I 43.6% I
Assume that the potential energy associated with the core repulsion of the
two ions of a diatomic molecule with ionic bonding can be represented by a potential
energy of the form Urep = Clr", so the total potential energy is
U= Ue + Urep + M, where Ue
=
-ke2 /r.
/)ill is the energy of the two ions at infinite
separation less the energy of the two neutral atoms at infinite separation (see Figure 3 71).
Use dUidr= 0 at r = ro to show that n
=
U I e(r o)1 Urep (ro)
-'--'--.:......!.
Picture the Problem U(r) is the potential energy of the two ions as a function of separation distance r. U(r) is chosen so U(oo) = -M, where M is the negative of the
energy required to form two ions at infinite separation from two neutral atoms also at
infinite separation. Urep(r) is the potential energy of the two ions due to the repulsion of
the two closed-shell cores. Ed is the disassociation energy, the energy required to separate the two ions plus the energy M required to form two neutral atoms from the two ions at
infinite separation. The net force acting on the ions is the sum of Frep and Fe . We can find
Frep from Urep a nd Fe from Coulomb ' s law and then use dUldr = Fnet = 0 at r= ro to solve for n .
(1)
Express the net force acting on the IOns: Find Frep from Urep :
Frep
=
dUrep dr nC
=
!!:..-[Cr-"l dr
=
-nCr-n-1
382 Chapter 37 The electrostatic potential energy of the two ions as a function of separation distance is given by: Find the electrostatic force of attraction Fe from Ue:
Substitute for Frep and Fe in equation
( 1 ) to obtain:
Because dU/dr
=
Ue =
Fe =
FlIet
_r
ke2
dUe dr
=
_
=
nC
r
11+1
�[_ ] ke2
dr
+
r
=
ke2 r2
ke2 r2
Fnet = 0 at
Multiply both sides of this equation
by ro to obtain:
Solve for n to obtain:
n=
lue h)1 U rep (ro)
Energy Levels of Spectra of Diatomic Molecules
*27
·
The separation of the two oxygen atoms in a molecule of O is actually
2
slightly greater than the 0. 1 nm used in Example 37-3, and the characteristic energy of rotation Eor is 1 .78x 1 0 -4 eV, rather than the result obtained in that example. Use this value to calculate the separation distance of the two oxygen atoms.
Picture the Problem We can relate the characteristic rotational energy Eor to the moment
of inertia of the molecule and model the moment of inertia of the O molecule as two
2
point objects separated by a distance r. The characteristic rotational energy of a molecule is given by: Express the moment of inertia of the molecule: Substitute for 1 to obtain:
li2 EOr = 21
Molecules
383
Solve for r:
Substitute numerical values and evaluate r: r
�
=
*30
(1.055X�O-" J ·s ) I O.121nm I
The equilibrium separation between the nuclei of the LiH molecule is 0 . 1 6 nm. Determine the energy separation between the = and rotational levels o f this ••
f! 3
diatomic molecule.
f! = 2
Picture the Problem We can use the expression for the rotational energy levels of the
diatomic molecule to express the energy separation M between the
f! = 3 and
f! = 2 rotational levels and model the moment of inertia of the LiH molecule as two point
objects separated by a distance roo
The energy separation between the
f! 3 and R 2 rotational levels of =
=
this diatomic molecule is given by:
Express the rotational energy levels Ee=3 and Ee= in terms of Eor:
2
EM = 3(3 + I)Eor = 12EOr
and
E =2 2(2 + I)Eor 6EOr t
=
=
Substitute for Ee= 3 and Et=2 to
obtain:
The characteristic rotational energy of a molecule is given by:
h2 = .L/1E EOr = 21 6
�
-
/1E = 3h2 1
Express the moment of inertia of the molecule:
where f.1 is the reduced mass of the
molecule.
384 Chapter
37
Substi tute for 1 to obtain:
M=
2
2
3tz = -3tz ---2 mLimH mLi + mH 3tz2 (mLi + mH ) mLimH rO 2 JL
r0
Substitute numerical values and evaluate M: M=
*31
3 1 .055x lO-34J·s
(6.94u+1u) (6.94u)(1 u)(0.16nmY 1.602x10-19 J/eV 1.660 x 1 0-27 kg/u ••
=
1 5 . 6 1meV I
Derive Equations 37-14 and 37- 1 5 for the moment of inertia in tenns of the
reduced mass of a diatomic molecule. Picture the Problem Let the origin of coordinates be at the point mass m l and point mass
m2 be at a dis tance ro from the origin. We can express the moment of inertia of a diatomic molecule with respect to its center of mass using the definitions of the center of mass and
the moment of inertia of point particles. Express the moment of inertia of a
(1)
diatomic molecule: The r coordinate of the center of mass IS:
The distances of
m
center of mass are:
I
and
m2 from the
Substitute for r l and r 2 in equation
(1) to obtain:
Simplifying this expression leads to: or
M olecules
11
=
/-t
whe re
ro2
385
36-14 36-15
ml and m2 are attached to a spring of force constant k roo (a) Show that when ml is moved a distance L'lrl from the center of mass, the force exerted by the spring is m,;,m, )",r, -k ( *36
··
Two objects of mass
and equilibrium length F
�
(b) Show that the frequency of oscillation is
f
=
_12:r_ �k/
I-l, where J.L is the reduced
mass. Picture the Problem For a two-mass and spring system on which no external forces are
acting, the center of mass must remain fixed. We can use this condition to express the net force acting on either obj ect. Because this force is a linear restoring force, we can conclude that the motion of the object whose mass is angular frequency given by
OJ �
given in (b). (a) If the particle whose mass is
�km'ffl
ml
ml will be simple harmonic with an
. Substitution for leo" will lead us to the result
I1r2 _mmI I1lj from (or toward) the center 2
L'lrl from (or toward) the cente r of mass, then the particle whose mass is m2 must
of mass.
Express the force exerted by the
F
moves a distance
=
move a distance:
sprIng: Substitute for L'lrz to obtain:
F
=
�
�
(b) A displacement in a restoring force:
L'lr of m results 1
1
F
=
-kl1r = -k(l1lj +I1r2)
-{ +�, ) m,;,m } -k ( m +m 2 ) -k -k ( m "'r,
"'r,
r,
1
2
11r.1
=
efT 11r.1
386 Chapter
37 where
keff
=
k
(
m1 +m2 m2
Because this is a linear restoring
J
force, we know that the motion will be simple harmonic with:
Substitute for ke ff and simplify to
obtain:
or, because f.1 =
mm I 2 ml + m2
is the reduced
mass of the two-particle system,
f = \llkl .
lEkJ
General Problems *40
··
The effective force constant for the HF molecule
frequency of vibration for this molecule.
IS
970 N/m. Find the
Picture the Problem We can use the result of Problem 36 to find the frequency of
vibration of the HF molecule. In Problem 36 it was established
that:
l fI f =_ 2n � -;
The reduced mass is:
Substitute for f.1 to obtain:
Substitute numerical values and evaluate f
l f =_ 2n
k = 1 ----
Mole cule s 387
*47
··
For a molecule such as CO, which has a permanent electric dipole moment,
= 1
radiative transitions obeying the selection rule 6,.e ± between two rotational energy levels of the same vibrational level are allowed. (That is, the selection rule 1'1 v = ± does
1
not hold.) (a) Find the moment of inertia of CO and calculate the characteristic rotational
Eo r £= 5 o for £ = O .
£= 0 E
(in eV). (b) Make an energy level diagram for the rotational levels for energy to for some vibrational level. Label the energies in electron volts starting with Indicate on your diagram transitions that obey 6,.£
= -1 and calculate the
=
energy of the photon emitted (c) Find the wavelength of the photons emitted for each transition in (b). In what region of the electromagnetic spectrum are these photons?
Picture the Problem We can find the reduced mass of CO and the moment of inertia of a
CO molecule from their definitions. The energy level diagram for the rotational levels for = Finally, we can find the wavelength to can be found using Me,t-I =
£ 0 £= 5
2 £EOr'
1 . . d lor . C each transItIon usmg o f the ph otons emItte /l,e , (-I .
1-
(a) Express the moment of inertia of CO:
he = --M
r,2
e t- I
he 2 £MOr
/I 0 r
where f.1 is the reduced mass of the CO
molecule. Find f.1:
In Problem 39 it was established that
ro = 0. 1 1 3 nm. Use this result to evaluate I:
1 = (6.86u)(1.66x 10-27 kg/u)(0.113nrnY = 11.45 x 10-46 kg ·m2 1
2
EOr = li1 2
The characteristic rotational energy
Eor is given by:
Substitute numerical values and evaluate
Eor:
EOr = -'----r--L.�--::---L = 1 0.239 meV I
388 Chapter 37 f = 5, E
(b) The energy level diagram is shown to the right. Note that Me,c-I>
=
2.14 meV
-
the energy difference between adj acent levels for t>"e = - 1 , is
1iEe,c_1 = 2 .eEOro
e
=
e
f.
e
=
=
=
4, E
3, E
2, E
1, E
liEu_I
2.86 meV----,--'--
=
=
=
e
(c) Express the energy difference between energy levels in
4.76 meV-
=
1.43 . meV --,--'----
0.476 mev
=
O,E
=
, -
0-'-
liEeH , , = hieH
tenus of the frequency of the emitted radiation: Be cause
e= ieHAeH
he e,H
Ae,e-I = liE
:
he
Substitute numerical values to obtain:
Ae,e-I =
(40136x10-15 eV os)(3x108m/s)= 2 596,urn .e 2 .e(0.2 39meV)
For JI. = 1 :
29 �,O= 5 ,urn = 1 2 596 Jl111 1
For.e = 2:
Az,1 = 2 59 ,urn = 1 1298,urn I
For.e
�,2 = 2 59 ,urn = I 865,urn I
=
3:
�
�
�
-
Mole cule s 389 For e Fore
=
=
4:
A4.3
5:
A5,4
=
=
2596f1I11 4 2596f1I11 5
=
=
\ 649 \ f-trl1
\ 5 1 9 f1I11 \ .
The s e wave le ngths fa lli n the microwave re gion of the s pe ctrum. *48
•••
Use the results of Problem 24 to calculate the vibrational frequency of the
LiCl molecule. The dissociation energy of LiCl is 4.86 eV, and the equilibrium separation is 0.202 nm. The electron affinity of chlorine is 3.62 eV, and the ionization energy of lithium is 5 .3 9 eV. To do this, expand the potential about r = ro, where ro i s the
equilibrium separation, in a Taylor series. Retain only the term proportional to (r - ro)2.
Recall that the potential energy of a simple harmonic oscillator is given by USHO = t m(ix2. What is the wavelength resulting from transitions between adj acent harmonic oscillator levels of this molecule?
Picture the Problem The wavelength resulting from transitions between adjacent
harmonic o scillator levels of a Liel molecule is given by A expression for
OJ by
=
2ne . We can find an (()
following the procedure outlined in the problem statement.
The wavelength resulting from transitions between adj acent harmonic oscillator levels of this
A
=
he he 2ne M n(() (() =
=
(1)
molecule is given by: From Problem 24 we have:
U(r)
=
-
ke2
r + r� , where M i s constant.
The Taylor expansion of U(r) about r
=
ro i s:
Because U(ro) i s a constant, it can be dropped without affecting the physical results and because
(2)
390
Chapter
(dU )
dr .
37
= 0:
'0
Differentiate U(r) twice to obtain: B ecause dU/dr = r= ro:
Fuet
=
0 at
c= ker2 a nrO 2
Solving for C yields:
( d2d U2 )
11+1 _
kera 2 II-I n
S ubstitute for C and evaluate
r
ro
to obtain:
Substitute for equation (2) :
(�:�1.
in
B ecause the potential energy o f a s imple harmonic oscillator is given by
/
USHO = tmo (r - ro Y :
Solve for
w to
obtain:
S ubst itute }1LiCI for m to obtain:
(n -1 )ke2
OJ ';::;
3
mrO
(n -1)ke2
OJ ';::;
mLimCl r O mLi +mCI
3
(3)
(n -l ) (mLi + mcl )ke2 3
mLimClrO
From Problem 24:
Urep is related to Ue, Ed, and M
n=
IUe(ro) I Urep (ro )
Urep -(Ue + Ed + M )
=
(4) (5)
Molecules
391
according to: The energy needed to form Li+ and cr from neutral lithium and
M
=
Substitute ro and evaluate Ue:
Substitute numerical values in
e2
Ue
= -k
Ue
=
equation (5) and evaluate Urep:
equation (4) and evaluate
n:
E
electronaffinity
n=1
1 .44 eV .run = 7.1 3eV 0.202nm
= -(- 7 . 1 3eV + 4.86 eV + 1 .77 eV) = 0.500 eV - 7.1 3 eV I = 14.3 0.500 eV
Substitute numerical values in equation (3) and evaluate
(j);::::
-
1 .44 eY·run
-
Urep
Substitute for Urep(ro) and Ue(ro) in
ionization
= 5.3geY - 3.62 eY = l.77 eY
chlorine atoms is:
Ue(ro) is given by:
E
co:
(1 4.3 -1)(6 . 941u + 35.453u)(1 .44 eV .run) 1 .60 x 10-19 l/ eV (6.941 u)(35.453u) 1 .66 x 10-27 kglU (0.202runY
= 1 1 .96 x 1 014 S-I
Substitute numerical values in
equation (1) and evaluate A:
I
Chapter 38 Solids and the Theory of Conduction Conceptual Problems *2
•
When the temperature of pure copper is lowered from 300 K to 4 K, its
resistivity drops by a much greater factor than that of brass when it is cooled in the same way. Why? Determine the Concept The resistivity of brass at 4 K is almost entirely due to the
"residual resistance, " the resistance due to impurities and other imperfections of the crystal lattice. In brass, the zinc ions act as impurities in copper. In pure copper, the
resistivity at 4 K is due to its residual resistance, which is very low if the copper is very pure. How does the change in the resistivity of copper compare with that of silicon *8 when the temperature increases? Determine the Concept The resistivity of copper increases with increasing temperature; the resistivity of (pure) silicon decreas es with increasing temperature b ecause the number of charge carriers increases.
The Structure of Solids *18
· Find the value of n in Equation 3 8 -6 that gives the measured dissociation energy of74 1 kJ/mol for LiCl, which has the same structure as NaCl and for which ro = 0 .257 nm.
Picture the Problem W e can solve Equation 3 8 -6 for n.
Equation 3 8-6 is: and
Solve for n to obtain:
1
393
394 Chapter 38 Substitute numerical values and evaluate n: n=
(1
1 14.641 eV/ion pair (741kJ/mol) (0.2 7nm) 6.47 kJ/mol 5 9 1 (1.7476)(1 .44eV . nm)
_______________
-
J
=
�--:-:-----:----
---
A Microscopic Picture of Conduction
*22 ·· Silicon has an atomic weight of28 .09 and a density of 2 .4 1 x1 0 3 Each atom of silicon has 2 valence electrons and the Fermi energy of the material is 4.88 eV. (a) Given that the electron mean free path at room temperature is A = 27.0 nm, estimate the resistivity. (b) The accepted value for the resistivity of Silicon is 640 n-m (at room temperature). How does this accepted value compare to the value calculated in part (a)?
kg/m3.
Picture the Problem We can use Equation 3 8-14 to estimate the resistivity of silicon.
(a) From Equation 38-14:
(1)
The speed of the electrons is given by: Substitute numerical values and evaluate Vav:
The electron density of Si is given by:
v
av
1
2(4.88 eV) 1.60 x 1 0-19 J (9. 1 x 10-31kg) 1 eV = 1 .31x106m1s =
ne =
MN
A
Nalam
where Nalam is the number of electrons per atom.
Substitute numerical values and evaluate ne: ne
=
2X 1023atoms J(�) (2.41X 1 03 mkg3 )( 6.00.0280 atom 9 kg
=
1.0 3x1 029 e/m3
Substitute numerical values in equation (1 ) and evaluate p:
(b) The accepted resistivity of 640 n·m is much greater than the calculated value. We assume that valence electrons will produce conduction in the material. Silicon is a
Solids and the Theory of Conduction 395 semiconductor and a gap between the valence band and conduction band exi sts. Only electrons with sufficient energies will be fo und in the conduction band. The Fermi Electron Gas *26
Calculate the Fermi temperature for (a) AI, (b) K, and (c) Sn.
•
Picture the Problem The Fermi temperature
Fermi energy.
TF is defined by kTF EF, where EF is the =
E F = kF
The Fermi temperature is given by:
T.
(a) For AI:
T.
(b) For K:
T.
(c) For Sn:
T.
F = 8.62x10-�) eV/K = 1 1.36x105K I 11.7 V
F = 8.62x10-�) eVIK = 1 2.45x104 K I 2.11 V
F = 8.62x10-)eVIK = 1 1.18x105 K I 10.2 �V
*31 ·· (a) Assuming that gold contributes one free electron per atom to the metal, calculate the electron density in gold knowing that its atomic wei ght is 196.97 and its mass density is 1 9 .3x1 03 kg/m3. (b) If the Fermi speed for gold is 1.3 9x1 06 mis, what is the Fermi energy in electron volts? (c) By what factor is the Fermi energy hi gher than the energy at room temperature? (d) Explain the difference between the Fermi energy and kTenergy.
kT
Picture the Problem We can use n e =
pV =
pN Na om A
m
t
, where Natom is the number of
electrons per atom, to calculate the electron density of gold. The Fermi energy is given 2 - 1 bY - z mevF•
EF
(a) The electron density of gold i s given by:
n e = pV =
pNANatom m
S ubstitute numerical values and evaluate n e:
ne =
(
19. 3X103
� ) (6.02x1023atoms)(1atIeom )
k
m
0.197 kg
(b) The Fermi energy is given by:
1
= 5.90x1028e/m3
1
39 6 Chapter 38 Substitute numerical val ues and evaluate
EF:
leV - = I s . so ev I EF =� (9. 1 1 XIO3- 1kg )(1 . 39XI06m/S) 1.60xlO 19J 2
)
(
(c) The factor by which the Fermi energy is higher than the kT energy at room temperature is:
E f= kTF
At room temperature kT= 0.026 eV. S ubstitute numerical val ues and evaluatef
S.SOeV f= 0.026 eV= @IJ
EF
(d) is 2 1 2 times kT at room temperature. There are so many free electrons present that most of them are crowded, as described by the Pauli exclusion principl e, up to energi es far higher than they would be according to the classical model . *32
••
particles by
The pressure of an ideal gas is related to the average energy of the gas PV t , where N is the number of particles and is the average
Eav
= NEav
energy. Use this to calculate the pressure of the Fermi electron gas in copper in newtons per square meter, and compare your result with atmospheric pressure, which is about 1 0 5 N/m 2. (Note: The units are most easily handled by using the conversion factors 1 N/m 2 = 1 J/m 3 and 1 eV = 1 .6 l 0 1 J.)
x 9 -
Picture the Problem We can solve
express P in terms of NIV and Solve
PV= t NEav
B ecause
for P:
Eav = t EF :
EF•
PV= t NEav for P and substitute for Eav in order to
p=� N Eav 3 V
( )
p= � N 'i EF = � N EF 3V S SV
( )( ) ( )
Substitute numerical values (see Table 38-1) and evaluate P:
p= �S (8.47 x1 022 electrons/cm3 )(7.04eV) (1.60x1 0-19 JleV) = 1 3.82xl0ION/m2 1. = 3.82xl0ION/m2 x 101 .32Slatm x1 03 N/m2 = 1 3.77xl05 atm I
Solids and the Theory of Conduction 397 Heat Capacity Due to Electrons in a Metal *35 ·· Gold has a Femli energy of 5 .53 eV. Determine the molar specific heat at constant volume and room temperature for gold.
Picture the Problem We can use Equation 3 8-29 to find the molar specific heat of gold at constant volume and room temperature.
The molar specific heat is given by Equation 3 8-29:
e 'y
The Fermi energy is given by:
Substitute for TF to obtain:
Substitute numerical values and evaluate e 'y :
Jr2 (8.3IJ/molK)(1.38x 10-23 J/mol)
e'
y
(
l eV 1.60x 10
-19
J
]
(300K)
= --------------------�----��--------�----=
1 0.192J/mol·K 1
2(5.53
eV )
Remarks: The value 0.192 J/mol K is for a mole of gold atoms. Since each gold atom contributes one electron to the metal, a mole of gold corresponds to a mole of electrons. Quantum Theory of Electrical Conduction
*37
··
The resistivity of pure copper is increased approximately 1 x 1 0-8 n·m by the
addition of 1 percent (by number of atoms) of an impurity throughout the metal. The mean free path depends on both the impurity and the oscillations of the lattice ions according to the equation 1/ A = 1/At + 1/ A,. (a) Estimate A; from data given in Table 38-1 . (b) If r i s the effective radius of an impurity lattice ion seen by an electron, the scatterin g cross section i s
Equation 3 8- 1 6.
7U"2. Estimate this area, using the fact that
r
is related to Ai by
Picture the Problem We can solve the resistivity equation for the mean free path and
then substitute the Fermi speed for the average speed to express the mean free path as a function of the Fermi energy.
398 Chapter 38 (a) In terms of the mean free path and the mean speed, the resistivity IS:
Solve for /L to obtain:
AI
= meuF2
ne Pi
Express the Fenni speed UF in
tenns of the Fenni energy
Er::
Substitute to obtain:
Substitute numerical values (see Table 38-1) and evaluate�:
�2(9.11x10-31 kg)(7.04eV)(1.60X1�-91 l/eV) = 66.1n mI 1 (8.47 x1028 electrons/m3)(1.60 x10-91 C) (10-8 Q. m)
Ai =
(b) From Equation 3 8 -1 6 we have:
Solve for 7r?:
Substitute numerical values and
evaluate ;rr2:
nA rcr
1 2- r 2 - \8.47X10 8 m-3)(66.1nm) xl 0--4-- nm-:--'2 1 = 1.79x10-22m2= lr--l .-79--
Band Theory of Solids *39
·
You are an electron sitting at the top of the valence band in a silicon atom,
longing to jump across the 1 . 1 4-eV energy gap that separates you from the bottom of the conduction band and all of the adventures that it may contain. What you need, of course, is a photon. What is the maximum photon wavelength that will get you across the gap? Picture the Problem We can relate the maximum photon wavelength to the energy gag using
DE = hf = he/A.
Express the energy gap as a function of the wavelength of the photon:
DE =hf = he A
Solids and the Theory of Conduction
399
Solve for A: Substitute numerical values and evaluate A:
A=
1 240eV ·nm 1 . 1 4eV
=
1 1.09 1 J11ll
Semiconductors *44
••
When a thin slab of semiconducting material i s illuminated with
monochromatic electromagnetic radiation, most of the radiation i s transmitted through the slab if the wavel ength i s greater than 1 . 85 J.lill. For wavel engths less than 1 .85 J.lill,
most of the incident radiation i s absorbed. Determine the energy gap of this semiconductor.
Picture the Problem We can use E= hf to find the energy gap of this semiconductor.
The energy gap of the semiconductor is given by:
E g=
hI" =
where Substitute numerical values and evaluate Eg:
*46
••
E 1=
'J
he
A
he = 1240 eV·nm o 1240eV ·nm E " = 1 .85,urn = 0.670eV
1
1
The ground-state energy of the hydrogen atom i s given by e 2 me
8 E o2 h 2
Modify this equation in the spirit of Problem 45 by replacing Eo by KEo and me by an effective mass for the electron to estimate the binding energy of the extra el ectron of an impurity arsenic atom in (a) silicon and (b) germanium. Picture the Problem We can make the same substitutions we made in Problem 45 in the expression for E1 (= 13.6 eV) to obtain an expression that we can use to estimate the binding energy of the extra electron of an impurity arsenic atom in silicon and germalll urn.
Make the indicated substitutions in the expression for E1 to o btain:
e2 memeff
8meK 2 E� h 2
400
Chapter 38
(a) For silicon:
(b) For gennanium:
Semiconductor Junctions and Devices *51
··
In Figure 3 8-27 for the pnp-transistor amplifier, suppose
Rb= 2 kO and RL=
1 0 kO . Suppose further that a 1 0-I'A ac base current generates a 0.5-mA ac collector current. What is the voltage gain of the amplifier? Picture the Problem We can use its definition to compute the voltage gain of the amplifier.
The voltage gain of the amplifier is given by: Substitute numerical values and evaluate the voltage gain:
IcRL . = -IbR b (0. 5 mA)(10 kQ) V o 1tage gam (1 0.uA) ( 2kQ) = 1 25 0 1 Voltage gam .
_
*54 ·· A "good" silicon diode has the current-voltage characteristic given in Problem 49. Let kT= 0 .025 eV (room temperature) and the saturation c urrent 10 = 1 nA. (a) Show that for small reverse-bias voltages, the resistance is 25 MO. (Hint: Do a Taylor expansion o/ the exponential/unction or use your calculator and enter small values/or V )b . (b) Find the dc resistance for a reverse bias of 0.5 V. (c) Find the dc resistance for a 0.5-V forward bias. What is the current in this case? (d) Calculate the ac resistance dVld1for a 0.5 -V forward bias.
Picture the Problem We can use Ohm's law and the expression for the c urrerit from Problem 49 to find the resistance for small reverse-and-forward bias voltages.
(a) Use Ohm ' s law to express the resistance:
R=Vb I
From Problem 47, the current across a pn j unction is given by:
(2)
For eVb« kT:
S ubstitute to obtain:
(1)
e b I= IoV kT
Solids and the Theory of Conduction 4 0 I Substitute for fin equation simplify:
(1) and
VIJ kT _ eVIJ elo 1o
R=
kT
Substitute numerical val ues and evaluate R:
1
1
= 25.0 MO (b) S ubstitute equation (2) in equation ( 1 ) to obtain:
R-
eVb
eVb
kT
kT
Evaluate -- for V b = - 0.5V: Evaluate equation (3) for
Vb=-0.5V: ev' (c) Evaluate _b for
Vb= + 0.5V:
kT
Evaluate equation (3) for
Vb= +0.5V: (d) Eval uate Rae = dVldlto obtain:
R=
eVb
kT
R=
b
(
(3)
)
) )
19 C -O.5V = l.60x 10= -19. 8 l. 38x lO-23 JIK 293K
(
I
-O.5V = 5 oOMO 10 9 A -,98. _1
)(e
(
)
( ) = 19.8 l . 38x 10-23 J/K (293K )
1
O.5V = l.26 0 1 0-9 A el98 -1
=
)
H
(
R ae _
1
l.60x lO-'9 C O.5V
=
=
Substitute numerical values and evaluate Rae:
V,
e 10 e vb/kT -1
dV dI
_
( dI )-1 dV
{:v [I, (
{ (
e -'.1" -1
}
1
ljf
e1o eVb/kT -1 kT -eVb/kT e = e e 10 kT
)
1
Rac = 25 MO e-19.8 = 0.0629 0
1
The BCS Theory *57
·
Repeat Problem 56 for lead 3 of 2.73xl O- eV.
(Tc = 7 . 1 9 K), which has a measured energy gap
Picture the Problem We can calculate Eg using
Eg = 3.5k�
and find the wavel ength of
a photon having sufficient energy to break up Cooper pairs in tin at
T = 0 using
402
Chapter38
(a) From Equation 3 8 -24 we have: Substitute numerical values and evaluate Eg:
Express the ratio of Eg to Eg,measured:
10-5 eV/K)(7.19K) = 1 2.17meV I
Eg = 3.5(8.62 x
Eg Eg,measured or
Eg
:::::
2.17meV = 0.795 2.73xl O-3eV
I 0.8Eg,measured
(b) The wavelength of a photon having sufficient energy to break up Cooper pairs in tin at T = 0 is given by: Substitute numerical values and evaluate A:
1
. V run = 4.54xl05 run = 1240e x 3 2.73 l O eV = I 0.454mmI
The Fermi-Dirac Distribution *60
··
(a) Use Equation 38-22a to calculate the Fermi energy for silver. (b) Determine the average energy of a free electron and (c) find the Fermi speed for silver.
Picture the Problem Equation 38-22a expresses the dependence of the Fermi energy EF on the number density of free electrons. Once we've determined the Fermi energy for silver, we can find the average electron energy from the Fermi energy for silver and then use the average electron energy to find the Fermi speed for silver.
(a) From Equation 38-22a we have:
2 3N 3/2 EF - h 8me "V _
Use Table 27- 1 to find the free electron number density NIV for silver:
-- (--)
= 5.86xl022 electrons V cm3 = 5.86xl 028 elect: ons m N
Solids and the Theory of Conduction 403 Substitute numerical values and evaluate EF:
3
6.63xlO-34J·s 2 [3(S.86xl028electrons/m3)]2/ [ leV EF = 89.11xlO-3Ikg 1.60xlO-19J 7[
= / S . S l eV /
(b) The average electron energy is given by:
3 Eav = -EF S
Substitute numerical values and evaluate Eav:
Eav =%(S.SleV)=/3.3lev /
]
(c) Express the Fermi energy in terms of the Fermi speed of the electrons: Solve for VF:
S ubstitute numerical val ues and evaluate VF:
*63
··
energy of
vF
2(3.3leV) [1.60X 10-19 J] 9.11xlO-3Ikg leV =11.08xl06m1s I -
What is the probability that a conduction electron in silver will have a kinetic
4.9 eV at T= 300
K?
Picture the Problem The probability that a conduction electron will have a given kinetic
energy is given by the Fermi factor. The Fermi factor is:
Because EF *70
•••
-
4.9 eV » 300k:
(a) Show that for E �
1_ = OJ f(4.geV)= _0+1 0, the Fermi factor may be written as
f(E) = l/(CeE/kT +1) (b) Show that if C» e-E1kT,f(E) Ae-E1kT « =
I; in other words,
A
show that the Fermi factor is a constant times the classical Boltzmann factor if « 1 . (c) Use
fn (E)dE = N and Equation 38-41 to determine the constant A. (d) Using the result
obtained in Part (c), show that the classical approximation is applicable when the electron concentration is very small and/or the temperature is very high.
(e) Most semiconductors
have impurities added in a process called doping, which increases the free electron
404
Chapter
38
concentration so that it is about 1 017/cm3 at room temperature. Show that for these systems, the classical distribution function is applicable. Picture the Problem We can follow the step-by-step procedure outl ined in the problem
statement to obtain the indicated results. (a) The Fermi factor is:
1
CeE/kT
=
provided C (b) If C»
e-E1kT:
1
e-EdkT
1 1 f ( E) - CeE/kT 1�CeE/kT - I Ae-E/kT 1 _
_
where A (c) The energy distribution function
=
+
+
=
l/C
n{E)dE g{E)dE f{E) =
IS:
S ubstitute for g(E)dE and j(E) in the expression for N to obtain: The definite integral has the value: S ubstitute to obtain:
Solve for A:
(d) Evaluate A at T= 300 K: A
.J2(6.63xl O-3 4J,sLn �4xl O-26n 2 - 81Z"3 /2�.l 1xl O-3 1 kgr/2[(1.38xl O-23JIK)3 { 00K)r/
-
where the units are S1.
Solids and the Theory of Conduction 4 05 The valence electron concentration is typically about 1039 m -3. To satisfy the condition that A « 1 at room temperature, n should be less than 1023 m -3, or
about one millionth of the valence electron concentration. Because A depends on T-3/2 , the electron concentration may be greater the higher the temperature.
(e)
1017 cm-3 = 1023 m-3. So, according to the criterion in
Cd), the classical
approximation is applicable.
General Problems *76 ·· Detennine the energy that has 1 0 percent free electron occupancy probability for manganese at T = 1 300 K.
Picture the Problem The Fenni factor gives the probability of an energy state being occupied as a function of the energy of the state E, the Fenni energy EF for the particular material, and the temperature T.
The Fenni factor is:
For 1 0 percent probability:
1 0.1 = (E-E )jkT F +1 e
Take the natural logarithm of both sides of the equation to obtain:
E -EF kT
Solve for E to obtain:
E = EF +kTln 9
From Table 37- 1 , EF(Mn)
(
=
1 1 .0 eV. Substitute numerical values and evaluate E:
)(
E = 11.0eV+ 1. 38X10-23JI K 1 300K *78
-,- = In 9
__
)
(
l eV 9
1. 60xlO-1 J
)
l
ln 9 = l1.2 eV
I
A 2 -cm2 wafer of pure silicon is irradiated with light having a wavelength of 775 nm. The intensity of the light beam is 4.0 W/m2 and every photon that strikes the •••
sample is absorbed and creates an electron-hole pair. (a) How many electron-hole pairs are produced in one second? (b) If the number of electron-hole pairs in the sample is 6.25x1 0" in the steady state, at what rate do the electron-hole pairs recombine? (c) If every recombination event results in the radiation of one photon, at what rate is energy radiated by the sample?
406
Chapter
38
Picture the Problem The rate of production of electron-hole pairs is the ratio of the
incident energy to the energy required to produce an electron-hole pair.
(a)The number of electron-hole pairs N produced in one second is:
N=
fA
he A
= fAA he
Substitute numerical values and evaluate N:
(b) In the steady state, the rate of recombination equals the rate of generation. Therefore: (c) The power radiated equals the power absorbed: Substitute numerical values and evaluate Prod:
Prad = 4( .0 W/m2 )(2 x 1 0-4 m2 ) = I 0.800mJ/s I
Chapter 39 Relativity Conceptual Problems *1
•
The approximate total energy of a particle of mass m moving at speed u « c
2 is (a) me2 +1mu2 . (b) 1mu • (c) emu. (d) 1me2 . (e) 1emu.
Picture the Problem The total relativistic energy E of a particle is defined to be the sum
of its kinetic and rest energies. The total relativistic energy of a particle is given by:
*2
•
and
I (a) is corr ect. I
A set of twins work in an office building. One twin works on the top floor
and the other twin works in the basement. Considering general relativity, which one will age more quickly? (a)
They will age at the same rate. (b) The twin who works on the
top floor will age more quickly. (c) The twin who works in the basement will age more quickly. (d) It depends on the speed of the office building. (e) None of these is correct. Determine the Concept The gravitational field of the earth is slightly greater in the
basement of the o ffice building than it is at the top floor. Because clocks run more slowly in regions of low gravitational potential, clocks in the basement will run more slowly than clocks on the top floor. Hence, the twin who works on the top floor will age more quickly.
I (b) is corr ect. I
Estimation and Approximation *7
The most distant galaxies which can be seen by the Hubble telescope are moving away from us with a redshift parameter of about z = 5 . (See Problem 30 for a definition of z) . (a) What is the relative velocity of these galaxies with respect to us (expressed as a fraction of the speed of light)? (b) Hubble's law states that the recession velocity is given by the expression v = Hx, where v is the velocity of recession, x is the distance, and H is the Hubble constant, H = 75 km/s/Mpc. ( 1 pc 3 .26 light-year). Estimate the distance of such a galaxy using the information given. ••
=
Picture the Problem We can use the result from Problem 30, for light that is Doppler
shifted with respect to an observer,
( : - 11)
v=c u
u- +
, where u = z + 1 and z is the red-shift
parameter, to find the ratio of v to c. In (b) we can solve Hubble's law for x and substitute our result from (a) to estimate the distance to the galaxy.
407
408 Chapter 39 (a) Use the result of Problem 30 to
express vic as a function o f z :
Substitute for z and evaluate vic:
� _ (z + lY - 1 (z + lY + l
c
: - 1 1 0.946 1 (5 + 1) + 1
� = (5 + 1)
c (b) Solve Hubble' s law for x:
Substitute numerical values and evaluate x:
=
v
X=H x
=
0.946c
H
=
(
0. 946 3 x 10 5
kmI s
kmI s )
75 --
Mp c
=
3.78 x 10 3 Mp c x
=
1 1 2 . 3 Gc . y 1
3.26 x 10 6 c · y
---- Mp c
Time Dilation and Length Contraction *10 ·· Unobtainium (Un) is an unstable particle that decays into Nonnalium (Nr) and Standardium (St) particles. (a) An accelerator produces a beam of Un which travel s t o a detector located 1 000 m away from the accelerator. The particles travel with a velocity of v = 0.866c. How long do the particles take (in the laboratory frame) to get to the detector? (b) By the time the particles get to the detector, half of them have decayed. What is the half-life of Un? JNote: Half-life as it would be measured in a frame moving with the particles). (c) A new detector is going to be used, which is located 10,000 m away from the accelerator. How fast should the particles be moving if half of them are to make it to the new detector?
Picture the Problem The time required for the particles to reach the detector, as measured in the laboratory frame of reference is the ratio of the distance they must travel to their speed. The half life of the particles is the trip time as measured in a frame traveling with the particles. We can find the speed at which the particles must move if they are to reach the more distant detector by equating their half life to the ratio of the distance to the detector in the particle' s frame of reference to their speed.
(a) The time required to reach the
detector is the ratio of the distance to the detector and the speed with which the particles are traveling:
Substitute numerical values and
evaluate /1{:
6t = & = v
&
0.866c
Relativity (b) The half life is the trip time as measured in a frame traveling with
6l
,
=
the particles: Substitute numerical values and evaluate 6.t' :
-;;
' 1- O.8 6C
( :
l1l' = 3.85,u s =
1 1. 93,u s 1
(c) In order for half the particles to reach the detector:
? Fffi) -
v
6t = 6l 1r
l1l ' = l1x ' rv
=
l1x '
)
Fffi v
where t:u' is the distance to the new detector.
Rewrite this expression to obtain:
fix '
v
Squaring both sides of the equation yields:
Substitute numerical values for t:u' and 6.t' and simplify to obtain:
_ _ = ( 104 2 V_
I-
m
'
m
93 s 1 ,u
J
2
= (17.3cY
Divide both sides of the equation by 2 c to obtain:
2 2
(17.3Y 1+ (17.3)-
Solve this equation for V /C :
Finally, solving for v yields:
v=
1
0.998c 1
?
=
0. 9967
409
410
Chapter
39
The Lorentz Transformation, Clock Synchronization, and Simultaneity *17
••
A spaceship of proper length Lp = 400 m moves past a transmitting station at
a speed ofO.76c. At the instant that the nose of the spaceship passes the transmitter, clocks at the transmitter and in the nose of the spaceship are synchronized to t = t'
=
O.
The instant that the tail o f the spaceship passes the transmitter a signal i s sent and subsequently detected by the receiver in the nose of the spaceship. (a) When, according to the clock in the spaceship, is the signal sent? (b) When, according to the clock at the transmitter, is the signal received by the spaceship? (c) When, according to the clock in the spaceship, is the signal received? (d) Where, according to an observer at the transmitter, is the nose of the spaceship when the signal is received? Picture the Problem Let S be the reference frame of the spaceship and S' be that of the
earth (transmitter station). Let event A be the emission of the light pulse and event B the
reception of the light pulse at the nose of the spaceship. In (a) and (c) we can use the
classical distance, rate, and time relationship and in (b) and (d) we can apply the inverse Lorentz transformations. (a) In both S and S' the pulse travels
at the speed c. Thus:
(c) The elapsed time, according to
L 400rn = 1 . 76,us 1 tA = -P = v 0.76c 1 tB
=
tpulse to travel length of ship
the clock on the ship is: Find the time of travel of the pulse to the nose of the ship:
tpulse to travel lcngth of ship
=
+ tA
400rn 2.998x1 08 rn1s
= 1 .33,us Substitute numerical values and evaluate tB : (b) The inverse time transformation is:
tB
= 1 .33 ,us + 1 .76,us = 1 3 . 09,us I
( ;)
tB ' = r t -
1 1 g = ---;= === = 1 .54 2 - (0.7 6c y -
where r
=
1- 2 c
1
c2
Relativity Substitute numerical values and evaluate t' B :
tB '
=
= =
(d) The inverse transfonnation for x
IS:
41 1
(1 .54 3. 0 9 JiS ( 0 . 7 6 (400m)
{
�
_
[
1
6) (400m) (1 . 5 4) 3.09 JiS - (- 0.7 3 x 08 m/s
1 6.3 2 JiS I
)
J
x ' = r (x - vt )
Substitute numerical values and evaluate x' :
*22
•••
An observer in frame S standing at the origin observes two flashes of colored
light separated spatially by &
=
2400 m. A blue flash occurs first, followed by a red flash
5 J..lS later. An observer in S' moving along the x axis at speed v relative to S also observes the flashes 5 f.1s apart and with a separation of 2400 m, but the red flash is observed first.
Find the magnitude and direction of v.
Picture the Problem We can use the inverse time dilation equation to derive an
expression for the elapsed time between the flashes in S' in tenns of the elapsed time
between the flashes in S, their separation in space, and the speed v with which S' is movmg.
From the inverse time transfonnation we have:
[
M' = r M -
;
&
]
where �t' is the time between the flashes in
S' and �t and & are the elapsed time
between the flashes and their separation in S.
Set �t'
=
-�t to obtain:
- �t v = l1t - - & 2 r
or - l1t
Square both sides of the equation:
c
1 -H c
2 2
v c2
= M --&
4 12 Chapter 39
Simplify to obtain:
Solve for v: v =
( 1L\�)2 L\t [ 1( J ]2 1 ( J 18 I I
1+
Substitute numerical values and evaluate v:
-
e
2
v =
+
2400m 5 f.1s
2400m
3 x 10' m ls
5 ps
= 2.697 x 0 m1s = 0.89ge Because v is positi ve, S' ism ovingin the positi ve x directi on . The Velocity Transformation
*24 ·· A spaceship, at rest in a certain reference frame S, is given a speed increment of O.50c (call this boost 1). Relative to its new rest frame, the spaceship is given a further 0.50c increment 1 0 seconds later (as measured in its new rest frame; call this boost 2). This process is continued indefinitely, at 1 0- second intervals, as measured in the rest frame of the ship. (Assume that the boost itself takes a very short time compared to 1 0 s .) (a) Using a spreadsheet program, calculate and graph the velocity of the spaceship in reference frame S as a function of the boost number for boost 1 to boost 1 0. (b) Graph the gamma factor the same way. (c) How many boosts does it take until the velocity of the ship in S is greater than 0.999c? (d) How far has the spaceship moved after 5 boosts, as measured in reference frame S? What is the average speed of the spaceship (between boost 1 and boost 5) as measured in S?
Picture the Problem We'll let the velocity (in S) of the spaceship after the ith boost be
and derive an expression for the ratio of v to c after the spaceship' s
Vi
( i + 1 )th boost a s a function o f N. W e can use the definition o f y, i n terms o f vic t o plot r as a function of N.
(a ) and (b) The velocity of the spaceship after the (i + 1 )th boost is given by relativistic velocity addition equation:
Vi+1
=
1
v. I
+
+ 0.5e 0.5e i
(
e2
)V
Relativity 4 1 3 Factor c from both the numerator and denominator to obtain:
v.
----'- + 0.5 C --== Vi+l Vi 1 + 0.5 c
1; is given by:
1
A spreadsheet program to calculate vic and r as functions of the number of boosts N is shown below. The formulas used to calculate the quantities in the columns are as follows: ContentlFormula 0 0 (B2+0.5)/( 1 +0.5 *B2) 1/(1 -B2/\2)/\0.5
Cell A3 B2 B3 C1
Algebraic Form N Vo Vi+ l r
,
B · A boost 1 vic 0 2 0.000 3 1 0.500 4 2 0.800 3 0.929 '.' " . 5 4 0.976 ;, 6 5 7 0.992 6 8 0.997 7 0 .999 9 :' 10 1 .000 8 ':' 1 1 9 1 .000 12 1 0 1 .000 '. A graph of vic as a functIOn of N IS shown below: . ;.
.. . ,.
rr c
gamma 1 .00 1.15 1 .67 2 .69 4.56 7.83 1 3 .52 23.39 40.5 1 70. 1 5 1 2 1 .50
1 .0 0.8 � �
0.6 0.4 0.2 0.0
a
2
4
N
6
8
10
4 14
Cha pter 39
A graph of y as a function of N is shown below: 1 20 1 00
'"
E E '"
0/)
80 60 40 20 0
(c)
0
2
4
N
6
8
10
E xa mi nati on of the sp read sheet or of the grap h of vic a s a fu ncti on of N i ndi cate s that, after 8b oost s , t he ve locity of t he sp a ce s hip i s greate rtha n 0.99g e .
(d) After 5 boosts, the spaceship has traveled a distance ill, measured in the earth frame of reference (S), given by:
!'!:,xH 2+ !'!:,x -+2 3 + !'!:,x 3-+4 + !'!:,x4-+S = (0. 5 e)(1 Os )r1-+ 2+ (0.8e)(1 Os )Y -+2 3 + (0.92ge)(1 Os)Y -+3 4 + (0.976e)(1 Os)Y4-+S + (0. 992e )(1 0S)YS-+ 6 = (0. 5 e)(1 Os )(1 1. 5 )+ (0. 8e)(1 Os )(1 .67) + (0.92ge)(1 Os )(2.69) + (0.976e)(1 0s) (4.5 6)+ (0. 992e)(1 Os )(7.83) = 1 1 66e .s I
!'!:,x =
The average speed of the spaceship, between boost 1 and boost 5 , as measured in S is given by:
!'!:,x vav = -
f..t
where !J.t is the travel time as measured in the earth frame of reference.
Express !J.t as the sum of the times the spaceship travels during each 1 0-s interval following a boost in its speed:
f..t = f..tl-+ 2+ f..t -+2 3 + f..t 3-+4 + f..t4-+S = ( 1 0S)YI-+ 2+ (1 0s)Y -+2 3 + (1 0S)Y 3-+4 + (1 0S)Y4-+S + (1 0S)YS-+ 6 = (l Os )( YH 2+ Y -+2 3 + Y 3-+4 + Y4-+S + YS-+ 6)
Substitute numerical values and evaluate !J.t: f..t =
(l Os )(1 1. 5 + 1 .67+ 2. 69+ 4.5 6+ 7.83) = 1 79s
Relativity 4 1 5 Substitute for t..,'( and /':,1 and evaluate Vav :
vav
=
1 66 c · s 1 79 s
= I O. 92 7c I
Remarks: This result seems to be reasonable. Relativistic time dilation implies that the spacecraft will be spending larger amounts of time at high speed (as seen in reference frame S). The Relativistic Doppler Shift *29
··
A clock i s placed in a satellite that orbits the earth with a period of
90 min. By what time interval will this clock differ from an identical clock on earth after 1 y? (Assume that special relativity applies and neglect general relativity.) Picture the Problem Due to its motion, the orbiting clock will run more slowly than the
earth-bound clock. We can use Kepler' s third law to find the radius of the satellite's orbit in terms of its period, the definition of speed to find the orbital speed of the satellite from
the radius of its orbit, and the time dilation equation to find the difference 0 in the
readings of the two clocks.
Express the time 0 lost by the clock:
(
!::"t = !::,.t 1 --I 0= !::"t - !::"t = !::,.t - P
r
Because v « c, we can use part (b)
r
J
of Problem 1 3 :
Substitute to obtain:
Express the square of the speed of the satellite in its orbit:
(1)
v'
=
e;r ) '
(2)
where T is its period and is the radius of
r
its (assumed) circular orbit.
Use Kepler' s third law to relate the period of the satellite to the radius of its orbit about the earth: Solve for r:
Substitute numerical values and evaluate r:
416
Chapter
39
4Jr2 (6.65 x 106mL (90min x 60s/min Y = 5.99 x 107 m 2 I S2
Substitute numerical values in equation (2) and evaluate v2 :
v2 =
Finally, substitute for v2 in equation ( 1 ) and evaluate 5:
(
)
1
5 .!. 5.99 x 107 m 2 1 S2 (1 y x 3 1 .56 Ms/ y ) = = 10 . 5ms 8 2 (3 x 10 m/sL
I
,
A particle moves with speed O.8c along the x" axis of frame S which moves with speed O.8c along the x ' axis relative to frame S '. Frame S ' moves with speed O.8c along the x "
axis relative to frame S. (a) Find the speed of the particle relative to frame S '. (b) Find
the speed of the particle relative to frame S. Picture the Problem We can apply the inverse velocity transformation equation to
express the speed of the particle relative to both frames of reference. (a) Express
u/ in terms of u/' :
u "+v ux ' = ---"x .....,.. vu " l + -xe2 _
where V of S ', relative to S', is O.8c. Substitute numerical values and evaluate u/ :
(b) Express
Ux in terms of u/ :
Substitute numerical values and evaluate
Ux:
u/ =
0 . 8e + 0.8e = 1 .6e = I 0.976c I (0.8e)2 1 .64 1+ e2
u '+v ux = ---"x,--_ vu ' l + --x e2 where v of S, relative to S ', is O.8c. 0.976e +0 . 8e = 1 .776e Ux 1 + (0.8e)(0 .976e) 1 .781 e2 = I O.997e I =
Relati vit y
41 7
Relativistic M omentum and Relativistic Energy *34
•
A proton (rest energy 938 MeV) has a total energy of 2200 MeV.
(a ) What is its speed? (b) What is its momentum? Picture the Problem We can use the relation for the total energy, momentum, and rest
energy to find the momentum of the proton and Equation 39-26 to relate the speed of the
proton to its energy and momentum. Relate the energy of the proton to its momentum: (b) Solve for p to obtain:
Substitute numerical values and evaluate p:
p= p=
�{2200 MeV Y - (938 MeV Y
1
= 1 .99 (a) From Equation 39-26 we have:
v
c
= pc
Solve for v to obtain: Substitute numerical values and evaluate v: *39
··
v=
;
GV
1
c
E
1 .99 GeV c = I O.905c I 2200 M eV
Two protons approach each other head-on at O.Sc relative to reference frame
S' . (a) Calculate the total kinetic energy of the two protons as seen in frame S' . (b)
Calculate the total kinetic energy of the protons as seen in reference frame S, which is moving with speed O.Sc relative to S' so that one of the protons is at rest.
Picture the Problem The total kinetic energy of the two protons in part (a) is the sum of
their kinetic energies and is given by K
= 2 {r - l)Eo . Part (b) differs from part (a) in
that we need to find the speed of the moving proton relative to frame S. (a) The total kinetic energy of the protons in frame S' is given by:
4 1 8 Chapte r 3 9 Substitute for r and Eo and evaluate
K:
K
1
=2 1-
(O.Scy c2
- 1 (93 8 . 2 8 MeV)
= 1 290 MeV I
(1)
(b) The kinetic energy of the moving proton in frame S is given by:
1
Express the speed u of the proton in frame S:
Substitute numerical values and evaluate
where
u:
Evaluate r
Substitute numerical values in equation (1) and evaluate K:
+ u = u 'vuv l + __2_x c
----'x', --_
.Sc + O. Sc = O. 00c 8 u = O(O .Sc)(O.Sc) + 1 c2 r
K
=
1
� 1-
(O . 8 c)(O 8 c) c2 .
= 1 . 67
= (1 .67 - 1)(93 8 . 2 8 MeV) = 1 62 9 MeV I
General Relativity *42 " Light traveling in the direction of increasing gravitational potential undergoes a frequency redshift. Calculate the shift in wavelength if a beam of light of wavelength A. = 63 2 . 8 nm is sent up a vertical shaft of height L 1 00 m. =
Picture the Problem Let m represent the mass equivalent of a photon. We can equate the change in the gravitational potential energy of a photon as it rises a distance L in the gravitational field to h;').j and then express the wavelength shift in terms of the frequency shift.
The speed of the photons in the light beam are related to their frequency and wavelength:
C = fA
=>
f = Ac
Relativity
41 9
e dl -2 = - ell = - 112 dll
Differentiate this expression with respect to A to obtain: Approximate dj1dA by f..j1f..1l and solve for f..J Divide both sides of this equation by
f..1 I
fto obtain:
e
Solve for f..1l :
(1) M = f.. U = mgL
The change in the energy of the photon as it rises a distance L in a gravitational field is given by: Because M = hf..f :
hf..1 = mgL
Letting m represent the mass
E = me2
Divide equation (2) by equation (3)
-- =
equivalent of the photon:
hf..1 hi
to obtain:
=
(2)
hi
mgL
--
me2
(3)
f..1 I
gL � - =e2
Substitute for f..j1fin equation ( 1 ):
(9.8 1 mJs2 )(1 00 m)(632. 8 nm) (3 x 1 08 mJs) = 1 - 6. 9 0x 10-1 2 I
Substitute numerical values and
f..1l =
evaluate f..1l :
_
Dm
General Problems *47
··
Frames S and S are moving relative to each other along the x and x' axes.
Observers in the two frames set their clocks to t
=
0 when the origins coincide. In frame
S, event 1 occurs at X I 1 .0 c·y and tl 1 y and event 2 occurs at X 2 2.0 c·y and t2 = 0.5 y. These events occur simultaneously in frame S . (a) Find the magnitude and =
=
=
direction of the velocity of S relative to S. (b) At what time do both these events occur as measured in S?
420
Chapter 3 9
Picture the Problem We can use Equation 39- 1 2, the inverse time transformation
equation, to relate the elapsed times and separations of the events in the two systems to the velocity of S' relative to S. We can use this same relationship in (b) to find the time at
which these events occur as measured in S' . (a) Use Equation 39-12 to obtain:
[(tz - tJ- ; (X2 J] = [ �t - :2 ]
�t' = t/ -t/
=
-x
&
r
Because the events occur
r
simultaneously in frame S' , /).t' = 0 and:
Solve for v to obtain:
Substitute for evaluate V: Because M
�t and /).x and
= t2
- tl
= -0.5 y :
e2 (0 . 5 y - l y) = 1 - 0.5e I V = 2.0e · y - 1 .0e · y
I
S' mo ves in
the negative xd ire ction .
I
(b) Use the inverse time transformation to obtain :
Substitute numerical values and evaluate tz' and tl ' :
*51
•••
t2 ' t ' = =
I
O
. 5 y (- 0.5e) (22.0e . y) e 1 - (- 0.52 e? e _
In a simple thought experiment, Einstein showed that there is mass
associated with electromagnetic radiation. Consider a box of length L and mass M resting
on a frictionless surface. At the left wall of the box is a light source that emits radiation of
energy E, which is absorbed at the right wall of the box. According to classical
electromagnetic theory, this radiation carries momentum of magnitude p
= Ele (Equation
30- 1 3). (a) Find the recoil velocity of the box so that momentum is conserved when the
light is emitted. (Since p is small and M is large, you may use classical mechanics.) (b)
Relativity 42 1 When the light is absorbed at the right wall of the box the box stops, so the total momentum remains zero. If we neglect the very small velocity of the box, the time it takes for the radiation to travel across the box is 111
Lie. Find the distance moved by the box in this time. (c) Show that if the center of mass of the system is to remain at the same place, the radiation must carry mass m = Ele2 • =
Picture the Problem We can use conservation of energy to express the recoil velocity of
the box and the relationship between distance, speed, and time to find the distance
traveled by the box in time 111 = Lie. Equating the initial and final locations of the center of mass will allow us to show that the radiation must carry mass m = Ele2 . (a) Apply conservation of momentum to obtain:
E -+MV = Pi = 0 e
Solve for v:
vL e
(b) The distance traveled by the box
d = v!::..t = -
Substitute for v from (a):
d=
in time 111 = Lie is:
(c) Let x = 0 be at the center of the box and let the mass of the photon be m. Then initially the center of
x
CM
( )I
L _� = LE e Me Me 2 �
= -M.l+mLm
I
----=2'----_
mass is at:
When the photon is absorbed at the other end of the box, the center of mass is at: Because no external forces act on the system, these expressions for
XCM must be equal: Solve for m to obtain:
-{mL M+m
[ -MeMEL2 + m("2 L EL )] 1
_
M+m
E
- ];k2
Chapter 3 9
422
Because and E
=
Mc2 is of the order of 1 0 16 J
hfis of the order of 1 J for
reasonable values off and: *55
EIMc2 «
1
When a projectile particle with kinetic energy greater than the threshold kinetic energy Kth strikes a stationary target particle, one or more particles may be created in the inelastic collision . Show that the threshold kinetic energy of the projectile is given by ...
Here Lmjn is the sum of the masses of the projectile and target particles, Lmfin is the sum of the masses of the final particles, and mtarget is the mass of the target particle. Use this
expression to determine the threshold kinetic energy of protons incident on a stationary proton target for the production of a proton-antiproton pair; compare your result with the result of Problem 40. Picture the Problem Let mj denote the mass of the incident (proj ectile) particle. Then Lmjn
=
j
m + mtarget and we can use this expression to determine the threshold kinetic
energy of protons incident on a stationary proton target for the production of a proton antiproton pair. Consider the situation in the center of mass reference frame. At threshold we have:
E 2 - p 2 2 = '"' � mfin c 2 C
Note that this is a relativistically invariant expreSSIOn.
In the laboratory frame, the target is at rest so: We can, therefore, write: For the incident particle:
E 2 - po2 2 E'20 .
1
I
C
=
I,
and
Ej = Ej O + Kth
where Kth is the threshold kinetic energy of the incident particle in the laboratory frame. Express Kth in terms of the rest energIes:
(Et,o + Ej•O ) 2 + 2Kth Et,0 = (I mfin c2 Y
where
Relativity 4 2 3 E1,0 + Ei. O =
and
L
n1fill c
2
Substitute to obtain: Solve for Kth to obtain:
mill +
For the creation of a proton - antiproton pair in a proton proton collision:
L mfin L mfin - L min
c2
2 m larget
L min = 2 mp L mfin = 4 mp
and
(
J(
J
2 mp + 4 m 4 mp - 2 m c2 Kth = --'------'-----'--'-----'----'--2 mp
Substitute to obtain:
�
(6rnp
��
)
rn, c' p
�
I 6rnpc' I
in agreement with Problem 40. *59
•••
For the special case of a particle moving with speed
u
along the y axis in
frame 5, show that its momentum and energy in frame S' , a frame that is moving along
the x axis with velocity v, are related to its momentum and energy in 5 by the transformation equations
( (
Px = r Px
,
E'
c
=r
VE
-
7
) )
E vPx
c
_
c
, Py
,
,
= Py ' pz = pz
.
Compare these equations with the Lorentz transformation for x ' , y' , Zl, and t' . These
equations show that the quantities px, Py, pz, and Elc transform in the same way as do x, y,
z,
and ct.
Picture the Problem We can use the expressions for
p and E in 5 together with the
relations we wish to verify and the inverse velocity transformation equations to establish
424
Chapter 3 9
the condition
U,2
=(
U/) 2 + (U/) 2 + (U/ ) 2
= �r2 v2 +
and then use this result to verify
the given expressions for px' , py' , p/ and Eric. In any inertial frame the momentum and energy are given by:
_
P=
g c2 mu
2
1 --
where Ii is the velocity of the particle and u
The components of
p in S are:
Because Ux = Uz = 0 and uy = u :
is its speed.
Px
= :-2 ' = -2 g 11c2 g c2
Px =
and
mu
pz =
m uy
Py
, and
0
mu
Substituting zeros for Px and pz in
the relations we are trying to show yields:
px'=r(o-::) = -r ::, p/ = pE/ = E E ' = r( c c )=r c g g 1-c2 1 - -c2 g 1 -c2 0 , and
_
In S' the momentum components are:
, Px =
P/ =
The inverse velocity transformations are:
mu x
0
,
U ,2
mu ' Z
PY '
U '2
' PY
,
=
mu y '
U, 2
, and
Relativity 425
u: = �
Uz
1-
Substitute obtain:
Ux Uz =
=
vu ,
c-,-
U = -v ' U y 1
0 and u) ' = u to
X
1
=
yu and
,
uz 1 = 0
U , 2 = (U/) 2 (U /) 2 + (u/Y u2 = v2 r2
Thus:
+
+
First we verify that
Next we verify that
pz' pz =
P; C
mu l -� �G1
= [£J Next we verify that
pz - m (O)1 2 = pz = fOl � u 1 c2
0:
, _
j
py' = py:
1 mu y I Py = 2 U 1 2
-
=
mu 2 v2 - u 1 r c 2 2c 2
r
g� C
V
( J ( J
u2 - v2 - v2 1-1 c2 c2 c2 = 2 PY � 1 - �2 - �2 1 - �2 1- 2 c c c c
(l -2 �)2(l - �J J -�( 1-� c2 c 2
p/ = r( Px - ::
� -::
1 mu U r 1 - 2 - u2 1 2 e2 e2 r 2-
)
:
426 Chapte r 3 9
Px - !n u x 2 U 1-C2
g I
I -
Y
�
-I 1 - u: y 2 c m cJ'V 2 - U2 c - 1 Uc 2 1 - c2 y 2c2 ' 2 - ;2 2 ;1-� 1 c2 C C J'V E 2c v 2 - �2 - v2 1 1-c2 c2 c2
zg
- mv v - U_2 1--C 2 Y 2 C2 ?
V
1 - :: J ( - J ( ( J ---,:: -;---- C 2 E v 2 u 2 v2 1 - 2 - 2 (1 - 2 ( J J = -;E 1 I vPx . E' = y( -E - _ E E' = yE : = y ) c c c c g U -I -' �1 - U 2' 1 y 2 2 2 yE v 2 c 2 E I = !n C I 2 = rm C 2 g ' � = U 1-u g - -2 - 1 - Uc 2 g 1 -1 2 2 c c y 2c2 C (1 - -Uc?-2 J(1 - �c2 J 1 - -vc22 - �c22 (1 - -cv22 J = yE I 2 2 v2 = yE v 2 2 v 2 u (1 - -J -�( --1 -� 1-2 c C2 c2 c2 c 2 1 c 2 J -
.
_
J'V
1
----,----
�
_ _
Fmally, we venfy that
,
or
Y
-
= 1 yE 1
The x, y, z, and t transformation equations are:
X ' = y {x - vt)
y' = y Z' = z
( ) = y ( x - � ct )
and
v t = Y t - 2x I
The x, y, are:
-
z,
and ct transformation equations
X'
y' = y z' = z
and
(
ct' = y ct -� x
)
, ( - -; -;;E) ,
The Px, PY' Pz, and £/c transfonnation equations are:
Px = r Px
v
Py = Py ' p PZ = z
and E' c
Note that the transfonnation equations for x, y, for Ph PY' pz, and £/c are identical. *62
000
Z,
=r
(
E c
Relativity 427
_'!.- Px c
)
and ct and the transfonnation equations
A long rod that is parallel to the x axis is in free fall with acceleration g
parallel to the -y axis. An observer in a rocket moving with speed v parallel to the x axis passes by and watches the rod falling. Using the Lorentz transfonnations, show that the observer will measure the rod to be bent into a parabolic shape. Is the parabola concave upward or concave downward? We can use the inverse Lorentz transfonnation for time to show that the observer will conclude that the rod is bent into a parabolic shape.
Picture the Problem
In frame S where the rod is not
moving along the x axis, the height of the rod at time t is:
The inverse Lorentz time transfonnation is: Express y' (t) in the moving frame of reference:
(1) B ecause equation (1)is th e equation of a parabola, w e'v e shown that th e moving obs erv er conclud e that th e rod is b ent into a parabolic shap e. B ecaus e th e co effici ent of x2 is n egativ e,th e parabola is concav e downward.
Evaluate y'(t) at t' = 0 to obtain:
will
Chapter 40 Nuclear Physics Conceptual Problems *4
The half-life of 14C is much less than the age of the universe, yet14C is found in nature. Why?
Determine the Concept14C
is found on earth because it is constantly being formed by cosmic rays in the upper atmosphere in the reaction 14N + n -7 14C + IH.
Write and balance equations for each of the following nuclear decays: *13 (a) beta decay of I�, (b) alpha decay of248Fm, (c) positron decay of 12N, (d) beta decay of81Se, (e) positron decay of6lCu, and if) alpha decay of228Th. Knowing the parent nucleus and one of the decay products, we can use the conservation of charge, the conservation of energy, and the conservation of the number of nucleons to identify the participants in the decay.
Determine the Concept
(a) beta decay of1�
214008Fm�244Cf+ 4He + Q I 98 2
(b) alpha decay of248Fm
(c) positron decay of 12N
I · + 013 + + Q 6291C U�61N I 28 1 + 1 0 22908Th�22884Ra+ 4He + Q .I 2
0 0813 S e�381B 5 r+_1 13 + 0 v � Q
(d) beta decay of81Se
4
(e) positron decay of61Cu if) alpha decay of 228Th
°
V
Write and balance reaction equations for each of the following: (a) 240pU undergoes spontaneous fission to form two fission fragments and three neutrons. One of the fission fragments is a 90Sr nucleus. (b) A 72Ge nucleus absorbs an alpha particle and ejects a photon. (c) A 1271 nucleus absorbs a deuteron and ejects a neutron. (d) A235U nucleus absorbs a slow neutron and fissions forming a113 Ag nucleus, two neutrons, and another fission fragment. (e) A 55Mn nucleus is struck with a high-energy 7Li nucleus, resulting in a triton, 3H, and a new nucleus. if)238U absorbs a slow neutron resulting in a compound nucleus that emits a beta particle. What is the resulting nucleus? *14
·
We can use the information regarding the daughter nuclei to write and balance equations for each of the reactions.
Determine the Concept
429
430
Chapter
40
Properties of Nuclei *17
238U.
40-1
·
Calculate the binding energy and the binding energy per nucleon from the masses given in Table for (a) 12C, (b) 56Fe, and (c) To find the binding energy of a nucleus we add the mass of its neutrons to the mass of its protons and then subtract the mass of the nucleus and multiply by c2 . To convert to MeV we multiply this result by 931.5 MeV/u. The binding energy per nucleon is the ratio of the binding energy to the mass number of the nucleus.
Picture the Problem
(a)For 12C, Z = 6 and N= 6. Add the mass of the neutrons to that of the protons:
6mp +6mn 6x1.007825u+6xl.008665u 12.098940u =
=
Subtract the mass of 12C from this result:
6mp +6m,J -m'2c 12.098940u -12u 0.098940u
(
=
=
Multiply the mass difference by c2 and convert to MeV:
Eb (�m )c2 (0.098940u)c2 x 931.5�ev/c2 I 92.2MeV I . . . Eb 92.2MeV I 7.68MeV I and the bmdmg energy per nucleon A 12 =
=
=
IS
=
=
(b)For56Fe, Z = 26 and N= 30. Add the mass of the neutrons to that of the
protons:
26mp +30mn 26x1.007825u+30x1.008665u 56.463400u =
Subtract the mass of 56Pe from this result:
=
Nuclear Physics 431 (26m p + 30m n ) -m" = 56.463400u -55.934942 u 0.528458u Multiply the mass difference by c2 and convert toMeV: (�m)c2 (0.528458u)c 2 931.51Meu V/ c2 = !. 492MeV ! and the binding energy per nucleon is = 49256MeV = 18.79Me · V I. (c)For238U, and Add the mass of the neutrons to that of the =
-c
E
b
=
=
x
.
E _b A
Z = 92
N= 146.
protons:
92mp +l46mn 92x1.007825u +146x1.008665u 239.984990 u =
=
Subtract the mass of238U from this result:
(92mp +l46mn)-mmu 239.984990u -238.050783u 1.934207u Multiply the mass difference by c2 and convert toMeV: (�m)c2 (1.934207u)c2 x 931.5�eV/ c2 11802MeV I 1802MeV I 7.57MeVI . and the binding energy per nucleon 238 · =
E b
=
=
=
=
.
1S
E _b A
=
=
e+
*21 ·· The neutron, when isolated from an atomic nucleus, decays into a proton, electron, and an antineutrino as follows: �n�:H+_� gv. The thermal energy of a neutron is of the order of kT, where k is the Boltzmann constant. (a) In both joules and electron volts, calculate the energy of a thermal neutron at 25°C. (b) What is the speed of this thermal neutron? A beam of monoenergetic thermal neutrons is produced at 25°C with intensity 1. After traveling 1350 lan, the beam has an intensity of I12. Using this information, estimate the half-life of the neutron. Express your answer in minutes.
(c)
Picture the Problem The speed of the neutrons can be found from their thermal energy. The time taken to reduce the intensity of the beam by one-half, from 1 to 112, is the half life of the neutron. Because the beam is monoenergetic, the neutrons all travel at the same speed.
432 Chapter 40 (a) The thermal energy of the
neutron is:
Ethen11"1 = = (1.38 x 10-23 11K)(25 273)K 1 4.11x10-21J I l eV 4.1lxlO-21Jx 1.60xlO-1 9J = 1 25.7m eV I Ethermal = 2 2Ethermal kT
+
=
=
Ethermal
(b) Equate and the kinetic energy of the neutron to obtain: Solve for v to obtain:
I
2: mn V
v=
Substitute numerical values and evaluate v: (c) Relate the half-life, t1/ ' to the 2 speed of the neutrons in the beam: Substitute numerical values and evaluate tl/ : 2
2 4.11xlO-2I J 7 1.67 x10-2 I 2.22 s I
v=
kg
=
kml
1350 = 608s x Imin /12 2.22 60s s 1 10.1min I km km1
=
t
=
*24 ·· In 1920, twelve years before the discovery of the neutron, Rutherford argued that proton-electron pairs might exist in the confines of the nucleus in order to explain the mass number, A, being greater than the nuclear charge, Z. He also used this argument to account for the source of beta particles in radioactive decay. Rutherford's scattering experiments in 1910 showed that the nucleus had a diameter of about 10 fm. Using this nuclear diameter, the uncertainty principle, and given that beta particles have an energy range of 0.02MeV to 3 MeV, show why electrons cannot be contained within the nucleus.
.40
Picture the Problem
1 2
TheHeisenberg uncertainty principle relates the uncertainty in
position, Llx, to the uncertainty in momentum, I1p, by Solve theHeisenberg equation for I1p:
I1.P>::;j
Substitute numerical values and evaluate I1p:
11.
� � .p =
&l1.p;:::
- ti.
ti 2&
1.05xlO-34J·s 2�OxlO-15m) 5.25 x10-21 . m/s kg
Nuclear Physics 433 The kinetic energy of the electron is given by: Substitute numerical values and evaluate K:
K =pc
5.25x l O 1 kg .m/s )(3xl08rn /s ) leV 1.58xlO-12Jx 1.60xlO-1 9 9.88MeV
K= = =
1 1000
(
-
2
J
This result contradicts experimental observations that show that the energy of electrons in unstable atoms is of the order of to eV. Radioactivity *31 ·· Plutonium is a highly hazardous and toxic material to the human body. Once it enters the body it collects primarily in the bones, although it can also be found in other organs. Red blood cells are synthesized within the marrow of the bones. The isotope 239pU is an alpha emitter with a half-life of years. Since alpha particles are an ionizing radiation, the blood-making ability of the marrow is, in time, destroyed by the presence oe39pu. In addition, many kinds of cancers will also be initiated in the surrounding tissues by the ionizing effects of the alpha particles. (a) If a person accidentally ingested Jig of239pU and it is absorbed by the bones of the victim, how many alpha particles are produced per second within the skeleton? (b) When, in years, will the activity be alpha particles per second?
24,360
2.0 1000
A,
Picture the Problem Each 239pU nucleus emits an alpha particle whose activity, depends on the decay constant of239pU and on the number N of nuclei present in the ingested239PU. We can find the decay constant from the half-life and the number of nuclei present from the mass ingested and the atomic mass of239pu. Finally, we can use the dependence of the activity on time to find the time at which the activity be alpha particles per second.
(a) The activity of the nuclei present in the ingested239pU is given by: Find the constant for the decay of 239pU:
A=AN
A
_
-
=
Express the number of nuclei present in the quantity of 239pU ingested: Substitute numerical values and evaluate N:
1000 (1)
1n 2 - 0.693 (24360y)(31.56Ms/y) tl/ 2 9.02xlO-13s-1 _
N= mPu
_ N
A_
M Pu
where Mpu is the atomic mass of 239PU. N= =
u lei/mol ] (2.0 j.Lg) ( 6.02xl239023gnlmol 5.04xlOl5nuclei c
434 Chapter 40 Substitute numerical values in
A
equa ti on(I) and evaluate A:
= =
(9.02xlO-13s-I)(5.04xlOI5a) 14.55xl03 a/s I
(b) The activity varies with time
according to:
Solve for t to obtain:
Substitute numerical values and evaluate t:
The fissile material 239pU is an alpha emitter. Write the equation of this reaction. Given that 239pU, 235U, and an alpha particle have respective masses of 239.052 156 u, 235.043 923 u, and 4.002 603 u, use the equations appearing inProblem 32 to calculate the kinetic energies of the alpha particle and the recoiling daughter nucleus. *33
··
A
We can write the equation of the decay process by using the fact that the post-decay sum of the Z and numbers must equal the pre-decay values of the parent nucleus. The Q value in the equations fromProblem 32 is given by Q = -(f1m)c2• Picture the Problem
239pU undergoes alpha decay
according to:
The Q value for the decay is given by: Substitute numerical values and evaluate Q:
Q [ (239.052156u) -(23S.043923u + 4.002603U)](93 1.::eV ) IS.24MeV I FromProblem the kinetic energy ( A�4 )Q of the alpha particle is given by: Substitute numerical values and =( 2��4 )S.24MeV) evaluate = I S.lSMeV I =
=
32,
Ka:
K K
a
a
=
From Problem 32, the kinetic energy of the J39U IS ' given b y:
-
.
K
U
=
Nuclear Physics 435
4Q A
Substitute numerical values and evaluate Ku: *36 ·· Radiation has long been used in medical therapy to control the development and growth of cancer cells. Cobalt-60, a gamma emitter of 1.17 and 1.33MeV energies, is used to irradiate and destroy deep-seated cancers. Small needles made of 60Co of a specified activity are encased in gold and used as body implants in tumors for time periods that are related to tumor size, tumor cell reproductive rate, and the activity of the needle. (a) A 1.00 f.1.g sample of 60 Co, of half-life of 5.27 y, is prepared in the cyclotron of a medical center to irradiate a small internal tumor with gamma rays. In curies, determine the initial activity of the sample. (b) What is the activity of the sample after 1.75 years?
R = Roe-A.t
Picture the Problem
We can use
Ro =
AN to find the initial activity of the sample and
to find the activity of the sample after 1.75 y.
(a) The initial activity of the sample is the product of the decay constant 60 It for Co and the number of atoms N of 60 Co initially present in the sample: Express N in terms of the mass m of the sample, the molar mass M of 60Co, and Avogadro's number NA:
R
-AN ° -
N
=.!!!..M
(1)
NA
Substitute numerical values and evaluate N: N
=(1.6000Xg/10mol-6g ](6.02x1023nuclei/mol) 1.00x1016 nuclei =
The decay constant is given by:
Substitute numerical values and evaluate It:
Ao:
Substitute numerical values in equation (1) and evaluate
0.693 (5.27 y )(31.56Ms /y ) =4.l7x10-9S-1 Ro =(4.17x10-9 )(1.00x1016nuclei ) 1Ci =4.l7x107s-IX 3.7x1010s -1 =! 1.13mCi !
A
=
S
-I
436
Ca e
h pt r 40
(b) The activity varies with time according t o: Evaluate R at
I
=
R
=
1.75 y:
R
= =
*41
··
Roe-A1
=
( a.69JI)
. R0e 5 27y
mCi) e I0.898mCi I (1.13
( a.693xI.75Y ) 5.27y
The counting rate from a radioactive source is measured every minute. The
resulting counts per second are 1000, 8 20, 673, 552, 453, 371, 305, and 250. Plot the
counting rate versus time on semilogarithmic graph paper, and use your graph to find the half-life of the source. Picture the Problem
The following graph was plotted using a spreadsheet program.
Excel's "Add Trendline" feature was used to determine the equation of the line.
7.0 6.8 6.6 6.4 6.2
� ..=
6.0 5.8 5.6 5.4 5.2 5.0
0
2
3
5
4 t
6
7
(min)
The linearity and negative slope of this graph tells us that it represents an exponential decay.
The decay rate equation is:
R - Rae-AI
Take the natural logarithm of both
lnR
sides of the equation to obtain: This equation is of the form:
= =
lne-AI + InRa -At+lnRa
y = mx+b where y = In R,
x
=
I, m
= -A, and
Nuclear Physics 437 b
The half-life of the radioisotope is:
= InRo.
tl/2 =
In 2 }L,
-
=
In 2 I 3.50mm I 0.198mm .
_I
=
.
Nuclear Reactions *47
for
I� N
••
(a) Use the atomic masses
m = 14.003 242 u for I:C and m = 14.003 074 u
to calculate the Q value(inMeV) for the fJ decay
14C�14N +fJ- +v 6
7
e
(b) Explain why you do not need to add the mass of the fJ- to that of atomic N for this calculation. Picture the Problem
)
We can use Q = -(L'1m c2 to find the Q values for this reaction.
(a) The masses of the atoms are:
14.003242u 14.003074u
m '4C = m'4N =
Calculate the increase in mass:
14.003074u -14.003242u -0.000168u -(L'1m) -(-0.000168 )C' (931.5Me�1 c' J I 0.156MeV I
= =
Calculate the Q value:
Q= = =
c2
U
The masses given are foratoms,not nuclei,so fornuclear masses the masses are too large bythe atomic numbertimes the mass of an electron. For the given nuclearreaction, the mass of the carbon atom is too large by 6me and the mass of the nitrogen atom is too large by 7me. Subtracting 6me from both sides of the reaction equation leaves an extra electron mass on the right. Not including the mass of the beta particle (electron)is mathematicallyequivalent to explicitlysubtracting 1me from the right side of the equation.
(b)
438 Chapter 40 Fission and Fusion *49
Assuming an average energy of 200MeV per fission, calculate the number of
•
fissions per second needed for a 500-MW reactor. Picture the Problem
The power output of the reactor is the product of the number of
fissions per second and energy liberated per fission. Express the required number N of
N=
fissions per second in terms of the power output P and the energy
released per fission Eper fission:
P ---
Eper fission
500MW N= 200 1 5x108 .:!.x 1.60xl019J 200 =1 1.56x1019 S-1
Substitute numerical values and evaluate N:
MeV
eV
s
MeV
1 9
*51
··
139L +2on1 + Q
Consider the following fission process:
2 + on --*425 M0+ 9235U
57
a
. The masses 0f the neutron, U,M0, and La are
1.008 665 u, 235.043 923 u, 94.905 842 u, and 138.906 348 u, respectively. Calculate the Q-value, inMeV, for this fission process. Compare the result to the result obtained in Problem 23. Picture the Problem
We can use Q
value. The Q value is given by:
=
-(t::.m)c2, where /)"m
Q= (
)
= mf - mi,
- �m c2
x
to calculate the Q
931.5 lu
MeV/c2
Calculate the change in mass /)"m:
=94.905842 u + 138.906348 u + 2(1.008665 u)-(235.043923 u 1.008665 u) =-0.223068 u Q =-(-0.223068 u)x 931.lu5 =1208 1
� m = mf - mi
+
Substitute for /)"m and evaluate Q:
MeV
MeV
Nuclear Physics
The ratio of Q to U found in Problem 23 is: *54
•••
Q U
=
208 MeV 1 88 236MeV =
.1%
439
1
The fusion reaction between 2H and 3H is
Using the conservation of momentum and the given Q value, find the final energies of both the 4He nucleus and the neutron, assuming the initial kinetic energy of the system is 1.00MeV and the initial momentum of the system is zero. We can use the conservation of momentum and the given Q value to find the final energies of both the 4He nucleus and the neutron, assuming that the Picture the Problem
initial momentum of the system is zero. Apply conservation of energy to obtain:
18.6MeV
tmHeV�e +tmnv� = KHe +Kn =
(1) (2 )
Apply conservation of momentum to obtain: Solve equation (2) for VHe:
Substitute for
v�e in equation (1):
Solve for Kn:
Kn
=
18.6MeV mn 1 +-mHe
Substitute numerical values for mn and mHe and evaluate Kn:
Kn
=
18.6MeV I 14.86MeV I 1 + 1.008665 4.002603u u
=
440
Chapter 40 KHe
Use equation (I) to find KHe:
=
18.6MeV - Kn
= 18.6MeV -14.86MeV =
1 3.74MeV I
General Problems *57
·
The counting rate from a radioactive source is 6400 counts/so The half-life of
the source is lO s.Make a plot of the counting rate as a function of time for times up to
1 min. What is the decay constant for this source? Picture the Problem
We can use the given information regarding the half-life of the
source to find its decay constant. We can then plot a graph of the counting rate as a function of time. The decay constant is related to the
A
half-life of the source:
=
In 2 In 2 I 0.0693s -1 I tl/2 lOs =
=
The activity of the source is given by: The following graph of R program.
=
6400Bq)e-0.0( 6 93S-I)t
(
was plotted using a spreadsheet
7000 6000 5000 4000 E9.. :: 3000 0'
;r..;
2000 1000 0
*61
··
0
10
20
30
t (5)
40
50
60
Show that the10 9 Ag nucleus is stable against alpha decay,
1��Ag��He+I��Rh + Q. The mass of the10 9 Ag nucleus is 108.904 756 u, and the
products of the decay are 4.002 603 u and 104.905 250 u, respectively.
Nuclear Physics 441 Picture the Problem We can show that 109Ag is stable against alpha decay by demonstrating that its Q value is negative.
The Q value for this reaction is:
Q=
-[(mRh +mJ -mAJ c2 (93l.5 M_ euV_ _Ic_2 J o
Substitute numerical values and evaluate Q:
Q= =
-[ (4.002603 u +104.905250 u) -108.904756 u](931.5MeV /u) 1 -2.88 MeV I
Remarks: Alpha decay occurs spontaneously and the Q value will equal the sum of the kinetic energies of the alpha particle and the recoiling daughter nucleus,
Q
=
Ka
+
K D Kinetic energy cannot be negative; hence, alpha decay cannot occur •
unless the mass of the parent nucleus is greater than the sum of the masses of the alpha particle and daughter nucleus, mp > ma + mD • Alpha decay cannot take place unless the total rest mass decreases.
*66
··
(a) Determine the closest distance of approach of an 8-MeV
a
particle in a
lO head-on collision with a nucleus of 197Au and a nucleus of B, neglecting the recoil of the
struck nuclei. (b) Repeat the calculation taking into account the recoil of the struck
nuclei. Picture the Problem
We can solve this problem in the center of mass reference frame for
the general case of an
a
particle in a head-on collision with a nucleus of atomic mass M u
lO and then substitute data for a nucleus of 197Au and a nucleus of B. In the eM frame, the kinetic energy IS:
KeM
m 1+� _ kqlq2 - k(2e)(Ze) - ke22Z 2 ke (1.44MeV ·fm)(2Z) =
Klab
M
At the point of closest approach:
KeM
R min
or, because
KeM Solve for Rmin to obtain:
_
_
-
=
R Olin
=
1.44MeV·fm,
. Rmm
. (1.44MeV .fm)(2Z)
R rrun
=
R min
KeM
(1)
442 Chapter 40 (a) Neglecting the recoil of the target nucleus is equivalent to replacing KCM by Klab. Evaluate equation (1) for 197Au: lO Evaluate equation (1) for B:
(b) Find KCM for the 197Au nucleus:
Substitute numerical values in equation (1) and evaluate Rmin:
(1.44MeV .fm)(2x 79) 8MeV =I 28.4fm I .fm)(2x 5) = (1.44MeV 8MeV =11.80fm I 8MeV =7.841MeV KCM = 4u 1+-197u .fm)(2x79) = (1.44MeV 7.84lMeV =129.0fm I .
R 01111
=
.
R mm
Rnun .
Note that this result is about 2% greater that Rmin calculated ignoring recoil. lO Find KCM for the B nucleus:
Substitute numerical values in equation (1) and evaluate Rmin :
KCM
.
R mill
= 8MeV 4u =5.7l4MeV 1+lOu .fm)(2x5) = (1.44MeV 5.714MeV =12.52fm I
Note that this result is about 40% greater that Rmin calculated ignoring recoil. The total energy consumed in the United States in 1 y is about 7.Ox 1019 J. How many kilograms of23SU would be needed to provide this amount of energy if we *70
··
assume that 200MeV of energy is released by each fissioning uranium nucleus, that all of the uranium atoms undergo fission, and that all of the energy-conversion mechanisms used are 100 percent efficient?
Picture the Problem
The mass of235U required is given by m235
=
N NA
M235, where
M23S is the molecular mass of235U and N is the number of fissions required to produce 7.0xl019 J.
Nuclear Physics 443 Relate the mass of 235U required to the number of fissions N required:
N 1n 35 = N M 35 2 2 A
(1)
where M235 is the molecular mass of mU. Determine N:
N=
Substitute numerical values and evaluate N:
N= =
Eonnua! Eperfission
7.0x1019J 19J 200MeVx 1.60x10eV 2.18x1030
Substitute numerical values in equation(1) and evaluate mm:
1n235 *73
•••
=
2.1823x1030 . (235 glmol)=I 8.51xl05 kg I 6.02x10 nucleI/mol
Assume that a neutron decays into a proton plus an electron without the
emission of a neutrino. The energy shared by the proton and electron is then O.782 MeV. In the rest frame of the neutron, the total momentum is zero, so the momentum of the
proton must be equal and opposite the momentum of the electron. This determines the
relative energies of the two particles, but because the electron is relativistic, the exact calculation of these relative energies is somewhat difficult. (a) Assume that the kinetic energy of the electron is O.782MeV and calculate the momentum p of the electron in units of MeVic. (Hin t: Use Equation 39-28.) (b) Using your result from Part (a) , calculate the kinetic energy p2/2mp of the proton. (c) Since the total energy of the electron plus the proton is 0.782 MeV, the calculation in Part (b) gives a correction to the assumption that the energy of the electron is 0.782 MeV. What percentage of 0.782MeV is this correction? Picture the Problem
E 2 = p2C2
+ E�
The momentum of the electron is related to its total energy through
and its total relativistic energy E is the sum of its kinetic and rest
energies.
(a) Relate the total energy of the
(1)
electron to its momentum and rest energy: The total relativistic energy E of the electron is the sum of its kinetic
E=K + Eo
444 Chapte 40 r
energy and its rest energy:
- c-
(K + Eo )? = p 2? + Eo2
Substitute for E in equation (1) to obtain: Solve for p:
Substitute numerical values and evaluate
p
=
p:
�(0.782MeV)(0.782MeV 2x 0.511MeV) = !1.188MeV/ c ! c
(b)Because Pp
+
=
-Pe:
Kp
Substitute numerical values (see Table
7-1
Kp
for the rest energy of a
proton) and evaluate Kp:
Kp
(c) The percent correction is:
K
*77
•••
2
2m = 1.188MeV/ cY = ! 752eV I 2938.28MeV/ c2 . . =� p
=
752eV =! 0.0962% I 0.782MeV
Frequently, the daughter of a radioactive parent is itself radioactive. Suppose
the parent, designated by A, has a decay constant AA; while the daughter, designatedB, has a decay constant
AB.
differential equation
dNBldt
= AANA
The number of nuclei ofB are then given by the solution to the
- AaNB
(a) Justify this differential equation.
(b) Show that the solution for this equation is
where NAO is the number of A nuclei present at t (c) Show that NBCt)
>0
a function of time when
whether
TB
AA > AB
= 3TA.
or AB
Picture the Problem We can differentiate
=
0
when there are noB nuclei.
> AA.
(d) Make a plot of NA(t) and NB(t) as
( )=
NA AA O -AA
NB t
As
to t to show that it is the solution to the differential equation
(e-A.At - e-A.st) with respect
Nuclear Physics 445
(a) The rate of change of
NB is the rate of generation ofB nuclei minus the rate of decay
ofB nuclei. The generation rate is equal to the decay rate of
AANA. The decay rate ofB nuclei is AsHB.
A
nuclei, which equals
dNB =AA NA _:l B N dt
(b) We're given that:
(1)
/"8
(3)
Differentiate equation
(2)
with respect to
Substitute this derivative in equation
Multiply both sides by
As - AA AsAA
(1)
t to obtain:
to get:
and simplify to obtain:
= NAO [- AAe-AA1 Ase-As1] As +
which is an identity and confirms that equation
is the solution to equation
(1).
AA As the denominator and the expression in the parentheses are both negative for t O. If AA As the denominator and the expression in the parentheses are both positive for t O. If
(c)
(2)
>
>
446 Chapter 40 (d)
The following graph was plotted using a spreadsheet program. 1.0 "1.
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-- _ :1.'��..�_.::
lotio.
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0.9 0.8
1'.1
.;o..,,�.
0.7
1,\
_;;�,.,.
0.6
1',1,
.,
0.5
I- 'I':" ,
0.4
I;\l..
0,3
I
0.2
1."" �1t
0.0
I
o
"
'\ .
,",., -;'; ..
\.-
'=
, ",or' '
'l'1umOt:rO(Uilllgmt:fl 0) nU(;It::l
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