Student Solutions Manual for Modern Physics Third Edition PDF

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Student Solutions Manual for Modern Physics Third Edition Raymond A. Serway Professor Emeritus, James Madison University Clement J. Moses Professor Emeritus, Utica College of Syracuse University Curt A. Moyer University ofNorth Carolina-Wilmington T H O M S O N *- BROOKS/COLE Australia • Canada • Me...

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CHAPT ER 21: Elect ric Charges and Elect ric Field Responses t o Quest ions Paola Cabay Elect ric Fields CHAPT ER OUT LINE Laura Orozco Sol. Serway Fisica. DannieL Morenoo, Francis Suasua

Student Solutions Manual for

Modern Physics Third Edition

Raymond A. Serway Professor Emeritus, James Madison University Clement J. Moses Professor Emeritus, Utica College of Syracuse University Curt A. Moyer University ofNorth Carolina-Wilmington

T H O M S O N

*BROOKS/COLE

Australia • Canada • Mexico • Singapore • Spain • United Kingdom • United States

COPYRIGHT © 2005 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license. ALL RIGHTS RESERVED. No part of this work covered by the copyright hereon may be reproduced or used in any form or by any means—graphic, electronic, or mechanical, including but not limited to photocopying, recording, taping, Web distribution, information networks, or information storage and retrieval systems—without the written permission of the publisher. Printed in the United States of America 1 2 3 4 5 6 7 08 07 06 05 04

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Contents Chapter 1: Relativity I Chapter 2: Relativity II Chapter 3: The Quantum Theory of Light Chapter 4: The Particle Nature of Matter Chapter 5: Matter Waves Chapter 6: Quantum Mechanics in One Dimension Chapter 7: Tunneling Phenomena Chapter 8: Quantum Mechanics in Three Dimensions Chapter 9: Atomic Structure Chapter 10: Statistical Physics Chapter 11: Molecular Structure Chapter 12: The Solid State Chapter 13: Nuclear Structure Chapter 14: Nuclear Physics Applications Chapter 15: Particle Physics

f

%

1 Relativity I

1-1

F =—. Consider the special case of constant mass. Then, this equation reduces to FA = maA at in the stationary reference system, and v B = v A + vBA where the subscript A indicates that the measurement is made in the laboratory frame, B the moving frame, and vBA is the velocity of B with respect to A. It is given that a : = —-^-. Therefore from differentiating the velocity at equation, we have aB = a A + aj. Assuming mass is invariant, and the forces are invariant as well, the Newton's law in frame B should be J]F = maA = maB -mav which is not simply ma B . So Newton's second law £ F = W 3 B is invalid in frame B. However, we can rewrite it as X F+maj =ma B , which compares to £ F+mg = m&B. It is as if there were a universal gravitational field g acting on everything. This is the basic idea of the equivalence principle (General Relativity) where an accelerated reference frame is equivalent to a reference frame with a universal gravitation field.

1-3

IN THE REST FRAME: In an elastic collision energy and momentum are conserved. m/s) + (0.2kg)(-3 m/s) = 0.9 kg m/s

jji=m1vli+m2v2i=(03kg)(5 Pt=m1vli+m2v2i

This equation has two unknowns, therefore, apply the conservation of kinetic energy Ei =E{ =—WjUji +—m2v2i =—m1Vif H—m2v2f and conservation of momentum one finds that Zi

£i

Jl

/U

u lf = -1.31 m/s and v^ = 6.47 m/s or vlf = -1.56 m/s and v^ = 6.38 m/s. The difference in values is due to the rounding off errors in the numerical calculations of the mathematical quantities. If these two values are averaged the values are v1{ = -1.4 m/s and o a = 6.6 m/s, p f =0.9 kg m/s. Thus, y>x =p{. IN THE MOVING FRAME: Make use of the Galilean velocity transformation equations. p{ -m^'^+m^o'^; where v '\\ ~ vn ~ v'= 5 m/s ~ ( - 2 m/s) = 7 m/s. Similarly, v'2i = -1 m/s and p[ =1.9 kg • m/s. To find p'f use v'lt = vVl - v and v'2i = v2i-v' because the prime system is now moving to the left. Using these results give p'{ = 1.9 kg • m/s.

1

CHAPTER 1

RELATIVITY I

This is a case of dilation. T = yT in this problem with the proper time T = T0 -1/2

121V2

L 0 /2

in this case T = 2T0, v = \ 1 •

21V2

Tn=>- =

1-1 f-

r

1 -1 - jl

therefore v = 0.866c.

The problem is solved by using time dilation. This is also a case of v « c so the binomial „ -,

expansion is used Af = yAt' = 1 + -

2c^

„

Af',Af-Af' =

2c2

,- ; i> =

2c 2 (Af-AO Af

-,1/9

A* = (24 h/ d a y)( 3 600 s/h) = 86 400 s; At = At' - 1 = 86 399 s; 1/2

2(86 400 s-86 399 s) v= 86399 s L

= 0.004 8c = 1.44 xlO 6 m/s.

- ^

^earth

_

^earth

-

2lV2

, L', the proper length so L earth =L = L[l -(0.9) 2 ] 1/2 = 0.436L.

^

At = yAt' A

„2A _ 1 /2

/

2 \

i+

At = At'

(4.0 xlO 2 m/s) 2

(3 600 s)

2(3.0 xlO 8 m/s)'

= (1 + 8.89 x 10"13 )(3 600 s) = (3 600 + 3.2 x 10~9) s At - At' ~ 3.2 ns. (Moving clocks run slower.) (a)

T=yr'=[l-(0.95)2]

(b)

Af•' =

1/2

(2.2/zs) = 7.05//s

d 3 x l 0 3 m , „ 1n _ 5s . , = = 1.05x10 s, therefore, 0.95c 0.95c N = N0 expf- —) = (5 x 104 muons) exp(-1.487) -1.128 x 104 muons.

(a)

For a receding source we replace v by —v in Equation 1.15 and obtain:

tic-vf'2 /ob

\[c + vf' V

[1-Vc]1/2

2

V

[ l + l;/c]V2

c

•"-3K>-

2 >

+ —=

1

Source

/source

—

1

l/i

2

4c

where we have used the binomial expansion and have neglected terms of second and higher order in - . Thus, -^— ^

/source

= iob

^source = - /source

**

MODERN PHYSICS

1-17

(b)

From the relations f =—,-*- = — T we find — = —'-—dX, or — = —— = — 1 X dX X2 f c/X X f c

(c)

Assuming v«c,

(a)

Galaxy A is approaching and as a consequence it exhibits blue shifted radiation. From 7, 21 - fi u (550nm) 2 -(450nm) 2 -^source "~o Example 1.6, p=- = 2| o u r c e 22°bs s o t h a t P: 2- = 0.198. Galaxy T A c I source + (550nmr + (450nm) z "" ""obs A is approaching at v = 0.198c.

(b)

For a red shift, B is receding, /3-

— B—— ,or VB\ —— c = c = 0.050c = 1.5 x 10 m/s. v c X \ X ) V397nm;

6

71

r

32

— ^'

source "obs i2 -i. I2 source "*" ^obs

SQ

t h a t

_ (700nm) 2 -(550nm) 2 „ „2 3„7 „ „ , „. ,. t 5= '-z— V2 = °• Galaxy 2 J B is receding b at v = 0.237c. (700nm) +(550nm) 1-19

KjM

= -u „ ; M ^ = 0.7c = .

Mx4

" " , ; 0.70c = 2

1-"XA"XBA '

"

2MX4 1 + ("XA/C)

or OJOwf^ - 2cuXA +0.7c2 = 0. 2

Solving this quadratic equation one finds w ^ = 0.41c therefore u^ = -Ux/i = ~0-41c. 1-21

u'x =

ux-v , . 2

l-uxv/c 1-23

(a)

0.50c-0.80c

. , =-0.50c l-(0.50c)(0.80c)/c 2

Let event 1 have coordinates x^ - yx = Zj = fj = 0 and event 2 have coordinates x2 =100 mm, y 2 =z 2 = t2 =0.1n S', x^ = y(x1 -vt1) = Q, y{ =y a = 0 , zj =z x =0,and ..2 "T 1 / 2

t{ = rh-

= 0,with y-

rn

systemS', x'2 = y(x2-vt2) t'2 = 7h

1-25

and so y = [l -(0.70) 2 ]" 1/2 = 1.40. In

= l'i0 m, y2 = z 2 = 0 , a n d

-\-j:\X2

(1.4)(-0.70)(100 m) = -0.33//s. 3.00xl0 8 m/s

(b)

Ax' = x 2 - ; t i = 1 4 0 m

(c)

Events are not simultaneous in S', event 2 occurs 0.33 jus earlier than event 1.

We find Carpenter's speed:

mGM

GM l1'2 R + h.

mv2

(6.67x10"" X5.98xlO M ) 6.37 x l 0 6 + 0 . 1 6 x l 0 6

1/2

= 7.82 km/s.

2x(R + h) 2^6.53 xlO 6 ) , '- = — ^ - — - ^ - = 5.25 x 103 s. Then the period of one orbit is T = =^ 7.82x10"

3

4

CHAPTER 1

RELATIVITY I .2 A -1/2

(a)

The time difference for 22 orbits is At - At' = (y -1)At' =

1-

c2J

- 1 (22)(T).

Using the binomial expansion one obtains '

(b)

lv2 ^ 1 + - 2- V - 1 (22)(7> 2- c

7.82 xlO 3 m/s 3 x10 s m/s

(22)(5.5xl0 3 s) = 39.2//s.

39.2 us For one orbit, At - At' = —'• = 1.78 //s » 2 //s. The press report is accurate to one significant figure.

1-27

For the pion to travel 10 m in time At in our frame, -1/2

10 m = vAt = v{yAt') = u(26 x 10"9 s) (3.85xl0 8 m/s)"

c2

1.46xlO17 m 2 / s 2 = u 2 ( l + 1.64) v = 2.37xl0 8 m/s = 0.789c 1-29

(a)

A spaceship, reference frame S', moves at speed v relative to the Earth, whose reference frame is S. The space ship then launches a shuttle craft with velocity v in the forward direction. The pilot of the shuttle craft then fires a probe with velocity v in the forward direction. Use the relativistic compounding of velocities as well as its inverse transformation: u'=- ur-v -, and its inverse u = - ur + v .The above

i-Mc2)

— - - i+(wxv/c2y

variables are defined as: v is the spaceship's velocity relative to S, u'x is the velocity of the shuttle craft relative to S', and ux is the velocity of the shuttle craft relative to S. Setting u'x equal to v, we find the velocity of the shuttle craft relative to the Earth to be: ur =2 l + (o/c) (b)

"

If we now take S to be the shuttle craft's frame of reference and S' to be that of the probe whose speed is v relative to the shuttle craft, then the speed of the probe 2v relative to the spacecraft will be, u'x = r-. Adding the speed relative to S yields: 1 + {v/c) \2 3v + v3/c3 3+(p/c)" . Using the Galilean transformation of velocities, we see «,= 2 l + 2v2/c2 l + 2(z;/c) that the spaceship's velocity relative to the Earth is v, the velocity of the shuttle craft relative to the space ship is v and therefore the velocity of the shuttle craft relative to the Earth must be 2v and finally the speed of the probe must be 3v. In the limit of low — , ux reduces to 3v. On the other hand, using relativistic addition of velocities, we find that ur = c when v —> c.

MODERN PHYSICS 1-31

In this case, the proper time is T0 (the time measured by the students using a clock at rest relative to them). The dilated time measured by the professor is: At = y T0 where At = T +t. Here T is the time she waits before sending a signal and t is the time required for the signal to reach the students. Thus we have: T +t = yT0. To determine travel time t, realize that the distance the students will have moved beyond the professor before the signal reaches them is: d v d = v(T +t). The time required for the signal to travel this distance is: t = — = —(T + t). Solving c c for t gives: t

—T 1—

. Substituting this into the above equation for (T +t) yields:

yT0, or T 1 —

= yT0. Using the expression for y this becomes: -1/2 V2 1+1-1^

-1/2

i-i^ 1-33

5

(a)

r0,orT:

*.U-f

'('-7

We in the spaceship moving past the hermit do not calculate the explosions to be simultaneous. We measure the distance we have traveled from the Sun as

L = LpJl-(-)

= (6.00 ly)Vl -(0.800) 2 = 3.60 ly.

We see the Sun flying away from us at 0.800c while the light from the Sun approaches at 1.00c. Thus, the gap between the Sun and its blast wave has opened at 1.80c, and the time we calculate to have elapsed since the Sun exploded is —

— = 2.00 yr. We

see Tau Ceti as moving toward us at 0.800c, while its light approaches at 1.00c, only 0.200c faster. We measure the gap between that star and its blast wave as 3.60 ly and growing at 0.200c. We calculate that it must have been opening for —

— = 18.0 yr

0.200c

and conclude that Tau Ceti exploded 16.0 years before the Sun (b)

1-35

Consider a hermit who lives on an asteroid halfway between the Sun and Tau Ceti, stationary with respect to both. Just as our spaceship is passing him, he also sees the blast waves from both explosions. Judging both stars to be stationary, this observer concludes that the two stars blew up simultaneously

In the Earth frame, Speedo's trip lasts for a time At = — = '•—-— = 21.05 Speedo's age v v v 6 v 0.950 ly/yr advances only by the proper time interval: At. - — = 21.05 yrVl - 0.95 2 = 6.574 yr during his 7 W trip. ° l y Vl-0.75 2 = 17.64 yr. While Speedo V K Similarly y for Goslo, Af_ v = — J l - \2 = v V c 0.750 ly/yr ' has landed on Planet X and is waiting for his brother, he ages by

20.0 ly 0.750 ly/yr

0.20 ly /, nnr2 z ,n,A —Vl-0.75 = 17.64 yr. 0.950 ly/yr

Then Goslo ends up older by 17.64 yr - (6.574 yr + 5.614 yr) = 5.45 yr.

6

CHAPTER 1

RELATIVITY I

1-37

Einstein's reasoning about lightning striking the ends of a train shows that the moving observer sees the event toward which she is moving, event B, as occurring first. We may take the S-frame coordinates of the events as (x = 0, y = 0, z = 0, t = 0 ) a n d ( x = 100 m, y = 0, z = 0, f = 0). Then the coordinates in S' are given by Equations 1.23 to 1.27. Event A is at (x' = 0, y' = 0,z' = 0,t' = 0). The time of event B is: 1

t' = r\t

vn0.8

l

0 - ^ ( 1 0 0 m) | = 1.667|

80 m = -4.44xl0"" 7 s. 3x10 s m/s

The time elapsing before A occurs is 444 ns. 1 M

1-39

, > (a)

- ., . „.. ^ c GMEm mv2 mflnrS? For the satellite 2, f = ma '• , — = =— r r r \ T J GMET'=4/rr

.2.3 1/3

6.67X10-11 N-m 2 (5.98xlO a * kg)(43080s)

= 2.66x10' m

2A~2

kg 4/r

(b) (c)

2^(2.66 xlO 7 m)

27tr v—

43080 s

= 3.87xl0 3 m/s

The small fractional decrease in frequency received is equal in magnitude to the fractional increase in period of the moving oscillator due to time dilation:

fractional change i n / = -(y-1)

= -

^-(S^xloVSxlO 8 )

f 3.87 xlO 3 ^ 2 1 = 1- (,1 2 — — \ 3x10°

I

(d)

= -8.34x10""

The orbit altitude is large compared to the radius of the Earth, so we must use GMEm U.

s

~

6.67 x l 0 ~ " N m ^ W S x l O 2 4 kg)m

6.67x10"" N m ^ S x l O 2 4 kg)m

kg22.66xl07m

kg 2 6.37xl0 6 m

= 4.76xl0 7 J/kgm A / _ A U g _ 4 . 7 6 x l 0 7 m 2 /s 2 / ~ rnc2 ~ ( 3 x i o 8 m/s) 2 (e)

,-1

= +5.29x10 -10

-8.34 x 10""" + 5.29 x 10""10 = +4.46 x 10 -10

2 Relativity II

2-1

p=

mv

2

[l-(, /c2)f2 (l.67xl0 _27 kg)(0.01c)

(a)

2 V2

:5.01xl0 -21 kg m/s

[l-(0.01c/c) ]

(b)

(l.67xl0 _27 kg)(0.5c) ^ — = 2.89xl0 _ w k g m / s 2 [l-(0.5c/c) ]

(c)

(l.67xl0- 27 kg)(0.9c) P r f w ^ =1-03x10-18 k g m / s [l-(0.9c/c) 2 ] 7

(d)

1.00 MeV

13

1.602x10"" J 2.998x10" m/s

5.34 x 10 ^ kg • m/s so for (a)

(5.01 xlO - 2 1 kgm/s)(100 MeV/c) 5.34 xlO - 2 2 kg m/s

= 9.38 MeV/c

SimUarly, for (b) p = 540 MeV/c and for (c) p = 1930 MeV/c. 2-3

As F is parallel to v, scalar equations are used. Relativistic momentum is given by p = ymv =

y=-, and relativistic force is given by

l-(v/cf]2lV2 '

mv dt

F=

dt

dp

dt

[[M'AOTJ m

cfo

[l-(t, 2 / c 2 )fHd£

7

10

CHAPTER 2

RELATIVITY II

2-17

Am = mRa - m ^ - m He (an atomic unit of mass, the u, is one-twelfth the mass of the 12C atom or 1.660 54xlCT27 kg) Am = (226.025 4 - 22.017 5 - 4.002 6) u = 0.005 3 u £ = (Am)(931 MeV/u) = (0.005 3 u)(931 MeV/u) = 4.9 MeV

2-19

Am = 6mp + 6m„ - m c = [6(1.007 276) + 6(1.008 665) -12] u = 0.095 646 u, AE = (931.49 MeV/u)(0.095 646 u) = 89.09 MeV. Therefore the energy per nucleon =

2-21

/E, e(-)

89 09 MeV = 7.42 MeV. 12

p

e(+)

O— O

K, p(e-) positron at rest

Conservation of mass-energy requires K + 2mc2 = 2E where K is the electron's kinetic energy, m is the electron's mass, and E is the gamma ray's energy. E = — + mc2 = (0.500 + 0.511) MeV = 1.011 MeV. 2 Conservation of momentum requires that pe = 2p cos 6 where p _ is the initial momentum of £

the electron and p is the gamma ray's momentum, — = 1.011 MeV/c. Using c E2. = p\.c2 + {mc2) where Ee_ is the electron's total energy, E - -K + mc2, yields

P

_=iVi^W ^ e

0 0 ) 2

c

^-OPXasii) MeV =1

m

c

Finally, cos 0 = ^ - = 0.703; 9 = 45.3°. 2p 2-23

In this problem, M is the mass of the initial particle, mx is the mass of the lighter fragment, vx is the speed of the lighter fragment, mh is the mass of the heavier fragment, and vh is the speed of the heavier fragment. Conservation of mass-energy leads to Mc 2 =-

^

•

^

^w ^W

MODERN PHYSICS : —-, = 7.04 x 10 -8 s. The current when electrons are The time of flight is At = — = 6 6 v 3.98 xlO m/s

2 8 c m a p a r t i s J = ^ = — = I 6 x l ° " I "-8C = 2 . 2 7 x l 0 - 1 2 A . t At 7.04xlO s

39

6 Quantum Mechanics in One Dimension 6-1

6-3

(a)

Not acceptable - diverges as x -» °°.

(b)

Acceptable.

(c)

Acceptable.

(d)

Not acceptable - not a single-valued function.

(e)

Not acceptable - the wave ...