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INSTRUCTOR SOLUTIONS MANUAL Preface and Acknowledgments This book contains solutions to the problems presented in Modern Physics for Scientists and Engineers, Fourth Edition by Thornton and Rex. It is intended for instructors only and is specifically not intended for general distribution to students...

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INSTRUCTOR SOLUTIONS MANUAL

Preface and Acknowledgments This book contains solutions to the problems presented in Modern Physics for Scientists and Engineers, Fourth Edition by Thornton and Rex. It is intended for instructors only and is specifically not intended for general distribution to students. Instructors may wish to share solutions to specific problems with students, but the copying of whole sections or chapters of this book for distribution to students is strongly discouraged. A Student Solutions Manual, which contains solutions to approximately one-fourth of the problems, is available. If you want your students to have that manual, either as a required or an optional text, you can order it through your bookstore (through Cengage Learning). We would like to thank Stephen T. Thornton who offered suggestions for solutions to many of the problems. We also especially thank Paul Weber of the University of Puget Sound and Thushara Parera of Illinois Wesleyan University who checked these solutions and offered many corrections and suggestions for clarification. For most physics problems, alternate solutions are possible. We welcome suggestions of alternate solutions and especially identification of any errors in this text, which are ours alone.

Allen P. Flora Department of Chemistry and Physics 401 Rosemont Avenue Hood College Frederick, MD 21701 [email protected] Andrew Rex Physics Department CMB 1031 University of Puget Sound Tacoma, WA 98416-1031 [email protected]

iii © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

iv © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Table of Contents CHAPTER 2 – Special Theory of Relativity

1

CHAPTER 3 – The Experimental Basis of Quantum Physics

28

CHAPTER 4 – Structure of the Atom

46

CHAPTER 5 – Wave Properties of Matter and Quantum Mechanics I

62

CHAPTER 6 – Quantum Mechanics II

78

CHAPTER 7 – The Hydrogen Atom

100

CHAPTER 8 – Atomic Physics

114

CHAPTER 9 – Statistical Physics

123

CHAPTER 10 – Molecules, Laser, and Solids

142

CHAPTER 11 – Semiconductor Theory and Devices

160

CHAPTER 12 – The Atomic Nucleus

169

CHAPTER 13 – Nuclear Interactions and Applications

187

CHAPTER 14 – Particle Physics

201

CHAPTER 15 – General Relativity

212

CHAPTER 16 – Cosmology and Modern Astrophysics – The Beginning and the End

221

v © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

vi © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 2

Special Theory of Relativity

1

Chapter 2  d 2 x ˆ d 2 y ˆ d 2 z ˆ 1. For a particle Newton’s second law says F  ma  m  2 i  2 j  2 k  . dt dt   dt Take the second derivative of each of the expressions in Equation (2.1): d 2 x d 2 x d 2 y d 2 y d 2 z d 2 z  2  2  2 . Substitution into the previous equation gives dt 2 dt dt 2 dt dt 2 dt  d 2 x d 2 y d 2 z  F  ma  m  2 iˆ  2 ˆj  2 kˆ   F . dt dt   dt

 dx dy ˆ dz ˆ  2. From Equation (2.1) p  m  iˆ  j  k . dt dt   dt dx dx dy dy dz dz  v   In a Galilean transformation . dt dt dt dt dt dt  dx  ˆ dy ˆ dz ˆ j  k  p .  v i  Substitution into Equation (2.1) gives p  m  dt dt  dt 

 dx dy ˆ dz ˆ  j  k  the same form is clearly retained, given However, because p  m  iˆ  dt dt   dt dx dx the velocity transformation  v. dt dt

3. Using the vector triangle shown, the speed of light coming toward the mirror is c 2  v 2 2 2 distance . and the same on the return trip. Therefore the total time is t2   speed c2  v2 Notice that sin  

v v , so   sin 1   . c c

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2

Chapter 2

Special Theory of Relativity

 0.350 m/s  4. As in Problem 3, sin   v1 / v2 , so   sin 1 (v1 / v2 )  sin 1    16.3 and  1.25 m/s 

v  v22  v12  (1.25 m/s)2  (0.35 m/s)2  1.20 m/s .

5. When the apparatus is rotated by 90°, the situation is equivalent, except that we have effectively interchanged 1 and 2 . Interchanging 1 and 2 in Equation (2.3) leads to Equation (2.4).

6. Let n = the number of fringes shifted; then n  n

c  t   t 



n 1

vc

7. Letting

1

2





v2 



c  1 2

2

.



. Because d  c  t   t  , we have

Solving for v and noting that

 0.005  589 109 m 

  3.00 108 m/s  1

d

22 m

1

+

2

= 22 m,

 3.47 km/s.

1   2 (where   v / c ) the text equation (not currently numbered) for

t1 becomes

t1 

2

1  2 2 1  c 1    c 1  2 1

which is identical to t2 when

1



2

so t  0 as required.

8. Since the Lorentz transformations depend on c (and the fact that c is the same constant for all inertial frames), different values of c would necessarily lead two observers to different conclusions about the order or positions of two spacetime events, in violation of postulate 1. 9. Let an observer in K send a light signal along the + x-axis with speed c. According to the Galilean transformations, an observer in K measures the speed of the signal to be dx dx   v  c  v . Therefore the speed of light cannot be constant under the Galilean dt dt transformations. 10. From the Principle of Relativity, we know the correct transformation must be of the form (assuming y  y and z  z ): x  ax  bt ; x  ax  bt . The spherical wave front equations (2.9a) and (2.9b) give us: ct  (ac  b)t ; ct   (ac  b)t .

Solve the second wave front equation for t  and substitute into the first: © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 2

Special Theory of Relativity

 (ac  b)(ac  b)t  or c2  (ac  b)(ac  b)  a 2c 2  b2 . ct    c   Now v is the speed of the origin of the x -axis. We can find that speed by setting x  0 which gives 0  ax  bt , or v  x / t  b / a , or equivalently b = av. Substituting this into

the equation above for c 2 yields c 2  a 2c 2  a 2v 2  a 2  c 2  v 2  . Solving for a:

 . / c2 This expression, along with b = av, can be substituted into the original expressions for x and x to obtain: x    x  vt  ; x    x  vt  a

1

1 v

2

which in turn can be solved for t and t  to complete the transformation. 11. When v  c we find 1   2  1 , so: x 

t 

x t

x   ct

 x   ct  x  vt ;

1  2

t x/c 1  2

x   ct  1  2

t x/c t ;

 x   ct   x  vt  ;

t    x / c 1  2

 t    x / c  t  .

  v / c  (26.39 m/s) / 3.00 108 m/s   8.8 108

12. (a) First we convert to SI units: 95 km/h = 26.39 m/s, so (b)   v / c  (240 m/s) /  3.00 108 m/s   8.0 107

(c) v  2.3 vsound   2.3 330 m/s  so

  v / c   2.3  330 m/s  /  3.00 108 m/s   2.5 106

  v / c  (7500 m/s) /  3.00 108 m/s   2.5 105

(d) Converting to SI units, 27,000 km/h = 7500 m/s, so (e) (25 cm)/(2 ns) = 1.25 108 m/s so   v / c  (1.25 108 m/s) /  3.00 108 m/s   0.42 (f) 11014 m  /  0.35 10 22 s   2.857 10 8 m/s , so

  v / c  (2.857 108 m/s) / 3.00 108 m/s   0.95

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3

4

Chapter 2

Special Theory of Relativity

13. From the Lorentz transformations t    t  vx / c 2  . But t   0 in this case, so solving for v we find v  c2 t / x . Inserting the values t  t2  t1  a / 2c and

c 2  a / 2c   c / 2 . We conclude that the frame K travels a at a speed c/2 in the  x -direction. Note that there is no motion in the transverse direction. 14. Try setting x  0    x  vt  . Thus 0  x  vt  a  va / 2c . Solving for v we find x  x2  x1  a , we find v 

v  2c , which is impossible. There is no such frame K .

15. For the smaller values of β we use the binomial expansion   1   2  (a)   1   2 / 2  1  3.87 1015

1/2

 1  2 / 2 .

(b)   1   2 / 2  1  3.2 1013 (c)   1   2 / 2  1  3.11012

(d)   1   2 / 2  1  3.11010 (e)   1   2 

(f)   1   2 

1/2

1/2

 1  0.422 

 1  0.952 

1/2

1/2

 1.10

 3.20

16. There is no motion in the transverse direction, so y  z  3.5 m.



1

1  2



x    x  vt   

1

1  0.82

 5/3

5  2m  0.8c  0   10 / 3 m 3

t    t   vx / c 2  

5 0   0.8c  2 m  / c 2   8.9 109 s  3

 3 m   (5 m)2  (10 m)2 x2  y 2  z 2   3.86 108 s 17. (a) t  8 3.00 10 m/s c (b) With   0.8 we find   5 / 3 . Then y  y  5 m, z  z  10 m, 2

5 x    x  vt   3m   2.40 108 m/s  (3.86  10-8 s)  10.4 m 3 2 5 t     t  vx / c 2    3.86 108 s    2.40 108 m/s   3 m  / 3.00 108 m/s    51.0 ns  3

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Chapter 2

Special Theory of Relativity

5

 10.4 m    5 m   10 m  x2  y2  z2   2.994 108 m/s which equals c (c) 9  t 51.0 10 s to within rounding errors. 2

2

2

18. At the point of reflection the light has traveled a distance L  vt1  ct1 . On the return trip it travels L  vt2  ct2 . Then the total time is t  t1  t2 

2 Lc 2L / c .  2 c  v 1  v2 / c2 But from time dilation we know (with t   proper time  2L0 / c ) that t   t  

2 L0 / c

1  v2 / c2

. Comparing these two results for t we get

which reduces to L  L0 1  v 2 / c 2 



L0

2

2 L0 / c 2L / c  2 v 1  v2 / c2 1 2 c

. This is Equation (2.21).

19. (a) With a contraction of 1%, L / L0  0.99  1  v 2 / c 2 . Thus 1   2  (0.99)2  0.9801. Solving for  , we find   0.14 or v  0.14c. (b) The time for the trip in the Earth-based frame is

d 5.00 106 m   1.19 101 s . With the relativistic factor   1.01 v 0.14  3.00 108 m/s (corresponding to a 1% shortening of the ship’s length), the elapsed time on the rocket ship is 1% less than the Earth-based time, or a difference of t 

 0.011.2 101s = 1.2 103 s.

20. The round-trip distance is d = 40 ly. Assume the same constant speed v   c for the

entire round trip. In the rocket’s reference frame the distance is only d   d 1   2 . Then

40ly 1   2 d distance    c 1  2 . in the rocket’s frame of reference v  time 40 y 40 y

v  1   2 . Solving for  we find   0.5 , or v  0.5 c  0.71c . c 1 To find the elapsed time t on Earth, we know t   40 y, so t   t   40y  56.6 y. 1  2 Rearranging  

21. In the muon’s frame T0  2.2 μs. In the lab frame the time is longer; see Equation (2.19): T    T0 . In the lab the distance traveled is 9.5cm  vT   v T0   c T0 , since v   c .

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6

Therefore  

9.5 cm



1  2

cT0

 , so   v  9.5 cm  c

Chapter 2

1  2

c 2.2μs 

Special Theory of Relativity

 . Now all quantities are

known except β. Solving for β we find   1.4 104 or v  1.4 104 c .

22. Converting the speed to m/s we find 25,000 mi/h = 11,176 m/s. From tables the distance is 3.84 108 m . In the earth’s frame of reference the time is the distance divided by speed, or t 

d 3.84 108 m   34,359 s . In the astronauts’ frame the time elapsed is v 11,176 m/s

t   t /   t 1   2 . The time difference is t  t  t   t  t 1   2  t 1  1   2  .  

2  11,176 m/s    5 Evaluating numerically t  34,359 s 1  1      2.4 10 s. 8 3.00 10 m/s       1 23. T    T0 , so we know that   5 / 3  . Solving for v we find v  4c / 5. 1  v2 / c2 1 and solving for v we find 24. L  L0 /  so clearly   2 in this case. Thus 2  1  v2 / c2

v

3c . 2

25. The clocks’ rates differ by a factor of   1/ 1  v 2 / c 2 . Because  is very small we will use the binomial theorem approximation   1   2 / 2 . Then the time difference is

t  t  t   t   t  t   1 . Using   1   2 / 2 and the fact that the time for the trip

equals distance divided by speed,

 375 m/s    6 8  8 10 m  3.00  10 m/s  t  t   2 / 2   375 m/s 2

2

t  1.67 108 s  16.7 ns.

26. (a) L  L /   L 1  v 2 / c 2   3.58 104 km  1  0.942  1.22 104 km (b) Earth’s frame: t  L / v 

3.58 107 m  0.127 s  0.94   3.00 108 m/s 

Golf ball’s frame: t   t /   0.127 s 1  0.942  0.0433 s

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Chapter 2

Special Theory of Relativity

7

27. Spacetime invariant (see Section 2.9): c2 t 2  x2  c2 t 2  x2 . We know x  4 km,

x2  x 2  5000 m    4000 m  t  0 , and x  5 km. Thus t     1.0 1010 s 2 2 8 c2  3.00 10 m/s  2

2

2

and t  1.0 105 s. 28. (a) Converting v = 120 km/h = 33.3 m/s. Now with c = 100 m/s, we have 1 1   v / c  0.333 and     1.061 . We conclude that the moving 2 1  1  0.3332 person ages 6.1% slower. (b) L  L /   (1 m) /(1.061)  0.942 m.

29. Converting v = 300 km/h = 83.3 m/s. Now with c =100 m/s, we have   v / c  0.833 and  

1

1 

2



1

1  0.8332

 1.81. So the length is L  L0 /   40 /1.81  22.1 m.

30. Let subscript 1 refer to firing and subscript 2 to striking the target. Therefore we can see that x1  1 m, x2  121 m, and t1  3 ns. t2  t1 

distance 120 m  3 ns +  3 ns + 408 ns = 411 ns. speed 0.98c To find the four primed quantities we can use the Lorentz transformations with the

known values of x1 , x2 , t1 , and t2 . Note that with v  0.8c ,   1  v 2 / c2  5 / 3 . t1    t1  vx1 / c 2   0.56 ns

t2    t2  vx2 / c 2   147 ns

x1    x1  vt1   0.47 m

x2    x2  vt2   37.3 m

31. Start from the formula for velocity addition, Equation (2.23a): ux  0.62c  0.84c 1.46c   0.96 c 2 1  (0.62c)(0.84c) / c 1.52 0.62c  0.84c 0.22c (b) ux    0.46 c 2 1  (0.62c)(0.84c) / c 0.48

(a) ux 

32. Velocity addition, Equation (2.24): ux  ux 

ux  v . 1  vux / c 2

ux  v with v  0.8 c and ux  0.8 c. 1  vux / c 2

0.8c  (0.8c) 1.6c   0.976 c 2 1  (0.8c)(0.8c) / c 1.64

© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8

Chapter 2

Special Theory of Relativity

33. Conversion: 110 km/h = 30.556 m/s and 140 km/h = 38.889 m/s. Let ux  30.556 m/s

and v  38.889 m/s. Our premise is that c  100 m/s. Then by velocity addition, 30.556 m/s   38.889 m/s  ux  v ux    62.1 m/s. By symmetry 2 1  vux / c 1   38.889 m/s  30.556 m/s  / 100 m/s 2

each observer sees the other one traveling at the same speed. 34. From Example 2.5 we have u  u 

c 1  nv / c  . For light traveling in opposite directions n 1  v / nc 

c 1  nv / c 1  nv / c  . Because v / c is very small, use the binomial expansion:  n 1  v / nc 1  v / nc 

1  nv / c 1  1  nv / c 1  v / nc   1  nv / c 1  v / nc   1  nv / c  v / nc , where we 1  v / nc 1  nv / c  1  nv / c  v / nc. Thus have dropped terms of order v 2 / c 2 . Similarly 1  v / nc c 2v u  1  nv / c  v / nc   1  nv / c  v / nc   1  1/ n   2v 1  1/ n2  . Evaluating n n 1   numerically we find u  2(5 m/s) 1   4.35 m/s. 2   1.33 

35. Clearly the speed of B is just 0.60c . To find the speed of C use ux  0.60 c and v  0.60 c : ux 

ux  v 0.60c  (0.60c)   0.88 c. 2 1  vux / c 1  (0.60c)(0.60c) / c 2

36. We can ignore the 400 km, which is small compared with the Earth-to-moon distance 3.84 108 m. The rotation rate is   2 rad 100 s1  2 102 rad/s. Then the speed across the moon’s surface is v   R   2 102 rad/s  3.84 108 m   2.411011 m/s.

37. Classical: t 

   ln 2  t  4205 m  1.43 105 s . Then N  N0 exp    14.6 t 0.98c 1/2  

or about 15 muons.

Relativistic: t   t /  

1.43 105 s  2.86 106 s so 5

   ln 2  t  N  N0 exp    2710 muons. Because of the exponential nature of the decay  t1/ 2  curve, a factor of five (shorter) in time results in many more muons surviving.

38. The circumference of the fixed point’s rotational path is 2 RE cos(39 ) , where RE  Earth’s radius = 6378 km. Thus the circumference of the path is 31,143 km. The © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 2

Special Theory of Relativity

9

rotational speed of that point is v   31,143 km  / 24 h  1298 km/h  360.5 m/s . The observatory clock runs slow by a factor of  

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