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
flora@hood.edu
Andrew Rex
Physics Department CMB 1031
University of Puget Sound
Tacoma, WA 98416-1031
rex@pugetsound.edu
iii
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publicly accessible website, in whole or in part.
iv
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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
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publicly accessible website, in whole or in part.
vi
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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
 v i 
j  k  p .
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

v2 
1
2

2
c 
.
. Because d  c  t   t  , we have
 0.005  589 109 m 
22 m
2
1

Solving for v and noting that
  3.00 108 m/s 

d
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:
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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:
1
 .
/ 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 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
t x/c t ;
1  2
x   ct 
1  2
 x   ct   x  vt  ;
t    x / c
1  2
 t    x / c  t  .
12. (a) First we convert to SI units: 95 km/h = 26.39 m/s, so
  v / c  (26.39 m/s) / 3.00 108 m/s   8.8 108
(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
(d) Converting to SI units, 27,000 km/h = 7500 m/s, so
  v / c  (7500 m/s) /  3.00 108 m/s   2.5 105
(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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publicly accessible website, in whole or in part.
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 
1/2
 1  2 / 2 .
(a)   1   2 / 2  1  3.87 1015
(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 
1/2
 1  0.422 
1/2
(f)   1   2 
1/2
 1  0.952 
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
c
3.00 10 m/s
(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
2 Lc
2L / c
.

2
c  v 1  v2 / c2
But from time dilation we know (with t   proper time  2L0 / c ) that
trip it travels L  vt2  ct2 . Then the total time is t  t1  t2 
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
distance
d


 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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publicly accessible website, in whole or in part.
6
Chapter 2
Therefore  
9.5 cm

1  2
cT0
 , so   v  9.5 cm 
c
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
24. L  L0 /  so clearly   2 in this case. Thus 2 
and solving for v we find
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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publicly accessible website, in whole or in part.
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
1
and  
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.
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
t2  t1 
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 
ux  v
.
1  vux / c 2
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 
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
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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
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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  
1
1 
2
 1   2 / 2  1  7.22 1013 . In
41.2 h the observatory clock is slow by  41.2 h   7.22 1013   2.9746 1011 h = 107 ns.
In 48.6 h it is slow by  48.6 h   7.22 1013   3.5089 1011 h = 126 ns. The Eastwardmoving clock has a ground speed of 31,143 km/41.2 h = 755.9 km/h = 210.0 m/s and thus
has a net speed of 210.0 m/s + 360.5 m/s = 570.5 m/s. For this clock
1

 1   2 / 2  1  1.811012 and in 41.2 hours it runs slow by
2
1 
 41.2 h  1.811012   7.4572 1011
h = 268 ns. The Westward-moving clock has a
ground speed of 31,143 km/48.6 h = 640.8 km/h = 178.0 m/s and thus has a net speed of
360.5 m/s − 178.0 m/s = 182.5 m/s. For this clock
1

 1   2 / 2  1  1.85 1013 and in 48.6 hours it runs slow by
2
1 
 48.6 h  1.85 1013   8.9911012
h = 32 ns. So our prediction is that the Eastward-
moving clock is off by 107 ns – 269 ns  162 ns, while the Westward-moving clock is
off by 126 ns − 32 ns = 94 ns. These results are correct for special relativity but do not
reconcile with those in the table in the text, because general relativistic effects are of the
same order of magnitude.
39. The derivations of Equations (2.31) and (2.32) in the beginning of Section 2.10 will
suffice. Mary receives signals at a rate f  for t1 and a rate f  for t2 . Frank receives
signals at a rate f  for t1 and a rate f  for t2 .
L L L L 2L
   
; Frank sends signals at rate f, so Mary receives
v c v c
v
f T  2 f L / v signals.
40. T  t1  t2 
T   t1  t2 
2L
; Mary sends signals at rate f , so Frank receives fT   2 f L /  v signals.
v
41. s 2  x2  y 2  z 2  c2t 2 ; Using the Lorentz transformation
s 2   2 ( x  vt ) 2  y2  z2  c 2 2 (t   vx / c 2 )2
 x2 2 1  v 2 / c 2   y2  z2  c 2t  2 2 1  v 2 / c 2 
 x 2  y  2  z  2  c 2 t  2  s  2
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10
Chapter 2
Special Theory of Relativity
42. For a timelike interval s 2  0 so x2  c 2 t 2 . We will prove by contradiction. Suppose
that there is a frame K is which the two events were simultaneous, so that t   0 . Then
by the spacetime invariant x2  c2 t 2  x2  c2t 2  x2 . But because x2  c 2t 2 ,
this implies x2  0 which is impossible because x is real.
43. As in Problem 42, we know that for a spacelike interval s 2  0 so x2  c2 t 2 . Then in
a frame K in which the two events occur in the same place, x  0 and
x2  c2 t 2  x2  c2t 2  c2t 2 . But because x 2  c 2 t 2 we have c 2 t 2  0 ,
which is impossible because t  is real.
44. In order for two events to be simultaneous in K , the two events must lie along the x
axis, or along a line parallel to the x axis. The slope of the x axis is   v / c , so
ct
. Solving for v, we find v  c2 t / x . Since the slope of the x axis
x
must be less than one, we see that x  ct so s 2  x2  c2 t 2  0 is required.
v / c  slope =
45. parts (a) and (b) To find the equation of the line
use the Lorentz transformation. With t   0 we
have t   0    t  vx / c 2  or, rearranging,
ct  vx / c   c . Thus the graph of ct vs. x is a
straight line with a slope  .
(c) Now with t  constant, the Lorentz transformation
gives t     t  vx / c 2  . Again we solve for ct :
ct   x  ct  /    x  constant . This line is
parallel to the t   0 line we found earlier but
shifted by the constant.
(d) Here both the x and ct  axes are shifted
from their normal (x, ct) orientation and they
are not perpendicular.
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Chapter 2
Special Theory of Relativity
11
46. The diagram is shown here. Note that there is only one worldline for light, and it bisects
both the x, ct axes and the x, ct  axes. The x and ct  axes are not perpendicular. This can
be seen as a result of the Lorentz transformations, since x  0 defines the ct  axis and
t   0 defines the x axis.
47. The diagram shows that the events A and B that occur at the same time in K occur at
different times in K .
48. The Doppler shift gives   0
1 
. With numerical values 0  650 nm and
1 
  540 nm, solving this equation for  gives   0.183 . The astronaut’s speed is
v   c  5.50 107 m/s. In addition to a red light violation, the astronaut gets a speeding
ticket.
49. According to the fixed source (K) the signal and receiver move at speeds c and v,
respectively, in opposite directions, so their relative speed is c  v . The time interval
t

between receipt of signals is t   / (c  v)  1/ f0 . By time dilation t  
.

  (c  v )
Using   c / v0 and  
f
1
1  v2 / c2
we find t  
1  2
c 1  v2 / c2

and
f 0 (c  v )
f 0 (1   )
f (1   )
1
1 
.
 0
 f0
2
t 
1 
1 
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12
Chapter 2
Special Theory of Relativity
50. For a fixed source and moving receiver, the length of the wave train is cT  vT . Since n
cn
cT  vT
waves are emitted during time T,  
and the frequency f  c /  is f 
.
cT  vT
n
As in the text n  f 0T0 and T0  T /  . Therefore f 
51. f  f 0
f 1  2
cf 0T / 
1 
.
 0
 f0
cT  vT
1 
1 
1 
1  0.95
 1400 kHz 
 224 kHz
1 
1  0.95
52. The Doppler shift function f   f 0
1 
1 
is the rate at which #1 and #2 receive signals from each other and the rate at which #2
and #3 receive signals from each other. But for signals between #1 and #3 the rate is
f   f 
1 
1 
.
 f0
1 
1 
53. The Doppler shift function f   f 0
1 
1 
is the rate at which #1 and #2 receive signals
from each other and the rate at which #2 and #3
receive signals from each other. As for #1 and
#3 we will assume that these plumbing vans are
non-relativistic  v  c  . Otherwise it would be
necessary to use the velocity addition law and apply
the transverse Doppler shift. From the figure we see
1
that f  
. Now f 0  1/ t0 and
t0  (t2  t1 )
2 x 2vt0 cos 

. With an angle of 45 , cos(45)  1/ 2 and
c
c
f0
f0
1
f


.
1/ f0  (2v cos  ) / cf 0 1   2v cos   / c 1  2 v / c
t2  t1 
54. The Doppler shift to higher wavelengths is (with 0  589 nm)   700 nm  0
1 
.
1 
8
v  0.171  3.00 10 m/s 
 1.75 106 s
Solving for  we find   0.171 . Then t  
a
29.4 m/s 2
which is 20.25 days. One problem with this analysis is that we have only computed the
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Chapter 2
Special Theory of Relativity
13
time as measured by Earth. We are not prepared to handle the non-inertial frame of the
spaceship.
55. Let the instantaneous momentum be in the x-direction and the force be in the y-direction.
dv
Then dp  F dt   mdv and dv is also in the y-direction. So we have F   m
  ma .
dt
v2
56. The magnitude of the centripetal force is  ma   m for circular motion. For a charged
r
v2
particle F  qvB , so qvB   m or, rearranging qBr   mv  p . Therefore
r
p
r
.
qB
When the speed increases the momentum increases, and thus for a given value of B the
radius must increase.
dp
mv
57. p   mv 
and F 
. The momentum is the product of two factors that
2
2
dt
1 v / c
contain the velocity, so we apply the product rule for derivatives:
F m

d 
mv


dt  1  v 2 / c 2 
 dv / dt
d 
1
 m
v 
dt  1  v 2 / c 2
 1  v 2 / c 2
 1   2v  dv
  ma  mv      2   3
 2   c  dt


 
 v2 
  ma   3 ma  2 
c 
 v2 v2 
  3 ma 1  2  2 
 c c 
  3 ma
58. From the preceding problem F   3ma . We have a  1019 m/s2 and m  1.67 1027 kg.

(a)
1
1  v2 / c2

1
1  0.012
 1.00005
F  1.00005 1.67 1027 kg 1019 m/s 2   1.67 108 N
3
(b) As in (a)   1.005 and F  1.70 108 N
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14
Chapter 2
Special Theory of Relativity
(c) As in (a)   2.294 and F  2.02 107 N
(d) As in (a)   7.0888 and F  5.95 106 N
59. p   mv with  
1
1  v2 / c2

1
1  0.922
 2.5516 ;
p
1016 kg·m / s
m

 1.42 1025 kg
8
 v  2.5516  0.92   3.00 10 m/s 
60. The initial momentum is p0   mv 
1
1   0.5
2
m  0.5c   0.57735mc .
(a) p / p0  1.01
1.01 
 mv
0.57735mc
 v  1.01 0.57735c   0.58312c

1
1
Substituting for  and solving for v, v  

2
2
 .58312c  c 

1
1
(b) Similarly v  

2
2
 .63509c  c 
1/ 2
 0.504c.
1/2
 0.536c
1/2

1
1
(c) Similarly v  
 2   0.756c
2
 1.1547c  c 
61. 6.3 GeV protons have K  6.3 103 MeV and E  K  E0  7238 MeV. Then
p
E 2  E02
c
 7177 MeV/c . Converting to SI units
 1.60 1013
p  7177 MeV/c 
MeV

J 
c

18

  3.83 10 kg·m/s
8
  3.00 10 m/s 
From Problem 56 we have B 
p
3.83 1018 kg  m/s

 1.57 T .
qr 1.60 1019 C  15.2 m 
62. Initially Mary throws her ball with velocity (primes showing the measurements are in
Mary’s frame): uM x  0 uM y  u0 . After the elastic collision, the signs on the above
expressions are reversed, so the change in momentum as measured by Mary is
mu0
mu0
2mu0
.
pM 


2
2
2
2
1  u0 / c
1  u0 / c
1  u02 / c 2
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Chapter 2
Special Theory of Relativity
15
Now for Frank’s ball, we know uFx  0 and uFy  u0 . The velocity transformations give
for Frank’s ball as measured by Mary: uF x  v
uF y  u0 1  v 2 / c 2 .
    u 
2
To find  for Frank’s ball, note that uF x

1
1  u / c
2
F
2

2
Fy
1
1  v 2 / c 2  u02 1  v 2 / c 2  / c 2

 v 2  u02 1  v 2 / c 2  . Then
1
1  u
2
0
/ c 2 1  v 2 / c 2 
. Using
p   mu along with the reversal of velocities in an elastic collision, we find
pF   m  u0  1  v 2 / c 2   mu0 1  v 2 / c 2  2 mu0 1  v 2 / c 2

2mu0 1  v 2 / c 2
1  u
2
0
/ c 2 1  v 2 / c 2 
Finally p  pF  pM 

2mu0
1  u
2
0
2mu0  2mu0
1  u02 / c2 
/ c2 
 0 as required.
1
63. To prove by contradiction, suppose that K   mv 2 . Then
2
1
K  E  E0   mc 2  mc 2    1 mc 2   mv 2 . This implies   1  v2 / 2c2 , or
2
2
2
  1  v / 2c , which is clearly false.
64. The source of the energy is the internal energy associated with the change of state,
commonly called that latent heat of fusion L f . Let m be the mass equivalent of 2 grams
and M be the mass of ice required. m 
E Lf M
 2 Rearranging
c2
c
8
mc 2  0.002 kg   3.00 10 m/s 
M

 5.39 108 kg
3
Lf
334 10 J/kg
2
65. In general K    1 mc 2 , so   1 
  1
K
. For 9 GeV electrons:
mc 2
9000 MeV
 1.76 104 Then from the definition of  we have
0.511 MeV
  1
1

2
 1
1
1.76 10 
4 2
 1  1.6 109 . Thus
v  1  1.6 109 c  0.9999999984c .
© 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.
16
Chapter 2
For 3.1 GeV positrons:   1 
Special Theory of Relativity
3100 MeV
1
 6068 and   1 
 1  1.4 108 .
2
0.511 MeV
 6068
Thus v  1  1.4 108 c  0.999999986c .
66. Note that the proton’s mass is 938 MeV/c2. In general K    1 mc 2 , so   1 
Then from the definition of  we have   1 
MeV, and   1 
1
2
K
.
mc 2
. For the first section K  0.750
0.750 MeV
1
1
 1.00080 with   1  2    1 
 0.040 .
2
938 MeV

1.00080 
Thus v = 0.04c at the end of the first stage. For the other stages the computations are
similar, and we tabulate the results:
K (GeV)


0.400
1.43
0.71
8
9.53
0.994
150
160.9
0.99998
1000
1067
0.9999996
511 keV/c   0.020c   10.22 keV/c ;
2
67. (a) p   mu 
1  0.0202
511 keV/c  (c )  511.102 keV ;

2
E   mc
2
2
1  0.022
K  E  E0  511.102 keV  511.00 keV  102 eV
The results for (b) and (c) follow with similar computations and are
tabulated:

p (keV/c) E (keV) K (keV)
0.20
104.3
521.5
10.5
0.90
1055
1172
661
68. E  2E0   E0 so   2 . Then   1 
1

2

3c
3
and v 
.
2
2
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publicly accessible website, in whole or in part.
Chapter 2
Special Theory of Relativity
17
69. For a constant force, work = change in kinetic energy = Fd  mc2 / 4, because
8
mc 2  80 kg   3.00 10 m/s 
d

 2.25 1017 m = 23.8 ly.
4F
4 8 N 
2
25%  1/ 4 .
70. E  K  E0  2E0  E0  3E0   E0 so   3 . Then
  1
1

2
   1
1 2 2

 0.943 . Thus v = 0.943c.
32
3
71. (a) E  K  E0  0.1E0  E0  1.1E0   E0 , so   1.1 . Then
  1
1

2
   1
1
 0.417 and v = 0.417c.
1.12
(b) As in (a)   2 and v  3 c / 2 .
(c) As in (a)   11 and v = 0.996c.
2
 E0 
72. K  E  E0 
. Rearranging 1    
 E0 so K  E0 
 which
1  2
1  2
 K  E0 
E0
E0
2
2
2
 E0 
 E0 
can be written as   1  
 .
 or   1  
 K  E0 
 K  E0 
2
73. Using E   mc 2 along with p   mv we see that   E / mc 2  p / mv . Solving for v/c we
find   v / c  pc / E .
74. The speed is the same for protons, electrons, or any particle.
1
1

K     1 mc 2  1.01 mv 2   0.505mc 2  2 so   1 
 1  0.505 2 .
2
2

1 
Rearranging and solving for  , we find   0.114 or v = 0.114c.
75. Converting 0.1 ounce = 2.835 103 kg.
E  mc 2   2.835 103 kg  3.00 108 m/s   2.55 1014 J .
2
Eating 10 ounces results in a factor of 100 greater mass-energy increase, or 2.55 1016 J.
This is a small increase compared with your original mass-energy, but it will tend to
increase your weight; depending on how they are prepared, peanuts generally contain
about 100 kcal of food energy per ounce.
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18
Chapter 2
Special Theory of Relativity
76. The energy needed equals the kinetic energy of the spaceship.
 1

K     1 mc 2  
 1 mc 2
 1  2





2
1

 1 104 kg  3.00 108 m/s   4.35 1019 J
2
 1  0.3

or 4.35% of 1021 J.
77. Up to Equation (2.57) the derivation in the text is complete. Then using the integration
 xdy  xy   ydx and noting that in this case
have  ud ( u)   u    udu . Thus
by parts formula,
x  u and y   u , we
2
u
K  m  ud ( u )   mu 2  m   udu
0
u
  mu 2  m 
1  u 2 / c2
du
Using integral tables or simple substitution:
K   mu 2  mc 2 1  u 2 / c 2
u
0
  mu 2  mc 2 1  u 2 / c 2  mc 2

mc 2  mc 2 1  u 2 / c 2 
 mc 2
1 u / c
  mc  mc 2  mc 2 (  1)
2
2
2
78. Converting 0.11 cal · g 1 ∙  C
1
= 460 J · kg 1 ∙  C 1  cV
(specific heat at constant volume). From thermodynamics the energy E used to change
the temperature by T is mcV T . Thus
E  1000 kg   460 J/(kg  C   0.5 C   2.30 105 J and
m 
E
2.30 105 J

 2.56 1012 kg . The source of this energy is the internal
2
2
8
c
3.00 10 m/s 
energy of the arrangement of atoms and molecules prior to the collision.
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Chapter 2
Special Theory of Relativity
19
79.
Eb   2m p  2mn  m  He   c 2
 931.494 MeV 
  2 1.007276 u   2(1.008665 u)  4.001505 u  c 2 
  28.3 MeV
c 2·u


80.
E   mn  m p  me  c 2
 931.494 MeV 
 1.008665 u  1.007276 u  0.000549 u  c 2 
  0.782 MeV
c 2·u


81. E  K  E0  1 TeV  938 MeV  1 TeV ;
1 TeV  938MeV    938 MeV 
E 2  E02
p

 1.000938 TeV/c
c
c
E  E0 1.000938 TeV
1


 1067 ;
 2  1  2  1  8.78 107
E0
0.000938 TeV

2
2
  1  8.78 107  1  4.39 107 ; and v   c  0.999999561c
82. (a) E 
p 2c 2  E02 
 40GeV    511 keV 
2
2
 40.0 GeV
K  E  E0  40.0 GeV
(b) E 
p 2c 2  E02 
 40 GeV    0.938 GeV 
2
2
 40.011GeV
K  E  E0  40.011 GeV  0.938 GeV  39.07 GeV
83. E  K  E0  200MeV  106MeV  306 MeV
 306 MeV   106 MeV 
E 2  E02
p

 287.05 MeV/c
c
c
E 306 MeV


 2.887
E0 106 MeV
2
  1
1
2
2
 0.938
so v  0.938c
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publicly accessible website, in whole or in part.
20
Chapter 2
Special Theory of Relativity
84. (a) The mass-energy imbalance occurs because the helium-3 (3 He) nucleus is more
tightly bound than the two separate deuterium nuclei (2 H) . (Masses from Appendix 8.)
E  [2m( 2 H)]  [mn  m(3 He)] c 2
 931.494 MeV 
  2(2.014102 u)  (1.008665 u  3.016029 u)  c 2 
  3.27 MeV
c2  u


 931.494 MeV 
(b) The initial rest energy is 2m(2 H) = 2(2.014102 u) 
 = 3752 MeV.
c2  u


Thus the answer in (a) is about 0.09% of the initial rest energy.
85. (a) The mass-energy imbalance occurs because the helium-4 (4 He) is more tightly bound
than the deuterium (2 H) and tritium nuclei (3 H) .
E  [m( 2 H)+m(3 H)]  [mn  m( 4 He)] c 2
 931.494 MeV 
  (2.014102 u  3.016029 u)  (1.008665 u  4.002603 u)  c 2 

c2  u


 17.6 MeV
 931.494 MeV 
(b) The initial rest energy is  m(2 H)+ m(3 H)   c 2  (5.030131 u)   c 2 
=
c2  u


Thus
the
answer
in
(a)
is
about
0.37%
of
the
initial
rest
energy.
4686 MeV.
86. (a) In the inertial frame moving with the negative charges in wire 1, the negative charges
in wire 2 are stationary, but the positive charges are moving. The density of the positive
charges in wire 2 is thus greater than the density of negative charges, and there is a net
attraction between the wires.
(b) By the same reasoning as in (a), note that the positive charges in wire 2 will be
stationary and have a normal density, but the negative charges are moving and have an
increased density, causing a net attraction between the wires.
(c) There are two facts to be considered. First, (a) and (b) are consistent with the physical
result being independent of inertial frame. Second, we know from classical physics that
two parallel wires carrying current in the same direction attract each other. That is, the
same result is achieved in the “lab” frame.
v d 1  2

where d is the length of the
c
ct 
87.
particle track and t  the particle’s lifetime in its rest frame. In this problem t   8.2 1011 s
As in the solution to Problem 21 we have  
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Chapter 2
Special Theory of Relativity
21
and d = 24 mm. Solving the above equation we find   0.698 . Then
E0
1672 MeV
E

 2330 MeV
1  2
1  0.6982
88.
1/2
dp d
d   v2  
F
   mv    mv 1  2  
dt dt
dt   c  


dv mv 3  2v  dv
 m

  2 
dt
2
 c  dt
dv  v 2 2 
dv 
v2 / c2 
1


m

1

dt 
c 2 
dt  1  v 2 / c 2 
dv 
1
dv
dv
1

 m
 m 3
m
2
2

dt 1  v / c 
dt
dt 1  v 2 / c 2 3/2
 m
89. (a) The number n received by Frank at f  is half the number sent by Mary at that rate, or
fL /  v . The detected time of turnaround is
t
n

f 
fL /  v

L 1 
1    / 1     v
1 

L 1    L L
  .
v
v c
(b) Similarly, the number n received by Mary at f  is
f L 1   
T
1  L
.
Her turnaround time is T  / 2  L /  v .
 f

2
1   v
v
(c) For Frank, the time t2 for the remainder of the trip is t2 = T − t1 = L/v − L/c.
n  f 
1 
fL
 L / v  L / c  .
1 
v
fL fL 2 fL
Total number received =
.


v v v
Total number received 2L
Mary’s age =
.

f
v
(d) For Mary, t2  T   t1  L /  v .
Number of signals = f t2  f
1  L
fL

1    .
1   v v
fL
2 fL
Total number received =
.
1    1    
v
v
Number of signals = f  t2  f
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publicly accessible website, in whole or in part.
22
Chapter 2
Frank’s age =
Special Theory of Relativity
Total number received 2L
.

f
v
90. In the fixed frame, the distance is Δx = 8 ly and the elapsed time is Δt = 10 y, so the
interval is s 2  x2  c2 t 2  64 ly2  100 ly2 =  36 ly2 . In the moving frame, Mary’s
clock is at rest, so x  0 , and the time interval is t   6 y. Thus the interval is
s2  x2  c2 t 2  0 ly2  36 ly2 =  36 ly2 . The results are the same, as they should
be, because the spacetime interval is the same for all inertial frames.
52y   4.3 ly  1  0.8  55.9  56
fL
91. (a) From Table 2.1 number 
1    


v
0.8c
1
(b) t1 
fL
L L 4.3 ly 4.3 ly
1   2  167.7  168
 

 9.68 y ; number  f  t1 
v
v c
0.8c
c
fL
1   2  168 so the total is 168 + 168 = 336.
v
Mary: number  2 f L / v  559
(c) Frank: f  t2 
(d) Frank: T  2L / v  10.75 y
Mary : T   2L /  v  6.45 y
(e) From part (c), 559 weeks = 10.75 years and 336 weeks = 6.46 years, which agrees
with part (d).
92. Notice that the radar is shifted twice, once upon receipt by the speeding car and again
upon reemission (reflection) of the beam. For this double shift, the received frequency f is
f
1  1  1 


f0
1  1  1 
For speeds much less than c, a Taylor series approximation gives an excellent result:
f 1 
1

 1   1     1  2 .
f0 1  
Converting 80 mph to 35.8 m/s gives β = 1.19 × 10−7, so the received frequency is
approximately
f  f0 1  2   10.4999975 GHz , which is 2.5 kHz less than the original frequency.
93. For this transformation v = 0.8c (so γ = 5/3), ux  0 and uy  0.8c. Applying the
transformations,
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Chapter 2
Special Theory of Relativity
ux 
23
ux  v 0  0.8c

 0.8c ;
vux
1

0
1 2
c
uy 
uy
0.8c

 0.48c
 vux  5 1  0
 
 1  2 
c  3

The speed is u  ux2  u y2  0.93c , safely under the speed of light.
94. (a) K  E  E0  250E0 . Thus K  249E0  249  511keV   127.2 MeV
(b)   250
  1
1
2
 0.999992
v  0.999992c
E 2  E02

c
(c) p 
 250  511 keV    511 keV 
2

 128 MeV/c
c
95. (a) For the proton: p   mu 
For the electron: E 
2
1
1  0.9
p 2c 2  E02 
938 MeV / c   0.9c   1940 MeV / c .
2
2
1940 MeV    0.511 MeV 
2
2
 1940 MeV .
E 1940 MeV

 3797
E0 0.511 MeV
  1
1

2
 1
1
 1  6.94 108  1  3.97 108
37972
v  1  3.97 108  c


1
 1  938 MeV   1214 MeV .
(b) For the proton: K    1 E0  
2
 1  0.9

K  E0 1214 MeV  0.511 MeV
For the electron:  

 2377 .
E0
0.511 MeV
  1
1

2
 1
1
 1  1.77 107  1  8.85 108
2
2377
v  1  8.85 108  c
96. In the frame of the decaying K 0 meson, the pi mesons must recoil with equal speeds in
opposite directions in order to conserve momentum. In that reference frame the available
kinetic energy is 498MeV  2 140 MeV   218MeV . The pi mesons share this equally,
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publicly accessible website, in whole or in part.
24
Chapter 2
Special Theory of Relativity
so each one has a kinetic energy of 109 MeV in that frame. The speed of each pi meson
can be found:

K  E0 109 MeV  140 MeV
1

 1.779 so u  1  2 c  0.827c .
E0
140 MeV

The greatest and least speeds in the lab frame are obtained when the pi mesons are
released in the forward and backward directions. Then by the velocity addition laws:
0.9c  0.827c
vmax 
 0.990c
1   0.9  0.827 
vmin 
0.9c  0.827c
 0.285c
1   0.9  0.827 
97. (a) The round-trip distance is L0  8.60 ly . Assume the same constant speed v   c for
the entire trip. In the rocket’s frame, the distance is only L  L0 1  L0 1   2 .
Mary will age in the rocket’s reference frame a total of 22 y, and in that frame
8.60 ly 1   2
distance
L
v
v


 0.39c 1   2 . Therefore,    0.39 1   2 .
time
22y
22y
c
0.392
Solving for  we find  
, or   0.36 so Mary’s speed is v  0.36c .
1.00  0.392
(b) To find the elapsed time on Earth, we know that T0  22y , so
T   T0 
1
1  2
22y  23.6 y . Frank will be 53.6 years old when Mary returns at the
age of 52 y.
98. With v  0.995c then   0.995 and   10.01 . The distance out to the star is
L0  5.98 ly . In the rocket’s reference frame, the distance is only
L  L0 1 
5.98 ly
 0.597 ly .
10.01
(a) The time out to the star in Mary’s frame is time 
distance 0.597 ly

 0.6 y. The
speed
0.995c
time for the return journey will be the same. When you include the 3 years she spends at
the star, her total journey will take 4.2 y.
(b) To find the elapsed time on Earth for the outbound journey, we know that T0  0.6y so
T   T0  (10.01)0.6 y  6.006 y. The return journey will take an equal time. The 3 years
the spaceship orbits the star will be equivalent for both observers. Therefore Frank will
measure a total elapsed time of 15.012 y, which makes him 10.8 years older than her.
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publicly accessible website, in whole or in part.
Chapter 2
Special Theory of Relativity
25
99. (a) The Earth to moon distance is 3.84 108 m. The rotation rate is
  2 rad  0.030 s1  1.885 101 rad/s. Then the speed across the moon’s surface is
v   R  1.885 101 rad/s  3.82 108 m   7.24 107 m/s .
(b) The required speed can be found using c   R  2 f R which requires the frequency
3.00 108 m/s

 0.124 Hz .
to be f 
2 R 2  3.84 108 m/s 
c
100. With the data given,   3.28 106 , which is very small. We will use the binomial
approximation theorem. From Equation (2.21), we know that
L  L0 1. 1  1   2  1   2 / 2 .
(a) The percentage of length contraction would be:
 L0  L0 1 
L0  L 

% change 
100% 
100%
L0
L0
 1   1  100%  1  (1   2 / 2)  100%

2
2
100%  5.37 1010
(b) The clocks’ rates differ by a factor   1/ 1   2 . The clock on the SR-71 measures
the proper time and Equation 2.19 tells us that T    T0 so the time difference is
t  T   T0   T0  T0  T0 (  1) . Using   1   2 / 2 and the fact that the time for
the trip in the SR-71 equals the distance divided by the speed
 983 m/s 
6
3.2 10 m  3.00 108 m/s 
t  t   2 / 2  
983 m/s
2
2
 1.75 108 s  17.5 ns
101. As the spaceship is approaching the observer, we will make use of Equation (2.32),
f
1 
1 
f 0 with the prime indicating the Doppler shifted frequency (or wavelength).
This equation indicates that the Doppler shifted frequency will be larger than the
frequency measured in a frame where the observer is at rest with respect to the source.
Since c   f , this means the Doppler shifted wavelengths will be lower. As given in the
problem, we see that the difference in wavelengths for an observer at rest with respect to
the source is 0  0.5974nm . We want to find a speed so that the Doppler shifted
difference is reduced to    0.55nm . We have
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publicly accessible website, in whole or in part.
26
Chapter 2
   2  1 
Special Theory of Relativity
 f  f  
c c
  c 1 2 
f 2 f1  f1f 2 
 1  
 1 
 
 f 0,1  
 1  
 1 
 c 
1 
f 0,1 f 0,2

1 



 f 0,2 
 1  



c


 1  


 1  
 

 1  
 c
c    1  

 
   
f
f
0,2
0,1
   1  

 1  
 

 1  


  0  


  f 0,1  f 0,2  
 
 
  f 0,1 f 0,2  


  0,2  0,1  


We can complete the algebra to show that  
02     
2
02     
2
 0.0825 so that
v  2.47 107 m/s .
102. As we know that the quasars are moving away at high speeds, we make use of Equation
(2.33) and the equation c   f . Using a prime to indicate the Doppler shifted frequency
(or wavelength), Equation (2.33) indicates that frequency is given by f  
1 
1 
f 0 ,or
1 
f0

so
f
1 
z
    0       1   c / f   1
0

 0
 
  c / f0



 f
  1 
  0  1  
 1


 f    1 

1 
2
Therefore  z  1 
. We can complete the algebra to show that
1 
2
  z  12  1 
z  1  1


and thus v  
 c. For the values of z given, v  0.787c for
2
2
 z  1  1
  z  1  1
z  1.9 and v  0.944c for z  4.9 .
103. (a) In the frame of the decaying K 0 meson, the pi mesons must recoil with equal
momenta in opposite directions in order to conserve momentum. In that reference frame
the available kinetic energy is 498 MeV  2 135 MeV   228 MeV .
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publicly accessible website, in whole or in part.
Chapter 2
Special Theory of Relativity
27
(b) The pi mesons share this equally, so each one has a kinetic energy of 114 MeV in that
frame. The energy of each pi meson is E  K  E0  114 MeV  135 MeV  249 MeV .
The momentum of each pi meson can be found:
 249 MeV   135 MeV 
E 2  E02
p

 209.2 MeV/c
c
c
104. (a) For each second, the energy used is 3.9 × 1026 J. The mass used in each second is
E
3.9 1026 J
m 2 
 4.3 109 kg
2
8
c
3.0 10 m/s 
2
2
(b) The time to use this mass is
t 
msun
2.0 1030 kg
 efficiency 
 0.007  3.3 1018 s  1.0 1011 y.
m / t
4.3 109 kg/s
That is 20 times longer than the expected lifetime of the sun.
2 r
 1.30 104 m/s . If the system’s center of mass
T
is at rest, conservation of momentum gives the star’s speed:
mp vp 1.90 1027

1.30 104 m/s = 12.4 m/s
mp vp  ms vs , so vs 
30
ms
1.99 10
105. a) The planet’s orbital speed is vp 
(b) The redshifted wavelength is   0
1 
 550 nm + 2.3 105 nm
1 
Similarly, the blueshifted wavelength is   0
1 
 550 nm  2.3 105 nm
1 
These small differences can be detected using modern spectroscopic tools.
106. To a good approximation, the shift is the same in each direction, so the redshift for the
approaching light is half the difference, or 0.0045 nm. Using this in the equation
1 
leads to a speed parameter β = 6.9 × 10−6, or v = βc = 2070 m/s. The speed
  0
1 
is equal to the circumference 2πR divided by the period, so the period is
8
2 R 2  6.96 10 m 
T

 2.11106 s = 24.5 days.
v
2070 m/s
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publicly accessible website, in whole or in part.
28
Chapter 3
The Experimental Basis of Quantum Theory
Chapter 3
1. The required field is homogeneous within the desired region and decreases in magnitude
to zero as rapidly as possible outside that region. The magnitude of the field is B  E / v0 .
The best design is an electromagnet with flat, parallel pole faces that are large compared
with the distance between them. But no matter what the design, it is impossible to
eliminate edge effects.
2. eE  evB
B
E (2.5 105 V/m)

 0.11T
v (2.2 106 m/s)
3. Assume the speed is exact. Non-relativistically use the energy obtained by accelerating
through a potential difference causes an increase in kinetic energy:
eV    12 mv 2 
31
7
mv 2  9.1094 10 kg 1.80 10 m/s 
V

 921.06 V
2e
2 1.6022 1019 C 
2
Relativistically eV  K  (  1)mc2 :





  511keV/c 2  2
1
V 
 1 
  c  923.59 V .
2
e
7




 1.80 10 m/s 
1





8
 2.9979 10 m/s 


The results differ by about 2.5 volts, or about 0.27%. Relativity is required only if that
level of precision is needed.
4. eE  evB so E  vB   4.0 106 m/s 1.2 102 T   4.8 104 V/m
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publicly accessible website, in whole or in part.
Chapter 3
The Experimental Basis of Quantum Physics
2
29
2
1 2 1  F    1  eE   
eE 2
y  at           
2
2  m   v0 
2  m   v0 
2mv02
1.602 10 C  4.8 10 V/m   0.02 m 

2  9.109 10 kg  4.0 10 m/s 
19
4
31
6
2
2
 1.0552 101 m  10.6 cm
5. The forces are shown below.
6. At terminal velocity the net force is zero, so Ff  fvt  mg and vt  mg / f .
7.
vt 
mg
mg

f
6  r
4
m    volume    r 3
3
2
 vt
4
  g  2g r
vt    r 3 

Solving for r: r  3

9
2g 
3
  6 r 
1.82 105 kg·m1  s 1 1.3 103 m/s 

 vt
3
 3.47μm
8. (a) r  3
2g 
2  9.80 m/s 2  900 kg/m3 
3
4
4
(b) m  V   r 3    900 kg/m3  3.47 106 m   1.58 1013 kg
3
3
13
2
mg 1.58 10 kg  9.80 m/s 

 1.19 109 kg/s
(c) f 
vt
1.3 103 m/s
1
1
 
1 
9. Lyman:    RH 1  2    RH1  1.096776 107 m1   91.2 nm
   
1
1
  1
1 
Balmer:    RH  2  2    4 RH1  4 1.096776 107 m1   364.7 nm
  2  
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publicly accessible website, in whole or in part.
30
Chapter 3

10. d  400 mm1

1
The Experimental Basis of Quantum Theory
 2.5μm  2500 nm and   d sin  in first order. Also tan   y / x so
y  x tan    3.0 m  tan sin 1 ( / d )  .

 656.5nm  
Red: y   3.0 m  tan sin 1 
   81.6 cm
 2500 nm  


 486.3nm  
Blue-green: y   3.0 m  tan sin 1 
   59.5 cm
 2500 nm  


 434.2nm  
Violet: y   3.0 m  tan sin 1 
   52.9 cm
 2500 nm  


11. d  420 mm1

1
 2.381 μm ;
  656.5nm for red. We know n  d sin  so
 n 
 . For n = 1 (first order) we find
 d 

 656.5 nm  
y  x tan    2.8 m  tan sin 1 
   80.3 cm .
 2381 nm  

Similarly for n = 2 we find y =185.1 cm, and for n = 3 we find y = 412.2 cm.
  sin 1 
Therefore the separations are:
between n =1 and n = 2, y  185.1cm  80.3cm  104.8cm ;
between n = 2 and n = 3, y  412.2cm  185.1cm  227.1cm .
12. Use Equation (3.13) with n = 4 for the Brackett series and n = 5 for the Pfund series. The
largest wavelengths occur for the smallest values of k.
1
1
  1 1 

 1 1 
Brackett:    RH  2  2    1.096776 107 m-1   2  2   .
 4 k 
  4 k 

For k = 5,   4.052 μm ; for k = 6,   2.626 μm ; for k =7,   2.166 μm ;
for k = 8,   1.945 μm .
1
1
  1 1 

 1 1 
Pfund:    RH  2  2    1.096776 107 m-1   2  2   .
 5 k 
  5 k 

For k = 6,   7.460 μm ; for k = 7,   4.654 μm ; for k = 8,   3.741 μm ;
for k = 9,   3.297 μm .
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publicly accessible website, in whole or in part.
Chapter 3
The Experimental Basis of Quantum Physics
31
13. Beginning with Equation (3.10) with n = 1, we have   d sin  with d  0.20mm .
d
 d cos  . Assuming the spectrum was viewed in the forward direction,
Therefore
d
d  

 d . An angle of 0.50 minutes of arc corresponds
then   0 and cos  1 so
d 
to 1.45 104 rad so
  d    0.20 103 m 1.45 104 rad  29 nm.
 1 1 
 RH  2  2  with RH  1.096776 107 m1 and n = 2 for

n k 
1
 1 1
the Balmer series. With k = 3 we find  1.096776 107 m1  2  2   656.47nm.

2 3 
This wavelength, found with the smallest value of k, is referred to as the hydrogen alpha
or H line. Using the equation above with k = 4 we find the hydrogen beta or H 
14. (a) Use Equation (3.13)
1
wavelength equals 486.27 nm. Similarly with k = 5 the hydrogen gamma or H
wavelength equals 434.17 nm, and finally with k = 6 the hydrogen delta or H
wavelength equals 410.29 nm.
(b) Because only three wavelengths are observed, the source must moving relative to the
detector. The wavelengths have increased, so the source must be moving away from the
detector. [See Equation (2.34) for example.] The H line at 656.47 nm is redshifted out
of the visible.
(c) Because the wavelengths are known and c   f , we will use the reciprocal of
Equation (2.33). We select this equation since the wavelengths are larger and thus the

1 
source is receding from the observer. observed 
. Using algebra to solve the
source
1 
equation for  and substituting the smallest wavelength gives
 observed 
 453.4nm 
 
 1 
 1
 source 
 410.29nm 


 0.0995 .
2
2
 observed 
 453.4nm 
 
  1  410.29nm   1


 source 
2
2
Therefore the speed is v  0.10c or v  3.0 107 m/s . Using other pairs of wavelengths
gives a similar result. If the object is rotating, then one side would be moving toward and
another side away from the detector so a range of wavelengths would be observed. This
effect can be used to determine the rotation speeds.
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publicly accessible website, in whole or in part.
32
Chapter 3
The Experimental Basis of Quantum Theory
 1 1 
 RH  2  2  with RH  1.096776 107 m1 and n = 3 for the

n k 
1
1 1
Paschen series. With k = 4 we find  1.096776 107 m1  2  2 ;   1875.63 nm .

3 4 
With k = 5,   1282.17 nm ; with k = 6,   1094.12 nm ;
with k = 7,   1005.22 nm; with k = 8,   954.86 nm.
15. (a) Use Equation (3.13)
1
(b) The observed spectral lines have been Doppler shifted. It might appear as if the
wavelengths have been blueshifted since the largest observed wavelength is smaller than
the largest expected wavelength using the Paschen series. However it is more likely that
the wavelengths have been redshifted and the calculated wavelength just below 1000 nm
in part (a) corresponds to the observed wavelength of 1046.1 nm.
(c) Using the formula from Problem 14 and noting from part (b) that the star is receding
2
 observed 
 1334.5nm 
 
 1 
 1
1282.17nm 


source 

 0.040 or v  1.20 107 m/s .
from the detector  
2
2
 observed 
 1334.5nm 
 
  1  1282.17nm   1


 source 
2
16. (a) To obtain a charge of +1 with three quarks requires two charges of +2e/3 and one of
charge  e / 3 . Three quarks with charge + e/3 would violate the Pauli Exclusion
Principle for spin 1/2 particles.
(b)\ To obtain a charge of zero we could have either two e / 3 and one 2e / 3 or one
2e / 3 and two e / 3 . At this point in the text there is no reason to prefer either choice.
(The latter turns out to be correct.)
2.898 103 m  K
 0.69 mm
17. (a) max 
4.2 K
2.898 103 m  K
 9.89 μm
(b) max 
293 K
2.898 103 m  K
 1.16 μm
(c) max 
2500 K
2.898 103 m  K
 0.322 μm
(d) max 
9000 K
2.898 103 m  K
 1.932 1011 K
18. (a) T 
14
1.50 10 m
2.898 103 m  K
 1.932 106 K
(b) T 
1.50 109 m
2.898 103 m  K
 4528 K
(c) T 
640 109 m
© 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 3
The Experimental Basis of Quantum Physics
33
2.898 103 m  K
 2.898 103 K
1m
2.898 103 m  K
 1.42 105 K
(e) T 
204 m
4
P1  T14
T14  2300 K 

19. (a)
; so P1  P0 4  
 P0  42.7 P0 ; The power increases by a factor
P0  T04
T0  900 K 
of 42.7.
(d) T 
(b) To double the power output, the ratio of temperatures to the fourth power must equal
4
 T 
2.  1   2 . Solving we find T1  1070 K .
 900 K 
2.898 103 m·K
 9.35 μm
310K
(b) At this temperature the power per unit area is
20. (a) max 
R   T 4   5.67 108 W·m2  K 4   310K   524W/m2 . The total surface area of a
4
cylinder is 2 r  r  h   2  0.13 m 1.75 m  0.13 m   1.54 m2 so the total power is
P   524 W/m2 1.54 m2   807 W.
(c) The total energy radiated in one day is the power multiplied by the time;
E  P  t  807 W  86400 s   6.97 107 J.
2000 kcal   2 106 cal    4.186 J/cal   8.37 106 J .
There are several assumptions. First, a cylinder may overestimate the total surface area;
second, radiation is minimized by hair covering and clothing.
21. max 
2.898 103 m  K
 1.447 108 m
200, 273 K
22. We know from Example 3.8 that in the long-wavelength limit, the two expressions are
the same. By comparing the expressions, we can also note that the Rayleigh-Jeans
spectral distribution will be larger than the Planck expression for a given  and T.
[Compare graphs or calculations using Equations (3.22) and (3.23) one can obtain using
Excel, Mathcad, or similar program.] So we want

 2 c 2 h  
2 ckT
1
 . Simplifying we find
 0.95  4    5   hc / kT
 1 
       e
 kT
 kT 
1
1
 x and this simplifies to  0.95 x  1/ x
. Let
. We
 0.95 
  hc /  kT
hc
e  1
 1
 hc  e
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publicly accessible website, in whole or in part.
34
Chapter 3
The Experimental Basis of Quantum Theory
can solve this transcendental equation to find x = 9.83 so 9.83 
 kT
hc
. Substituting
numerical values for the constants we find
9.83 1.986 1025 J  m 
9.83hc


 2.44 105 m  24.4 μm.
23
1
kT
1.38 10 J  K  5800K 
2.898 103 m  K
 966 nm which is in the near infrared.
3000 K
24. The graph is a characteristic Planck law curve with a maximum at   966 nm (see
Problem 23).
(a) Numerical integration of the ( , T ) function shows that approximately 8.1% of the
radiated power is between 400 nm and 700 nm. Details of the calculation are:
710
710
 4.798  106  5
 hc  5
16
2 c 2 h 
exp  

d


3.74

10
exp



410      d  That is
410
  kT 
23. max 
7
7
7
7
 3.71105 W/m2
the power per unit area emitted over visible wavelengths. Over all wavelengths we
know the power per unit area is R   T 4  4.59 106 W/m2 . Therefore the fraction
3.7128 105 W/m2
 0.081 .
emitted in the visible is
4.593 106 W/m2
(400 nm, T )
 0.073 ;
(966nm, T)
(b) Using computed numerical intensity values
(700 nm, T )
 0.754 .
(966nm, T)
2 c 2 h
 hc 
exp  
 . The exponential goes

  kT 
to zero faster than  5 , so the intensity approaches zero in this limit.
25. In this limit exp  hc /  kT   1 , so
( , T ) 
5
26. (a) At this temperature the power per unit area is
R   T 4   5.67 108 W  m2  K 4   293 K   419 W/m2 . For the basketball, a sphere
4
of radius r = 12.5 cm, we obtain P  R  4 r 2    419 W/m2   4  0.125 m   82.3 W .
2
(b) At this temperature the power per unit area is
R   T 4   5.67 108 W  m2  K 4   310 K   524 W/m2 . Assume the body is roughly
4
cylindrical with a radius of 13 cm and a height of 1.65 m. The total surface area of a
cylinder is 2 r  r  h   2  0.13 m 1.78 m   1.45 m2 so the total power is
P   524 W/m2 1.45 m2   760 W. Numerical values will vary depending on estimates
of the human body size.
© 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 3
The Experimental Basis of Quantum Physics
35
2.898 103 m  K
 9.35 μm
310 K
2
 2  2 a
n 2 2
 n x   
 n x 
28. Taking derivatives
;

sin


a
sin 


 . Substituting these
2
2
2
2
t
t
L
 L  x
 L 
27. max 
values into the wave equation produces
1  2 a  n x   n2 2
 n x  
sin 
   a 2 sin 
  0
2
2
c t
L
 L  
 L 
n c
2L
2a
n2 2c 2


a
  2 a where  
. Since  
for this
2
2
L
n
t
L
system and c   f , then   2 f .
which simplifies to
29. Let r  nx2  ny2  nz2 be the radius of a three-dimensional number space with the ni the
three components of that space. Then let dN be the number of allowed states between r
and r  dr . This corresponds to the number of points in a spherical shell of number
1
space, which is dN   4 r 2 dr where we have used the fact that 4 r 2 dr is the
8
“volume” of the shell (area 4 r 2 by thickness dr ), and the 1/8 is due to the fact that only
positive numbers ni are allowed, so only 1/8 of the space is available. Also
r 2  nx2  ny2  nz2 
2Lf
2 L2 4 L2 f 2
 2L 

or r 
. Then from this dr  
 df . Putting
2 2
2
c
 c
c
 c 
1

  2 Lf  2 L
4 L3 2
everything together: dN   4 r 2 dr  r 2 dr  
df

f df

8
2
2 c  c
c3
2
30. From the diagram at right we see that the
average x-component of the velocity c of

electromagnetic radiation within the cavity is
 /2
cx
  ccos 2 r sin  d .


 2 r sin  d
2
0
/2
2
0
1
Letting u  cos  we have cx 
c  udu
0
1
 du

c
.
2
0
On average only one-half of the photons are
traveling to the right. Thus the mean velocity of photons traveling to the right is c/4.
c
Therefore power =  intensity  area   dU  A .
4
© 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.
36
Chapter 3
The Experimental Basis of Quantum Theory
31. For classical oscillators the Maxwell-Boltzmann distribution gives
n( E )  A exp   E / kT   A exp   E  where E  nhf and   1/ kT . The mean energy

is E 
 En( E )
n 0

 n( E )
n 0


 nhf exp   nhf 
n 0

 exp   nhf 
. Notice that E 


ln  exp    nhf  . Now
 n0
n 0
letting x  exp   hf  we see that by Taylor series

 exp   nhf   1  x  x
n 0
E
2
 ...  1  x  for x 2  1.
1
hf exp    hf 


1
ln 1  x    ln 1  x  


1  exp    hf 
hf
. Using the result of Problem 29 (along with a factor of 2 for two
exp  hf / kT   1
photon polarizations) we can see that
4 2
hf
8 hf 3 / c3
U ( f ,T )  2 3 f

.
c
exp  hf / kT   1 exp  hf / kT   1
To change from U to requires the factor c/4 (Problem 30), and changing from a
frequency distribution requires a factor c /  2 , because with f  c /  we have
E
df   c /  2 d  . Putting these together
( , T ) 
2
8 h /  3
1
 c  c  2 c h

.
 2  
5
exp  hf / kT   1    4 
 exp  hc /  kT   1
32. Energy per photon  hf   6.626 1034 J  s  98.1106 s1   6.50 1026 J
5.0 10
4
J/s 
1 photon
 7.69 1029 photons/s
26
6.50 10 J
33. (a) Energy per photon  hf   6.626 1034 J  s 1100 103 s1   7.29 1028 J
180 J/s 
1 photon
 2.47 1029 photons/s
7.29 1028 J
(b) energy per photon  h
 3.00 108 m/s 
17
  6.626 1034 J  s  
  2.48 10 J
9

8

10
m


c
1 photon
 7.26 1018 photons/s
17
2.48 10 J
1 photon 1 MeV
 2.811014 photons/s
(c) 180 J/s 
13
4 MeV 1.60 10 J
180 J/s 
© 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 3
The Experimental Basis of Quantum Physics

37
2.93 eV
hc
1  hc

 7.08 1014 Hz : eV0    so V0 =     ;
15
h 4.136 10 eV  s

e

1 1240 eV  nm

V0  
 2.93 eV   0.333 V
e  380 nm

34. ft 
35. t 

hc


1240 eV  nm
 267.2 nm .
4.64 eV
If the wavelength is halved (to   133.6 nm), then
K
hc

 
36. Notice that
1240 eV  nm
 4.64 eV  4.64 eV
133.6 nm
hc


1240eV  nm
 2.33eV   , so photoelectrons will be produced.
532 nm
19
 105 electrons   1 photon  
1 eV
  1.60 10 C 
3
2

10
J/s

  1 photon   2.33eV  1.60 1019 J   electron   8.58 nA





hc 1240 eV  nm
37.  

 4.59 eV ; K  2.0 eV  hf   ;
t
270 nm
K 
2.0 eV  4.59eV
f 

 1.59 1015 Hz
15
h
4.136 10 eV  s
1240 eV  nm
 hc 
38. E  100    100
 203 eV
610 nm
 
39. eV01  hc / 1   and eV02  hc / 2   . Subtracting these equations and rearranging we
find
h
e V02  V01 
1 1
c  
 2 1 

e  2.3 V  1.0 V 
3.00 10
8
1
 1

m/s  


 207 nm 260 nm 
 4.40 1015 eV  s .
This is about 6% from the accepted value. For the work function we use the first set of
data (the second set should give the same result):
 4.40 1015 eV  s 3.00 108 m/s   1.0 eV  4.1 eV
hc
   eV01 
1
260 109 m
© 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.
38
Chapter 3
40. hf    K or
hc

The Experimental Basis of Quantum Theory
2
   K    12 mvmax
; Using 4.63 eV as the work function for tungsten
(see table 3.3) and 9.111031 kg as the mass of the electron, one can solve for  .




hc
1240 eV  nm



2
  12 mvmax
  4.63eV  12  9.111031 kg 1.4 106 m/s 2 1eV19 
1.6 10 J 

2
 1.2110 nm  121nm
41. min
1.24 106 V  m 1.24 106 V  m


 0.0413 nm
V
30 kV
hc 1240 eV  nm

 2.48 1017 m. A photon produced by bremsstrahlung is still an
K
5 1010 eV
x ray, even though this falls outside the normal range for x rays.
42.  
43.  
1240 eV  nm
 0.0496 nm
2.5 104 eV
44. Because the electron is accelerated through a potential difference, it has kinetic energy
equal to K  eV0 . As described in the text, Equation (3.37) gives the minimum
wavelength (assuming the work function is small). So
6
 hc   1  1.240 10 V·m
min      
 3.54 1011 m  3.54 102 nm .
4
e
V
3.5

10
V
  0 
The work function for tungsten from Table 3.3 is 4.63 eV. If we include the work
function, the energy available for the photon and the wavelength of the photon will
change by 4.63 parts out of 35000 or about 0.013% which is very small.
45. From Figure (3.19) we observe that the two characteristic spectral lines occur at
wavelengths of 6.4 1011 m and 7.2 1011 m . Rearrange Equation (3.37) to solve for
hc 1
1.24 106 V  m

 17.2 kV and we have used the larger of
e min
7.2 1011 m
the two wavelengths in order to determine the minimum potential difference.
the potential V0 ; V0 
h
1  cos   so at maximum cos  1 and
mc
2 1240 eV  nm 

2h
2hc



 1.01105 . This corresponds to
2
  mc mc   511.0 keV  480 nm 
46.  
  4.85 1012 m and therefore is not easily observed.
© 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 3
The Experimental Basis of Quantum Physics
39
47. The maximum change in the photon’s energy is obtained in backscattering (   180 ), so
2h
 4.853 1012 m. The photon's original wavelength was
1  cos  2 and  
mc
hc 1240 eV  nm
 
 0.0310 nm  3.10 1011 m and the new wavelength is
E
40000 eV

      3.585 1011 m. The electron's recoil energy equals the change in the
photon's energy, or
K
hc


hc 1240 eV  nm
1240 eV  nm


 5411 eV  5.41 keV
2
  3.10 10 nm 3.585 102 nm
48. Use the Compton scattering formula but with the proton's mass and   90 :
h
h
hc 1240 eV  nm
 
 1.32fm .
1  cos    2 
mc
mc mc
938.27 MeV
49. We find the Compton wavelength using c 
photon energy is E 
hc
c
requirements are high.
50.


 0.004 
h
hc 1240 eV  nm
 2
 1.32 fm. The
mc mc
938.27 MeV
 938 MeV . In principle this could be observed, but the energy
c
1  cos 

so
  250c 1  cos   ;
(a)   250  2.43 1012 m 1  cos30   8.14 1011 m
(b)   250  2.43 1012 m 1  cos90   6.08 1010 m
(c)   250  2.43 1012 m 1  cos170   1.21109 m
51. By conservation of energy we know the electron’s recoil energy equals the energy lost by
hc hc hc       hc 

the photon: K   
.
 
 
 
  /   hf
hc 
hf 
hf 



Using       we have K 
.
          1   /   1  

h
h
Conservation of px : pe cos   cos   ;


h h
pe cos    cos  .
(1)
 
Conservation of p y : pe sin  
h
sin   0 ;

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publicly accessible website, in whole or in part.
40
Chapter 3
pe sin  
h
sin 

The Experimental Basis of Quantum Theory
(2)
h
sin 


Dividing equation (2) by equation (1), tan  
.
h h
 cos 
 
h sin 
h

1  cos  
h
mc
tan


1

cos

Using     
we
have:
.


h
h cos 
mc

   h 1  cos 
mc


h

Multiplying numerator and denominator by   
1  cos   , we have:
mc


 h sin 
 sin 
tan  

.
h2
   h  1  cos 

h 
1  cos     h cos  

mc 

mc
sin 
 
Using the trig identity:
 cot   , we find:
1  cos  
2

1
1
 
 
 
tan  
cot   
cot   
cot   .
h
 2  1 h
 2  1  hf
 2

mc
mc
mc 2
hf 

 
Inverting the equation gives cot   1  2  tan   .
 mc 
2
52.      c 1  cos   
hc
 c 1  cos   ;
E
1240 eV  nm
  2.43 103 nm 1  cos110   5.17 pm
3
650 10 eV
hc 1240 eV  nm
E 

 2.40 105 eV  240 keV
3

 5.17 10 nm
 
By conservation of energy we find: Ke  E  E  650keV  240keV  410keV which
agrees with the K formula in the previous problem. Also from the previous
 110 
hf 

    650 keV 
cot   1  2  tan    1 
tan
problem we have:

  3.245 so
 mc 
 2   511 keV 
 2 
  17.1 .
53. For   90 we know      c  2.00243nm ;

c 2.43 103 nm
 
 1.22 103


2 nm

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publicly accessible website, in whole or in part.
Chapter 3
The Experimental Basis of Quantum Physics
41
54. E  2mc2  2  938.3 MeV   1877 MeV . This energy could come from a particle
accelerator.
55. a) To find the minimum energy consider the zero-momentum frame. Let Ee be the energy
of the electron in that frame, and E0 is the rest energy of the electron. From conservation
of energy and momentum:
Ee2  E02
hf
 pe 
or hf  Ee2  E02 .
c
c
energy: hf  Ee  3E0 or hf  3E0  Ee .
momentum:
Squaring and subtracting these two equations gives: 0  10E02  6Ee E0 which simplifies
5
to Ee  E0 . This tells us that for the transformation from the lab frame to the zero3
momentum frame,   5 / 3 and v  0.8c . Then from the momentum equation we have in
25E02
4
 E02  E0 .
9
3
In the lab the photon’s energy is obtained using a Doppler shift:
1  4
1  0.8
hf lab  hf
 E0
 4 E0  2.04 MeV
1  3
1  0.8
the zero-momentum frame: hf 
b) The proton’s rest energy is Mc 2 . Now as in (a) we let the proton’s energy in the lab
frame be Ep and conservation of momentum and energy give
momentum: hf  E p2   Mc 2  ;
energy: hf  E p  2E0  Mc2 .
Squaring and subtracting, we find E p
 Mc 

2
2 2
 2 E02  2 E0 Mc 2
2 E0  Mc 2
. This is very close to
E p  Mc 2 . Therefore the zero-momentum and lab frames are equivalent, and we
conclude hflab  2E0  1.02 MeV .
56. The maximum energy transfer occurs when   180 so that
  (h / mc)(1  cos )  2h / mc . By conservation of energy the kinetic energy of the
hc hc hc
hc
electron is K  E  E    
. Multiplying through by  (   ) we
      
find  (   ) K  hc(   )  hc  hc or  2 K    K  hc  0 . This is a
quadratic equation that with numerical values can be solved for  to find
hc 1.24 keV  nm
 104 keV .
  1.20 1011 m . Then E  
 1.20 102 nm
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publicly accessible website, in whole or in part.
42
Chapter 3
The Experimental Basis of Quantum Theory
57. The power of the light received is intensity times area, or
2
P  IA   4.0 1011 W/m2    4.5 103 m    2.5 1015 W . The energy of each


hc
 6.626 10

34
J  s  3.00 108 m/s 
 3.6 1019 J . Thus, the number
5.50 107 m
1 photon
 6900 photons .
of photons per second is 2.5 1015 J/s 
3.6 1019 J
photon is E 
58. (a) max 

2.898 103 m  K
 2.13 106 m
1358 K
(b) The peak is well into the infrared part of the spectrum. However, the shape of the
“tail” of the distribution implies that more red photons are emitted than from any other
part of the visible spectrum, and so we see red.
59. To find the asteroid mass m, note that Earth (matter) would supply an equal mass m to the
GM E2
process, so 2mc 2 
. Solving for m we find:
2 RE
11
3
1
2
24
GM E2  6.67 10 m  kg  s  5.98 10 kg 
m

 1.04 1015 kg .
2
3
8
4 RE c 2
4  6378 10 m  3.00 10 m/s 
2
 3 1.04 1015 kg  


 4  5000 kg/m3  


Evaluating the energy:
1/3
1/3
 3m 
Then r  

 4 
GM E2  6.67 10
E

2 Re
11
 3.68 km which is relatively small.
m3  kg 1  s 2  5.98 1024 kg 
2  6378 10 m 
3
2
 1.87 1032 J ;
E
1.87 1032 J

 8.9 1012 . There is a lot of energy in the
nuclear arsenals 5000  4.2 1015 J 
annihilation process!
60. For maximum recoil energy the scattering angle is   180 and   0 . Then as usual
2h
 
. Using the result of Problem 56,
mc
 / 
2h / mc
2hf / mc 2
K
hf 
hf 
hf .
1   / 
1  2h / mc
1  2hf / mc 2
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publicly accessible website, in whole or in part.
Chapter 3
The Experimental Basis of Quantum Physics
43
 2hf  2  hf 
We can rearrange this equation: K 1  2  
or
mc 2
 mc 
2
 2 
 2K 
 2   hf    2   hf   K  0 . This constitutes a quadratic equation in hf which
 mc 
 mc 
can be solved with the given value of K  100 keV numerically to yield hf  217 keV.
2
61. See graph below.
62. (a) For   180 we have     2h / mc . Therefore E  
(b) With  
hc


hc

2h
.
mc
hc
we find
E
1


1
2
E 



  71.9 keV .
5
hc 2h
1
2
 110 eV 511000 eV 


2
hc
E
mc
1
E
mc
63. (a) The energy per second detection capability of the detector in the Hubble Space
Telescope would be  2.0 1020 W/m2 0.30 m2  6.0 1021 J/s . Each photon at 486 nm
would have energy E  hf 
hc
 6.626 10

34
J  s  3.0 108 m/s 

486 10 m
Therefore the average number of photons per second would be
6.0 1021 J/s
n
 0.015 photons/s
4.09 1019 J/photon
and long exposure times would be required.
9
 4.09 1019 J .
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publicly accessible website, in whole or in part.
44
Chapter 3
The Experimental Basis of Quantum Theory
(b) If 30th magnitude corresponds to 2 1020 W/m2 and each change in magnitude by one
represents a change in brightness of 1001/ 5  2.5119 then we must increase by a factor of
100 
1/ 5 24
 (2.5119)24  3.98 109 . So 6 th magnitude corresponds to 7.96 1011 W/m2 .
If the diameter of the pupil is 6.5 mm, then the power reaching the retina is
2
P  7.96 1011 W/m2   3.25 103 m    2.64 1015 J/s. With each photon carrying


 photons 
energy 4.09 1019 J , then 2.64 1015 J/s  4.09 1019 J  n
 and
s


3
n  6.45 10 photons each second.
64. The laser’s power is 25 kW = 2.5 × 104 J/s. The energy of a single photon is
34
8
hc  6.626 10 J  s  3.00 10 m/s 
E

 1.88 1019 J . Thus,

1.06 106 m
1 photon
2.5 104 J/s
 1.3 1023 photons/s .
1.88 1019 J
65. The laser’s power is 25 kW = 2.5 × 104 J/s. The energy of a single photon is
34
8
hc  6.626 10 J  s  3.00 10 m/s 
E

 1.88 1019 J . Thus,
6

1.06 10 m
1 photon
2.5 104 J/s
 1.3 1023 photons/s
19
1.88 10 J
66. From Example 3.5, the sun uses 3.91 × 1026 J/s. At that rate, the sun’s lifetime is
5 106 J/kg
2.0 1030 kg
= 2.6 1010 s or about 820 years! This estimate is off, because
3.911026 J/s
the sun uses nuclear energy, not chemical.
67. (a) max 
2.898 103 m  K
 3.01107 m = 301 nm
9600 K
This is actually in the ultraviolet part of the spectrum, but the shape of the distribution
function guarantees that most of the visible photons will be in the violet-blue region,
which explains the star’s blue appearance.
(b) Assuming a perfect blackbody,


2
4
P   T 4 A  5.67 108 W/  m2  K 4   9600 K   4 1.6  6.96 108 m  


27
 7.5110 W
This is about 19 times the value for the sun, reported in Example 3.5.
© 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 3
The Experimental Basis of Quantum Physics
45
68. Kinetic energy is related to momentum (nonrelativistically) by K  p 2 / 2m , so an
electron of mass m and kinetic energy K has momentum p  2mK . Thus the magnitude
of the electron’s change in momentum is p  2mKf  2mKi  7.5 1024 (SI units)
By conservation, this is the momentum of the recoiling nucleus. The nucleus then has
kinetic energy
 p 
K
2
 9.2 1023 J = 5.8 104 eV This is negligible, compared with
2m
the energy (5 keV) of the x ray produced in the process.
69. For the 10-keV gamma ray, the electron absorbs half this amount, so its kinetic energy is
5 keV. That is less than 1 percent of its rest energy, so we’ll use a non-relativistic
1
calculation K  mv 2
so
2
v  2K / m  2  5000 eV  1.60 1019 J/eV  / 9.111031 kg  4.2 107 m/s .
Half of the 300-MeV gamma ray’s energy is 150 MeV, which is highly relativistic. In
K
150 MeV
this case K    1 E0 so   1 

 294 and   295 . Solving for
E0 0.511 MeV
1
speed v,   295 
, which yields β = 0.999994 . Thus v = 0.999994 c.
1  2
70. If n gamma rays strike the target, the energy deposited is Q = nE0, where E0 = 100 MeV
is the energy of each gamma ray. Converting to SI units,
1.60 1013 J
E0  100 MeV 
 1.60 1011 J . The relationship between thermal energy
1 MeV
and temperature is Q  mcV T , where cV is the specific heat. Solving for n:
Q  nE0  mcV T ; n 
mcV T  2.5 kg  430 J/(kg  K)  0.010 K 

 6.7 1011
11
E0
1.60 10 J
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46
Chapter 4
Structure of the Atom
Chapter 4
1. With more than one electron we are almost forced into some kind of Bohr-like orbits.
This was the dilemma faced by physicists in the early 20th Century.
2  7.7 MeV 
1 2
2K
mv and v 

 6.4281102 c .
2
m
3727 MeV/c 2
7.7 MeV
 1.002066
Relativistically K    1mc 2 so   1  K / mc 2 and   1 
3727 MeV
1
which gives   1  2  6.4181102 c . The difference is about 10-4 c or about 0.16%

of the velocity.
2. Non-relativistically K 
3. Conserving momentum and energy:
M  v  M  v   me ve
(1)
(2)
M v2  M v  2  meve 2
m
From (1) we see v   v  e ve which, when substituted into (2) gives
M
2


m
M  v  M  v  e ve   meve 2 .
M 


m 
This can be solved to find ve 1  e   2v .
 M 
2
But with me  M  we have ve  2v .
 802 
4. P( )  exp   2   3 102780
 1 
Therefore multiple scattering does not provide an adequate explanation.
5. (a) With Z1  2 , Z 2  79 , and   1 we have
9
Z1Z 2e2
    2  79  1.44 10 eV  m 
b
cot   
cot  0.5   1.69 1012 m
6
8 0 K
2
2  7.7 10 eV 
 
9
Z1Z 2e2
    2  79  1.44 10 eV  m 
cot   
cot  45   1.48 1014 m
(b) For   90 b 
6
8 0 K
2  7.7 10 eV 
2

2
 Z1Z 2e2 
2  
6. f   nt 
 cot   . For the two different angles everything is the same except
2
 8 0 K 
f (1 ) cot 2 (0.5 )

 4.00.
the angles, so the ratio is
f (2 ) cot 2 1.0 
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Chapter 4
Structure of the Atom
47
7. The fraction f is proportional to nt and to Z 2 from Equation (4.12). However, the
question states that the number of scattering nuclei per unit area (equal to nt) is the same
N (Au) n(Au)t (79) 2 792
for either target. Therefore


 36.93 .
N (Al) n(Al)t (13) 2 132
8. From Example 4.2 we know n  5.90 1028 m3 . Thus
2
 Z Z e2 
 
f   nt  1 2  cot 2  
2
 8 0 K 
  2  79  1.44 109 eV  m  
 cot 2 (4 )
m 
6


2
5

10
eV




2
   5.90 1028 m 3 108
 1.96 104
9. (a) With all other parameters equal the number depends only on the scattering angles, so
2

f (90 ) cot 45 

 0.217 so the number scattered through angles greater than 90 is
f (50 ) cot 2 25 
(10000) ( 0.217) = 2170.
2

2

f (70 ) cot 35 
f (80 ) cot 40 

 0.4435 and

 0.3088 . The
(b) Similarly:
f (50 ) cot 2 25 
f (50 ) cot 2 25 
numbers for the two angles are thus 4435 and 3088 and the number scattered between
70 and 80 is 4435  3088  1347 .
10. From the Rutherford scattering result, the number detected through a small angle is
n(50 ) sin 4 (3 )
 

 2.35 104 and if they
inversely proportional to sin 4   . Thus

4

n
(6
)
2
sin
25
 
 
count 2000 at 6 the number counted at 50 will be  2000   2.35 104   0.47 which is
insufficient.
11. In each case all the kinetic energy is changed to potential energy:
Z1Z 2e2
where r1 and r2 are the radii of the particular particles (as
K  V  V 
4 0  r1  r2 
the problem indicates the particles just touch).
(a) Z1  2 and Z 2  13 for Al, Z 2  79 for Au;
Al: K 
Au: K 
(2)(13) 1.44  109 eV  m
2.6 1015 m  3.6  1015 m
(2)(79) 1.44  109 eV  m
2.6 1015 m  7.0  1015 m
 6.04 MeV
 23.7 MeV
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publicly accessible website, in whole or in part.
48
Chapter 4
Structure of the Atom
(b) Now Z1  1 and for the two different values of Z 2 :
(1)(13) 1.44  109 eV  m
Al: K 
 3.82 MeV
1.3 1015 m  3.6  1015 m
(1)(79) 1.44  109 eV  m
Au: K 
 13.7 MeV
1.3 1015 m  7.0  1015 m
12. (a) The maximum Coulomb force is at the surface and equal to 2Z 2e2 / 4 0 R 2 . Then
2Z 2 e 2 2 R 4Z 2 e 2
p  F t 

. For maximum deflection
4 0 R 2 v
4 0 Rv
2Z 2 e2
p 1 4Z 2e2
  tan  


, for small angles.
p mv 4 0 Rv 4 0 KR
 2  79 1.44 eV  nm   2.11104 rad
8 MeV  0.135 nm 
 2  79 1.44 eV  nm   2.28 104 rad
13. (a)  
10 MeV  0.1 nm 
(b)  
 0.012
 0.013
(b) These results are comparable in magnitude with those obtained by electron
scattering (Example 4.1).
14. (a) v 
e
4 0 mr

ec
4 0 mc r
2

1.44 eV  nm
 511000 eV  1.2 106 nm 
c  1.53 c
which is not an allowed speed.
e2
1.44 eV  nm

 600 keV
(b) E  
8 0 r
2 1.2 106 nm 
(c) Clearly (a) is not allowed and (b) is too much energy.
15. (a) v 
e
4 0 mr

ec
4 0 mc r
2

1.44 eV  nm
938 10
6
eV   0.05 nm 
c  1.75 104 c
or v  5.25 104 m/s .
1.44 eV  nm
 14.4 eV
8 0 r
2  0.05 nm 
(c) The “nucleus”' is too light to be fixed, and there is no way to reconcile this model
with the results of Rutherford scattering.
(b) E  
e2

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Chapter 4
Structure of the Atom
49
16. For hydrogen:
e
ec
v


4 0 mr
4 0 mc 2 r
1.44 eV  nm
 51110
3
eV   0.0529 nm 
c  7.30 103 c
 2.19 106 m/s
6
v 2 2.19 10 m/s 
a 
 9.07 1022 m/s 2
11
r
5.29 10 m
2
For the hydrogen-like Li  we know: F 
Ze2
mv 2
Z e2
2

v

or
. We also
4 0 r 2
r
4 0 r m
4 0 2 a0
 .
Zme2
Z
2 2
32 1.44 eV  nm 
Z e
v2 

 4.79 104 c 2 or
11
3
2
4 0 a0 m  5.29 10 m  51110 eV/c 
know r 
v  2.19 102 c  6.57 106 m/s . This is a factor of 3 greater than the speed for
6.57 106 m/s 

v2
 2.45 1024 m/s 2
hydrogen. The acceleration is: a  
r 5.29 1011 m / 3
2
17. Both forces are attractive. The magnitude of the gravitational force:


6.67 1011 Nkgm2  9.111031 kg 1.67 1027 kg 
Gm1m2
Fg 

 3.6 1047 N
2
2

11
r
5.3 10 m 
2
The magnitude of the electrical force:
28
1 q1q2
e2 1  2.3110 J  m 
Fe 


 8.2 108 N
2
2
2

1
1
4 0 r
4 0 r
5.3 10 m 
The electrical force is much larger. The ratio is Fe / Fg 
8.2 108 N
 2.28 1039
3.6 1047 N
18. The total energy of the atom is e2 / 8 0 r  . Differentiating with respect to time:
dE
e2 dr

. Equating this result with the given equation from electromagnetic
dt 8 0 r 2 dt
2
2
e2 dr
2e2  d 2 r 
dr
1 2e2  d 2 r 
theory gives:
.
This
simplifies
to





 .


2r 2 dt
3c3  dt 2 
8 0 r 2 dt
4 0 3c3  dt 2 
e2
In a circular orbit
a
d 2r
is just the centripetal acceleration, which is also given by
dt 2
F
e2

.
m 4 0 mr 2
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50
Chapter 4
2

dr
e2 dr
2e2  e2

Substituting: 2
or
 3 
2 
2r dt
3c  4 0 mr 
dt
 4
Solving by separation of variables: dt    4
3m2c3
0
4e4
t  1.55 1011 s.
t    4
2

0
a0
r 2 dr   4
0

2
0
2
4e4
0
2
3m2c3r 2
Structure of the Atom
.
3m2c3 2
r dr which can be integrated:
4e4
m2c3 3
a0 . Inserting numerical values we find
4e4
 1 1 
 1
 1
 Z 2 Rc  2  2  ; f ( K )  Z 2 Rc 1   and f ( K  )  Z 2 Rc 1   and

 4
 9
 nl nu 
1 1
f ( L )  Z 2 Rc    . With 1  1/ 4  (1/ 4  1/ 9)  1  1/ 9 we see
4 9
f ( K )  f ( L )  f ( K  ) .
19. f 
c
20. (a) As in Problem 16, v  2.19 106 m/s and
L  mvr   9.111031 kg  2.19 106 m/s 5.29 1011 m   1.0554 1034 kg  m 2/s
Notice that L  .
(b) Using Equation (4.31) for v and Equation (4.24) for rn we find
1
 2
L  mv2 r2  m 
 n a0  n  2 as expected.
n
ma
0 

15
21. hc   4.135669 10 eV  s   299792458 m/s   1239.8 eV  nm ;
9
1.6021733 1019 C 

e2
1 eV

 10 nm

 1.4400 eV  nm


4 0 4 8.8541878 1012 F/m   1.6021733 1019 N  m  m
2
mc 2  510.99906 keV/c 2 c 2  511.00 keV ;
12
34
4 0 2 4  8.8541878  10 F/m1.05457  10 J  s
a0 

2
me2
 9.1093897 1031 kg1.6021733 1019 C
2
 5.2918  1011 m  5.2918  102 nm
E0 
e2
8 0 a0
1.6021733 10 C

8  8.8541878 10 F/m  5.2917725 10
19
2
12
11
m
 2.179874 1018 J  13.606 eV
22. For a hydrogen-like atom E 
 Z 2 E0
  Z 2 E0 for n = 1.
2
n
H: E   E0  13.6 eV
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publicly accessible website, in whole or in part.
Chapter 4
Structure of the Atom
51
He : E  4E0  54.4 eV
Li  : E  9E0  122.5 eV
The binding energy is larger for atoms with larger Z values due to the greater attractive
force between the nucleus and electron.
1240 eV  nm
 3.02 eV . This is the energy difference

410 nm
between the two states in hydrogen. From Figure 4.16, we see E3  1.51 eV so the
initial state must be n = 2 in order to have a large enough difference. This energy
difference exists between n = 2 (with E2  3.40 eV) and n = 6 (with E6  0.38 eV).
23. The photon energy is E 
hc

1240 eV  nm
 13.05 eV . This can only be a transition to

95 nm
n = 1 ( E1  13.6 eV ) and because of the energy difference it comes from n = 5 with
E5  0.54 eV .
24. The photon energy is E 
hc

25. In general the ground state energy is Z 2 E0 .
(a) E  12 E0  13.6 eV . Using the reduced mass does not change this result to three
significant digits.
(b) E  22 E0  54.4 eV
(c) E  42 E0  218 eV
26. Following the strategy of Example 4.8, we use Equation (4.37) to determine appropriate
Rydberg constants for atomic hydrogen, deuterium, and tritium. The Balmer series has n
lower equal to 2. The  refers to nu  3 ;  refers to nu  4 , etc. Then we use Equation
 1 1 
 R  2  2 

 n nu 
The example with n upper equal to 3 is completed in Example 4.8; the results are
repeated here:
  H ,hydrogen   656.47 nm ;   H ,deuterium   656.29 nm ;
(4.30) to calculate the “isotope shifted” wavelengths.
1
  H ,tritium   656.23 nm
  H  ,hydrogen   486.27 nm ;   H  ,deuterium   486.14 nm ;
  H ,tritium   486.10 nm
  H ,hydrogen   434.17 nm ;   H ,deuterium  434.05 nm
  H ,tritium   434.02 nm
  H ,hydrogen   410.29 nm ;   H ,hydrogen   410.18 nm
  H ,hydrogen   410.15 nm
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52
Chapter 4
Structure of the Atom
27. We know from Equations (4.25) and (4.29) that E1  hcR . We must adjust the Rydberg
constant to account for the finite mass of the nucleus. Use the calculations of example
4.8 that provide the constants for deuterium RD  0.99973R and tritium
RT  0.99982 R . Then we see
E1, D  hc  0.99973 R  1239.8 eV  nm  0.99973 1.097373 10 2 nm 1   13.602 eV
E1,T  hc  0.99982  R  1239.8 eV  nm  0.99982  1.097373 10 2 nm 1   13.603 eV In
Problem 21 we found that E1,H  13.606 eV .
28. (a) It is only the first four lines of the Balmer series, with wavelengths 656.5 nm, 486.3
nm, 434.2 nm, and 410.2 nm.
(b)To determine the energy levels in ionized helium, perform the same analysis as in the
text but with e 2 replaced by Ze2  2e2 . This results in an extra factor of Z 2  4 in the
energy, so the revised Rydberg-Ritz equation is
1 4 E0  1 1 
1 
7
1  1

 2  2   4.377 10 m  2  2  . We need the wavelength to be
 hc  nl nu 
 nl nu 
between 400 and 700 nm. The combinations of nl and nu that work are tabulated below:
nl nu  (nm)
comments
3
4
470
4
6
658
4
4
7
8
543
487
4 13
5 12
5 13
5 
404
691
670
571
etc. with nl  4 to nu  13
but nl  4 and nu  13 gives   400 nm
etc. with nl  5 all the way to...
a series limit
29. From Equation (4.31), vn  (1/ n)  / ma0  which gives the speed in the n = 3 state as
v  7.30 105 m/s. The radius of the orbit is n2 a0  9a0 . Then from kinematics:
7.30 105 m/s 108 s 

vt
number of revolutions 

 2.44 106.
11
2 r
2  9   5.29 10 m 
 1239.8 eV  nm 
30. The energy of each photon is hc /   
  3.12 eV . Looking at the
397 nm


energy difference between levels in hydrogen (with En   E0 / n2 ) we see that
E7  E2  3.12 eV, which matches the photon energy to three significant figures.
Therefore this laser can promote the atom to the n = 7 level.
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publicly accessible website, in whole or in part.
Chapter 4
Structure of the Atom
53
31. We must use the reduced mass for the muon:
106 MeV/c 2  938 MeV/c2 

mM


 95.2 MeV/c2
m  M 106 MeV/c 2  938 MeV / c 2
(a) a0 
(b) E 
4
2
0
2
e
e2
8 0 a0
 6.58 10 eV  s 
3.00 10

c
1.44 10 eV  m 95.2 10 eV/c 
1.44 10 eV  m   2535 eV

2  2.84 10 m 
16
9
2
8
6
2
2
m/s 
2
 2.84 1013 m
9
13
hc 1240 eV  nm

 0.49 nm
E
2535 eV
4hc 4 1240 eV  nm 
Second series:  

 1.96 nm
E
2535 eV
9hc 9 1240 eV  nm 
Third series:  

 4.40 nm
E
2535 eV
(c) First series:  
mm
 m / 2 where m is the mass of each
mm
E
e2
e2
 2a0 and E  

  0  6.8 eV .
8 0 r
8 0  2a0 
2
32. The reduced mass for this system is  
particle. Then r 
4
2
0
2
e
33. (a) As shown in Problem 32, the radius of the orbit is 2a0 .
(b) With E0  6.8 eV (see Problem 32) we have
E0  E0  3E0
hc 1240 eV  nm

 243 nm .
 1  
 5.1 eV . Then  
2
E
5.1 eV
2  1 
4
34. (a) r2  r1  4a0  a0  3a0  1.59 1010 m
E  E2  E1  
(b) r6  r5  36a0  25a0  11a0  5.83 1010 m
(c) r11  r10  121a0  100a0  21a0  1.11109 m
(d) Successive radii (different by 1) would have a difference given by:
2
rm  rn   n  1  n2  a0   n2  2n  1  n2  a0   2n  1 a0 . For large values of n this


is approximately rm  rn  2n  1a0  2na0 .
1 1 
 1 1 
1 4 
1
 RH   2  ; helium:  Z 2 RHe  2  2   RHe   2 


 4 nu 
 4 nu 
 4 nu 
We see that the lines match very well when nu is even for helium but not when it is odd.
For example, when nu  6 , for the ionized helium, then the nu  3 state for hydrogen will
be very similar. In the same fashion, the nu  8 level for the ionized helium will be very
similar to the nu  4 level for hydrogen. Also, all the “matched'” lines differ slightly
35. hydrogen:
1
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54
Chapter 4
Structure of the Atom
because of the different Rydberg constant (which is due to the different reduced masses).
They differ by a factor of RHe / RH  0.99986 / 0.99946  1.0004 .
36. As mentioned in the text below Equation (4.38), if each atom can be treated as singleelectron atoms (and the problem states we can make this assumption), then
1
R
R where R  1.0973731534 107 m1 and m  0.0005485799 u.
m 
1
M
4
He (M = 4.0026 u), R  0.999863 R  1.097223 107 m1 (off by 0.14%).
39
K (M = 38.963708 u), R  0.9999859 R  1.097358 107 m1 (off by 0.0014%)
238
U (M = 238.05078 u), R  0.9999977 R  1.097371107 m1 (off by 0.00023%)
37. The derivation of the Rydberg equation follows as in the text. Because He is hydrogenlike it works with R  Z 2 RHe and RHe  1.097223 107 m1 as in Problem 36. Then
R  4 1.097223 107 m1  4.38889 107 m1 .
38. For L we have  
Z  43 :  
Z  61:  
Z  75 :  
c
36
.

f 5R Z  7.4 2
36
5R 43 - 7.4 
2
36
5R 61  7.4 
2
36
5R 75  7.4 
2
 0.52 nm ;
 0.23 nm ;
 0.14 nm
39. For Pb Z  82 and for Bi Z  83 . For the K lines  
4
3R Z  1
2
.
4
4
 18.52 pm ;
Bi :  
 18.07 pm
2
3R(81)
3R(82)2
Therefore   0.45 pm, which is less than the specified resolution, so it will not work.
Pb:  
c
 1 1  8cR
 cR( Z  1)2  2  2  
( Z  1) 2 This is higher than the K
 (K  )
1
3
9


frequency by a factor of (8 / 9)(3 / 4)  32 / 27 , which seems to be in agreement with
Figure 4.19.
40. f (K  ) 
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Chapter 4
Structure of the Atom
41. Helium:  (K ) 
Lithium:  (K ) 
55
4
3R  Z  1
2
4
3R  Z  1
2
 122 nm ;  (K  ) 
9
 103 nm
8R( Z  1)2
 30.4 nm ;  (K  ) 
9
 25.6 nm
8R( Z  1)2
42. Refer to Figure 4.19 and Equation (4.41).  is inversely proportional to ( Z  1)2 for the
K series and to ( Z  7.4)2 for the L series.
 (U) (6  1)2

 0.0030
(a)
 (C) (92  1)2
 (Pt) (20  7.4)2
(b)

 0.032
 (Ca) (78  7.4)2
43. The longest wavelengths are produced for an electron vacancy in the K-shell. We begin
with Equation (4.43) and notice that the longest wavelengths will occur when the
expression on the right of the equal sign is as small as possible; thus we choose n equal to
2, 3, and 4. The series limit occurs for n   . Molybdenum (Mo) has Z = 42. Then
1
1 
1
2
2
 R  Z  1 1  2   1.09737 107 m1   41 1  2   72.28 pm .
K
 n 
 2 
In a similar fashion, we find   K    60.99 pm ;   K   57.82 pm and the K series
limit is   54.21 pm .
44. The longest wavelengths occur for an electron vacancy in K shell. The two longest
wavelengths correspond to the K and the K  . Use Equation (4.41) and rearrange it to
solve for Z  1 . Then
2
 Z  1
2

4 1
4

 783.89 . This gives
7
3 R K 3 1.09737 10 m1  0.155 109 m 
Z  1  28 or Z = 29. Using the expression for K

from problem 40, we have
 Z  1
2

9 1
. Using the second wavelength given, 0.131 nm, we find
8 R K 
 Z  1
2

9 1
9
1

 782.58 . This yields
7
1
8 R K 8 1.09737 10 m  0.131109 m 
Z  1  27.97
or Z = 29. Therefore we can conclude the target must be copper.
45. Non-relativistically 45eV 
2 45 eV 
1 2
mv1 so v1 
 0.0133c  3.98 106 m/s .
2
2
511 keV / c
From elementary physics
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56
Chapter 4
Structure of the Atom
2 0.0005486 u 
2m1
v1 
3.98 106 m/s  21.8 m/s where we have

m1  m2
0.0005486 u  200.59 u
used the molar mass of mercury. Then


1
1
c2
2
K 2  m2v22   200.59 u   931.49MeV/u·c2  
21.7 m/s 
16
2
2 
2
2
 9 10 m / s 
v2 
 4.89 104 eV
46. Without the negative potential an electron with any energy, no matter how small, could
drift into the collector plate. As a result the electron could give up its kinetic energy to a
Hg atom and still contribute to the plate current. The Franck-Hertz curve would not show
the distinguishing periodic drops, but rather would rise monotonically.
47. Using E  hc /  we find h 
E

 254 nm  4.88 eV   4.13 1015 eV  s .
c
3.00 108 m/s
48. 3.6 eV, 4.6 eV, 2(3.6 eV) = 7.2 eV, 3.6 eV + 4.6 eV = 8.2 eV, etc. with other
combinations giving 10.8 eV, 11.8 eV, 12.6 eV, 14.4 eV, 15.4 eV, 16.2 eV, 16.4 eV,
17.2 eV, 18.0 eV.
49. Magnesium:  (K ) 
Iron:  (K ) 
4
3R  Z  1
4
3R  Z  1
2
2
 1.00 nm ;  (L ) 
 0.194 nm ;  (L ) 
36
5R Z  7.4 
2
36
5R Z  7.4 
2
 31.0 nm
 1.90 nm
50. K is a transition from n = 2 to n = 1 and K  is from n = 3 to n = 1. We know those
wavelengths in the Lyman series are 121.6 nm and 102.6 nm, respectively. The redshift
1 
1  1/ 6
factor  / 0  is (with   1/ 6 )

 1.183 . Then the redshifted
1 
1  1/ 6
wavelengths are higher by 18.3% in each case. The observed wavelengths are:
K :   1.183121.6 nm   143.9 nm and K  :   1.183102.6 nm   121.4 nm .
2
 Z Z e2 
 
51. f   nt  1 2  cot 2  
2
 8 0 K 
  2  79  1.44 109 eV  m  
 cot 2  0.5 
m
6


2  8 10 eV 


2
(a)
f (1 )    5.90 1028 m 3  0.32 106
 0.157
  2  79  1.44 109 eV  m  
 cot 2 1 
m 
6


2
8

10
eV




2
f (2 )    5.90 1028 m 3  0.32 106
 0.0394
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Chapter 4
Structure of the Atom
57
The fraction scattered between 1 and 2 is 0.157  0.0394  0.118 .
2

2

f (1 ) cot 0.5 
f (1 ) cot 0.5 
(b)

 100.5 ;

 1.31104
f (10 )
f (90 ) cot 2 45 
cot 2 5 
52. From classical mechanics we know
that L is conserved for central forces.
L  L  r  p  rp0 sin  . But
r sin   b so L  p0b  mv0b .
53. If the positions of the electron and proton are respectively (along a line) re and rM then
mr  MrM
putting the center of mass at R  0 we have R  0  e
mM
M
m
 m 

m 
r.
or rM    re  . r  re  rM  re    re   re 1   which gives re 
M m
 M 
 M
M 
 m 
Similarly we can show that rM  
 r . From these two expressions we can see
M m
r
m
that M 
. Now the centripetal force on the electron and the nucleus must be equal (in
re M
magnitude) by Newton's third law so we have
mve2 MvM2
v 2 mr
v
m2
m

or M2  M  2 so M  . The total angular momentum of the
re
rM
ve
Mre M
ve M
atom is the sum of the electron and nuclear orbital angular momenta, so
m
 m  m 

L  mve re  MvM rM  mve re  M  ve  re   mve re 1    mve r where the last
 M  M 
 M
substitution comes from using the expression for re from above. Because the total L must
n
equal n we find ve 
. This is the equivalent of Equation (4.22b). Now the
mr
centripetal force depends on the distance of the electron from the center of mass while the
Coulomb force depends on r, the distance between the electron and the nucleus. So
2
 1   e2  mve2
 e2   1   re   n 
2
or ve  
Equation (4.18) becomes: 
 2  
 2    
.
re
 4 0   r 
 4 0   r   m   mr 
This can be solved to give mre 
4 0 n2
e2
2
4 0 n2
 M 
r

or m 

e2
M m
2
. Finally this can
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58
Chapter 4
Structure of the Atom
be solved for r and the subscript added to account for allowed orbits. We find
mM
m
4 0 n 2 2 4 0 n2 2
with  
.

rn 

2
mM 2
M

m
1

m
/
M

e
e
M m
This is the modification to Equation (4.24).
9.111031 kg 1.60 1019 C 

me4
54. (a) f  2 3 
 6.55 1015 Hz
2
3
12
34
40h
4  8.85 10 F/m   6.626 10 J  s 
4
(b) As determined in previous problems v  2.19 106 m/s and r  a0  5.29 1011 m.
f 
v
2 r

2.19 106 m/s
 6.59 1015 Hz
11
2 5.29 10 m 
(c) We know E  
e2
e2
 E.
and K 
8 0 a0
8 0 a0
(d) Because K  nhforb / 2  hforb / 2 for n = 1 we have
2K
2(13.6 eV)
f orb 

 6.58 1015 Hz which agrees with (a) and (b) to within
15
h
4.136 10 eV  s
rounding errors.
55. We start with K  nhf orb / 2 . From classical mechanics we have for a circular orbit
 v   mv 2   1 
1 2
L

mvr

mv
f  v / 2 r , or r  v / 2 f :



 . Using K  mv ,
2
 2 f   2    f 
K
n h f nh
L


n
 f 2 f 2
56. From problem 32 we know that the reduced mass of the system   m / 2 where m is the
mass of either particle. Then from Equation (4.37), we see that the effective Rydberg
constant would be Reff  R / 2 . We know that
En
1239.8 eV  nm    12 1.09737 102 nm1  6.803 eV
hc  Reff 




.
Therefore
n2
n2
n2
E1  6.803 eV , E2  1.701 eV , E3  0.756 eV , and E4  0.425 eV . As in other
problems, we know the K wavelength occurs for an nu  2 to an n  1 and the L
wavelength occurs for an nu  3 to an n  2 transition, etc. Further, we know that
1239.8 eV  nm
hc
 243.0 nm . In

. For example, we have   K  
1.701 eV   6.803 eV 
Eu  E
a similar fashion we can show that   K    205.0nm ,   L   1312nm , and
  L   971.6nm .
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Chapter 4
Structure of the Atom
59
57. The longest wavelength in the Lyman series is 121.57 nm, using the value of RH. Then
 137.15nm
1 
from the Doppler-shift formula,
. We can solve to find


0 121.57 nm
1 
  0.12 or v = 3.6 × 107 m/s.
58. (a) The assumptions identified in the statement of the problem lead to the following
 1   Ze2   1   e2 
expression for the force on each electron: 
  Fnet ,
 2   

2
 4 0   r   4 0    2r  
where the terms are of opposite sign since the force of the nucleus is attractive whereas
the force of the other electron is repulsive. The above expression simplifies to:
 e2 
Z  14   Fnet .

2 
4

r
0


 e2 
 e2 
mv 2
2
1
1
(b) 
or
Z


F




Z  4   v .
net
4
2 
4

m
r
4

r
r
0


0


(c) Bohr’s rule for angular moment: L  mvr  n  1 in the ground state. Rearrange
this equation to solve for v and square that expression. v  / mr which gives
2
v 2   / mr  . This is substituted into the expression from part (b). Then solve for r to
2
4 0 2
0h
find r  2
.

me  Z  14   me2  Z  14 
(d) The total energy of the atom is the sum of the kinetic energy term for each of the
electrons (easily found using the answer to part (b) and the sum of the potential energy
of each electron, with respect to the nucleus, and the potential energy of the electron pair.
Each can be found using the expression for r found in part (c). The general form of the
 1  Ze2
potential energy is given by V   
, using the appropriate r and Z and the

 4 0  r
appropriate sign.
 1  Ze 2  1  e 2
2
1
E  K  V  2  2 mv   2 



 4 0  r  4 0  2r
  e2 

 e2 
1
1
 m
  Z  4    
  2Z  2  
 4 0 mr 
  4 0 r 

 e2 
1
1

  Z  4  2Z  2 
4

r
0 

 e2 
1
 
Z  4 
 4 0 r 
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60
Chapter 4
Structure of the Atom
Now substitute the expression for r and simplify:
2


 e2 
 e2 
0h
1
1
E  
Z



Z




4
4 
2
  me  Z  1  
 4 0 r 
 4 0 
4 

1
 me4 
   2 2   2Z  12  Z  14 
8 0h 
You may notice the first parenthesis in the answer is E0 . See Equation (4.26).
(d) The calculation for helium, Z = 2, is shown.
 me 4 
EHe   2 2   2 Z  12  Z  14 
8 0h 
  E0   4  12  2  14 
 13.6 eV  3.5 1.75 
 83.3 eV
The calculation for lithium (Z = 3) follows in a similar fashion and gives
ELi  205.7 eV . These are fairly close to the measured values but not precise enough
to rank with the Bohr model. Thus, our statement that the Bohr model cannot be
applied to multi-electron atoms is fair.
59. The result from dimensional analysis is the same as the exact result [Equation (4.24)]
gives for the ground state. In general, dimensional analysis might be expected to give the
correct answer to within a multiplicative constant.
60. (a) Performing a binomial expansion on the expression in parentheses for a transition
from the n + 1 level to the n level gives:
2
1
1
1 1  1
1 1  2


 2  2 1    2  2 1   ... 
2
2
n n  1 n n  n 
n n  n

or
1
1
1 1 2
2

 2  2  3  3 . Therefore using Equation (4.30)   n3 / 2R . (b)
2
2
n n  1 n n n
n
Performing the derivative gives the same result. E  
E0
dE 2E0
 3 . With
so
2
n
dn
n
dE E 2E0

 3 and E  hc /  , E0 / hc   R , and n  1 , we have   n3 / 2R .
dn n
n
(c)   100  / 2 1.097373 107 m1   4.556 cm using the approximation and
3
1 
 1
 1.097373 107 m1 

 21.6226 m1 which gives   4.625cm using the
2
2 

 100 101 
exact formula.
1
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publicly accessible website, in whole or in part.
Chapter 4
Structure of the Atom
61
3cR
2
42  1  4.15  1018 Hz , so the
4
wavelength is λ = c/f = 7.2 × 10−11 m. Using Equation (4.43) with n = 3 for the Kβ x ray,
the wavelength is λ = 6.1 × 10−11 m.
61. Using Equation (4.40) for the Kα x ray, f 
Both of these results compare favorably with the graph. We can compute the frequency
of the Lα x ray using Equation (4.44) and then use that frequency to find the wavelength,
which is about 5.5 × 10−10 m. This value does not place the wavelength on the graph. You
would want to use a less energetic electron beam to generate the x rays in order to
observe the L series.
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62
Chapter 5
Wave Properties of Matter and Quantum Mechanics I
Chapter 5
1. Starting with Equation (5.1), with n = 1 and 1  12.5 , sin 1 

2d
 0.216 .
2
 2sin 1 ; 2  sin 1  2sin 1   sin 1  2(0.216)   25.6 ;
2d
1
3  sin  3sin 1   40.4
Second order: sin  2 
2. Use   0.207 nm and we know from the text (Example 5.1) that d = 0.282 nm for NaCl.
n 

 0.367 ;   21.3
n  1: sin  
2d 2d
n 
  0.734 ;   47.2
n  2 : sin  
2d d

  47.2  21.3  25.9
3. For n = 1 we have   2d sin   2  0.314nm   sin12.8   0.139 nm.
1240 eV  nm
 8.92 keV. The largest order n is the largest integer for which

0.139 nm
n
2d
 1 or n 
 4.52 so we can observe up through n = 4.
2d

E
hc

4. As in Davisson-Germer scattering n  D sin  , so
 1240 eV  nm 

1  hc 
1
  3.0
  sin 
  sin  5

D
 ED 
 10 eV   0.24 nm  
  sin 1 
h
6.626 1034 J·s
 9.2 1035 m . No, the wavelength of the water waves
5.   
p 1.2 kg  6.0 m/s 
depends on the medium; they are strictly mechanical waves.
6. Using the mean speed from kinetic theory (also in Chapter 9)
4
v
2

kT
4

m
2
1.38 10 J/K  310.15 K   484.2 m/s ;
28 1.66 10 kg 
23
27
h
6.626 1034 J  s

 2.94 1011 m which is roughly 3% of the size
27
p 28 1.66 10 kg   484.2 m/s 
of the molecule.
7. Using the approach of Example 5.2, and with K  eV we have, assuming relativistic
h
h
hc
effects are small,   
. If we do not assume that relativistic

p
2mK
2  mc 2  K
effects are small, then  
h

p
hc
K 2  2  mc 2  K
. See problem 11 for details. When
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
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Chapter 5
Wave Properties of Matter and Quantum Mechanics I
63
K  40keV , if we use the non-relativistic formula then
1240eV  nm

 6.13 103 nm  6.13 pm. Using the relativistic
6
3
2  0.51110 eV  40 10 eV 
1240eV  nm
formula, we find  
 40 10 eV 
3
2
 2  0.51110 eV  40 10 eV 
6
 6.02 pm. This
3
represents a 2% difference. Clearly when K  100keV , we must use the relativistic
approach and we find   3.70 103 nm  3.70 pm.
8. The resolution will be comparable to the de Broglie wavelength. The energy of the
microscope requires a relativistic treatment, so
h
hc
 
p
K 2  2  mc 2  K
1240eV  nm

 3.0 10 eV 
6
2
 2  0.511106 eV  3.0 106 eV 
 3.57 104 nm  0.357 pm
9.
 50 eV  1.602 1019 J/eV   8.011018 J ;

h
6.626 1034 J  s

 1.73 1010 m.
31
18
2mK
2  9.109 10 kg 8.0110 J 
This is the same result as in the textbook’s example.
10. When E  E0 then E  pc for the particle; E  pc for a photon. Therefore the electron's
energy is approximately equal to the photon energy.
If E  2 E0 then we cannot use E  pc . The exact expression is p 
for the electron's momentum. Then the de Broglie wavelength is  
photon has the same wavelength, its energy is E 
11. (a) Relativistically p 

h

p
E 2  E02

c
hc

c
c

3E0
c
h
hc

. If the
p
3E0
 3E0 .
 K  mc    mc 
2 2
E 2  E02
2 2

K 2  2 Kmc 2
;
c
hc
K  2 Kmc 2
2
(b) Non-relativistically, as in the text  
h
hc

.
2mK
2mc 2 K
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64
Chapter 5
Wave Properties of Matter and Quantum Mechanics I
12. The rest energy of the electron is very small compared to the total energy.
p
E 2  E02
c
 50 GeV/c ;  
h 1240 eV  nm

 2.48 1017 m ;
p
50 GeV
2.48 1017 m
 0.012.
2 1015 m
13. (a) For photons, kinetic energy equals total energy and de Broglie wavelength is
hc 1240 eV  nm

 9.54 keV.
wavelength: K  E 

0.13 nm
(b) The energy is low enough that we can use the non-relativistic formula:
fraction 
 hc  
1240 eV  nm 
h2

 89.0 eV
2
2
2 2
2m
2mc 
2  511103 eV   0.13 nm 
2
K
2
 hc  
1240 eV  nm 
h2
(c) K 

 0.048 eV
2
2
2 2
2m
2mc 
2  939 106 eV   0.13 nm 
2
2
 hc  
1240 eV  nm 
h2
(d) K 

 1.22 102 eV
2
2
2 2
6
2m
2mc 
2  3727 10 eV   0.13 nm 
2
2
14. (a) As in Problem 6, we use the mean speed formula from kinetic theory:
v
4
2
kT
4

m
2
1.38 10 J/K  5 K   324 m/s ;
1.675 10 kg 
23
27
h
6.626 1034 J  s
 
 1.22 nm .
p 1.675 1027 kg   324 m/s 
4
(b) v 
2

kT
4

m
2
1.38 10 J/K   0.01 K   14.5 m/s ;
1.675 10 kg 
23
27
h
6.626 1034 J  s

 27.3 nm
p 1.675 1027 kg  14.5 m/s 
15. From the accelerating potential we know K  eV  3.00keV so E  K  E0  514 keV .
 514 keV    511 keV 
E 2  E02
p

 55.4 keV/c ;
c
c
h hc 1240 eV  nm
 

 22.4 pm
p pc 55.4 103 eV
2
2
16. We use the relativistic formula derived in Problem 11:  
(a)   0.194nm
(b)   6.13 102 nm
h

p
hc
.
K  2 Kmc 2
(c)   1.94 102 nm
2
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publicly accessible website, in whole or in part.
Chapter 5
Wave Properties of Matter and Quantum Mechanics I
(d)   6.02 103 nm
65
(f)   2.77 104 nm
(e)   1.64 103 nm
From Example 5.1 in the text, we know the spacing of the NaCl lattice is 0.282 nm. Thus
even the lowest energy electrons here could be used to probe the crystal structure.
h
6.626 1034 J  s
 2.60 1011 m
17. (a)   
27
p 32 1.66110 kg   480 m/s 
(b)  
h
6.626 1034 J  s

 1.02 1016 m
13
5
p  6.5 10 kg 10 m/s 
18. (a) Using the relativistic formula from Problem 11,
h
hc
1240 eV  nm
 

 1.24 1018 m
2
2
2
p
12
12
6
K  2 Kmc
10 eV   2 10 eV 938 10 eV 
(b)

h

p
hc
K  2 Kmc
2
2

1240 eV  nm
 7.0 10
12
eV   2  7.0 1012 eV  938 106 eV 
2
 1.77 1019 m
Note: For both parts (a) and (b), the same answers (to three significant figures) may be
obtained using the relativistic approximation K = pc.
19. d  D sin( / 2)   0.23 nm  sin16  0.063nm ;
  D sin    0.23 nm  sin 32  0.122 nm ; p 
E
p 2c 2  E02 
10.2 keV    511 keV 
2
2
h


hc 1240 eV  nm

 10.2 keV/c ;
c  0.122 nm  c
 511.102 keV ; K  E  E0  102 eV
20. Beginning with Equation (5.7) and with V0  48 V and D  0.215nm for nickel, we have
1.226 nm V 1/ 2

 0.177 nm 

 0.177nm ; and   sin 1    sin 1 
  55.4 .
48 V
D
 0.215 nm 
1/ 2
1.226 nm V

 0.153nm 

At 64 eV:  
 0.153nm ; and   sin 1    sin 1 
  45.4 .
64 V
D
 0.215 nm 
21. First we compute the wavelength of the electrons:
h
hc
1240 eV  nm
 

 2.74 102 nm.
2
2
2
2
p
E  E0
 513 keV    511keV 

From Figure 5.7(a) we see that 21  tan 1  2.1cm/35cm   3.434 or 1  1.717 . Now

2.74 102 nm

 0.457 nm ;
because   2d sin  we have: d1 
2sin 1 2sin 1.717 
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publicly accessible website, in whole or in part.
66
Chapter 5

1
2
2  tan 1  2.3 cm/35 cm   1.880 ; d2 
1
2
3  tan 1  3.2 cm/35 cm   2.612 ; d3 
22.  
2sin  2
Wave Properties of Matter and Quantum Mechanics I

2.74 102 nm
 0.412 nm .
2sin 1.880 

2.74 102 nm

 0.301 nm
2sin 3 2sin  2.612 
h
hc
1240 eV  nm


 0.181 nm ;
2mK
2mc 2 K
2  939 106 eV   0.025 eV 
  D sin  ;   sin 1 

1  0.181 nm 

  sin 
  23.7
D
 0.45 nm 
23. We begin with Equation (5.23)  t  2 . Since   2 f the first equation is
1
1
equivalent to f t  1 . Then t 

 3.33s . If we want the time to be onef 0.3Hz
half of that found from the bandwidth relation, then we must monitor the frequency of the
system every 1.67 s.
24. Refer to Problem 11. The 6-eV electrons are nonrelativistic. We must treat the 600 keV
electrons relativistically.
For the 6 eV electrons,  
hc
2mc 2 K

1240 eV  nm
2  0.511106 eV   6eV 
electron speed can be found using p  mv 
h

 0.501 nm . The
. Therefore
v
6.626 1034 J  s
v
 1.45 106 m/s and    4.83 103 .
31
9
c
9.1110 kg  0.50110 m 
1 2
mv to find v.
2
The phase velocity is found using a modification of Equation (5.32). We must note that E
represents the total energy of the particle, so
6
19
 E/
E 0.51110 eV 1.602 10 J/eV 
vph  
 
 6.20 1010 m/s.
31
6
k p/
p
9.11

10
kg
1.45

10
m/s



Alternatively for this nonrelativistic case you could use K  6eV 
This speed exceeds the speed of light but is not associated with the transmission of
information. In Problem 28 we will show that a simpler approach to find the phase
c
velocity is to use vph 
but either approach yields the same answer. We find the group

velocity from Equation (5.31), namely, ugr 
pc 2
. In problem 73 from chapter 2, we
E
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publicly accessible website, in whole or in part.
Chapter 5
Wave Properties of Matter and Quantum Mechanics I
67
pc
so ugr   c  1.45 106 m/s which is the same as the particle speed.
E
For the 600 keV electrons,
hc
1240 eV  nm
so


2
2
2
3
6
3
K  2  mc  K
600

10
eV

2
0.511

10
eV
600

10
eV

 


showed that  
  1.26 103 nm . Using the same approach as before p 
h

 5.26 1022 kg  m/s .
However, this is a relativistic momentum so p   mv . We find  
therefore v  2.66 108 m/s  0.885c . The phase velocity is vph 
c

K  E0
 2.174 and
E0
 1.13c and again
the group velocity is the same as the particle velocity, ugr  0.885 c .
25. (a) f 
v


5.0cm/s
 0.714 Hz
7.0cm
(b) From the initial conditions given, the phase in Equation (5.18) is    / 2 rad  90 .
This is equivalent to representing the wave with a cosine function and zero phase:
 2 

 2 

  A cos 
 10 cm   5.0 cm/s 13s    2.49 cm
  x  vt     4.0 cm  cos 
  

 7.0cm 

4.2 cm/s
 1.05 Hz
 4.0 cm
b) T  1/ f  0.95 s
26. (a) f 
v

c) k  2 /    / 2cm1
d)   2 / T  2.1 rad/s
27. (a)   1   2  0.003 sin(6.0 x  300t )  sin  7.0 x  250t  We can use a trig identity
 A B 
 A B 
sin A  sin B  2sin 
 cos 
 so   0.006sin(6.5x  275t )cos(0.5x  25t )
 2 
 2 
or because cosine is an even function   0.006sin(6.5x  275t )cos(0.5x  25t ).
 50 rad/s
  2 550 rad/s

 50 m/s
(b) v ph  1

 42.3 m/s and ugr 
1
k
1 m1
k1  k2
13m
(c) As in Equation (5.22) x   2 / k   2 m is the separation between zeros.
(d) k x  1 m1   2 m   2
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68
Chapter 5
28. ugr 
1/2
d dE d

  p 2c 2  E02  
dk dp dp
pc 2
p 2c 2  E02
Wave Properties of Matter and Quantum Mechanics I

pc 2
 c ;
E
h  E pc 2 / v c 2 c
v ph   v 
 
  . The particle and its “signal”' are associated
p 2 p
p
v 
with the group velocity, not the phase velocity.
29. As in Example 5.6 we start with v ph  c n where c is a constant. We also know from
dv ph  d  

 . Since
dk
dk
d   dk 
2
dv ph
2
d  2  2
 2  dv ph   

then
. Therefore ug  v ph  
.

v


 2 




ph

d
k
dk
k
2
   d   2 
dv ph
Equation (5.33) that ug  v ph  k
. We note further that
dv ph

Setting ug  v ph  c n , we find c n  c n  cn n . This can be satisfied only if n = 0, so
v ph is independent of  . This is consistent with the idea that when a medium is nondispersive, the phase and group velocities are equal and the speed independent of
wavelength.
K  E0 948.27 MeV
1
 0.145 ;
30. Protons:  

 1.01066 ;   1 
1.010662
E0
938.27 MeV
c
ugr   c  0.145 c and vph   6.90c .

K  E0 10.511 MeV
1
 0.9988 ;

 20.57 ;   1 
20.572
E0
0.511 MeV
c
ugr   c  0.9988c and vph   1.001c
Electrons:  

31.
 ( x, 0)   A(k ) cos(kx) dk  A0 
k0 k /2
k0 k /2
cos(kx) dk
k k /2
A
sin(kx) 0
 A0
 0  sin  k0  k / 2  x  sin  k0  k / 2  x 
x k0 k /2 x

2 A0
 kx 
sin 
 cos  k0 x 
x
 2 
2
 kx  1
See the diagrams below. At the half-width of ( x, 0) , we have sin 2 
  , so
 2  2
k x 


 and x 
. Then x  2 x 
, and k x   .
2
4
k
 2k 
© 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 5
Wave Properties of Matter and Quantum Mechanics I
32. Relativistically: u 
Classically: u 
69
pc 2
1/2
dE d
  p 2c 2  E02  
dp dp
p 2c 2  E02

pc 2
;
E
dE d  p 2  p



dp dp  2m  m
33. For a double slit, the amplitude of E is doubled, and hence the intensity (proportional to
E 2 ) is higher by a factor of four for the double slit.

h/ p
hc
1240 eV  nm
34. d 



 2.22 nm
2
2

sin  sin 
E 2  E 2 sin 
512
keV

511
keV
sin1

 


0



35. We make use of the small angle approximations: sin   tan  and sin    .
0.3 mm
sin   tan  
 3.75 104 ;
0.8m
  d sin   d   2000 nm   3.75 104   0.75 nm ;
p
h


hc 1240 eV·nm

 1.653 keV/c ;
c  0.75 nm c
K  E  E0  p 2c 2  E02  E0 
1.653 keV   511 keV 
2
2
 511 keV  2.67eV .
Such low energies will present problems, because low-energy electrons take longer to
move through the region of the field and thus the deflection by stray electric fields is
greater.
36. By the uncertainty principle px 
Emin
 p 

2
at a minimum. Non-relativistically (with x  d )
197.3 eV  nm 
( c) 2



 25.0 keV
2
2
2 2
2m
8md
8mc d
8  938.27 106 eV 1.44 105 nm 
2
2
2
37. For the n = 1 energy level:
1240 eV  nm 
h2
h2c 2
E


 21.3MeV .
2
2
2 2
8md
8mc d
8  939.57 106 eV  3.1106 nm 
2
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publicly accessible website, in whole or in part.
70
Chapter 5
By the uncertainty principle px 
 p 

2
at a minimum. Non-relativistically (with x  d )
2
 0.539 MeV .
2m
8md 2
The two answers differ by the factor of (2 ) 2 in using
Emin

2
Wave Properties of Matter and Quantum Mechanics I
or h in the energy formula.
38. The uncertainty ratio is the same for any mass and independent of the box length.
px  m v x 
2
or v 
2m x

2mL
h2
h
1
h2
v


or
E  mv 2 
2 2
2
4m L
2mL
2
8mL
v
/ 2mL 1


v h / 2mL 2
39. For circular motion L  rp and so L  r p . Along the circle x  r and x  r .
L
Thus px 
 r    L  . For complete uncertainty   2 and
r
2
/2
L 

.
2
4
40. E t 
2
so E 
2t

1.054 1034 J  s
 1.67 1078 J
36
7
2 110 y  3.16 10 s/y 
41. If we use Equation (5.44),  t 
42. px  m vx 
v 
2m x
43. (a) E t 

2
2
1
1
1

 2.1105 Hz .
we find  
2t 2  2.4μs 
2
so at minimum uncertainty
1.054 1034 J  s
 1.76 1014 m/s
15
6
2  3.0 10 kg 1.0 10 m 
so E 
2t

6.582 1016 eV  s
 3.29 103 eV
13
2 1.0 10 s 
(b) Using the photon relation E 
hc

and taking a derivative dE  
Then letting   d  and E  dE we have
E
3.29 103 eV
  hc 2  1240 eV  nm 
 0.185nm .
2
E
 4.7 eV 
hc

d  
2
E2
d
hc
© 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 5
Wave Properties of Matter and Quantum Mechanics I
71
44. The wavelength of the electrons should be 0.14 nm or less. For this wavelength
h hc 1240 eV  nm
p 

 8.86 keV/c ;
 c  0.14 nm  c
K  E  E0  p 2c2  E02  E0 
8.86 keV 
45. For the angle  R we find  R  tan  R 
2
 (511 keV)2  511 keV  77 eV.
4000 nm
 2 105. The wavelength is
20 cm
5
d  R  0.05m   2 10 


 820 nm .
1.22
1.22
hc 1240 eV  nm

 1.51 eV
(a) E 

820 nm
(b) For non-relativistic electrons
1240 eV  nm 
p2
h2
h2c 2
K



 2.24 106 eV .
2
2
2 2
3
2m 2m
2mc 
2  51110 eV   820 nm 
2
Check with the uncertainty principle:
1.055 1034 J  s
p 

 1.32 1029 kg  m/s. The actual momentum is
9
2x 2  4000 10 m 
6.626 1034 J  s
p 
 8.08 1028 kg  m/s . This is allowed because p  p .
9

820 10 m
h
46.  
h
h
1240 eV  nm


 6.12 fm. The minimum kinetic energy
p
2mK
2  3727 MeV  5.5 MeV 
according to the uncertainty principle is
 p 
K
2
2m
 c

2
8mc 2  x 
2
197.33 eV  nm 

2
8  3727 MeV  16 106 nm 
2
 5.10 keV . Because the kinetic
energy exceeds the minimum, it is allowed.
47. The proof is completed in Example 6.11. With   k / m we have a minimum energy

k 1.055 1034 J  s 8.2 N/m

 9.03 1034 J
2
2 m
2
0.028 kg
48. We can determine the value using the normalization condition:
L
2
2
2 x 
2 L
0 A sin  L  dx  1  A 2 or A  L .
E

No answers are provided for Problems 49 through 51
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publicly accessible website, in whole or in part.
72
Chapter 5
Wave Properties of Matter and Quantum Mechanics I
x 
 2 x 
 3 x 
52.  1  A sin 
 2  A sin 
;
 ;  3  A sin 
;
 L 
 L 
 L 
where A  2 / L . Refer to problem 48.
2L
nh
53. As before  
so pn 
. At high energies we must use relativistic formulas, so
n
2L
E
p 2c 2  E02  E0
p 2c 2
 1 . The ratios follow:
E02
2 2
2 2
2 2
2 2
E3 1  9h c /  4 L E0  
E2  1  h c /  L E0  
 ;
 ; and


E1 1  h 2 c 2 /  4 L2 E02  
E1  1  h 2 c 2 /  4 L2 E02  




1/2
1/2
2 2
2 2
E4 1  4h c /  L E0  
 .

E1 1  h 2 c 2 /  4 L2 E02  


These are quite different from the non-relativistic results, as one might expect. They do
reduce to the non-relativistic results in the low-energy limit.
p

54. At time t = 0 the velocity is uncertain by at least v0 
. After a time
m 2mx
1/2
m  x 
x

 m  x 
we have x   v0  t  
which means the uncertainty


2
 2mx 
equals half the distance of travel.
t
2
2
55. Both the spatial distribution  ( x) and the wavenumber distribution  (k ) should have the


x2 
k2 

(
k
)

exp

same Gaussian form:  ( x)  exp  
and
. For


2
2
  2x  
  2k  
conjugate variables ( x, k ) it is possible to obtain one distribution by taking a Fourier
transform of the other. Letting A be a normalization constant for  (k ) we have

k2 
dk
exp

exp  ikx  .

2

  2k  
The integral is evaluated by completing the square:
 k2

A 
 ( x) 
dk
exp

 ikx  x 2 k 2  x 2 k 2 

2

2 
 4k

 ( x) 
 ( x) 
A
2


A

1 
2 2
exp   x 2 k 2   dk exp  
k

2
ix

k



2

2
 4k 

© 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 5
Wave Properties of Matter and Quantum Mechanics I
 k  2ixk  we have  ( x)  A
Letting u 
2
2
73
k exp   x 2 k 2   exp  u 2 du . The


2k

integral has a value  so  ( x)  2 Ak exp   x 2 k 2  . Now comparing with the

1
x2 
1
2
Gaussian form  ( x)  exp  
we see that  k  
or k x  .
2
2
2
 2x 
  2x  
56. (a) These electrons have a kinetic energy that is a significant fraction of the rest energy,
so they must be considered relativistically. These electrons have a total energy
E  K  E0  390 keV + 511 keV = 901 keV.
h hc
hc
1.240 keV  nm
 


 0.00167 nm  1.67 pm.
2
2
2
2
p pc
E  E0
 901 keV    511 keV 
(b) Using the Davisson-Germer result Equation (5.5),
n 1 0.00167 nm 
sin  

 0.00777 which gives an angle   0.45 . This small
D
0.215 nm
angle is quite difficult to measure, which is why less energetic electrons are better for this
type of scattering experiment.
57. For such highly relativistic electrons p  E / c, so the wavelength is
h
h
hc 1240 eV  nm
 


 2.1107 nm  2.11016 m. Typically only gamma
9
p E/c E
6.0 10 eV
ray photons have such high energies.
58. (a) This is a fairly low energy, but to be safe we should use relativistic calculations. The
electron’s total energy is E  K  E0  5.7 keV + 511.0 keV = 516.7 keV.
h hc
hc
1.24 keV  nm
 


 0.0162 nm  16.2 pm.
2
2
2
2
p pc
E  E0
 516.7 keV    511.0 keV 
(b) For this analysis we will find the minimum energy of an electron confined in a onedimensional box of length 3.4 fm. We can use either the uncertainty principle result
Equation (5.43) or the exact particle-in-a-box result Equation (5.51). The two results are
the same order of magnitude. Using the latter,
1240 eV  nm 
h2
h2c 2
E1 


 3.3 1010 eV = 33 GeV.
2
2
2 2
5

6
8m
8mc
8  5.1110 eV  3.4 10 nm 
2
This is much larger than the energy of the emitted electron, which implies that it did not
exist in the nucleus prior to emission. Further, note that the electron’s wavelength in part
(a) is much larger than the nucleus, also implying that it could not have been confined
there.
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74
Chapter 5
Wave Properties of Matter and Quantum Mechanics I
59. From Chapter 4, r  n2 a0 where a0  5.29 1011 m is the Bohr radius. In this case we
have n = 10, so r  100a0  5.29 109 m . Using Equation (5.51), the electron’s energy is
1240 eV  nm 
h2
h2c 2
E1 


 3.36 103 eV.
2
2
2 2
5
8m
8mc
8  5.1110 eV   2  5.29 nm 
2
This is a very small energy compared with the electron’s actual kinetic energy, even in
such a high orbital state.
60. Using Equation (5.23), the range of angular frequencies is
2
  2 / t 
 2.51 s 1. Because   2 f , the range of frequencies is
2.5 s
2.51 s 1
f   / 2 
 0.40 Hz. Thus if the central frequency is 440 Hz, the range is
2
from about 439.8 Hz to 440.2 Hz.
61. (a) Because of conversation of energy and momentum, each photon gets an energy equal
to the electron’s rest energy, so its wavelength is given by E = hc/λ, or
hc 1240 eV  nm
 
 2.43 103 nm = 2.43 pm.
5
E 5.1110 eV
(b) Now the total energy of each electron is (Equation 2.65):
mc 2
5.11105 eV
E

 5.357 105 eV. This is the new energy of each photon, so
2
2
1 
1  0.30
hc 1240 eV  nm

 2.31103 nm = 2.31 pm.
E 5.357 105 eV
Because of the additional kinetic energy in the initial configuration, the photons
generated in the process have more energy and hence shorter wavelength.
the photon’s wavelength is:  
62. (a) The uncertainty principle (Equation 5.45) provides the relationship between
uncertainly in energy and time as, Et  / 2 or
E  / 2(t )   6.58211016 eV  s  /  2  8.4 1017 s   3.92eV.
(b) The uncertainty in mass is given by the energy-mass relation as m  E / c2 .
With a mass of 135 MeV/c2, the relative uncertainty is
m E
3.92 eV
 2
 2.9 108 .
6
m mc 135 10 eV
d 1.2 1015 m
6.58211016 eV  s
24

E


 82 MeV .

4.0

10
s
63. t  
and
c 3.0 108 m/s
2t
2  4.0 1024 s 
This “lower bound” estimate of the rest mass is E / c 2 which is within a factor of two of
the rest energy.
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publicly accessible website, in whole or in part.
Chapter 5
Wave Properties of Matter and Quantum Mechanics I
64. (a) In general n  2d sin  so sin  
75
n
0.5 nm
n
 0.3125n . It is required that
2d
2  0.8 nm 
sin   1 so the allowed values of n are 1, 2, 3. Substituting we find   18.2 for n = 1,
  38.7 for n = 2, and   69.6 for n =3.
h hc 1240 eV  nm

 2480 eV/c and
(b) For electrons p  
 c
 0.5 nm c
K  E  E0 
p 2c 2  E02  E0 
 2.480 keV 
2
 (511 keV)2  511 keV  6.0 eV.
65. The uncertainty of the strike zone width is x  0.38 m. There is a second uncertainty,
in the x-component of the ball’s velocity, given by px  mvx 
. Due to the
2x
uncertainty in velocity the ball’s x position will be uncertain by the time it reaches home
t
plate in an amount vxt 
 0.38 m where we have added the inequality because
2mx
this is the condition for a strike. Since this inequality must be satisfied simultaneously
with the first one x  0.38 m, we multiply the two inequalities together to find
t
2
  0.38 m   0.144 m2 . Rearranging we find
2m
2m  0.144 m2  2  0.145 kg   0.144 m2 


 0.081 J  s.
t
18 m  /  35 m/s 
66. Assume that the given angle corresponds to the first order reflection. We have:
  2d sin   2  0.156nm  sin 26  0.1368nm. Next we find the energy:
p
h


hc 1240 eV  nm

 9.06 103 eV/c .
c  0.1368 nm c
K  E  E0 
p 2c 2  E02  E0 
 9.06keV 
2
 (939.57MeV)2  939.57MeV
 4.36 102 eV.
The Oak Ridge Electron Linear Accelerator Pulsed Neutron Source (ORELA) produces
intense, nanosecond bursts of neutrons, each burst containing neutrons with energies
from 103 to108 eV.
67. (a) Using the given information, the wave packet will equal:
15
1
1
1
1
   n  cos(18 x)  cos(20 x)  cos(22 x)  cos(24 x)  cos(26 x)
4
3
2
2
n 9
1
1
 cos(28 x)  cos(30 x).
3
4
A graph of the first two terms and a graph of the entire packet are shown below.  9 has a
frequency of 9 and an amplitude of 0.25.  10 has a frequency of 10 and an amplitude of
0.33. Other terms would have a frequency of 11 and amplitude of 0.5, etc. The peak
value of the packet is approximately 3.16.
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76
Chapter 5
Wave Properties of Matter and Quantum Mechanics I
(b) The packet is centered about x = 0 but extends to  . The packet repeats every unit
along the x axis.
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publicly accessible website, in whole or in part.
Chapter 5
Wave Properties of Matter and Quantum Mechanics I
68. (a) Starting from Equation (5.45), E t 
77

and substituting, we have    .
2
2
2
Therefore   .
(b) We can find the minimum value for  from the equation above. Using the data for the
6.58211016 eV  s
neutron, for example,  neutron 
 7.42 1019 eV. The other values
887s
follow in a similar fashion. pion  2.53 108 eV and upsilon  65.8keV.
69. (a) With E = hf, the uncertainty in energy is ΔE = hΔf. With this the uncertainty relation
Et  / 2 becomes h f t  / 2 . This simplifies to f t  1/ 4 , so the minimum
uncertainty in frequency can be computed:
1
1
f 

 7.96 1012 Hz.
15
4t 4 10 10 s 
The original frequency is f = c/λ, so the relative uncertainty is
9
12
f f  532 10 m  7.96 10 Hz 


 0.014. That’s a fairly large uncertainty,
f
c
3.00 108 m/s
over 1%!
(b) With f = c/λ, f /   c /  2 , so the absolute value of the wavelength range is
 
 2 f
532 10

9
m   7.96 1012 Hz 
2
 7.5 109 m = 7.5 nm.
c
3.00 10 m/s
(c) This is an appreciable fraction of the wavelength (also 0.014) and of the original
pulse, which has a length L  ct   3.00 108 m/s 10 1015 s   3.0 106 m = 3000 nm.
8
The wavelength spread is 0.0025 of the pulse length.
70. As in the preceding problem, the uncertainty principle leads to a frequency spread
f  1/  4 t  . In this case the wavelengths (and frequencies) differ by enough that it
would be risky to use the approximation f /   c /  2 . Instead we will compute the
value of Δf directly:
c c 3.00 108 m/s 3.00 108 m/s
f  f 2  f1   

 3.211014 Hz.
2 1 400 109 m 700 109 m
Solving for the pulse duration Δt:
1
1
t 

 2.5 1016 s = 0.25 fs.
4 f 4  3.211014 Hz 
This is just possible with today’s technology.
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publicly accessible website, in whole or in part.
78
Chapter 6
Quantum Mechanics II
Chapter 6
1. The function’s output does not approach zero in the limits  and  , so it cannot be
normalized over these limits.
2. (a) Given the placement of the  sign, the wave moves in the +x-direction. (One always
assumes t increases starting from 0. For the argument of the sine function to remain
constant x must increase as t increases. Therefore a value of constant phase with have
increasing x as t increases.)
(b) By the same reasoning as in (a), it moves in the  x -direction.
(c) It is a complex number. Euler’s formula eix  cos x  i sin x shows that an expression
written as ei ( kx t ) will have a real and complex part. Therefore the expression is complex.
(d) It moves in the +x-direction. Looking at a particular phase kx  t , x must increase as
t increases in order to keep the phase constant.
3. The derivatives are  / t  i and  2  / x2  k 2  . Substituting in the timedependent Schrödinger equation, we find: i
 i   
2
 k    V  .
2m
2
The term on
k2
p2

  K
2m
2m
where K is the kinetic energy. The result is E  K  V , which is a statement of
conservation of mechanical energy.
the left reduces to   E and the first term on the right is
4.
*  A2 exp  i(kx  t )  i(kx  t )  A2 . The condition for normalization becomes

a
0
5.
2
a
*dx  A2  dx  A2 a  1 so A 
0
1
1
and  
exp i(kx  t ) .
a
a
 2r 
*  A2 r 2 exp 
 . The condition for normalization becomes
  
 2  A2 3


 2r 
*
2
2
2


dr

A
r
exp
dr

A

 1 . Therefore



3
0
0
4
  
  2 /   
4
A
 2 3/2 .
3

6. In order for a particle to have a greater probability of being at a given point than at an
  
adjacent point, it would need to have infinite speed. This is seen as p  i 
   at a
 x 
discontinuity. Another problem is that the second derivative must exist in order to satisfy
the Schrödinger equation.
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publicly accessible website, in whole or in part.
Chapter 6
Quantum Mechanics II
79
7. (a) The wave function does not satisfy condition 3. The derivative of the wave function is
not continuous at x = 0.
(b) Based on (a) the wave function cannot be realized physically.
(c) Very close to x = 0 we could modify the function so that its derivative is continuous.
If we do so just in the neighborhood of x = 0, we need not change the function elsewhere.
8.
3.4  3.9  5.2  4.7  4.1  3.8  3.9  4.7  4.1  4.5  3.8  4.5  4.8  3.9  4.4
15
 4.247
x2 
x
3.42  3.92  5.22  4.7 2  4.12  3.82  3.92  4.7 2  4.12  4.52  3.82  4.52  4.82  3.92  4.42
15
x
 18.254
2
The standard deviation is
 x  x 

 x
2
2
i

i
 2 xi x  x 2 

x
2
i
 2x
x  x
2
i
. Look at the three
N
N
N
N
N
terms in the sum. The first is just x 2 . The second is 2 x  x   2 x 2 . The third term is
x
N
2

Nx 2
 x 2 . Putting the results together  
N
the data given we have  
x2  2x 2  x 2 
x 2  x 2 . For
x 2  x 2  18.254   4.247   0.466 .
2
9. If V is independent of time, then we can use the time-independent Schrödinger equation.
Then by Equation (6.15) *   * ( x) ( x)eit eit   * ( x) ( x) . Then
  xdx  
*
*
( x) ( x) xdx which is independent of time.
10. Using the Euler relations between exponential and trig functions, we find
  A  eix  eix   2 A cos  x  .
Normalization:


    dx  4 A   cos  x dx  4 A   1 .
*
2

2
2

Thus A 
(a) The probability of being in the interval [0,  / 8] is
 /8
P     dx 
0

*
1


 /8
0
1
2 
.
 /8
1x 1

cos  x dx    sin(2 x) 
 2 4
0
2
1
1

 0.119 .
16 4 2
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80
Chapter 6
Quantum Mechanics II
(b) The probability of being in the interval [0,  / 4] is
 /4
1
0

P    * dx 

 /4
cos 2  x dx 
0
 /4
1x 1

  sin(2 x) 
 2 4
0
1 1
 
 0.205 .
8 4
11.




 * dx  A2  sin 2  x dx  A2
 1 so A 
2
.

2
(a) The probability of being in the interval [0,  / 4] is
0
0
 /4
P     dx 
*
0
2

 /4

0
 /4
2x 1

sin  x dx    sin(2 x) 
 2 4
0
2
2  1 1 1

 
 0.091
  8 4  4 2
(b) The probability of being in the interval [0,  / 2] is

P
 /2
  dx 
0

*
2


 /2
0
 /2
2x 1

sin  x dx    sin(2 x) 
 2 4
0
2
2 
 1
  0    0.50.
4
 2
n 2 2 2
12. En 
;
2mL2
En 1
2
n  1  2


2
;
2mL2
2 2
 n  12  n 2   
2n  1 .
 2mL2 
2mL2 
Computing specific values:
En  En1  En 
E1 
2
2
3 ;
2  
2mL
2
E8 
2
2
2
17  ;
2 
E800 
2mL
13. The wave function for the nth level is  n ( x) 
2
2
2mL2
1601
2
 n x 
sin 
 so the average value of the
L
 L 
square of the wave function is
L
 ( x)
2
n
   dx  1   dx  2


L 
 dx L
*
L
0
L
0
*
2
L
0
2 L 1
 n x 
sin 2 
 dx  2  . This result for the
L 2 L
 L 
0
average value of the square of the wave function is independent of n and is the same as
the classical probability. The classical probability is uniform throughout the box, but this
2
is not so in the quantum mechanical case which is sin 2  kn x  .
L
14. (a) We know the energy values from Equation (6.35). The energy value En is
proportional to n 2 where n is the quantum number. If the ground state energy is 4.3 eV ,
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publicly accessible website, in whole or in part.
Chapter 6
Quantum Mechanics II
81
then the next three levels correspond to: 4E1  17.2 eV for n = 2; 9E1  38.7 eV for n =
3; and 16E1  68.8 eV for n = 4.
(c) The wave functions and energy levels will be like those shown in Figure 6.3.
15. (a) Starting with Equation (6.35) and using the electron mass and the length given, we
have:
2
2
2
2 2
 2 197.3 eV  nm 
2 
2   c
2
En  n
n
n
2
2mL2
2  mc 2  L2
2  5.11105 eV   2000 nm 
 n 2  9.40 108 eV 
Then the three lowest energy levels are: E1  9.40 108 eV ; E2  1.88 107 eV ; and
E3  2.82 107 eV .


3
3
1 eV
kT  1.3811023 J/K  
13K 
19   
2
2
1.602

10
J


3
which equals 1.68 10 eV . Substitute this value into the equation above as En and solve
for n. We find n = 134.
(b) Average kinetic energy equals
16. For this non-relativistic speed we have
2
1
1
1
E  mv 2   mc 2   2   511000eV/c 2 1.25 104 c   0.003992eV .
2
2
2
2 2
nh
Using E 
, we find
8mL2
2
3
8mL2 E 8mc 2 L2 E 8  51110 eV   48.5 nm   0.003992 eV 
2
n 


 24.97 and
2
h2
h2c 2
1240 eV  nm 
therefore n = 5.
2
x 
17. The ground-state wave function is  1 
sin 
.
L
 L 
L /3
2 L /3
2x L
x 
 2 x  
 dx   sin 2   dx   
sin 

0
L
L  2 4
 L 
 L  0
L /3
P1  
0
2
1
1
3
 2  
  0.1955
 6 8 
2 L /3
P2  
2x L
 2 x  
 dx   
sin 

L  2 4
 L   L /3
2 L /3
L /3
2
1
1
3
 2  
  0.6090
 6 4 
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82
Chapter 6
Quantum Mechanics II
L
2x L
 2 x  
P3    dx   
sin 

2 L /3
L  2 4
 L   2 L /3
L
2
1
1
3
 2  
  0.1955
6
8



Notice that P1  P2  P3  1 as required.
2
2
sin  kx  with k 
.
L
L
1
3
 2  
  0.402
 6 16 
18. The first excited state has a wave function  2 
L /3
P1  
2x 1

 dx    sin  2kx  
L  2 4k
0
L /3
0
2
2
1
3
Similarly P2  2  
  0.196 and P3  P1  0.402 . Notice that P1  P2  P3  1 .
 6 8 
1240 eV  nm 
h2
h2c 2
19. E1 


 3.76 GeV.
2
2 2
8mL 8mc L 8  511103 eV 105 nm 2
2
Then E2  4E1  15.05 GeV and E  E2  E1  11.3 GeV. This is much larger than any
energy observed in nuclear process; the electron is not in the nucleus.
20. (a) As in the previous problem
1240 eV  nm 
h2c 2
E1 

 935 keV
2 2
8mc L 8  938.27 106 eV 1.48 105 nm 2
2
1240 eV  nm 
h2c 2

 235 keV
(b) E1 
8mc 2 L2 8  3727 106 eV 1.48 105 nm 2
2
21. As in previous problems the ground state energy is
1240 eV  nm 
h2
h2c 2
E1 


 0.7676 eV. The other energy levels
2
2 2
8mL 8mc L 8  511103 eV   0.70 nm 2
2
are En  n2 E1 : E2  4E1  3.07eV ; E3  9E1  6.91eV ; E4  16E1  12.28 eV . The
allowed photon energies are:
E4  E3  5.37 eV
E4  E2  9.21eV
E4  E1  11.5 eV
E3  E2  3.84eV
E3  E1  6.14eV
E2  E1  2.30 eV
22. Lacking an explicit equation for finite square well energies, we will approximate using
the infinite square well formula. In order to contain three energy levels the depth of the
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publicly accessible website, in whole or in part.
Chapter 6
Quantum Mechanics II
well must be at least E 
83
n2 h2
9h 2
. Evaluating numerically with the given mass

8mL2 8mL2
9 1240 eV  nm 
9h 2
9h 2 c 2
E


 102 MeV.
2
2 2
8mL 8mc L 8 1.88 109 eV  3.00 106 nm 2
2
23. (a) The wavelengths are longer for the finite well, because the wave functions can leak
outside the box.
(b) Generally shorter wavelengths correspond to higher energies, so we expect energies
to the lower for the finite well.
(c) Generally the number of bound states is limited by the depth of the well. We expect
no bound states for E  V0 .
24. Using the same notation as the text, from the boundary condition  I ( x  0)   II ( x  0)
we have Ae0  Ce0  De0 or A  C  D . From the condition  I( x  0)   II ( x  0) we
have  A  ikC  ikD . Solving this last expression for A and combining with the first
ik
ik
C ik  
boundary condition gives C  D  C  D or after rearranging
.



D ik  
25. As in the previous problem matching the wave function at the boundary gives
CeikL  DeikL  Be L and matching the first derivative gives
ikCeikL  ikDeikL   Be L . Eliminating B from these two equations gives
1
C ik   2ikL
.
CeikL  DeikL   ikDeikL  ikCeikL  . Thus

e

D ik  
26. E 
2
2
n
2
1
 n22  n32   E0  n12  n22  n32  where E0 
2mL2
fourth, and fifth levels are:
E2   22  12  12  E0  6E0
E3   22  22  12  E0  9E0
E4   32  12  12  E0  11E0
E5   22  22  22  E0  12E0
2
2
2mL2
. Then the second, third,
(degenerate)
(degenerate)
(degenerate)
(not degenerate)
 nx   n  y   n z 
27. In general we have  ( x)  A sin  1  sin  2  sin  3  .
 L   L   L 
For  2 ( x) we can have  n1 , n2 , n3   1,1, 2  or 1, 2,1 or  2,1,1 .
For  3 ( x) we can have  n1 , n2 , n3   1, 2, 2  or  2, 2,1 or  2,1, 2  .
For  4 ( x) we can have  n1 , n2 , n3   1,1,3 or 1,3,1 or 3,1,1 .
For  5 ( x) we can have  n1 , n2 , n3    2, 2, 2  .
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a
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84
Chapter 6
28. We must normalize by evaluating the triple integral of  * :
Quantum Mechanics II
  dxdydz  1with
*
 ( x, y, z) given by Equation (6.47) in the text. We can evaluate the iterated triple integral
2
A

L
0
L
L
x 
2  y 
2 z 
2 L
sin 
 dx 0 sin 
 dy 0 sin 
 dz  A    1 . Solving for A we find
 L 
 L 
 L 
2
3
2
3/ 2
2
A  .
L
29. Taking the derivatives we find 2    k12  k22  k32  so the Schrödinger equation
becomes 
2
2m
2 
k
2m
2
1
 k22  k32   E . From the boundary conditions ki 
ni
Li
n22 n32 
  . The three quantum numbers come directly from the three

2m  L12 L22 L23 
boundary conditions.
so E 
 2  n12
2
2

30. One possible solution is n1 = 2, n2 = 1, n3 = 1. Letting either of the other two quantum
numbers be equal to 2 (with the remaining two equal to 1) gives the same result for A.
For the choice given here, the wave function becomes
 2 x    y    z 
2
  A sin 
 sin 
 sin 
 . To normalize, integrate  *   over the entire
 L   L   L 
L
L
L
 2 x 
 y 
z 
L
box A2  sin 2 
dx  sin 2 
dy  sin 2 
dz  A2    1 Solving for A, we



0
0
 L  0
 L 
 L 
2
3
3/ 2
2
find A    . It is interesting that this is the same result as for the ground state
L
(Problem 28).
n32 
 2 2  n12 n22


31. From Equation (6.49), E 

 . For the ground state:
2m  L2 4 L2 16 L2 
2 2
 1 1  21
. The least amount of energy
1





2mL2  4 16  32mL2
 2 2  1 1  3 2 2
we can add is accomplished by making n3 = 2. Then E1 
. This
1    
2mL2  4 4  4mL2
state is not degenerate. For the next energy level, make n3 = 3:
 2 2  1 9  29 2 2
, also non-degenerate. For the next level, take n1 = 1,
E2 
1    
2mL2  4 16  32mL2
2 2 
1  33 2 2
1

1

n2 = 2, and n3 = 1: E3 
. This state is also non-degenerate.


2mL2 
16  32mL2
Notice that this energy is less than we would obtain with n1 = 1, n2 = 1, and n3 = 4.
n1 = n2 = n3 = 1. Thus Egs 
2
2
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publicly accessible website, in whole or in part.
Chapter 6
Quantum Mechanics II
85
32.  0 ( x) has these features: it is symmetric about x = 0; it has a maximum at x = 0 because
the wave function must tend toward zero for x   ; there is no node in the ground
state; the wave function decreases exponentially where V  E .
1
1


33. En  En1  En   n  1      n      for all n. This is true for all n, and
2
2


there is no restriction on the number of levels.



2
2
1 
34. Normalization 1    * dx  A2  x 2e x dx  2 A2  x 2e x dx  2 A2
. Solving


0
4 
for A we find A  2 1/4 3/4 .
over symmetric limits.
x 
x2  x
2


x  A2  x3e x dx  0 because the integrand is odd
2


x 2  A2  x 4e x dx 
2

3 2 1/2 5/2
3
;
A 

4
2
3
2
35. The energy is related to the frequency of oscillation by:
1
1
1



E   n      n   hf   4.136 1015 eV·s 11013 s 1   n  
2
2
2



1

  4.136 102 eV   n  
2

k
For the harmonic oscillator  2  so
m
k   2 m  4 2 f 2 m  4 2 11013 s 1   2.32 1026 kg   91.6 N/m.
2
36. Taking the second derivative of  for the Schrödinger equation:
2
2
d
 5 A xe x /2  2 2 Ax3e x /2 ;
dx
2
d 2
  5  5 2 x 2  6 2 x 2  2 3 x 4  Ae x /2 .
2
dx
Starting with Equation (6.56), and with the wave function given in the problem, we have:
2
d 2
  2 x 2      2 x 2    A  2 x 2  1 e  x /2
2
dx


 2 3 x 4  x 2  2   2    Ae x
2
/2
Matching our two values of d 2 / dx 2 we see that the Schrödinger equation can only be
satisfied if   5 . Then
2mE
2
5
mk
2
or E 
5
 . This is the expected result,
2
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86
Chapter 6
Quantum Mechanics II
because the wave function contains a second-order polynomial (in x), and with n = 2 we
1
5

expect E   n    
.
2
2

37. By symmetry p  0 . Setting the ground state energy equal to
p2
p2
2m
we find
1
 which gives p 2  m . Note, however, that a detailed calculation gives
2m 2
1
p2 
 m . The factor of one-half is evidently the difference between the kinetic
2
energy and the total energy, which if taken into account does give the correct result.

38. Taking the derivatives for the Schrödinger equation:
d 2
d
 x2 /2
2 3  x2 /2
 x 2 /2
2  x2 /2
;


3
A

xe

A

xe
  2 x 2  3  .
 Ae
 A x e
2
dx
dx
d 2
Combining Equation (6.56) with this, we see
  2 x 2      2 x 2  3  . Thus
2
dx
2mE
mk
 3 2 . Solving for the energy, we find
we see that   3 or
2
E
3
2
k 3
 .
m 2
39. The classical frequency for a two-particle oscillator is [see Chapter 10, Equation (10.4)]
  k /   k  m1  m2  / m1m2  2k / m since the masses are equal in this case. The
energies of the ground state ( E0 ) and the first three excited states are given by
1

En   n    so the possible transitions (from E3 to E2 , E3 to E1 , etc. are E   ,
2

2  , and 3  . Specifically these calculations give:
2 1.1103 N/m 
2k
16

  6.582 10 eV·s 
 0.755 eV with a wavelength,
m
1.673 1027 kg
hc 1240 eV  nm


 1640 nm .
E
0.755 eV
2   2  6.582 10

16
eV  s 
2 1.1103 N/m 
1.673 1027 kg
 1.51 eV
hc 1240 eV  nm

 821 nm
E
1.51 eV
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publicly accessible website, in whole or in part.
Chapter 6
Quantum Mechanics II
3   3  6.582 10
16
87
2 1.1103 N/m 
eV  s 
1.673 1027 kg
 2.26 eV
hc 1240 eV  nm

 549 nm
E
2.26 eV
40. The kinetic energy is (see Chapter 9)
3
3
K  kT  8.617 105 eV/K  12000 K   1.551 eV . Assume a square-top potential of
2
2
6 1.440 eV  nm 
q1q2
6e2


 3.145 MeV where we have used the
height V0 
4 0 r 4 0 r 1.2 106 nm 12 1/ 3



fact that the radius of a nucleus is approximately 1.2 A1/3 fm (see Chapter 12). For the
width of the potential barrier use twice the radius or
L  2 1.2 106 nm  12
1/3

2mc 2 V0  E 
 5.49 106 nm . Then
2  938.27 106 eV  3.145 106 eV  1.551 eV 

197.3 eV  nm
c
 3.89 105 nm 1
 L   3.89 105 nm1  5.49 106 nm   2.135
1
2


3.145 106 eV  sinh 2  2.135 


  1.14 107
T  1
 4 1.551 eV   3.145 106 eV  1.551 eV  


41. (a) p  2m  E  V0  ;  
(b) p  2m  E  V0  ;  
42. In each case  L

h
h

; K  E  V0
p
2m  E  V0 
h
h

; K  E  V0
p
2m  E  V0 
1so we can use T  16
2mc 2 V0  E 
 2 3727 10

6
E
E  2 L
where
1   e
V0  V0 
eV 10 106 eV 

1/ 2
 1.38 1015 m1 .
c
197.4 eV  nm
(a) With L  1.3 1014 m,
5 MeV 
5 MeV  21.381015 m1 1.31014 m
Ta  16
 9.3 1016 .
1 
e
15 MeV  15 MeV 
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publicly accessible website, in whole or in part.
88
Chapter 6
Quantum Mechanics II
(b) With V0  30 MeV,

2mc 2 V0  E 
 2 3727 10

6
eV  25 106 eV 

1/ 2
 2.19 1015 m1 .
c
197.4 eV  nm
5 MeV 
5 MeV  2 2.191015 m1 1.31014 m
Tb  16
 4.2 1025 .
1 
e
30 MeV  30 MeV 
(c) With V0  15 MeV we return to the original value of  , but now L  2.6 1014 m and
5 MeV 
5 MeV  21.381015 m1  2.61014 m
 2.4 1031 .
1 
e
15 MeV  15 MeV 
By comparison Ta  Tb  Tc .
Tc  16
43. When E  V the wave function is oscillating, with a longer wavelength as E  V
decreases. Then when E  V the wave function decays.
1
 V02 sin 2  k2 L  
44. In general for E  V0 we have R  1  T  1  1 
 . If E V0 then
4 E  E  V0  

1
4E  E  V0   4E 2 . From the binomial theorem 1  x   1  x for small x and
 V02 sin 2  k2 L   V02 sin 2  k2 L 
 V0 sin  k2 L  
R

R  1  1 

which
can
be
written
as


 .
2E
4E 2
4E 2




2
1
 V 2 sin 2  k2 L  
45. T  1  0

4 E  E  V0  

(a) To obtain T  1 we require sin(k2 L)  0 . Except for the trivial solution L = 0, this
occurs whenever k2 L  n with n an integer. Letting n = 1we find


1
hc
1
1240 eV  nm
L 


 0.220 nm
k2
2m  E  V0  2 2mc 2  E  V0  2 2  511103 eV   7.8 eV 
Any integer multiple of this value will work.
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publicly accessible website, in whole or in part.
Chapter 6
Quantum Mechanics II
89
n
for any odd integer n. From the
2k 2
result of (a) we see that the first maximum is with L equal to half the value of L for the
first minimum, or L = 0.110 nm.
(b) For maximum reflection sin 2  k2 L   1or L 
 2  511103 eV  1.5 eV  
  6.27 nm1

46.  
c
197.4 eV  nm
With a probability of 2 104 we know  L 1and we can use
E E
1 
1  2 L
T  16 1   e2 L  16
 3.84e2 L  2 104 .
1 
e
V0  V0 
2.5  2.5 
Solving for L:
ln 1.92 104 
L
 7.86 1010 m. Now using the proton mass:
9
1
2  6.27 10 m 
2mc 2 V0  E 
1/ 2
 2  938.27 106 eV  1.5 eV 
  268.8 nm1 .


c
197.4 eV  nm
9
1

2
268.8

10
m

7.861010 m  1.2 10183 .
T  3.84e2 L  3.84e
The proton's probability is much lower!
2mc 2 V0  E 
1/ 2
47. The sketch of the potential will be like Figure 6.12. Since the question mentions
tunneling, the assumption is that the total energy is less than the potential. Therefore the
wave function will be sinusoidal on either side of the barrier with an exponential decay in
the barrier region. Each sketch will be similar to Figure 6.15. From Equation (6.70) we
can see that in part (b) with a barrier twice as wide, the exponential factor will be
markedly smaller (a ratio of e2  0.135 while doubling the barrier height in c) will
reduce the transmission coefficient as compared to (a) by less than 1/2.
(a)
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publicly accessible website, in whole or in part.
90
Chapter 6
Quantum Mechanics II
(b)
(c)
48. Starting with T  2e
(a)  
and  
2m V0  E 
2  511103 eV   6.4eV  1.4eV 
197.33 eV  nm
Then T  e2 L  2e
(b)  
2 L

2 11.5nm1  2.8nm 

2mc 2 V0  E 
c
, we find:
 11.5 nm1
  2e64.4  2.11028
2  3727 106 eV 19.2 106 eV  4.4 106 eV 
197.33 eV  nm
 1.683 106 nm1
21.6831015 m1  7.41015 m  
Then T  e2 L  2e
 2e22.6  3.04 1011 . For values of  L 1,
Equation (6.70) is a good approximation for Equation (6.67) and Equation (6.73) gives at
least an estimate of the transmission that is to within an order of magnitude.
49. The second derivative of the wave function with respect to x is
d 2
 k 2 A sin  kx   k 2 B sin  kx   k 2 . Then the time-independent Schrödinger
2
dx
2 2
 2k 2
k
equation [Equation (6.13)] becomes 

  E . We know that for a free
2m
2m
particle E  p 2 / 2m  2 k 2 / 2m, which verifies that this is a suitable wave function.
50. (a) We apply the time-dependent Schrödinger equation [Equation (6.1)] with

V = 0. The left side is i
 i   A cos  kx  t    B sin  kx  t   . The right side is
t
2
2
2 2
2
k
2
2






k
A
sin
kx


t

k
B
cos
kx


t

 . Because of the




2


2m x
2m
2m
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Chapter 6
Quantum Mechanics II
91
complex number i and the extra minus sign on the left side, there is no way the two
expressions can be equal. This is not a valid wave function.
(b) Using the Euler relation ei  cos  i sin  , this wave function is equivalent to
( x, t )  A cos  kx  t   iA sin  kx  t   Aei kxt  . Example 6.2 verified that this is a
valid wave function. You can also substitute the original wave function (containing sine
and cosine) into Equation (6.1) to prove that it is valid.
n22 n32 
 2  2  2  and substituting the given values of L we
2m  L1 L2 L3 
2 2
n32 
  2
2
2 2
2
find E 
n

2
n

 . Letting E0   / 2mL we have:
2
2  1
2mL 
4
51. As in the text we find E 
 2  n12
2


22 
32  21
1  13

E1  E0 1  2    E0 ; E2  E0 1  2    4 E0 ; E3  E0 1  2    E0 ;
4
4 4
4 4





42 
22 
1  25

E4  E0  22  2   
E0 ; E5  E0 1  2    E0  22  2    7 E0 .
4
4
4 4



Of those listed, only E5 is degenerate.
3
52. Recognizing this as the infinite square well wave function we see that k 
and

p
k
9 
.


2m 2m 2m 2
53. (a) In general inside the box we have a superposition of sine and cosine functions, but
only the sine function satisfies the boundary condition  (0)  0 , and thus   A sin(kx) .
E
2
2
2
2
2
2 2
p2
k
2mE
With V = 0 inside the well, E 
or k 
. Outside the well the decaying

2m 2m
2 2
k
exponential is required as explained in section 6.4 of the text, with E 
 V0 which
2m
2m V0  E 
reduces to   ik 
.
(b) Equating the wavefunctions and first derivatives at x = L:
A sin(kL)  Be L and kA cos(kL)   Be L .
tan(kL)
1
Dividing the first equation by the second gives
  or  tan(kL)  k .
k

54. From the previous problem  tan(kL)  k or  L tan(kL)  kL . Let   kL and    L
so that
(1)
 tan    .
Now from the value of V0 given in the problem and the definitions of  and  we have:
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92
Chapter 6
1
2mV0 L2
2

2mEL2

2
2m V0  E  L2
2
 k 2 L2   2 L2   2   2 .
Quantum Mechanics II
(2)
Solving Equations (1) and (2) numerically we find   0.20 and   0.98 . Then E is
2
k2
given by E 

2m
2
 / L 
2m
2
0.04 2
.
2mL2

55. Referring to the solution to the previous problem, we see that only a finite number of
solutions to Equation (1) exist up to any particular (finite) value of V0 . Therefore for any
finite V0 only a finite number of combinations of  and  will satisfy both equations,
and the number of bound states is finite.
56. Using the nomenclature of Problem 53
2m V0  E  L
L 
 1.14 where we
197.4 eV  nm
have used the mass of one nucleon, because one nucleon is “bound” by the other. Now
 L     / tan  so   2.07  kL . Then
2
k2
E

2m
2
 / L 

2  939 106 eV  2.2 106 eV   3.5 106 nm 
2

2m
2
 2.07 / L 
2m
2
 2.07  197.4 eV  nm 
2

2
2  939 106 eV  3.5 106 nm 
2
 7.26 MeV
This means that V0  2.2 MeV  E  9.46 MeV . The next solution of the equation
   / tan  is at   4.94 , a value that will put E  V0 . Therefore there are no excited
states.
57. For the one-dimensional well [Equation (6.35)],
E1 
2
2
2
2mc 2 L2

 2 197.3 eV  nm 
2
 33.6 eV.
2
2  5.11105 eV   2  5.292 102 nm 
This is the same order of magnitude but larger than the electron’s ground-state kinetic
energy. The one-dimensional model is a fair but inaccurate representation of the atom.
2mL2

 2  c
58. For the cubical well, the ground-state energy is given by Equation (6.51):
2
3 2 197.3 eV  nm 
3 2 2 3  c 
E


 101 eV.
2
2mL2
2mc 2 L2
2  5.11105 eV   2  5.292 102 nm 
This is about an order of magnitude higher than the electron’s kinetic energy in the Bohr
atom. The cubical well is not a good approximation. Also, the electron’s spatial
distribution goes beyond the Bohr radius, as you will see in Chapter 7. A less-confined
electron would correspond to a larger well, leading to a lower ground-state energy.
1
59. (a) From introductory physics, the oscillator’s energy is E  kA2 where A is the
2
amplitude. Equating this to the quantum oscillator energy [Equation (6.58)],
2
2
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Chapter 6
Quantum Mechanics II
93
1
1 2

 n     kA . The classical frequency is   k / m . Solving for the quantum
2
2

1  kA2 kA2 m A2 km

number,  n   


2 2  2
k
2

1  A2 km  0.10 m  12 N/m  0.15 kg 

n


 6.4 1031 .


34
2
2
2 1.055 10 J  s 


2
Thus n  6.4 1031 . This is to be expected, because the classical system is exceedingly
large as compared to a quantum oscillator.
1
1
1

(b) For the ground state n = 0, so  n       kA2 . Again using   k / m
2
2
2

k
 kA2 , so
and solving for amplitude A,
m
A 
2
2
km

1.055 10
34
J s
2
12 N/m  0.15 kg 
 7.86 1035 m2 , so A  8.9 1018 m.
The first quantum state is too small to be observed.
(c) The difference between adjacent states is E   
k / m  1.055 1034 J  s 
values, E 
k / m . Inserting numerical
12 N/m
 9.4 1034 J. Again, this energy is
0.15 kg
too small to be observed.
60. (a) For the one-dimensional potential well [Equation (6.35)], the ground-state energy is
E1 
2
2
2mL2

 2 1.055 1034 J  s 
2
2  2.0 10 kg 1.0 10 m 
6
3
2
 2.75 1056 J. If this is the classical kinetic
2  2.75 1056 J 
2K

 1.7 1025 m/s,
energy K = ½ mv , the resulting speed is v 
m
2.0 106 kg
which is much too small to be measured.
2
(b) Now the kinetic energy is K = ½ mv2 = 6.25 × 10−11 J. Solving Equation (6.35) for the
quantum number n:
 2  2.0 106 kg 1.0 103 m 2  6.25 1011 J  
 2mL E 
  1.511021.
n 2 2  
2
2
34





 1.055 10 J  s 


This extremely large value of n is an indication of how small the quantum states are.
2
1/ 2
1/ 2
61. (a) This was calculated in Problem 36.
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publicly accessible website, in whole or in part.
94
Chapter 6
Quantum Mechanics II

(b) x   x * dx  0 because the integrand is odd over symmetric limits. To find

2
x we first need to normalize:
A2 


1  2 x  e
2 2  x 2
dx  2 A2  1  2 x 2  e  x dx  1

2
2
0
 2 A2  1  4 x 2  4 2 x 4  e  x dx  1

2
2
0
 2 A2
Thus A2 
 1
3
 1    1
 2
2
1 
and
2 
x 2   x 2 * dx  2 A2  x 2 1  4 x 2  4 2 x 4  e  x dx 



0
2
62. Using the known functions for  1 and  2 we see:
1
2
  1 
3
1 2
3
x 
2 
sin 

2
2 L
 L  2
2
1  1 3 15  5
 

.
  4 2 4  2
2
 2 x 
sin 

L
 L 
1
3
x 
 2 x 
sin 
sin 


2L
2L
 L 
 L 
For normalization:
L
L 1
3
3
  x   2 x  
*
2 x 
2  2 x 
0   dx  0  2L sin  L   2L sin  L   2L sin  L  sin  L  dx .


The third term vanishes because of the orthogonality of the trigonometric functions,
leaving
L
L 1
3
*
2 x 
2  2 x  


dx

0
0  2L sin  L   2L sin  L  dx


1 L 2 x 
3 L 2  2 x 
sin
dx

sin 


 dx
2 L 0
2 L 0
 L 
 L 

1 L 3  L
 
   1 as required.
2L  2  2L  2 
63. Using the Taylor approximation for the exponential e x  1  x for small x, we have

V (r )  D 1  e a ( r re )

2
 D 1  1  a(r  re )    D  a(r  re )   Da 2  r  re  which is
2
2
2
quadratic in  r  re  .
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publicly accessible website, in whole or in part.
Chapter 6
Quantum Mechanics II
95
64. We will solve for numerical values of the factors in front of the quantum numbers:
 a
2D

2 D  m1  m2 
m1m2
 a
   6.582 1016 eV·s  7.8 109 m 1 

2  4.42 eV  39.10 u  35.45 u 
 2.998 108 m/s 
1u


931.5 106 eV/c 2 
c
 39.10 u  35.45 u 

2
2
  0.03477 eV and
 0.03477 eV 

4  4.42 eV 
4D
Evaluating for specific energy levels:
2
 6.838 105 eV
2
1
1


E0   0    0.03477 eV    0    6.838 105 eV   0.017 eV
2
2


2
 1
 1
E1  1    0.03477 eV   1    6.838 105 eV   0.052 eV
 2
 2
2
1
1


E2   2    0.03477 eV    2    6.838 105 eV   0.086 eV
2
2


2
1
1


E3   3    0.03477 eV    3    6.838 105 eV   0.121 eV
2
2


Note that for these low quantum numbers the second-order correction is small.
65. The solution is identical to the presentation in the text for the three-dimensional box but
without the z dimension. Briefly, we assume a trial function for the form
 ( x, y)  A sin(k1 x)sin(k2 y) . Assuming that one corner is at the origin, applying the
n
n
boundary conditions leads to k1  x and k2  y and substituting into the Schrödinger
L
L
equation leads to E 
L
L
L
0
0
0

 * dxdy  A2 
so A 
2
2
2
2mL

L
0
n
2
x
 ny2  . To normalize, solve the iterated double integral
 n y y 
n x
2  L  L 
sin 2  x  sin 2 
 dxdy  A     1
 2  2 
 L 
 L 
2
. Now to find the energy levels use the energy equation with different values of
L
the quantum numbers. Letting E0 
n1  1, n2  1 ;
2
2
2
2mL
we have: E1  E0 12  12   2E0 with
E2  E0  22  12   5E0 with n1  2, n2  1 or vice versa ;
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publicly accessible website, in whole or in part.
96
Chapter 6
Quantum Mechanics II
E3  E0  22  22   8E0 with n1  2, n2  2 ;
E4  E0  32  12   10E0 with n1  3, n2  1 or vice versa ;
E5  E0  32  22   13E0 with n1  3, n2  2 or vice versa ;
E6  E0  42  12   17 E0 with n1  4, n2  1 or vice versa .
66. (a) For the infinite square well, the energies are given by Equation (6.35) and increase as
n 2 . The energies for the finite square well are expressed in a transcendental equation of
 2m V  E  L 
V E
0
 0
the form tan 
where L is the width of the well. (See a


E


Quantum Mechanics text such as Griffiths.) This equation always has at least one
solution, no matter how weak the well.
(b) For the infinite square well, the wave functions are given by Equation (6.34) and the
2
 can be determined easily. Note that for the finite well, the wave functions extend
beyond the boundaries.
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publicly accessible website, in whole or in part.
Chapter 6
Quantum Mechanics II
97
(c) For the infinite square well, the wave functions equal zero exactly at the boundaries.
This is not true for the finite square well. The determination of the exact energies is rather
difficult and is often treated in quantum mechanics texts. The number of available
energies depends on the value of the potential. We assume that four states exist in this
problem.
67. This tunneling problem is similar to Example 6.14. Because of the information given in
the problem we will use Equation (6.73) to approximate the transmission probability. We
know  
2m V0  E 

2mc 2 V0  E 
c
so  
Then from Equation (6.73), T  2e2 L  2e

2  5.11105   0.9 eV 
197.33 eV  nm

2 4.86nm1 1.3nm 
 4.86 nm1.
 6.51106 .
68. (a) Note that this is an approximate procedure for one-dimensional problems with a
gradually varying potential, V ( x) . We begin with Equation (6.62b) which was derived
for a scenario where E  V but with V constant. We found k 
2m  E  V0 
for a
constant V0 . Because our potential varies slowly, we approximate the wave number by
2m  E  V ( x ) 
h
. Combining all of the
p
h
above, we find a position-dependent wavelength  ( x) 
.
2m  E  V ( x ) 
k
. We also know that p  k and  
(b) If we neglect barrier penetration, then the wave function must be zero at the turning
points. From the particle in a box example, we know that the number of wavelengths that
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publicly accessible website, in whole or in part.
98
Chapter 6
Quantum Mechanics II
1
3
, or 1, or , etc., which equals the distance divided by
2
2
the wavelength. By analogy, the number of wavelengths that can fit inside our potential
dx
n
well with a slowly varying wavelength is 
 where n is an integer. Substituting
 ( x) 2
fit between the turning points is
from above and rearranging, we have 2 2m  E  V ( x) dx  nh where n is an integer.
69. (a) Strictly speaking this approach is valid only when the potential varies slowly. From
the previous problem, we know 2 2m  E  V ( x) dx  nh , where n is an integer. For
the potential energy V ( x)   for x  0 and V ( x)  Ax for x  0 , we know the classical
turning points occur at x = 0 and where E  V  x   0  E  Ax. Let us call this point
b; then E  Ab  0 so b  E / A . Now we must evaluate 2
b
0
b  E / A . We know that
2 2m 
b
0
  a  qx 
 E  Ax) dx  2
1/2
dx 
2
3q
 a  qx 
3
2m  E  Ax) dx  nh with
. Therefore
 2 
3/2 b 
2m    E  Ax 
  nh . Simplifying, we find
0
 3 A 

 3 Anh 
3 Anh
4 2m
  E 3/2   nh so E 3/2 
or E  5/3 1/3 .
2 m
3A
4 2m
2/3
(b) A sketch is shown below. The wave functions are oscillating with the lowest state
being one-half a cycle, the second state a full cycle, etc. As mentioned in the answer for
previous problem, part (b), these sketches ignore barrier penetration in the region where
the potential is finite. No barrier penetration occurs where x = 0 where the potential is
infinite.
70. Equation (6.70) gives the transmission coefficient for tunneling: T  16
E
E  2 L
.
1   e
V0  V0 
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publicly accessible website, in whole or in part.
Chapter 6
Quantum Mechanics II
99
The tunneling current is proportional to the transmission coefficient. Suppose there is
initially a tunneling current I0 with a barrier width L0. Then the barrier width is increased
by 0.4 nm to L, resulting in a current decrease to I. The ratio of currents is the ratio of the
transmission coefficients. All the factors in the ratio cancel except the exponential,
I
e2 L
leaving
 2 L0  e2 L
I0 e
where ΔL = 0.4 nm. We wish to solve for V0 − E, the difference between the particle
energy and barrier height. This is contained in the factor κ in the exponential. Taking the
natural logarithm:
 I 
ln    2 L . Notice that in this case I < I0, so the logarithm is negative, and the
 I0 
equation is correct with ΔL > 0. Thus   
E,
2
V0  E 
V0  E 
 ln  I / I  
2
0
2
8m  L 
197.3 eV  nm 
 c   ln  I / I 0  

2
8  mc 2   L 
2
2
 ln 10 
4
8  5.11105 eV   0.4 nm 
2m V0  E 
ln  I / I 0 
. Solving for V0 −

2L
2
2
2
 5.0 eV.
This is entirely reasonable, depending on the materials used for the sample and probe.
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100
Chapter 7
The Hydrogen Atom
Chapter 7
1. Starting with Equation (7.7), let the electron move in a circle of radius a in the xy plane,
so sin   sin  / 2   1. With both r and  constant, R and f are also constant. Let
R  f  1 . Then g   and the derivatives of R and f are zero. With this Equation (7.7)
reduces to 
2
2
a 2 E  V  
1 d 2
. In uniform circular motion with an inverse-square
 d 2
force, we know from the planetary model that E  V / 2 , and E  V 
1 d 2
1 d 2 2
or
 2 E  0.
2
 d 2
a 2 d 2
2. This is a simple harmonic oscillator equation. Assume a standard trial solution
Thus 
2
V
V
V    E .
2
2
a2 E 
  A exp iB  . With this trial solution d 2 / d 2   B2 . Substituting this into the
1
2
 B 2   2 E   0 . Solving for B,
2 
a
equation from the previous problem we find:
B
2 E a
. To find A, normalize

2
0
2
 * d  1  A2  d  2 A2 so A  1/ 2 .
0
Note that B must be an integer (let B  n ) so that  will be single-valued
 (0)   (2 ). With B  n we have n2 
2
2
E a 2 and E 
n2 2
. For circular
2 a 2
L2
motion E 
where rotational inertia I   a 2 for a particle of mass  . Thus
2I
n2 2
L  2 I E  2 a
 n2
2
2 a
2
2
2
or L  n , which is the Bohr condition.
3. Assuming a trial solution g    Aeik (which is easily verified by direct substitution),
and using the boundary condition g (0)  g (2 ) , we find Ae0  Ae2 ik which is only true if
k is an integer.
4. Using the transformations it can be shown that for any vector A

1  ˆ 1 
  rˆ


and
r
r 
r sin  
1  2
1

1 A
. Because 2  · we can
r Ar  
 sin  A  

2
r r
r sin  
r sin  
combine our results to find
·A 
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publicly accessible website, in whole or in part.
Chapter 7
The Hydrogen Atom
101
1   2  
1
 

r
 2
 sin 
2
r r  r  r sin   

rearrangement gives the desired result.
2 
1
 2

and from this a simple

 2 2
2
 r sin  

r 
5. Letting the constants in the front of R be called A we have R  A  2   e r /2 a0 ,
a0 

 2
 3
dR
r 
d 2R
r 
 A    2  e r /2 a0 and
 A  2  3  e r /2 a0 . Substituting these into
2
dr
dr
 a0 2a0 
 2a0 4a0 
Equation (7.13) we have
 1 2 E   5
4 E
2 e2   4
4 e2  1


r






 0 . To satisfy the

 2
3
2 
2
2  
2 
4
a
a
2
a
4

a
a
4

r
0
0
0 0
0

  0
  0

equation, each of the expressions in parentheses must equal zero. From the 1/ r term we
find a0 
4
2
which is correct. From the r term we obtain E  
0
2
e
2
8 a
2
0

E0
which is
4
consistent with the Bohr result. The other expression in parentheses also leads directly to
E   E0 / 4 , so the solution is verified.
6. As in the previous problem R  Are r /2a0 ,

dR
r   r /2 a0
 A 1 
,
e
dr
 2a0 


dR
r 3   r /2 a0
d  2 dR 
2r 2 r 3   r /2 a0
 A r2 
e
r

A
2
r

 2 e
,
and
. Substituting




dr
2
a
dr
dr
a
4
a


0 
0
0 


these into Equation (7.10) with  1 and after substituting the Coulomb potential, we
r2
 1
2 E   1
2 e2 
1

r



 2  2   0 . The 1/ r term vanishes, and the
have  2

2 
2  
r
 4a0
  2a0 4 0 
middle expression (without r) reduces to a0 
we get E  
7. R 
2
8 a
2
0
e r /2 a0
3 2a0 
3/2

4
2
0
2
e
which is correct. From the r term
E0
which is consistent with the Bohr result.
4
e r / a0  r 
r
so R* R 

3 
a0
3 2a0   a0 
2
To normalize, we integrate over all r space:

1  4  r / a0
1
4!
2 *
r
R
Rdr

r e
dr 
 1 , so the wave function R21 was
5 0
5
0
24a0
24a0 1/ a0 5
normalized.
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102
Chapter 7
The Hydrogen Atom
8. The wave function given is  100 r , ,    Ae r / a0 so * is given by  100* 100  A2e2 r / a0 .
To normalize the wave function, compute the triple integral over all space
2


0
0
0
  dV  A   
*
2
r 2 sin  e2 r / a0 drd d . The  integral yields 2 , and the 
integral yields 2. This leaves
  dV  4 A 
*
2

0
r 2 e2 r / a0 dr  4 A2
2
2 / a0 
3
  a03 A2
This integral must equal 1 due to normalization which leads to  a03 A2  1 so A 
 a03
 5 and m  .
9. It is required that
 4: m  0, 1, 2, 3, 4 ;
 3: m  0, 1, 2, 3 ;
 1: m  0, 1
 2: m  0, 1, 2
10. n = 3 and
1
 0: m  0
 1 , so m  0 or 1 . Thus Lz  0 or  . L 

 1)   2
Ly and Lx are unrestricted except for the constraint L2x  L2y  L2  L2z .
11. For 3p state n = 3 and
 310  R31Y10 
 1 and m  0, 1 .
1 2 3/2 
r  r 
a0  6    e r /3a0 cos 
81 
a0  a0 

 311  R31Y11 

1
r  r 
a03/2  6    e r /3a0 sin  e i
a0  a0 
81 

x
12. The sum is of the form
y
y  x
x
2
which by symmetry is equivalent to 2 y 2 . Let us first
y 1
consider (as a lemma) the sum:
x
  y  1  y
y 1
3
x
3
   3 y 2  3 y  1

 y 1 
 23  13  33  23  ...  x  1  x3
3
 x  1  13  x3  3x 2  3x
3
x
x
x
x
Now let us write 3 y 2   1  y   y 3   3 y  1 . The first of these sums is

 y 1
y 1
y 1
y 1
3
x
given by our lemma above. The others are
y
y 1
1
x x  1 and
2
x
1  x .
Combining
y 1
x
3
1
these results 3 y 2  x3  3x 2  3x  x x  1  x  x 2 x  1x  1 . Therefore
2
2
y 1
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publicly accessible website, in whole or in part.
.
Chapter 7
The Hydrogen Atom
103
x
1
y  x 2 x  1x  1 and

6
y 1
2
L2  3 L2z 
3
m2

2  1 m l
2
x
y
2
y  x


1
 x 2 x  1x  1 . Then
3
 1
2
.
13. As in Example 7.4 the degeneracy is n2  36 .
14. There are five possible orientations,
corresponding to the five different
values of m  0, 1, 2 .
For the m  1 component we have (with
Lz  m   with L2x  L2y  L2  L2z  6
15. cos  
Lz

L
could have
m

 1
 2 ), L 
2

2

 1
 6
and
 5 2.
(See diagram in the preceding problem.) For this extreme case we
 m so cos 10 

 1
or cos 2 10 
2

 1

2
2

. Rearranging
1
we find


1
  32.16 and we have to round up in order to get within 10 ,


1
2

 cos 10  


so  33 .
16. There is one possible m value for
 0 , three values of m for
 1 , five values of m
 2 , and so on, so that the degeneracy of the nth level is
1  3  5  ...(2  1)  1  3  5  ...(2(n 1)  1)  n2 .
for
17.  211  R21Y11 
 r   r /2 a0
1
sin  ei ;
e
3/2 
8  a0  a0 
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104
Chapter 7
 210  R21Y10 
The Hydrogen Atom
 r   r /2 a0
1
cos  ;
e
3/2 
4 2 a0  a0 
 r 2   r /3a0
e
sin  cos  ei
3/2  2 
a
81  a0  0 
18. We must calculate the triple integral over all space. The definite integrals can be
evaluated from a handbook or from Appendix 3.
1
 321  R32Y21 
2

r  r /a 2
  200 dV  32 a03 0 0 0  2  a0  e 0 r sin  drd d
The  integral yields 2 , and the  integral yields 2. This leaves
1
*
200


2
 2 4r 3 r 4   r / a0
0
0  4r  a0  a02 e dr

1 
4
1
 3  4 2!a03  3!a04  2 4!a05 
8a0 
a0
a0

1
1
  dV  8a03
*
2


r  2  r / a0
1
dr  3
2  r e
a0 
8a0


as required.
Repeat the process with the second wave function:
2
 r  2 3 r /a
  211 dV  64 a03 0 0 0  a0  r sin  e 0 drd d .
The  integral yields 2 as before. The  integral can be found in integral tables and
*
211
1
2


2

1   r  2  r / a0
1 1
equals 4/3. This leaves   dV 
dr 
4!a05   1
r e
3 0 
3  2 
24a0  a0 
24a0  a0

as required.
19. The value of n sets the maximum value of . However the letter p identifies the value of
for this problem. With  1 we have m  0, 1 and Lz  m  0,  .
*
20. The maximum difference is between the m  2 and m  2 levels, so m  4 . Then
V  B m  B  5.788 105 eV/T 4 3.5 T   8.10 104 eV .
21. Differentiating E 
hc

we find: dE  
hc

2
d  or E 
hc
2
 . In the normal Zeeman
 hc 
 2 B
effect, between adjacent m states E  B B so B B   2   or   0 B .
hc
 0 
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Chapter 7
The Hydrogen Atom
105
22. See the solution to problem 14 for the sketch. To compute the angles with  2 :
L
m
m
. There are five different values of  , corresponding to the
cos   z 

L
  1 6
different m values 0, 1, 2 :
 2 

  35.3
 6
  cos1 
 1 

  114.1
 6
  cos 1 
23. With
 1 

  65.9
 6
  cos1 
  cos 1 0   90
 2 

  144.7
 6
  cos1 
 3 we have (as in the previous problem) cos  
Lz

L
m

 1

m
. For the
12
 3
 3 
1

minimum angle m   3 and   cos 1 
  cos  2   30 .
 12 


24. From Problem 21,  
02 B B
so the magnetic field is
hc
1240 eV  nm  0.04 nm   1.99 T.
hc 
B 2

0 B  656.5 nm 2  5.788 105 eV/T 
25. There are seven different states, corresponding to m  0, 1, 2, 3 . In the absence of a
magnetic field E  
E0
13.606 eV

 0.544 eV . The Zeeman splitting is given by
2
n
25
E  B B m   5.788 105 eV/T 3.00 T m  1.7364 104 eV m . For m  0
we have E  0 . For the other m states:
m  1: E  1.7364 104 eV 1  1.74 104 eV ;
m  2: E  1.7364 104 eV 2   3.47 104 eV ;
m  3: E  1.7364 104 eV 3  5.21104 eV .
26. From the text the magnitude of the spin magnetic moment is
2 B S
2 B
3
 3B .
2
The z-component of the magnetic moment is (see Figure 7.9)

1/ 2
 z  s cos   s
 s  B .
3/2
3
s 

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106
Chapter 7
The Hydrogen Atom
The potential energy is V    B  z Bz and so the vertical component of force is
Fz  dV / dz  z dBz / dz  . From mechanics the acceleration is az 
Fz  z dBz
and

m m dz
1 2
az t . With the
2
time equal to the horizontal distance divided by incoming speed, or t  x / vx , we have
with constant acceleration the vertical deflection of each beam is z 
2
1 2 1   z dBz   x  1  9.27 1024 J/T 
 0.071 m 
4
z  az t  
 2000 T/m 
  3.017 10 m
   
25
2
2  m dz   vx 
2  1.8110 kg 
 925 m/s 
The separation between the two silver beams is twice this amount, 6.035 104 m.
2
27. The kinetic energy of the atoms is
3
3
K  kT  1.38 1023 J/K 1273 K   2.64 1020 J . From the previous problem,
2
2
2
  dBz   x 
we see that the separation of the beams is (remember  z  B ) s  2 z   B
  .
 m dz   vx 
20
dBz smv 2 2sK 2  0.01 m   2.64 10 J 



 57.0 T  m .
Rearranging we see that x
dz
B
B
9.27 1024 J/T
2
The magnet should be designed so that the product of its length squared and its vertical
magnetic field gradient is 57 T  m.
28. As shown in Figure 7.9 the electron spin vector cannot point in the direction of B ,
because its magnitude is S  s(s  1)  3 / 4
and its z-component is S z  ms  / 2 .
If the z-component of a vector is less than the vector's magnitude, the vector does not lie
along the z-axis.
29. For the 4f state n = 4 and
 3 . The possible m values are 0, 1, 2, and 3 with
ms  1/ 2 for each possible m value. The degeneracy of the 4f state is then (with 2 spin
states per m ) equal to 2(7) = 14.
30. For the 5d state n = 5 and
 2 . The possible m values are 0, 1, and 2, with
ms  1/ 2 for each possible m value. The degeneracy of the 5d state is then (with 2 spin
states per m ) equal to 2(5) = 10.
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Chapter 7
The Hydrogen Atom
107
31. If we determine the thermal energy that equals the energy required for the spin-flip
3
3
transition, we have 5.9 106 eV  kT  8.617 105 eV/K T . This gives
2
2
T  0.0456 K .
32. The spin degeneracy is 2 and the n 2 is shown in Problem 16.
33. The selection rule m  0, 1 gives three lines in each case.
34. (a)   0 is forbidden
(b) allowed but with n  0 there is no energy difference unless an external magnetic
field is present
(c)   2 is forbidden
 1 1
(d) allowed with absorbed photon of energy E  E0  2  2   2.55 eV
2 4 
35. We must find the maxima and minima of the following function.
2

r  2
4r 3 r 4   r / a0
2
2
P(r )  r R(r )  A e
 e
 2   r  A  4r 
a0 
a0 a02 


dP
To find the extremes set
 0:
dr
2
2
2  r / a0
0
1  2 4r 3 r 4   r / a0 
12r 2 4r 3   r / a0
which simplifies to
4
r


e

8
r

 2 e



a0 
a0 a02 
a0
a0 

0
r 3 8r 2 16r


8
a03 a02
a0
Letting x 
r
the equation above can be factored into x  2 x 2  6 x  4  0 . From the
a0
first factor we obtain x = 2 (or r  2a0 ), which from Figure 7.12 we can see is a
minimum. The second parenthesis gives a quadratic equation with solutions x  3  5 ,


so r  3  5 a0 . These are both maxima.


36. In the previous problem we found that the two maxima are at r  3  5 a0 . From


Figure 7.12 it is clear that the peak at r  3  5 a0 is higher. This can be verified by
substituting the two values for r and computing P(r ) . The most probable location is
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108
Chapter 7

The Hydrogen Atom

therefore at r  3  5 a0  5.24a0 , which is significantly further from the nucleus than
the 2p peak at r  4a0 .
37. From the solution to Problem 35 we see that P(r )  0 at r  2a0 . Note that P(0)  0 also.
A sketch is found in Figure 7.12.
38. The radial probability distribution for the ground state is P(r )  r 2 R(r ) 
2
With r
4 2 2 r / a0
r e
.
a03
a0 throughout this interval we can approximate the exponential term as
e2 r / a0  1 . Therefore the probability of being inside a radius 1.2 1015 m is

1.21015
0
4
P(r ) dr  3
a0

1.21015
0
4r 3
r 2 dr  3
3a0
1.21015
 1.55 1014
0
39. As in the previous problem, P(r )  r 2 R(r ) 
2
4 2 2 r / a0
re
. To find the desired
a03
probability, integrate P(r ) over the appropriate limits:
Letting x  r / a0 which implies dx  dr / a0 . Thus


1.05 a0
0.95 a0
1.05 a0
0.95 a0
P(r ) dr 
P(r ) dr  4
4
a03
1.05
0.95

1.05 a0
0.95 a0
r 2e2 r / a0 dr.
x 2e2 x dx  0.056
where the definite integral was evaluated using Mathcad.


40. In general r   r P(r ) dr   r 3 R(r ) dr . For the 2s state:
0
r 
1
8a03


0
2
0

4r r 
r 3  4   2  e r / a0 dr. Using integral tables
a0 a0 

2


0
r n e r / a0 dr  n !a0 
n 1

a0
4
1
4
5
6
 4 3!a0  4!a0  2 5!a0   24  96  120   6a0
a0
a0

 8
For the 2p state:
r 
1
8a03
2
 r  r /a
1
0 r  a0  e 0 dr  24a05
120a0
1

5! a 6 
 5a0
5   0 
24a0
24
1
r 
24a03

3


0
r 5e r / a0 dr
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Chapter 7
The Hydrogen Atom
109
2

1
r 
41. For the 2s state: P(r )  r R(r )  3 r 2  2   e  r / a0 . As in Problem 38 for r
8a0 
a0 
2
2
a0
we can say e r / a0  1 , so the probability is given by the integral

1.21015
0
1
P(r ) dr  3
8a0

1.21015
0
2

r 
1
r  2   dr  3
a0 
8a0

2

1.21015
0
 2 4r 3 r 4 
  dr
 4r 
a0 a02 

1.21015
1  4 r3 r4 r5 
 
 

8  3 a03 a04 5a05  0
 1.9 1015
Similarly for the 2p state: P(r )  r 2 R(r ) 
2

1.21015
0
1
P(r ) dr 
24a05

1.21015
0
r5
r 4 dr 
120a05
1 4  r / a0
re
.
24a05
1.21015
 5.0 1026
0
42. To find the most probable radial position in the 3d state:
2
2 2 r /3a0
P(r )  r R(r )  A e
2
2
To find the extrema, set
6
 r2  2
2  r  2 r /3 a0
r

A
 2
 4 e
 a0 
 a0 
1  6r 5 2r 6 
dP
 0 : 0   3  4  e2 r /3a0 which is equivalent to
a0  a0 3a0 
dr
6r 5 2r 6
0  3  4 . r  9a0 and r  0 are the two solutions. From Figure (7.12) you can see
a0 3a0
that r  0 is a minimum where P(r )  0 and r  9a0 is a maximum.
43. For the 3d state:

  r 2 2 2 r /3a 
  r 6  2 r /3a
16
8
0
0
e

e




 812 30    a02 
 812 15   a07 




To find that probability that the electron in the 3d state is location at a position greater
  r 6  2 r /3a


8
0
dr .
 7 e
than a0 , we must evaluate  P(r ) dr   
2
a0
a0 

a
81
15




0




(Alternatively, we could evaluate the integral from 0 to a0 and subtract this answer from 1
1
P(r )  r R(r )  r  
 a0 
2
2
3
2
since we know that the wave functions are normalized.) From integral tables, we find
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110
Chapter 7
m
integrals of the form
m bx
bx
 x e dx  e  1
i 0
i
The Hydrogen Atom
2
m ! x m i
where m = 6 and b 
for our
i 1
3a0
m  i !b
example. Evaluation of the summation gives a probability of 0.9999935 that the 3d
electron will be at a position greater than a0 . That means it is almost certain!
e2
1.44 109 eV  m
44. R 

 2.82 1015 m . From the angular momentum
2
3
4 0 mc
51110 eV
equation v 
3 197.33 eV  nm 
3
3 c

c
c  103 c . A speed of
2
4mR 4mc R
4  511103 eV  2.82 106 nm 
103c is prohibited by the postulates of relativity.

45. The electron radius would be c  1.211012 m. As in the previous problem
2
3 197.33 eV  nm 
3
3 c
v

c
c  0.24 c . This result is allowed
2
4mR 4mc R
4  511103 eV 1.21103 nm 
by relativity. However, in order to obtain this allowed result, we had to assume an
unreasonably large size for the electron (one thousand times larger in radius than a
proton!).
Ze2
46. (a) The only change in Equation (7.3) is in the potential energy term, with V  
so
4 0 r
the Schrödinger equation becomes:
1   2 
r
r 2 r  r
1
 


 2
 sin 

 r sin   
1
 2 2 
Ze2 



E

  0 .
 2 2
2
2 
4 0 r 
 r sin  

(b) Because V occurs only in the radial part, there is no change in the separation of
variables.
(c) Yes, from Equation (7.10) is it clear that the radial wave functions will change.
(d) No, there is no change in the  or  dependence.
47. Carrying Z through the derivation in the text [Equations (7.12) through (7.14)] we find
 100
1 Z

 
  a0 
3/2
e Zr / a0 .

r   r /3a0
48. Use the wave function R31  Ar 1 
where A is constant. Then
e
 6a0 

r3
r 4  2 r /3a0
dP
2
P(r )  r 2 R(r )  A2  r 2 

e
. To find the extrema, set
 0 . Doing
2 
3a0 36a0 
dr

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publicly accessible website, in whole or in part.
Chapter 7
The Hydrogen Atom
111
so and factoring out A2 re2 r /3a0 gives 2 
5r
r2
r3
r
 2
 0 . Letting x  and
3
a0
3a0 3a0 54a0
multiplying both sides by - 54 we obtain
x3  18x 2  90 x  108  0  x  6 x 2  12 x  18.
(a) The minimum is at x = 6, or r  6a0 , and we find P(6a0 )  0 . Clearly P(0)  0 also.
(b) The two maxima come from the quadratic equation in parentheses, with x  6  3 2


or r  6  3 2 a0 .
(c) Y11  constant sin  ei Then Y *Y is proportional to sin 2  , and the probability is
zero at   0 and   180 .
49. The ground state energy can be obtained using the standard Rydberg formula with the
reduced mass,  , of the muonic atom.
E0 
e2
8 0 a0

 e4
2
2  4 0 
2
Computing the reduced mass:
mm
1 938.27 MeV 105.66 MeV 
 p   2
 94.966 MeV/c 2
m p  m c 938.27 MeV  105.66MeV 
Thus
 e4
E0 
2
2  4 0 
1.44 eV  nm   94.966 106 eV 
 e2   c 2


 2.53 keV

2
2 2
4

2
c
2
197.33
eV

nm




0 

2
2
2
50. The interaction between the magnetic moment of the proton and magnetic moment of the
electron causes hyperfine splitting. The transition between the two states causes emission
hc 1240eV  nm 

 5.9 106 eV. From the uncertainty
of a photon with energy of E 
7

2110 nm
principle, we know E t  . With a lifetime of t  1107 y then
2
E 
6.58211016 eV
 1.0411030 eV.
7
7
2 110 y  3.16 10 s/y 
51. Break the vector r into its components r  xi  y j  zk . Because the two factors R and
Y in the hydrogen wave functions are given in spherical coordinates, it is best to express
each of the Cartesian components in spherical coordinates using the standard
transformations x  r sin  cos  , y  r sin  sin  , and z  r cos as shown in Figure 7.1.
We know that in spherical coordinates, the volume element d  r 2 sin  drd d .
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112
Chapter 7
The Hydrogen Atom
1
.
4
For simplicity we will write each component of the integral separately. Ignoring constant
In this case we have Y = Y00 = 1/ 2  for both the initial and final states, so Y f*Yi 


2
0
0
0
factors, the x component is  *f x i d   r 3 R*f Ri dr  sin 2  d  cos  d.
The integral over  is zero, so the product is zero. Similarly, the y and z components of
the integral are zero, and so the transition probability is zero. (For the z component, it is
the integral over θ that vanishes.)
52. The solution is similar to the preceding problem, but now the spherical harmonics give
1 1 5
5
3cos2   1 
3cos 2   1


4

8

2 
Ignoring constant factors, the x component of the integral is
Y f*Yi 
*
3 *
2
2
 f x i d   r R f Ri dr  sin  3cos  1d  cos  d.


2
0
0
0
As in the preceding problem, the integral over  is zero, so the product is zero. Similarly,
the y and z components of the integral are zero, and so the transition probability is zero.
(For the z component, it’s the integral over θ that vanishes.)
53. In this case we want to show that the transition integral is not zero. From Table 7.2,
Y f*Yi 
1 1 3
3
cos  
cos  . Ignoring constant factors, the x component of the
4
2  2 


2
0
0
0
integral is  *f x i d   r 3 R*f Ri dr  sin  cos  d  cos  d which is still zero, as is
the y component. However, in this case the z component is

*
f


2
0
0
0
z i d   r 3 R*f Ri dr  sin  cos2  d  d .
All three factors in this integral are nonzero, so the net result is nonzero, implying that the
transition may occur. This is consistent with the selection rules described in Section 7.6.
54. The probability density is
2
1  4  6 2 r / 3a0
P  r R32 (r )  7 
re
a0  81 30 
2
2
To find the peak, set the derivative of this function equal to zero:
dP 1  4 

dr a07  81 30 
2
 5 2 r / 3a0
 2r  2 r / 3a0 
 r6  
 6r e
0
e
3
a
0 



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Chapter 7
The Hydrogen Atom
113

2r 
Therefore 0  6 
 which leads to r  9a0 . This matches the Bohr result.
 3a0 
55. (a) As in Example 7.14, r3 p
Table 7.1 R3 p 
r3 p
2
r 3 R3 p (r ) dr . Using the value from
rP(r )dr
0
0
1
4 
r  r  r / 3a0
, so
6  e
3/ 2
a0 81 6 
a0  a0
1
4
3
a0 81 6
2
3
r 6
0
r
a0
2
r2
e
a02
2 r / 3 a0
dr
25a0
.
2
This is 12.5a0, which is substantially larger than the Bohr result 9a0. Our answer seems
reasonable, given the shape of the graph of P(r) in Figure 7.12.
(b) The probability is
Integration using a computer program yields r3 p 
P
2
P(r )dr
9 a0
9 a0
2
r R3 p (r ) dr
1
4
3
a0 81 6
2
2
r 6
9 a0
r
a0
2
r2
e
a02
2 r / 3 a0
dr.
Because the lower limit is no longer zero, we will use numerical integration on a
computer. The result to three significant figures is P = 0.820. That may seem rather high,
but again it is reasonable, considering the shape of the graph of P(r) in Figure 7.12.
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114
Chapter 8
Atomic Physics
Chapter 8
1. The first two electrons are in the 1s subshell and have
electron is in the 2s subshell and has
 0 , with ms  1/ 2 . The third
 0 , with either ms  1/ 2 or 1/ 2 . With four
particles there are six possible interactions: the nucleus with electrons 1, 2, 3; electron 1
with electron 2; electron 1 with electron 3; or electron 2 with electron 3. In each case it is
possible to have a Coulomb interaction and a magnetic moment interaction.
2. H: 1s1 , He: 1s 2 , Li: 1s 2 2s1 , Be: 1s 2 2s 2 , B: 1s 2 2s 2 2 p1 , C: 1s 2 2s 2 2 p 2 , N: 1s 2 2s 2 2 p3 ,
O: 1s 2 2s 2 2 p 4 , F: 1s 2 2s 2 2 p5 , Ne: 1s 2 2s 2 2 p6
3. L: n = 2, so there are two (2s and 2p)
N: n = 4, so there are four (4s, 4p, 4d, and 4f)
O: n = 5, so there are five (5s, 5p, 5d, 5f, and 5g)
4. In the first excited state, go to the next higher level. In neon one of the 2p electrons is
promoted to 3s, so the configuration is 2 p5 3s1 . By the same reasoning the first excited
state of xenon is 5 p5 6s1 .
5. K:  Ar  4s1 , As:  Ar  4s 2 3d 10 4 p3 , Nb:  Kr  5s1 4d 4 , Pd:  Kr  4d 10 , Sm:  Xe 6s 2 4 f 6 ,
Po: [Xe]6s 2 4 f 14 6s 2 5d 10 6 p 4 U:  Rn  7s 2 6d 1 5 f 3 where the bracket represents a closed
inner shell. For example,  Ar  represents 1s 2 2s 2 2 p6 3s 2 3 p6 .
6.
(a)
(b)
(c)
(d)
He, Ne, Ar, Kr, Xe, Rn
Li, Na, K, Rb, Cs, Fr
F, Cl, Br, I, At
Be, Mg, Ca, Sr, Ba, Ra
7. From Figure 8.4 we see that the radius of Na is about 0.16 nm. We know that for singleelectron atoms E  
Ze2
. Therefore
8 0 r
2  0.16 nm  5.14 eV 
8 0 rE
4
e  2 2 0 rEe  
e  1.14e .
2
e
e
1.44 eV  nm
8. Using the  n, , m , ms  notation, the first four electrons (in the 1s and 2s orbitals) have
Ze  
quantum numbers 1,0,0, 1/ 2  and  2,0,0, 1/ 2  . The remaining two electrons are in
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Chapter 8
Atomic Physics
115
the 2p orbital with n = 2 and  1 . By Hund's rules, we want the total spin angular
momentum, S , to be maximized, so ms  1/ 2 or ms  1/ 2 , and the total angular
momentum, L, should be maximized, so the possibilities are  2,1,1, 1/ 2  ,  2,1,0, 1/ 2 
or  2,1,1, 1/ 2  ,  2,1, 0, 1/ 2  .
9. (a) count the number of electrons, 9; F, or use Table 8.2
(b) count the number of electrons, 12: Mg
(c) filled 3p shell: Ar
10. Ag:  Kr  4d 10 5s1 , Hf:  Xe 4 f 14 5d 2 6s 2 , Sb:  Kr  4d 10 5s 2 5 p3 where the bracket
represents a closed inner shell. For example,  Kr  represents
1s 2 2s 2 2 p6 3s 2 3 p6 4s 2 3d 10 4 p6 .
11. (a) filled 3d: Se
(b) filled 4p and 4d: Ag
(c) The 4f shell fills in the lanthanides: Er
12. J ranges from L  S to L  S or 2,3, 4 . Then in spectroscopic notation
2 S 1
LJ , we
have three possibilities: 3 F2 , 3 F3 , or 3 F4 . The ground state has the lowest J value, or 3 F2 .
With n = 4 the full notation is 4 3 F2 .
13. For indium (In) there is a single electron in the 5p subshell. Therefore S  12 and
2S  1  2 . Because that electron is in the p subshell we have L  1 . The possible J values
range from L  S to L  S , which in this case gives J = 1/2 or J = 3/2. Therefore the
possible states are 2 P1/2 and 2 P3/2 , with 2 P1/2 being the ground state. The full notation is
5 2 P1/2 .
14. No spin-orbit interaction requires either a closed subshell (S = 0), which occur in He, Be,
Ne, Mg, Ar, and Ca or L = 0 which we have in H, Li, Na, and K. (It is the 1st excited
state of the alkali metals which accounts for the splitting in the alkali metal spectra with
the visible transition begin from the 2nd excited state to the first in Na for example.)
15. In the 4d state  2 and s  1/ 2 , so j = 5/2 or 3/2. As usual m  0, 1, 2 . The value of
m j ranges from  j to j so its possible values are 1/ 2 , 3 / 2 , and 5 / 2 . As always
ms  1/ 2 . The two possible term notations are 4D5/2 and 4D3/2 .
16. 1 S0 : S  0, L  0, J  0
2
D5/2 : S  1/ 2, L  2, J  5 / 2
F1 : S  2, L  3, J  1
3
F4 : S  1, L  3, J  4
5
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116
Chapter 8
Atomic Physics
17. The quantum number mJ ranges from  J to J , or 7 / 2 to+7/2. Then J z  mJ   / 2 ,
3 / 2 , 5 / 2 , 7 / 2 .
18. (a) The quantum number mJ ranges from  J to J , or 7 / 2 to 7 / 2 . Then
J z  mJ   / 2 , 3 / 2 , 5 / 2 , 7 / 2 .
(b) The minimum angle occurs when mJ is at its maximum value, which is 7 / 2 . Then
J z  7 / 2 and cos  
Jz

J
mJ
J  J  1

7/2
 7 / 2  9 / 2 

7
so   28.1 .
3
19. The 1s subshell is filled, and there is one 2s electron. Therefore L = 0 (which gives the S
in the middle of the symbol). The single unpaired electron gives S = 1/2 or 2S  1  2 .
With L = 0 and S = 1/2 the only possible value for J is J = 1/2.
20. Ga has a ground state configuration with a filled 3d shell and 4s 2 4 p1 . The single
unpaired electron gives S = 1/2, so 2S  1  2 . The unpaired electron is in a p subshell, so
L = 1, for which the symbol is P. Then J  L  S which is 1/ 2 or 3/2, but the 1/2 state has
a slightly lower energy. Therefore the symbol is 4 2 P1/2 .
1
1


3
 1240eV  nm  

  7.334 10 eV.
1 2
766.41
nm
769.90
nm


As in Example 8.8 the internal magnetic field is
21. E 
hc

hc
9.109 1031 kg  7.334 103 eV 

mE
B

 63.4 T
e
1.602 1019 C 6.582 1016 eV  s 
22. By the selection rules L  1 , S  0 , J  0, 1 , no transitions are allowed between
the pictured levels. The selection rule S  0 prohibits singlet  triplet transitions.
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Chapter 8
Atomic Physics
117
23. The 2s to 1s transition is forbidden by the L  1 selection rule. The two lines result
from the transitions from the two 2p levels to the 1s level.
24. As in Example 8.8
E 
e B 1.602 10

m
19
C  6.582 1016 eV  s   2.55 T 
9.109 10
31
kg
 2.95 104 eV .
25. The ground state in He is a singlet, so an excited state may be a singlet or triplet, and the
behavior of Ca ( 4s 2 ) and Sr ( 5s 2 ) must be the same. Al has a 3s 2 3 p1 configuration. The
single unpaired electron gives S  1/ 2 and 2S  1  2 , which is a doublet.
26. The minimum angle corresponds to the maximum value of J z and hence the maximum
value of m j , which is m j  j . Then cos  
j we find j 
27.
1
1
1
cos2 
Jz

J
mj
j  j  1

j
j  j  1
. Solving for
 2.50  5 / 2 .
 I  Zev
dq
Ze
Zev


From the Biot-Savart Law, B  0  0 2 . With the
dt 2 r / v 2 r
2r
4 r
L
assumption that the orbit is circular, the angular momentum is L  mvr , so v 
and
mr
 ZeL
ZeL
1
where we have used the fact that 0  2 . The directions of
B 0 3 
2 3
4 mr
4 0 mc r
c 0
I
the vectors follow from the right-hand rule.
28. Using the fact that g = 2 we have s  g
Ze2 S  L
eS eS
V


·
B

and
thus
.

s
s
4 0 m2c 2 r 3
2m m
29. (a) In order to use the result of the previous problem, we need to know the directions of
S and L . The electron has S  1/ 2 , so S  3 / 4
and the angle S makes with the
 1/ 2 


+z-axis is   cos 1 
  54.7 or 125.3 .
 3/ 4 
With L  1 we have L  2 and the vector L can have three possible orientations,
corresponding to m  0, 1 . If we choose m  0 , then the orientation of the L vector is
in the xy plane. Therefore for either spin state the angle between L and S is 35.3  and
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118
Chapter 8
3
2 cos  35.3  
4
Problem 40), we have
S L 
2
Atomic Physics
. Then using r  5a0 for a 2p electron (see Chapter 7
1.44 eV  nm 1240 eV  nm 
e2 2
V

 1.2 105 eV . The difference
2
2 2 3
3
4 0 m c r
4 2  5.11105 eV   5  0.0529 nm 
2
between spin-up and spin-down states is twice this amount, or 2.4 105 eV, which is just
over half the measured value.
(b) The two possibilities are j = 1/2 and j = 3/2. The difference between these two is




Z
2
2
2


V 
mc

2n 3
 2 1  1 2 3  1 
  
 2
2 
  
4
4


 1 




2
2
137 

5

  4.53 105 eV

5.1110 eV 

3 
 2 1  1 2 3  1 
2  2
  
 2
2 
  
which gives a more accurate result.
4
30. In Example 8.10, it was shown that the normal Zeeman effect is expected when the total
spin S = 0. The normal Zeeman effect is due to the interaction of the external magnetic
field and the orbital angular momentum, which is true for all levels. If S is not zero, then
the anomalous effect is expected. Thus the two effects are not expected in the same atom.
31.   
av  

B
B
 L  2S  and J  L  S
 L  2S    L  S  J

J2
Using the equation in the problem, we have

Now L  2S  L  S  L2  2S 2  3L  S and using J  L  S we have
J  J  L2  S 2  2L  S so L·S 
Therefore av  
V   av  B 
J 2  L2  S 2
and
2
B 3J 2  S 2  L2
2J 2
B B 3J 2  S 2  L2
2J 2
 L  2S    L  S   3J
2
 S 2  L2
.
2
J . With B defined to be in the +z-direction,
J z . As usual J z  mJ , so V  B BmJ
Now the vector magnitudes are J 2  j ( j  1)
2
, S 2  s(s  1)
2
3J 2  S 2  L2
.
2J 2
, and L2  (  1)
2
so we
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Chapter 8
Atomic Physics
119
obtain V  B BmJ g where
g
3 j ( j  1)  s( s  1)  (  1)
j ( j  1)  s( s  1)  (  1)
.
 1
2 j ( j  1)
2 j ( j  1)
32. E 
hc

hc

1240 eV  nm 1240 eV  nm

 1.166 104 eV
422.7 nm
422.7168 nm
1 2
[Note: The same answer to four significant figures is obtained using the approximation
E  hc  /  2 as in Example 8.3.] Also E  B Bm   e / 2m  B with m  1 .
4
e
2E 2 1.166 10 eV  1.602 1019 J
Thus


 1.868 1023 J/T . The accepted
m
B
2.00 T
eV
e
 2B  2  9.274 1024 J/T   1.855 1023 J/T which is less than 0.2% lower
value is
m
than the experimental value. Using the experimental data with the known value of ,
e  e  1 1.868 1023 J/T
  
 1.77 1011 C/kg, which compares favorably to the known
34
m m
1.055 10 J  s
value of
e 1.6022 1019 C

 1.76 1011 C/kg .
m 9.1094 1031 kg
1/ 2  3 / 2   1/ 2  3 / 2   1  1  2
2 1/ 2  3 / 2 
 3 / 2 5 / 2  1/ 2 3 / 2  1(2)  1  1  4
S  1/ 2 , and J  3 / 2 ; g  1 
2  3 / 2  5 / 2 
3 3
 5 / 2 7 / 2  1/ 2 3 / 2  2(3)  1  1  6
S  1/ 2 , and J  5 / 2 ; g  1 
2  5 / 2  7 / 2 
5 5
33. (a) L  0 , S  J  1/ 2 ; g  1 
(b) L  1 ,
(c) L  2 ,
34. Computing the g factors: 2 G9/2 : L  4 , S  1/ 2 , and J  9 / 2 ;
g  1
2
 9 / 211/ 2  1/ 2 3 / 2  4(5)  1  1  10
2  9 / 2 11/ 2 
9 9
H11/2 : L  5 , S  1/ 2 , and J  11/ 2 ;
g  1
11/ 213 / 2  1/ 2 3 / 2  5(6)  1  1  12
2 11/ 2 13 / 2 
11 11
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120
Chapter 8
Atomic Physics
1 2   1 2   1(2)  1  1  3
2 1 2 
2 2
 2  3  1 2  2(3)  1  1  7
g  1
2  2  3
6 6
35. 3 P1 : L  1 , S  1 , and J  1 ; g  1 
3
D2 : L  2 , S  1 , and J  2 ;
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Chapter 8
Atomic Physics
121
36. From Equation (8.22) we have an energy splitting of V  B Bg mJ .
For the 2 S1/2 state g  2 , so V  2B BmJ . Now mJ can vary from  J to  J , so
mJ  1/ 2 and we find V   B B .
For the 2 P1/2 state g  2 / 3 , so V  23 B BmJ . Now mJ can vary from  J to  J , so
mJ  1/ 2 and we find V   13 B B .
1
1
The maximum difference is from  B B to  B B or from  B B to  B B . In each
3
3
case the magnitude of the energy difference is 43 B B for a total difference of 83 B B .
Calling this difference E , it will show up as the usual wavelength difference  with
E   hc /  2   . Thus
7
24
8  5.8976 10 m   9.274 10 J/T   2.50 T 
 

B B 
hc
hc 3
3
 6.626 1034 J  s  2.997 108 m/s 
 2 E
2 8
2
 1.083 1010 m
37. As in the preceding problem we have g  4 / 3 for the 2 P3/2 state and V  43 B BmJ . The
states are split by mJ  1 and so the energy difference between two adjacent states is
E  43 B B .
4
4
B B   9.274 1024 J/T  1.20 T   6.18 1024 J  9.26 105 eV.
3
3
38. (a) Calcium has two 4s subshell electrons outside a closed 3 p 6 subshell. So the
Then E 
electronic configuration is 1s 2 2s 2 2 p6 3s 2 3 p6 4s 2 . Aluminum has one 3p electron outside
the closed 3s level of magnesium, so its electronic configuration is 1s 2 2s 2 2 p6 3s 2 3 p1 .
(b) The LS coupling for calcium can be determined since both outer shell electrons have
 0 , then L  0 and S  0 since one electron will have spin up and one will have spin
down. Then J  L  S will also be 0. Therefore in spectroscopic notation, n 2 S 1LJ ,
calcium will be 4 1S0 . The 3p electron for aluminum will have
 1 . Therefore L  1
and S  1/ 2 . Now J  L  S , so J  3 / 2 or J  1/ 2 . The state with lower J has lower
energy, so in spectroscopic notation, aluminum is 3 2 P1/2 .
39. (a) Yttrium (Y) has one 4d and two 5s electrons outside a closed 4 p 6 subshell. So Y 
would have one additional electron in the 4d subshell. So the electronic configuration is
1s 2 2s 2 2 p6 3s 2 3 p6 4s 2 3d 10 4 p6 5s 2 4d 2 . Aluminum has one 3p electron outside the closed
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122
Chapter 8
Atomic Physics
3s subshell. So Al  would have one additional electron in the 3p subshell. So its
electronic configuration is 1s 2 2s 2 2 p6 3s 2 3 p 2 .
(b) We will use the LS coupling model and Hund's rules to find the lowest energy state.
For Y  we must consider the two 4d electrons. According to Hund's rules, the triplet state
S  1 will be lowest in energy. Additionally, we attempt to maximize L as long as we do
not violate the exclusion principle. Both 4d electrons have  2 , so we can have
L  0,1, 2,3,or 4 . As discussed in Example 8.9 of the text, the antisymmetrization of the
wave function requires that the m values for the electrons forming the S  1 state must
be unequal. The largest allowed L is thus 3. Finally, J should be minimized, so we have
J = 2. Therefore in spectroscopic notation, n 2 S 1LJ , we have 5 3 F2 .
For Al  we consider the two 3p electrons outside the closed subshell. As above, the
triplet state S  1 will have lowest energy. The 3p electrons have  1 so we can have
L  0,1,or 2 . Again using the antisymmetrization of the wave function requires the m
values for the electrons forming the S  1 state to be unequal. The largest allowed L is
therefore 1. Finally the smallest J will be 0. So in spectroscopic notation we have 33 P0 .
40. (a)
2 S 1
LJ is standard spectroscopic notation. For 2 L 3/2 , reading from the notation, J =
3/2 and 2S + 1 = 2, so S = 1/2. The “D” implies that L = 2. This is permitted, because in
this case J = L − S. With J = 3/2, the possible values of Jz are  / 2 and 3 / 2.
(b) With J = 3/2, the quantum number mJ ranges from −3/2 to +3/2. The minimum angle
is achieved when mJ = 3/2. In this case
cos  
Jz

J
mJ
J  J  1

3/ 2
 3/ 2  5 / 2 
 3/ 5
Thus θ = 39.2°.
41. Co has a ground state configuration 3d74s2. By Hund’s rules there are two pairs of delectrons and three unpaired ones, so S = 3/2 and 2S + 1 = 4. The unpaired electrons have
m  0, 1, 2, so the total angular momentum is L = 3, indicating the symbol F. The
maximum value of J is L + S = 3 + 3/2 = 9/2. Therefore the full term symbol is 4F9/2.
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Chapter 9
Statistical Physics
123
Chapter 9
1. (a)
 m 
v   v g (vx ) dvx  


 2 

2
x
1/ 2
2
x
 m 
 2

 2 
1/ 2


0



 1

vx2 exp    mvx2 dvx
 2

 1

 m 
v exp    mvx2  dvx  2 

 2

 2 
1/ 2
2
x
 2 


4  m 
3/ 2

1
m
Therefore
 
vx rms  v
2
x
1/2
1/2
 1 


 m 
1/2
 kT 


m
 m 
 1
2
g (v x )  
 exp    mvx 
 2 
 2

1/2
(b)
and from (a) we see that   m 
1/2
g (vx ) dvx 
1
, so
 vxrms
 1 v2 
1 1
vx rms exp   2 x dvx
 2 vx rms 
2


 1 v2 
2. (a) With vx = 0.01vx rms we have exp   2 x   1 .
 2 vx rms 


 1 vx2 
1 1
1 1
g (vx ) dvx 
vx rms exp   2 dvx 
vx rms 1  0.002vx rms   7.98 104


2
2
 2 vx rms 
This is the probability that a given molecule will be in this range, so in one mole the
number is
N   7.98 104  N A   7.98 104  6.022 1023   4.811020
 1  0.20vx rms 2 
(b) With vx = 0.20vx rms we have exp  
  0.980 . Continuing as in (a) we find
2
 2  vx rms  
1 1
g (vx ) dvx 
vx rms  0.98 0.002vx rms  7.82 104 and therefore N  4.711020.
2
(c) N = 2.91 × 1020


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124
Chapter 9
Statistical Physics
(d) N = 1.79 × 1015
(e) In this case g (vx )dvx   7.98 104  exp  5 103  which is on the order of 10−2175 .
Therefore we conclude that no molecules travel at that speed.
 v 
 v
3. (a) f  f 0 1    f 0 1  x   f 0 1  0   f 0
c
 c

(b)

1/2
 f  f  
2
1/2
0
  f v 2 
  0 x  
 c  


1/2
 2 vx2 
  f0 2 
 c 


1/2
 f 02 kT 
But we know that v  kT / m , so    2

c m
2
x
(c) From (b) we have
H 2 at T  293 K:

f0


f0


f0
c
kT
.
m
1 kT
.
c m
1.3811023 J/K  293 K 
1
 3.66 106
8
27
3.00 10 m/s
2 1.674 10 kg 
1.3811023 J/K  5500 K 
1
H at T  5500 K:

 2.25 105
8
27
f 0 3.00 10 m/s
1.674 10 kg 

This is how we could deduce the surface temperature of a star.
4. (a) Letting d be the distance between the two atoms, we have
2
2
16 1.66 1027 kg 1.211010 m 
 d  md
I x  2  mr   2m   

2
2
2
 1.94 1046 kg  m 2
2
2
(b)
2
2
 4
I z  2  mR 2   mR 2  0.8 16  1.66 1027 kg  3.0 1015 m 
5
 5
55
 1.9110 kg  m2
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Chapter 9
Statistical Physics
125
(c) The rigid rotor is quantized (see Chapter 10) with an energy

2
E
 1
1.055 10 J  s  1 2   5.74 10

2 1.94 10 kg  m 
2
34
46
2I
2
23
J = 3.58 104 eV
(d) Rearranging the energy equation in (c) and using the value of Iz to find the
we find
(  1) 
2 IE
2

2 1.911055 kg  m 2  5.74 1023 J 
1.055 10
34
J  s
2
value,
 1.97 109
This shows that a much larger energy (larger Erot) is required to have =1 for the rotation
about the z axis. Therefore the diatomic molecule acts as if there were only two
degrees of rotational freedom.
5. (a)


c
F (v) dv  4 C 

c
 m 
 1

v exp    mv 2  dv with T = 293 K and C  

 2 
 2

2
3/2
.
(b) For example H 2 gas at T = 293 K we have
1 2   938 106 eV 
1
2
 mc 
 3.7 1010 .
5
2
2 8.62 10 eV/K   293 K 
The exponential of the negative of this value, exp  3.7 1010  , is almost 0.
6.
Computations depend on the software but should yield numbers very close to zero.
4
7. (a) v 
2
2kT
v 

m
*
(b) v 
v* 
4
2
2kT

m
1.38110 J/K  300 K   2510 m/s
1.675 10 kg 
2 1.38110 J/K   300 K 
 2220 m/s
1.675 10 kg 
1.38110 J/K   630 K   3640 m/s
kT
4

m
2
1.675 10 kg 
2 1.38110 J/K   630 K 
 3220 m/s
1.675 10 kg 
23
kT
4

m
2
27
23
27
23
27
23
27
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126
Chapter 9
Statistical Physics
 1

8. F (v)  4 C exp    mv 2  v 2 . In the limit as v  0 the exponential is approximately
 2

2
0
e  1 and v approaches 0, so lim F (v)  0 .
v 0
The other limit is lim F (v)  4 C lim
v 
v 
v2
.
1
2
exp   mv 
2

Applying L’Hopital’s rule,
2v
2
 1

lim F (v)  lim
 lim
exp    mv 2   0 .
v 
v 
1
 v  m
 2

 mv exp   mv 2 
2

2 1.3811023 J/K   258 K 
2kT
9. (a) v 

 391 m/s
m
28 1.6605 1027 kg 
*
2 1.3811023 J/K   308 K 
2kT
(b) v 

 428 m/s
m
28 1.6605 1027 kg 
*
 1

 1
 2kT 1
10. The equation to be satisfied is 2v 2 exp    mv 2   v*2 exp    mv*2  
e
m
 2

 2

2kT
 1
 kT 1
. Thus v 2 exp    mv 2  
e  28000
m
 2
 m
which can be solved graphically to yield v = 188 m/s and v = 639 m/s. The lower of
these is closer to v*  390 m/s , which follows from the shape of the distribution curve.
where we have used the fact that v* 
11. Various software packages should all give results very close to 1.
12. The calculations start from Equation (9.14) and are of the form:
 m 
I  4 

 2 
3/2

b
a
 1

v 2 exp    mv 2 dv
 2

The limits a and b are given in each part. Values are, as a fraction of the total number of
molecules,
(a) 2.0 × 10−7 (b) 2.0 × 10−4 (c) 0.156
(d) 0.494 (e) 0.350
(f) 0.99987
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Chapter 9
Statistical Physics
127
13. (a) From Equation (9.20) we have vrms 
3kT
. The mass of H2  2 1.008 u   2.02 u
m
and the mass of N2  2 14.003 u   28.0 u.
H2 :
N2 :
vrms 
 3 1.38 1023 J/K   293K 
 1902 m/s
 2.02 u  1.66 1027 kg/u 
vrms 
 3 1.38 1023 J/K   293K 
 511 m/s
 28.0 u  1.66 1027 kg/u 
(b) From classical physics we know the escape velocity vesc 
2GM
. Using the mass
R
and radius of the earth we find vesc  1.12 104 m/s if the object starts at the surface.
Neither H 2 nor N 2 has vrms  vesc ; however, a very small percentage of molecules in
the exponential tail of the distribution may have speeds greater than vesc and will
escape. Since H 2 has a larger vrms , a larger fraction will eventually escape.
14. (a) Use Equation (9.8) for the translational kinetic energy of one atom and multiply by
Avogadro’s number for one mole.
3 
3

K  N A  kT    6.022 1023   1.3811023 J/K   273K    3406 J.
2 
2

(b) Since the translational kinetic energy depends only on temperature and one mole of
anything contains the same number of objects, the answer is the same for argon or
oxygen.

15. (a) E   EF ( E ) dE 
0
8 C
2m3/2


0
E 3/2 exp   E dE 
8 C (5 / 2)
5/2
2m3/2 
3
3 
3/2
Using (5 / 2)  (3 / 2) 
and C    m / 2  we find
2
4
E
8   m 


2m3/ 2  2 
3/ 2
3 
3
3

 kT .
5/ 2
4
2 2
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128
Chapter 9
(b) As we know from the text E 
which is a bit less than
1 2
1
4
mv and by Equation (9.17) mv 2  kT  1.27kT
2
2

3
kT .
2
16. Starting with the distribution F ( E ) 
0
Statistical Physics
8 C 1/2
dF
E exp   E  and setting
 0 , we obtain
3/2
dE
2m
d
1
 E1/2 exp   E   E 1/2 exp    E    E1/2 exp    E  .
dE
2
Therefore 0  E 1/2  2 E1/2 and solving for E gives the desired result E * 
kT
.
2
17. The ratio of the numbers on the two levels is
n2 ( E ) 8exp    E2 

 4exp     E2  E1    105
n1 ( E ) 2exp    E1 
Therefore exp    E2  E1    2.5 106 .
Taking the logarithm of each side gives:
E  E1
  E2  E1    2
 ln  2.5 106   12.90 . For atomic hydrogen
kT
3
E2  E1  E0  10.20eV . Finally
4
E2  E1
10.20 eV
T 

 9175 K.
k  12.90 
8.618 105 eV/K   12.90
18. (a) With E 
(b) We have 
h
3mkT
p2
h
h
h
3
and the mean energy E  kT we obtain   
.

2m
2
p
2mK
3mkT
d . Using λ from part (a) and d  V / N  we obtain
1/3
1/3
N
h3
V 
If
we
cube
both
sides
and
rearrange
 
V  3mkT 3/2
N
(c) For any ideal gas
1.
N
6.022 1023

 2.69 1025 m3 .
3
3
V 22.4 10 m
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Chapter 9
Statistical Physics
129
For argon gas (a monatomic gas) at room temperature
6.626 1034 J  s 

N
h3
25
3
  2.69 10 m 
V  3mkT 3/ 2
3  40  1.66 1027 kg 1.38 1023 J/K   293 K 
3


3/ 2
 3 107
so Maxwell-Boltzmann statistics are fine. However, for electrons in silver,
N / V  5.86 1028 m3 and
 6.626 10 J  s 
N
h3
28
3

5.86

10
m


V  3mkT 3/ 2
3  9.111031 kg 1.38 1023 J/K   293 K 
34

3

3/ 2
 1.5 104
so in this case Maxwell-Boltzmann statistics fail.
 1

19. F (v) dv  4 Cv 2 exp    mv 2 dv  F ( E ) dE
 2

dE
dE
1
With E  mv 2 we differentiate to obtain dE  mvdv or dv 
. Then

mv
2
2mE
2E
dE
E1/ 2
F ( E ) dE  4 C
exp    E 
 8 C
exp    E dE
m
2mE
2m3/ 2
8 C 1/ 2

E exp    E dE.
2m3/ 2
20. (a) We assume that the magnetic moment is due to spin alone. In general
n  E   g  E  FMB . There is no reason to prefer one spin state or the other, so the two
g ( E ) are the same. Thus the ratio of the numbers in the two spin states is governed by the
Maxwell-Boltzmann distribution:
n( E2 ) FMB ( E2 ) exp    E2 


 exp    E1  E2   .
n( E1 ) FMB ( E1 ) exp    E1 
The energy of a magnetic moment  in a magnetic field B is E    B . We know from
e
e
e
S  B  Sz B  
B   B B .
m
m
2m
Then E1  B B is the energy of an electron aligned with the field, and E2   B B is the
Chapter 7 that this is equal to E 
energy of the spin opposed to the field. Therefore
n( E2 )
   B  B B 
 2B B 
 exp    E1  E2    exp  B
  exp 
.
n( E1 )
kT


 kT 
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130
Chapter 9
(b) At T = 77 K:
Statistical Physics
 2  9.274 1024 J/T   5.0 T  
n( E2 )
 2B B 
  0.916
 exp 
  exp 
23

n( E1 )
1.381

10
J/K
77
K
 kT 






 2  9.274 1024 J/T   5.0T  
n( E2 )
 2B B 
  0.976
At T = 273 K:
 exp 
  exp 
23

n( E1 )
1.381

10
J/K
273
K
 kT 






 2  9.274 1024 J/T   5.0T  
n( E2 )
 2B B 
  0.993
At T = 600 K:
 exp 
  exp 
23

n( E1 )
1.381

10
J/K
900
K
 kT 






As temperature is increased, the probabilities for spin aligned with the magnetic field
(lower energy) and anti-aligned (higher energy) become essentially equal, as the energy
difference between them becomes much less than kT.
21. Starting with Equation (9.30) and setting FFD  0.5 when E  EF , we have
0.5 
1
BFD exp   EF   1
. Solving for BFD we find BFD exp   EF   1  2 so
BFD exp   EF   1 and BFD  exp   EF  . Therefore in general
FFD 
1
1
1


.
BFD exp   E   1 exp   EF  exp   E   1 exp    E  EF    1
22. At first one might think it should be 0.5, but this is not quite true due to the asymmetric
shape of the distribution. Starting with Equation (9.43) for g ( E ) and using the fact that
FFD  1 in this range, we have
N  E  EF    g ( E )(1)dE 
E
0
E
3
3
NEF3/2  E1/2 dE  NEF3/2 E 3/2 . Recall that E  EF , and
0
5
2
3
we see that N  E  EF   N  
5
3/ 2
 0.465 N .
23. (a) From dimensional analysis
 1 mol   6.022 1023 
28
3
1.05 104 kg/m3 
  5.86 10 m .

0.10787
kg
mol



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Chapter 9
Statistical Physics
131
(b) For electrons an extra factor of 2 is required due to the Pauli principle:
N 2A
3/ 2
 3  2 mkT 
V
h
so
2/3
2/3
2
 5.86 1028 m 3 
34
 N 
2
6.626

10
J

s
h






2(1)
2 AV 

T

 5.28 104 K .
31
23
2 mk
2  9.109 10 kg 1.38110 J/K 
2/3
28
3
2/3
 N  h 2  5.86 10 m   6.626 1034 J  s 2




2(0.001) 
2 AV 


(c) T 

 5.28 106 K.
31
23
2 mk
2  9.109 10 kg 1.38110 J/K 
24.

The number is given by
EF
0.90 EF
EF
EF
3
NEF3/ 2 
E1/ 2 dE  NEF3/ 2 E 3/ 2
0.90
EF
0.90
E
F
2
 N 1.003/ 2  0.903/ 2   0.146 N
g ( E ) dE 
We see that about 14.6% of the electrons are in this range, which is about what one
would expect from the shape of the distribution.
25. (a) As in Problem 23, N / V  5.86 1028 m3 . Then
h2  3 N 
EF 


8m   V 
2/3
 6.626 10

8  9.109 10
2/3
J  s  3
28
3 
  5.86 10 m  
31
kg   

34
2
 8.811019 J  5.50 eV.
2 8.811019 J 
2 EF

 1.39 106 m/s.
(b) uF 
m
9.109 1031 kg
26. (a) Note: the term   kT  / EF is a small fraction of one eV and can be ignored. Then
2
3
3
E  EF   5.51 eV   3.31 eV
5
5
2  3.31 eV 
2E
3

 2.56 104 K
(b) With E  kT we have T 
5
3k 3 8.617 10 eV/K 
2
(c) As discussed in the text, thermal energies are small compared with the Fermi energy,
except at high temperatures.
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132
Chapter 9
Statistical Physics

  6.022 1023 
1 mol
28
3
27. 8.92 103 kg/m3 
  8.45 10 m .

0.063546
kg
mol



This represents a difference of 0.2% from the experimental value so within rounding
errors there is one conduction electron per atom.
 1 mol   6.022 1023 
28
3
28. (a) 2.70 103 kg/m3 
  6.03 10 m

0.02698
kg
mol



h2  3 N 
(b) EF 


8m   V 
2/3
N   8mEF 
 

V 3  h2 
3/ 2
31
19


  8  9.109 10 kg  11.63% eV  1.602 10 J/eV  

 6.626 10
3

34
J  s
2
3/ 2


 1.80 1029 m 3
(c) Dividing the conduction electron density by the number density we obtain almost
exactly 3; from this we conclude that the valence number is three.
29. In general EF 
1
muF2 so uF  2EF / m .
2
2  4.69 eV  1.602 1019 J/eV 
2 EF

 1.28 106 m/s
(a) uF 
m
9.109 1031 kg
2 14.3 eV  1.602 1019 J/eV 
2 EF

 2.24 106 m/s
(b) uF 
m
9.109 1031 kg
30. Beginning with Equation (9.34) consider the following cases as T  0 :
E  EF
  so FFD  0
kT
E  EF
E  EF :
  so FFD  1
kT
E  EF
1
E  EF :
 0 so FFD 
kT
2
E  EF :
31. In general n  E   g  E  FFD . Using Equation (9.43) for g ( E ) and the result of Problem 21
for FFD , we can substitute to find n( E ) 
3N 3/ 2
E1/ 2
EF
.
2
exp    E  EF    1
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publicly accessible website, in whole or in part.
Chapter 9
Statistical Physics
133
32. The graphs will resemble those of Figure 9.11 (b). The T = 0 K line will match the
dashed line and the T = 300 K line will match the solid line. The T = 1500 K line will
deviate a bit more from the dashed line.
33. Numerical integration should yield accurate results with kT = 0.02586 eV. However, in
Mathcad, for example, the upper limit cannot exceed 20 as the engine calculates e raised
to the power before taking the reciprocal and thus rejects the problem. As you adjust the
upper limit above 10, the integral equals 1 within rounding error.
1.5  7 
3/2


0
E1/2
dE  1
exp   E  7  /  0.02586    1
34. Setting up the numerical integration in Mathcad we have with kT = 0.02525 eV,
1.5  7 
3/2

7
6
E1/2
dE  0.203 .
exp   E  7  /  0.02525   1
We see that about one-fifth of the electrons are within 1 eV of the Fermi energy, which
makes sense given the shape of the distribution.
h 2  3N 
35. We can use Equation (9.42) EF 


8m   L3 
dimensional analysis
h 2  3N 
EF 


8m   L3 
2/3
2/ 3
. We use the neutron mass and from
N 4.50 1030 kg 1 (neutron)

 6.411044 m3 . Then
L3 4  104 m 3 1.675 1027 kg


3
 6.626 10

8 1.675 10
2/3
J  s  3
44
3 
 6.4110 m 
27
kg   
34
2
 2.36 1011 J  147 MeV
The close packing of the neutrons makes the Fermi energy large compared with Fermi
energies in normal matter.
36. The probability that a state will be occupied can be determined from the Fermi-Dirac
factor. With kT = 0.02525 eV:
1
1

 0.981  98.1%
(a) FFD 
  E  EF  
  0.1 
exp 
  1 exp 
 1
kT
0.02525




(b) When E  EF , then FFD 
1
 50% .
2
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134
Chapter 9
(c) FFD 
Statistical Physics
1
1

 0.0187  1.9%
  E  EF  
  0.1 
exp 
  1 exp 
 1
 kT 
 0.02525 
Therefore a state with energy less than the Fermi energy is almost certainly occupied
and one above the Fermi energy has a very small probability of being occupied.
37. We assume that the collection of fermions behaves like an ideal gas. Using MaxwellBoltzmann statistics, we know that E  32 kT or  E  3 2 . We note that
exp   E   exp  3 2   4.4817 . Now the MB factor is exp   E  and we want this to be
within 1 % of the FD factor: FFD 
1
. So we want
  E  EF  
exp 
 1
 kT 
exp    E  EF   1  1.01   4.4817   4.5265 or    E  EF   ln  3.5265 . At room
temperature     2.525 102 eV so the expression above can be solved to give
1
 E  EF   3.2 102 eV. This small value is possible at room temperature.
38. (a) Computing carefully:
N /V 
1.93 × 104 kg/m3
 5.90 × 1028 /m3
197.0 1.66× 1027 kg
2/3
h2  3N 
EF 

  5.53 eV. This is an exact match with Table 9.4.
8m   L3 
(b) Thermal energy is 1.5kT = 0.04 eV. The result shows that electrons aren’t classical
and the energy levels stack up in the Fermi distribution, forcing the electrons into higher
energy states.
39. (a) Use the same method as in the text for helium. The liquid density of 1200 kg/m3
translates to a number density of about 3.6 × 1028/m3. Plugging this into the usual formula
(9.65) gives T > 0.87 K.
(b) Neon is not a liquid at that temperature, so it cannot be a superfluid.
40. (a) To find N / V integrate n( E )dE over the whole range of energies:

N
8
  n( E ) dE  3 3
0
V
hc


0
E2
dE . From integral tables we have the following:
exp( E / kT )  1
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publicly accessible website, in whole or in part.
Chapter 9
Statistical Physics


0
135
x n1
1
, (3)  2!  2 , and from numerical
dx  m n (n) (n) . For our integral m 
mx
e 1
kT
tables  (3)  1.20 . Therefore
N 8
8 k 3T 3
3
 3 3  kT   2 1.20   3 3 (2.40) .
V hc
hc
(b) With T = 500 K:


1.3811023 J/K   500 K 

N 8 k 3T 3

 3 3 (2.40)  8 (2.40) 
  6.626 1034 J  s  2.998 108 m/s  
V
hc


15
3
 2.53 10 m .
(c) With T = 5500 K :
3


1.3811023 J/K   5500 K 

N 8 k 3T 3

 3 3 (2.40)  8  (2.40) 
  6.626 1034 J  s  2.998 108 m/s  
V
hc


18
3
 3.37 10 m .
3
u1/2
41. Evaluating the following integral in Mathcad we find  u du  2.315 .
0 e 1
42. Take the expressions in Equation (9.67) to be in the “normal” order (1,2). That is,
1
S 1, 2  
  a (1)b (2)   a (2)b (1) . Reversing the order for the symmetric wave
2
1
function yields S (2,1) 
  a (2)b (1)   a (1)b (2), which is the same as
2

S 1, 2  with the order of terms reversed. Thus, S 1, 2   S  2,1 .
Following the same procedure for the antisymmertic wave function,
1
 A (1, 2) 
  a (1)b (2)   a (2)b (1) and
2
1
 A (2,1) 
  a (2)b (1)   a (1)b (2) .
2
Comparison of these two expressions verifies that  A (1, 2)   A (2,1).
p2
43. E  K  V 
 mgz :
2m
  p2

  p2 
exp(  E )  exp    
 mgz    exp  
 exp    mgz 

 2m 
  2m
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136
Chapter 9
Statistical Physics
Absorbing the (assumed constant) first exponential factor into the normalization constant
C z , f ( z )dz  Cz exp   mgz dz . To find C z normalize the integral:


1
kT
.

 mg mg
44. For air we will use an average m = 29 u = 4.82 × 10−26 kg and T = 273 K. In general
 (h) exp    mgh 

 exp    mgh  .
 (0) exp    mg  0  

0
f ( z )dz  Cz  exp   mgz dz  Cz   mg  and therefore Cz 
0
  4.82 1026 kg  9.80 m/s 2  176 m  
  (0)  0.978 (0)
For Chicago:  (h)  exp  

1.3811023 J/K   273 K  

  4.82 1026 kg  9.80 m/s 2  1610 m  
  (0)  0.817  (0)
For Denver:  (h)  exp  
23


1.381

10
J/K
273
K






  4.82 1026 kg  9.80 m/s 2   4390 m  
  (0)  0.577  (0)
For Mt. Rainier:  (h)  exp  
23


1.381

10
J/K
273
K






45. In equilibrium a fluid layer of density  , mass M, thickness h, and surface area A has a
force F2  P2 A acting downward on its upper surface and a force F1  P1 A acting upward
on its lower surface. The difference between these forces equals the weight of the fluid
layer. F2  F1   P1  P2  A  Mg   gAh . With dP  P  P2  P1 and h  z  dz , we
have dP    gdz . With N particles of mass m, the mass density is   Nm / V . Putting
these together: dP    gdz  
Nmg
dz . From the ideal gas law N / V  P / kT , so
V
mgP
dz . Using separation of variables we can solve this differential equation for P
kT
dP
mg
mgz
as a function of z:

dz
ln P  
 constant    mgz  constant .
P
kT
kT
P   constant  exp   mgz   P0 exp   mgz 
dP  
46. (a)
dN
nv
NvA
; Rearranging this expression we find the following

A
dt
4
4V
vA
dN
vA

dt or ln N   t  constant . This can be rewritten as
4V
N
4V
N 1
 vA 
 vA 
N   constant  exp   t   N 0 exp   t  . Setting
 at t  t1/2 , we find
N0 2
 4V 
 4V 
differential equation:
4V
1
 vA 
ln 2 .
 exp   t1/2  or t1/2 
vA
2
 4V 
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publicly accessible website, in whole or in part.
Chapter 9
Statistical Physics
(b) V 
 D3
6

137
  0.4 m 
6
3
 0.0335 m ; A 
3
d2
4

  0.001 m 
4
2
 7.85 107 m2 ;
1.38110 J/K   293 K   462.6 m/s
29 1.66 10 kg 
4  0.0335 m 
4V

ln 2 
ln 2  256 s
vA
 462.6 m/s   7.85 10 m 
4
v
2
23
kT
4

m
2
27
3
t1/2
7
2
47. The number of molecules with speed v that hit the wall per unit time is proportional to v
and F(v), so that the distribution W(v) of the escaping molecules is by proportion
 1

W (v) ~ vF (v) ~ v3 exp    mv 2  . Let the normalization constant for W(v) be C  , so
 2

2
 1

 1   m 

2 2
C  v exp    mv 2 dv  1  C   
 or C   m / 2 . The mean kinetic
0
2
2
2


 

energy of the escaping molecules is


3
3

1
1
1   2 m2    m 
2
 1

E  mv 2  mC   v5 exp    mv 2 dv  m 

   2kT
0
2
2
2  2  2 

 2

3kT
mv 2
so T  rms  19,500 K
m
3k
(b) With the shape of the Maxwell-Boltzmann distribution, there are always some atoms
that are moving fast enough to escape. Whenever equilibrium is reestablished, more
helium atoms fill that part of the distribution, so more will continue to escape.
48. (a) vrms =
3kT
= 1.20 × 104 m/s. This is much less than escape speed, so we do not expect
m
many hydrogen atoms to escape from the sun.
N A
3/2
50. From Example 9.9 we have
 3  2 mkT  .
V h
(a) Letting m be the electron mass and inserting a factor of 2 for the Pauli principle,
N 2A
3/2
 3  2 mkT 
V
h
3/2
2(1)
31
23



2

9.109

10
kg
1.381

10
J/K
293
K





3

 6.626 1034 J·s  
49. vrms =
 2.42 1025 m 3
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publicly accessible website, in whole or in part.
138
Chapter 9
Statistical Physics
This is quite a bit less than the density of conduction electrons in a metal (such as
copper), which indicates that Fermi-Dirac statistics should be used.
(b) For He gas the Pauli principle does not apply, so

1
 6.626 10
J  s
3
34
N A
3/ 2
 3  2 mkT 
V h
 2 (4) 1.66 1027 kg 1.3811023 J/K   293 K  


3/ 2
 7.54 1030 m3 .
This is much larger than the density of He gas, which explains why MB statistics work
for helium.
N 2A
3/ 2
(c) For the neutron star
 3  2 mkT 
V
h
3/ 2
N
2(1)
 2 1.6749 1027 kg 1.3811023 J/K 106 K   1.90 1035 m 3

3


V
6.626 1034 J  s






(d) With mass density on the order of 1017 kg/m3, the number density of the neutron star
is on the order of 1043/m3, which is why Fermi-Dirac statistics are required there.
51. EF 
h 2  3N 


8m   L3 
2/3
Rearrange to solve for N/L3:
h 2  3N  N

3/ 2
EF 
 3  8mEF   1.811029
 3
3
8m   L  L 3h
where the numerical value comes from the Fermi energy in the table. We need to
compare this with the number density for aluminum atoms, which is
2/3
2.70 × 103 kg/m3
 6.0 × 1028 /m3
27
27.0 1.66× 10 kg
Note that the conduction electron density is exactly three times the number density, so
there are three conduction electrons per atom. This makes sense, because aluminum has
two 3s electrons and one 3p electron.
52. Expressions for I1 and I 2 are known from Appendix 6:
N /V 
I3  
dI1
1
 1 
  2   2
da
 2a  2a
dI 2
  3  5/2 3  5/2

a
 a 
da
4  2
8
dI
1
I 5   3    2a 3   a 3
da
2
53. For the harmonic oscillator the position and velocity are x  x0 cos( t ) and
I4  
v
dx
1
1
  x0 sin( t ) respectively. V  kx 2  k x02 cos 2 ( t ) ;
dt
2
2
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publicly accessible website, in whole or in part.
Chapter 9
Statistical Physics
139
1
1
1
K  mv 2  m  2 x02 sin 2 ( t )  k x02 sin 2 ( t ) where we have used the fact that
2
2
2
2
 m  k . Over one cycle the average of the square of the sine or cosine function is one1
1
1 E
half. Also the total energy is E  k x02 . Therefore K  V  k x02    .
2
2
2 2
54. (a) For these elements the number of electrons is half the number of nucleons. The
number of nucleons is
m(sun)
 1.1916 1057
m(nucleon)
The number of electrons is half this, or 5.96 × 1056.
5.96 1056
(b) Number density = number/volume =
 4.22 1035 m 3 .
3
4
  6.96 106 m 
3
2/3
h2  3N 
14
(c) From text EF 

  3.28 10 J = 205 keV
8m   L3 
This is huge! The high density of electrons gives them an enormous Fermi energy.
55. (a) They have different wave functions, so when you stack up the energies the protons
and neutrons go in separate categories. Because they are fundamentally different
particles, which must have different wave functions, it’s not a violation of Pauli to have
two of them in the same energy state.
(b) i) For protons m = 1.673 × 10−27 kg, and the number density is
N
N
92
4 3
 5.42 1043 m3
3
4

15
V 3r
  7.4 10 m 
3
Then the Fermi energy is
h 2  3N 
EF 


8m   L3 
2/ 3
 6.626 10

8 1.673 10
2/ 3
J  s  3
43
3 
  5.42 10 m  
27
kg   

34
2
or EF = 4.53 × 10−12 J = 28.3 MeV.
ii) For neutrons m = 1.675 × 10−27 kg, and the number density is
N
N
146
4 3
 8.60 1043 m3
3
4
V 3r
  7.4 1015 m 
3
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publicly accessible website, in whole or in part.
140
Chapter 9
h 2  3N 
The Fermi energy is EF 


8m   L3 
2/ 3
 6.626 10

8 1.675 10
Statistical Physics
2/ 3
J  s  3
43
3 
8.60

10
m


27
kg   
34
2
or EF = 6.19 × 10−12 J = 38.7 MeV.
These energies are enormous, but still small compared with the binding energy of the
uranium nucleus.
56. The mass of gas molecules would differ slightly depending on the isotope of uranium.
235
UF6 would have a molecular mass of 235 + 6(19) = 349 while 238 UF6 would have a
molecular mass of $352$. Find the rms velocity for each molecule.
235
238
UF6 :
UF6
vrms
3 1.3807 1023 J/K   293K 
3kT


 1.447 102 m/s
27
m
349 1.6605 10 kg 
vrms
3 1.3807 1023 J/K   293K 
3kT


 1.441102 m/s
27
m
352 1.6605 10 kg 
The difference is about 0.6 m/s which represents a 0.4% change from one isotope to the
other.
57. Rearranging Equation (9.64) and with m = 4 u, we have
N 2  2.315 
3/ 2

 2mkT 
3
V
h
3/ 2
2  2.315
 2(4) 1.6605 1027 kg 1.3811023 J/K  (293) 

3


 6.626 1034 J·s 
 1.97 1031 m 3 .
The number density of an ideal gas at STP is
N P
1.0135 105 Pa


.
V kT 1.3811023 J/K   293 K 
N
 2.50 1025 m 3 . As you would expect, the condensate has a number
V
density nearly one million times greater than the ideal gas.
Therefore
58. From Appendix 8, the 85Rb isotope has atomic mass of 84.9 u:
vrms
3 1.38 1023 J/K  20 109 K 
3kT


 2.42 103 m/s
27
m
84.9 1.66 10 kg 
For sodium, the mass is 23.0 u, so
vrms
3 1.38 1023 J/K  450 1012 K 
3kT


 6.99 104 m/s
m
23.0 1.66 1027 kg 
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Chapter 9
Statistical Physics
141
which is even slower!
59. (a) Beginning with Equation (9.65) and with

h2 
1
T


2mk  2  2.315  
2/3
N
 
V 
 6.626 10
34
N
 2.5 1028 m3 we have
V
2/3
J/K 
2


1



2  40  1.6605 1027 kg 1.38110 23 J/K   2 )  2.315  
2/3
 2.5 10
28
m 3 
2/3
 3.43 101 K  0.343 K.
(b) The temperature found in (a) is far below the freezing point of 84 K so no condensate
is possible.
60. Beginning with Equation (9.65)

h2 
N
T


2mk  2 V  2.315  
2/3
 6.626 10
J/K 


2000



2  87  1.6605 1027 kg 1.3811023 J/K   2 11015 m3   2.315  
 2.93 108 K  29.3 nK.
34
2
2/3
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142
Chapter 10
Molecules, Lasers, and Solids
Chapter 10
1.
2
(a) For each state the energy is given by Erot 
2
E 
2I
 2(3)  1(2) 
I

2 1.055 1034 J  s 
10
46
kg  m
2
2
 2.2 1022 J  1.4 103 eV.
31 2
 1.73 1021 J  1.08 102 eV.
2I
2I
This is still in the infrared part of the spectrum.
(b) As in part (a), E 
2
2
2
(  1)
, so the transition energy is
2I
10(11)  9(10)  
2. From Table 10.1   1860 N/m and f  6.42 1013 Hz.
(a) E  hf   4.136 1015 eV  s  6.42 1013 Hz   0.266 eV
(b) Set E  kT with two degrees of freedom in the vibrational mode. Then
E
0.266 eV
T

 3090 K
k
8.617 105 eV/K
1
1
1
1

3. In the ground state E   n    
  k A2    2 A2 . Solving for A:
2
2
2
2

A

. Using the
35
Cl  
m1m2
35

u  1.614 1027 kg .
m1  m2 36
From Table 10.1 f  8.66 1013 Hz. With   2  f we have
A


1.055 1034 J  s
 1.10 1011 m.
27
13 1
1.614

10
kg
2

8.66

10
s

 

4.
(a) Using the result of Problem 8 for the rotational inertia, we have
I   R2  1.614 1027 kg 1.28 1010 m   2.64 1047 kg·m2 .
2
For
 1 we have Erot 
2 2


I2
(  1)
2I
2

2 1.055 1034 J  s 
 2.64 10
47
kg  m
2
I

1 2
I  . Therefore
2
2

2 2
 5.65 1012 rad/s .
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publicly accessible website, in whole or in part.
Chapter 10
Molecules, Lasers, and Solids
 5 we have Erot 
For
110

I2
2
143
(  1)
2I
2

30 1.055 1034 J  s 

 2.64 10
47
kg  m

15
I
2

1 2
I  . Therefore
2
2
2 2
 2.19 1013 rad/s .
(b) The distance of the center of mass from the H atom is
m x
35
xcm  Cl Cl 
R  1.24 1010 m . Then from the rotational kinematics we have
M
36
 1 : v   xcm   5.65 1012 rad/s 1.24 1010 m   700 m/s and for
for
 5:
v   xcm   2.19 1013 rad/s 1.24 1010 m   2720 m/s.
v
0.1c 2.998 107 m/s
(c)  


 2.42 1017 rad/s . Then  
10
xcm
xcm
1.24 10 m
(  1)
I2
2
and
  2.42 1017 rad/s  2.64 1047 kg  m2  
  3.67 109
(  1)  2  


1.055 1034 J  s


4
from which it follows that  6.110 .
2
2I 2
2
(d) Erot  kT 
2
T

 1
2I
. Thus
1.055 1034 J  s  3.67 109   5.611010 K.
 1

2 Ik
2  2.64 1047 kg  m2 1.38 1023 J/K 
2

L2 n2 2

5. With Bohr’s condition L  n we find Erot 
. The Bohr version and the
2I
2I
correct version become similar for large values of quantum number n or , but they are
quite different for small .
6. E  E1  E0  E1 
2
I

hc

.
34
3
 1.055 10 J  s  2.60 10 m 
(a) I 


 1.46 1046 kg  m2
8
hc 2  c
2  2.998 10 m/s 
2

(b) The minimum energy in a vibrational transition E  hf . From Table 10.1
f  6.42 1013 Hz, which corresponds to a photon of wavelength   c / f  4.67 m . A
photon of this wavelength or less is required to excite the vibrational mode, so the 2.60mm photon is too weak.
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144
Chapter 10
Molecules, Lasers, and Solids
7. E  hf   6.626 1034 J·s  6.42 1013 Hz   4.25 1020 J . Using this energy for a
2
rotational transition from the ground state to the th state we have E 

2 I E
 1 
2

2  7.28 1047 kg  m2  4.25 1020 J 
1.055 10
34
J  s
2
 556 so

 1
2I
so
 24 .
This is prohibited by the   1 selection rule.
8. Combining R  r1  r2 with m1r1  m2 r2 we find that r1 
m2
m1
R and r2 
R.
m1  m2
m1  m2
Therefore
2
 m2

 m1

I  m r  m r  m1 
R   m2 
R
 m1  m2 
 m1  m2 
m m m  m 
 1 2 1 2 2 R2   R2
 m1  m2 
2
1 1
9. (a) E 
2
2
2 2
2
3
3  4   2  3  

2I
I
2
so
3 1.055 1034 J  s 
3 2
I

 1.46 1046 kg  m2
3
19
E 1.43 10 eV 1.602 10 J/eV 
2
12 16  u  1.139 1026 kg . Then I   R2 becomes
m1m2

m1  m2
28
(b)  
R
I


1.46 1046 kg  m2
 1.13 1010 m , which is a reasonable answer.
26
1.139 10 kg
10.
(a) The distance of each H atom from the line is
d   0.0958nm   sin 52.5   7.60 102 nm .
Then I  2mH d 2  2 1.67 1027 kg  7.60 1011 m   1.93 1047 kg  m2 .
2
(b) E1 
E2 
2
3
I
2
I

1.055 10

1.93 10
34
47
J  s
kg  m
3 1.055 1034 J  s 
1.93 10
47
kg  m
2
2
2
 5.77 1022 J  3.61 MeV
2
 1.73 1021 J  10.81 MeV
© 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 10
Molecules, Lasers, and Solids
(c)  
145
34
8
hc  6.626 10 J  s  2.998 10 m/s 

 344 μm
E1
5.77 1022 J
11. First we need to compute the rotational inertia. Including both the nucleus and electrons:
2
2
2
I  m rnuc
  2me  a02
5
5
2
2
2
  6.64 1027 kg 1.9 1015 m   2  9.109 1031 kg  5.29 1011 m 
5
 2.04 1051 kg  m 2


Notice that the nuclear contribution is negligible.
2
(a) Erot 
1.055 10

34
J  s
51
2
 5.46 1018 J  34.1 eV.
I
2.04 10 kg  m
(b) This is greater than the ionization energy for helium and note likely to be observed.
12.
f f 
2 I
2
2
 3 with f  c /   1.00 1013 Hz. Thus
f   1.00 1013 Hz 
1.055 1034 J  s
 2  3 or
2 1.30 1045 kg  m2 


f   1.00 1013 Hz   2.58 1010 Hz   3.87 1010 Hz where
usual  
is an integer and as
c
.
f
13. E  hf 
2
  1  2  
2I 

 1  . Therefore we can say that for a particular
transition f  C /  where C is a constant, because I   R 2 . Then
absolute values
df
C
  2 ; taking
d

d f
f
f 
 C /  2  . Thus

as required.
d

f

14. (a) The Maxwell-Boltzmann factor is FMB  A exp   E / kT  . The energies of the three
rotational levels are E0  0 ; E1 
are
2
I
; E2 
2
3
I
. Thus the Maxwell-Boltzmann factors
 0 : FMB  A .
For  1:
FMB
2


2
1.055 1034 J  s 




  0.981A .
 A exp  
  A exp  
46
2
23
IkT
1.4

10
kg

m
1.381

10
J/K
293K
 






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146
Chapter 10
Molecules, Lasers, and Solids
 2:
For
FMB
2


3 1.055 1034 J  s 
 3 2

  0.943 A .
 A exp  
  A exp  
46
2
23

IkT
1.4

10
kg

m
1.381

10
J/K
293
K









(b) The degeneracy factor g  E  is 1 for
 0 , 3 for
 1 , and 5 for
 2 . Therefore
the level populations n( E )
 0:
n( E)  g ( E)FMB  A
 1:
n( E )  g ( E )FMB =3  0.981A   2.94A
 2:
n( E)  g ( E)FMB  5  0.943A  4.72 A
(c) For the lower rotational states the degeneracy factor causes the state population to
increase with increasing . However as increases and the rotational energy
increases, the exponential factor begins to take over and decrease the state
populations.
15. The gap between adjacent lines is h  f  
(a) I 
2
h  f 

1.055 10
 6.626 10
34
34
J  s
2
/I.
2
J  s  7 10 Hz 
11
 2.4 1047 kg  m2
(b)   2 f   /  with f  8.65 1013 Hz. The reduced mass is (using the
isotope)  
35
Cl
m1m2
35

u  1.614 1027 kg . Solving for  we find
m1  m2 36
  4 2 f 2   4 2 8.65 1013 Hz  1.614 1027 kg   478 N/m in good agreement
2
with Table 10.1.
16. (a) P  qx  1.602 1019 C 2.67 1010 m   4.28 1029 C  m
(b) The fractional ionic character is the ratio
17. E  hf 
4.28 1029 C  m
 0.79
5.411029 C  m
hc
. Using the data from Table 10.1, the wavelength is

c 2.998 108 m/s
 
 4.48 μm which corresponds to an energy of 0.277eV . This
f
6.69 1013 Hz
photon is in the infrared.
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publicly accessible website, in whole or in part.
Chapter 10
Molecules, Lasers, and Solids
147
18. (a) Using dimensional analysis and the fact that the energy of each photon is
hc /   3.14 10
19
N
5 103 J/s

 1.59 1016 photon/s .
J , then
19
t 3.14 10 J/photon
(b) 0.02 mole is equivalent to 0.02 N A  1.20 1022 atoms . The fraction participating is
1.59 1016
 1.33 106 .
22
1.20 10
(c) The transitions involved have a fairly low probability, even with stimulated emission.
We are saved by the large number of atoms available.
1.15 eV 1.602 1019 J
19. (a) The energy of each photon is E2  E1 
·
 1.84 1019 J/photon .
photon
1eV
Since P  E / t then we have P  1.84 1019 J/photon  5.50 1018 photons/s   1.00W .
(b)  
hc 1.986 1025 J  m

 1.08 106 m  1.08 μm
19
E
1.84 10 J
20. (a) With the given wavelength, we know that each photon has an energy of
E
hc


1.986 1025 J  m
 5.658 1019 J .
9
35110 m
Therefore the number of transitions required is
40 103 J
N
 7.07 1022 photons .
19
5.658 10 J/photon
(b) Average power is energy divided by time: P 
40 103 J
 1.14 1013 W
9
3.5 10 s
or about 11 TW!
21. (a) From Chapter 9 we have f 
f0
c
kT
. We also have, in general, f  c /  so,
m
calculating the differential we have f   c /  2   . Combining the two expressions we
have  
 2 f
c

 2 f 0 kT
c
2
m
6.328 107 m
calculate  
2.998 108 m/s

 kT
c
m
. Using the mass of neon as 3.32 1026 kg we
1.38110
23
J/K   293 K 
3.32 1026 kg
 7.37 1013 m .
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148
Chapter 10
(b) E 
hc
2
 
Molecules, Lasers, and Solids
2
 6.328 10 m 
2
2
 


 1.06 1019 m.
8
3
2hc 4 c 4  2.998 10 m/s 10 s 
7
2
The Doppler broadening is much more significant than the Heisenberg broadening.
2 1 m 
 6.67 109 s
22. (a) t 
8
2.998 10 m/s
(b) For the 16-km round trip
t 
16 103 m
16 103 m

 1.6 108 s .
8
4
8
2.998 10 m/s 1  3 10  2.998 10 m/s
Because this result is larger than the desired uncertainty in timing, it is important to take
atmospheric effects into account.
23. In the three-level system, the population of the upper-level must exceed the population of
the ground state. This is not necessary in a four-level system.
24. Because the 3s state has two possible configurations and the n = 2 level has eight, let us
assign a density of states g = 2 to the excited state and g = 8 to the ground state.
f3 2 exp    E3 
 E  E2 

 0.25exp   3

f 2 8exp    E2 
kT 

(a)


16.6 eV
  7.2 10287
 0.25exp  
  8.617 105 eV/K   293 K  


(b)


f3
16.6 eV
  4.4 10559
 0.25exp  
5
 8.617 10 eV/K  150 K  
f2




f3
16.6 eV
  9.110141
 0.25exp  
5


f2
 8.617 10 eV/K   600 K  
(d) Thermal excitations can be neglected.
(c)
25. Using dimensional analysis the number density is
1980 kg/m3
1 mol 2  6.022 10
0.07455 kg
mol
d   3.20 1028 m3 
1/3
23
  3.20 10
28
m3 . Therefore the distance is
 3.15 1010 m  0.315 nm .
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Chapter 10
Molecules, Lasers, and Solids
149
26. Each charge has two unlike charges a distance r away, two like charges a distance 2r
2e2  1 1 1

away, and so on: V  
1     ...  . The bracketed expression is the Taylor
4 0 r  2 3 4

series expansion for ln 2, so V  
2e2
 e2
and we see that   2ln 2 .
ln 2  
4 0 r
4 0 r
27. Using the central positive
charge as a guide, we find:
V
e2  4
4
4
8
4

 
 

 ... 

4 0 r  r
2r 2r
5r
8r
,
so by definition of  we have   4 
4
8
4
2

 ...
2
5
8
2
dV
 e2
 r / 
r0 /    e
28. F  
.

 e . From Equation 10.20a we have 1  e
dr
4 0 r 2 
4 0 r02
Multiplying this factor of 1 by the last term in the force equation, we obtain
 e2
  r /  r0 /   e2
F 

e e
4 0 r 2 
4 0 r02

 e2  r02  r r0  /  
 e

4 0 r02  r 2

29. Inserting r  r0   r into the force equation from Problem 28:
F

r02
 e2 
 r / 


e

 . Factoring the first term in parentheses leaves

4 0 r02   r0   r 2





 e2 
1
 r /  
F

e
 . Applying the binomial theorem
4 0 r02    r 2
 1


 

r
0 
 

2
 r 
2
3
2
1    1   r  2  r   ... and we end the series at that point for small  r . The
r0 
r0
r0

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150
Chapter 10
Molecules, Lasers, and Solids
 r  r 

 ... Combining these

2 2
2
Taylor series for the exponential term is e r /   1 
 e2 
2
3
 r  r 
2
two series approximations gives: F 

1


r


r

1





4 0 r02 
r0
r02

2 2
F
Collecting terms we have:

.


 3
 e2   2 1 
1 
2


r




r









2 
2
2
4 0 r0   r0  
 r0 2  

 K1  r   K 2  r 
 e2  2 1 
where K1 
  
4 0 r02  r0  
2
and
2
 e2  3
1 
K2 
 2 2
2 
4 0 r0  r0 2  
30. (a) Looking at the result of the preceding problem we see that the spring constant is
   K1 , and we know that for the harmonic oscillator    /  . For NaCl

m1m2
 2.32 1026 kg . Recall that for NaCl we know   1.7476 ,
m1  m2
r0  0.282 nm, and   0.0316 nm . Substituting these values gives
 K1    
f 
 e2  2 1 
    124.7 N/m . Then the oscillation frequency is
4 0 r02  r0  

1

2 2
(b)  

1

 2
124.7 N/m
 1.17 1013 Hz .
2.32 1026 kg
c 2.998 108 m/s

 25.6 μm which is about one-half the observed value.
f
1.17 1013 Hz
31. (a) F  0  K1 r  K 2  r  and therefore  r  
2
K2
2
 r  .
K1
(b) From the equipartition theorem K1  r   kT so with a)  r 
2
K2
kT . The
K12
coefficient of thermal expansion α comes from L  L  T , or  
our nomenclature  
 
1 L
. Therefore in
L T
1 d r
1 k K2

.
r0 dT
r0 K12
(c) Evaluating with K1  124.7 N / m and K2  2.35 1012 N/m2 , we find
  7.4 106 K 1 , which is the right order of magnitude.
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Chapter 10
Molecules, Lasers, and Solids
151
32. We begin with Equation (10.32): L 
K 4k 2
. Solving for the thermal conductivity,

 T  e2
K, we find
7
1
1
23
1
4 k 2T 4  6.30 10   m 1.38 10 J  K  293K
K

 175 W  K 1  m 1 .
2
2
e
 1.60 1019 C 
2
As mentioned in the text, the Wiedemann-Franz Law was derived using classical
expressions for the mean speed and the molar heat capacity. Quantum mechanical
3
 2.58 . With this correction, our answer
12
would be 451 W  K 1  m1 which is much closer to the measured value.
corrections give an additional factor of
33. We begin with Equation (10.26) and compare that equation to the graph given in Figure
10.23. We notice that in the figure the mean displacement is plotted on the vertical axis
and the absolute temperature is on the horizontal axis. Equation (10.26) is of the form
y  mx  b where the mean displacement represents the vertical or y quantity and the
absolute temperature is the horizontal or x quantity. Thus the slope, m is the constant
3bk
multiplying the temperature, or in this case, m  2 . We can estimate the slope from
4a
the graph. In the region between temperatures of about 40 K and 90 K, the graph is
0.546nm  0.534nm
nearly linear. The slope is approximately m 
 2.67 104 nm  K 1 .
85K  40K
4
1
b 4m 4  2.67 10 nm  K 
From above we also know that 2 

 2.58 1010 N 1 .
a
3k 3 1.3807 1023 J  K 1 
34. (a) Substituting the results from Equations (10.24) and (10.25) into Equation (10.23), we
3  5/2 3/2
ba 
3
3bkT
4
have
.
 ba 2  1 
1/2 1/2 1/2
 a 
4
4a 2
3bk
 C0 . Then x  C0T .
4a 2
 x
C0 
  x  1.67 105 K 1 8.47 1028 m3
T
the number density of copper from Table 9.3.
(b) Let



1/3
 3.80 1015 m/K where we used
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152
Chapter 10
Molecules, Lasers, and Solids
35. The potential energy of a harmonic oscillator would have only the quadratic term and be
1
of the form  x 2 . Examine Equation (10.22) and notice that a is the multiplicative
2
constant for the x 2 term. Thus a  1/ 2   . From Table 10.1 we see that an average
value might be   1000 N/m . Using one-half this value for α along with the definition
of C0 from the preceding problem we find
2
15
4a 2C0 4  5 10 N/m   3.80 10 m/K 
b

 9.2 1013 N/m2 .
23
3k
3 1.38110 J/K 
2
2 E 2 NE
2E
3
and the ideal gas law becomes PV  NkT  Nk
.

kT so T 
3k
3
3k
2
N
3
(b) From Chapter 9 we know
 8.47 1028 m3 and we know E  EF  6.76 1019 J .
V
5
Thus in SI units we have
2 NE 2
P
 8.47 1028 m3 6.76 1019 J  3.82 1010 N/m2 which is quite high. The
3V
3
ideal gas law may not be the best assumption for conduction electrons.
2 NEF
2 NE 2 N 3

EF 
37. From the previous problem P 
. We must be careful because
3V
3V 5
5V
36. (a) E 



h 2  3N 
EF depends on the volume as: EF 


8m   V 
2/3
2 N h 2  3N 
3
P

  
5V 8 m   V 
 
The Bulk modulus is:
B  V
P
3
 V  
dV
 
5 3 
  
3 
2/3
2/3
2/3
2/3
so
N 5/3h2 5/3
V .
20 m
N 5/3h 2  5  8/3
  V
20 m  3 
N 5/3 h 2 5/3 5P
V

20 m
3
Using the fact from above that P 
5 2 NEF 2 NEF
2 NEF
we find B 
.

3 5V
3V
5V
38. For silver N / V  5.86 1028 m3 and EF  5.49eV .
(a) B 
2 NEF 2
  5.86 1028 m-3   5.49 eV  1.602 1019 J/eV   3.44 1010 N/m2 .
3V
3
(b) The computed result is about one-third of the measured value.
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Chapter 10
Molecules, Lasers, and Solids
153
39. (a) Following the arguments in the text, regardless of the sense of rotation of the
electron, as the magnetic field increases from zero, the flux upward through the loop
increases. Then, as in the text, the tangential electric field is still directed clockwise. The
torque does not depend on the sense of rotation either, so the torque is directed out of the
e
e2 r 2 B
page. Thus the vector  L is out of the page which means that   
,
L
2m
4m
and   is directed into the page and thus opposite the direction of B as before. (If the
reader is uncertain that the sense of rotation is irrelevant, recall that  B   B·da . You
use the right-hand rule to determine the sense of da . While the direction of da will
depend on the sense of rotation, the other side of the equation for Faraday's law contains
the expression
 E·d
. The sense of transit along the increment d will also change.)
(b) With the field directed into the paper, now the magnetic flux increases downward
through the loop. Thus Faraday's law will indicate that the electric field is tangent to the
orbit but now directed counterclockwise. Thus the torque will directed into the page and
e
thus  L will be directed into the page. Finally the negative sign in    
 L will
2m
e2 r 2 B
mean that  
, with   directed out of the page.
4m
40. We use the result of the previous problem and the semi-classical model that m
represents a projection of the angular momentum vector on the z-axis and thus represents
a sense of rotation. If we say that m  1 represents a clockwise rotation, then m  1
represents a counterclockwise rotation. Regardless of the sense of rotation, there is a
diamagnetic effect for both. The electron with m  0 will not contribute to the change in
the magnetic moment. The angular momentum of this electron is perpendicular to the
magnetic field. There is not magnetic flux through the orbit so Faraday's law indicates
there will be no tangential electric field. So the total induced magnetic moment is twice
 e2 r 2 B  e2 r 2 B
the effect of the previous problem. That is   2 
.

2m
 4m 
41. This is the same as the high-field limit. With  B / kT
1 we have tanh   B / kT   1 so
 .
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154
Chapter 10
Molecules, Lasers, and Solids
42. (a) See graph:
(b)    tanh(5)  0.99991 and the approximate result of Problem 41 is off by only
0.009%.
(c)    tanh(0.10)  0.0997 . The approximate result is off by just 0.3%.
43. The magnetic dipole moment has units of A  m2 , so M has units of A  m2 m3  A/m .
0 has units of T  m/A and B has units T, so  
0 M
B
has units
 T  m/A  A / m 
T
which
reduces to no unit.
44. (a) If we assume that every atom’s magnetic moment is a Bohr magneton aligned in the
same direction, M  nB where n is the number density, we have
n
7.87 103 kg 1 mol  6.022 1023 
28
3

  8.49 10 m . Thus
m3
0.05585 kg  1mol 
M  nB  8.49 1028 m3  9.274 1024 J/T   7.87 105 A/m .
(b) The computed value is almost exactly one-half the measured value.
(c) This implies that there are two unpaired spins per atom.
  T 2 
2
45. Bc  Bc (0) 1      0.25Bc (0) . Thus T / Tc   0.75 and T  0.75Tc  0.87Tc .
  Tc  


Similarly for a ratio of 0.50 we find T  0.50Tc  0.71Tc ; for a ratio of 0.75 we find
T  0.25Tc  0.50Tc .
46. The energy gap at T = 2 K is Eg (2 K) 
hc


1240 eV  nm
 2.18 103 eV . Inserting
5.68 105 nm
1/2
 T
Equation (10.46) into Equation (10.47) gives Eg (T )  1.74  3.54 kTc  1   . Using
 Tc 
Eg (2K)  2.18 103 eV and k  8.62 105 eV/K we obtain
1/2
1/2
 T

2
4.11 K  Tc 1    Tc 1   . Solving this equation using MathCAD we obtain
 Tc 
 Tc 
Tc  5.23K which is closest to vanadium.
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Chapter 10
Molecules, Lasers, and Solids
155
47. Using the value given in the text just below Equation (10.43), Tc  4.146 K for a mass of
203.4 u, we find M 0.5Tc  constant  59.1296 u 0.5  K . For a mass of 201 u we find
Tc  4.171K and for mass of 204 u we find Tc  4.140 K .
48. With
16
O the molar mass in grams is
88.906  2 137.34  3  63.546   7 16.00   666.224 . Replacing all of the
16
O atoms
with 18 O atoms adds 14 grams per mole, changing the mass to 680.224. Using the BCS
formula for the isotope effect M10.5Tc1  M 20.5Tc 2 and assuming Tc  93K (exactly) for the
1/2
M 
 666.224 
first sample Tc 2   1  Tc1  
  93 K   92.0 K , a change of 1.0 K.
 680.224 
 M2 
49. Extrapolating on the graph in Figure 10.37, it could be at about 130 K.
1/2
50. B  0 I n   4 107 N/A2   4.5 A   2500 m1   14.1 mT
  0.032 m 
B d 2
  BA 
 14.1103 T 
 1.13 105 T  m2
4
4
2

1.13 105 T  m2

 5.5 109 flux quanta .
 0 2.068 1015 T  m2
This large number shows how small the flux quantum is.
51. We know that for niobium Bc  0.206 T . Then the diameter (twice the radius) is
0 I  4 10 N/A   2.5 A 

 4.9106 m which is quite small.
B
  0.206 T 
7
D  2R 
52. P  I 2 R 
2
I 2L
where ρ is resistivity, L is length, and A is area. Since A   r 2 we have
A
I 2L
. The surface area is 2 rL so the power per unit area is
 r2
P
I 2L
I 2
 2
 2 3  80 W/m2 . Using r  3.75 104 m we find
area  r  2 rL  2 r
P
2 2  3.75 104 m  80 W/m2 
3
I2 
1.72 108   m
 4.842 A 2 so I = 2.2 A.
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156
Chapter 10
Molecules, Lasers, and Solids
53. (a) From the BCS theory we have
  T 2 
  4.2 2 
Bc (4.2 K)  Bc (0) 1       0.206 T  1  
 0.1635 T . From the result of
  9.25  
  Tc  




 Bc D   0.1635 T  1.29 10 m 
Problem 51 we know that we know that I 

 527A.
0
4 107 T  m/A
(b) This current is more than 200 times the current that the copper wire can carry.
54. Because f j is directly proportional to V, the frequency is known to within one part in
3


1010 or (1010 ) 483.6 109 Hz  48.36 Hz, which is pretty fine tuning.
55. By conversion 420 km/h = 119.44 m/s. Then from kinematics v 2  2ax , so
v 2 119.44 m/s 
a

 2.38 m/s 2 . This is about g/4 which would certainly be
2x
2  3000 m 
2
noticeable.
56. (a) To compute the escape speed use conservation of energy with
vesc
1 2
GMm
:
mvesc 
2
Re
2  6.673 1011 m3  kg1  s1  5.98  1024 kg 
2GM


 11.1km/s.
Re
6.378 106 m
(b) From Chapter 9 we know
4
v
2
kT
4

m
2
1.38110 J/K   293 K   1245 m/s.
4 1.66110 kg 
23
27
(c) There are always enough helium atoms on the (high-speed) tail of the MaxwellBoltzmann distribution that a significant number can escape, given enough time.
mv 2
 qvB or
57. Equating the centripetal force with the Lorentz (magnetic) force we have
R
mv  p  qBR . The formula p  qBR is also correct relativistically and note that for
these extremely high energies E  pc  qBRc . Therefore the energy is
 27000 m 
8
E  qBRc  1.602 1019 C  13.5T  
  2.998 10 m/s 
 2

6
 2.786 10 J  17.4 TeV.
58. (a) Let the atoms’ distances from the molecule’s center of mass be r1 and r2, respectively,
for the atoms with masses m1 and m2. Thus R = r1 + r2. The rotational inertia is
I  m1r12  m2 r22 , which can also be written
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publicly accessible website, in whole or in part.
Chapter 10
Molecules, Lasers, and Solids
157
m1  m2  m1r12   m1  m2  m2 r22

I
m1  m2

m12 r12  m22 r22  m1m2 r12  m1m2 r22
m1  m2
From the definition of center of mass, m1r1 = m2r2. Applying this to the last expression
above,
2m m r r  m1m2 r12  m1m2 r22 m1m2  r1  r2 
I 1 212

m1  m2
m1  m2
2
which reduces to I = μR2.
(b) Looking up the masses in Appendix 8, m1 = 23.0 u and m1 = 35.0 u. Then
I   R2 
 23 u  35 u  1.66 1027 kg
23 u + 35 u
u
 2.36 10
10
m   1.28 1045 kg  m2
2
This answer is in agreement with published values.
59. (a) Using the result of the preceding problem, the rotational inertias of the two isotopes
are:
I   R2 
I   R2 
1 u  35 u  1.66 1027 kg
1 u + 35 u
u
1 u  37 u  1.66 1027 kg
1 u + 37 u
u
1.27 10
1.27 10
m   2.603 1047 kg  m2
2
10
10
m   2.607 1047 kg  m2
2
Thus the change is only about 4 1050 kg  m2 , a fractional difference of 1.5 × 10−3.
(b) Using the two different rotational inertia values from (a),
Erot 
Erot 
2
I
2
I
1.0546 10

34
J  s
2
47
kg  m
1.0546 10

J  s
2.603 10
2.607 10
34
47
2
 4.2727 1022 J
2
kg  m
2
 4.26611022 J
The difference is 6.6 × 10−25 J = 4.1 × 10−6 eV, a very small difference.
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158
Chapter 10

2
60. (a) With Erot 
 1
2I
Molecules, Lasers, and Solids
2
, for this transition we have Erot 
 6  2  2
2I
2
I
. Solving
for I,
2 1.055 1034 J  s 
2 2
I

 1.46 1046 kg  m2
4
19
Erot  9.55 10 eV 1.602 10 J/eV 
2
(b) From Problem 58 I = μR2, The reduced mass is  
12 u 16 u   6.86 u ,
m1m2

m1  m2 12 u + 16 u
so the bond length is
1/ 2
I
R 

1/ 2


1.46 1046 kg  m2
 

27
  6.86 u  (1.66 10 kg/u) 
 1.13 1010 m
E 1.8 106 J

 4.5 1014 W or 150 times the average US power consumption.
61. (a) P 
9
t 4.0 10 s
(b) The energy of each photon is E = hc/λ = 5.68 × 10−18 J, so the number of photons is
1.8 106 J
 3.2 1023 photons.
18
5.68 10 J/photon
2
T 
B (T )
 1    , so
62. According to BCS theory, c
Bc (0)
 Tc 
1/ 2
1/ 2

T  Bc (T ) 
5 105 T 
 1 

1


  0.9994

2
Tc  Bc (0) 
 4.1110 T 
For mercury Tc = 4.15 K, so T ≈ 4.148 K, which is probably not noticeable.
63. From Table 10.5, Tc = 3.72 K. Take this to be the transition temperature associated with
the weighted atomic mass from the periodic table, 118.7 u. The BCS theory gives the
isotope effect for the two isotopes, relative to this average.
0.5
112 c112
For m = 112 u, M T
M T
0.5
av c,av
Similarly, for m = 124 u, Tc124 
, so Tc112 
M av0.5Tc,av
0.5
M124
M av0.5Tc,av
0.5
M112
 118.7 


 124 
0.5
 118.7 


 112 
0.5
 3.72 K   3.83 K
 3.72 K   3.64 K
The transition temperatures differ by 0.19 K.
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publicly accessible website, in whole or in part.
Chapter 10
Molecules, Lasers, and Solids
159
64. (a) In an RL circuit the current is I  I 0e Rt / L . For small values of R let us approximate
the exponential with the Taylor expansion 1 – Rt/L. Then 109  1 
I
Rt
 1  e Rt / L  ;
I0
L


L
3.14 108 H
  4.0  1025  .
 109 
 2.5 y  3.16  107 s/y  
t


0.1
(b) For a 10% loss t  9  2.5 y   2.5 108 y .
10
  T 2 
65. From the BCS theory B  Bc (0) 1     . Then
  Tc  


R  109
3
S
  B 2  2 Bc2 (0)  T  T  
     . For numerical values use T = 6 K,



V
T  20 
0Tc  Tc  Tc  


Tc  9.25K , and Bc (0)  0.206T . Substituting, we have
2
3
 6
2  0.206 T 
S
 6  
3
1



   2743 J  m  K .
7
V
 4 10 T  m/A  9.25 K   9.25  9.25  
The volume of one mole of niobium is V 
92.91 g
 10.84 cm3  1.084 105 m3 and
3
8.57 g/cm
thus S   2743 J·m3  K 1 1.084 105 m3   2.97 102 J/K for one mole of niobium.
The superconducting state has a lower entropy than the normal state.
66. Provided the supercomputing wire is in its superconducting state and remains in that
state, the resistance is zero. Therefore, the resistance of the circuit is determined by the
copper wire.
V 12V
(a) The current, from Ohm’s law is I  
 8.0A .
R 1.5 
(b) The potential difference across the copper is 12 V. The superconducting wire will
carry the current but will have zero potential difference across it.
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160
Chapter 11
Semiconductor Theory and Devices
Chapter 11
1. The Clement-Quinnell equation, log R 
K
B
 A  , can be rearranged to find
log R
T
B

  A    log R   K  0 which is a quadratic equation that can be solved
T

numerically at any given temperature.
a) T  77 K : log R  2.372 so R  236 
 log R 
b)
2
T  20K : log R  2.786 so R  610 
c) T  1K : log R  7.739 so R  5.48 107 
2. log R 
K
B
 A  ; inserting each of the R and T values, we have three equations in
log R
T
three unknowns  A, B, K  which can be solved to yield A = 2.09, B = 1.96, and K = 1.10.
3. Positive charges drift to the right, so the right side of the strip is at a higher potential and
the voltmeter reads a positive value.
4. (a) Starting from Equation (11.6) and with A = yz, we have
 0.10 A  0.050 T 
IB
n

 1.811022 m3
19
3
4
eVH z 1.602 10 C 11.5 10 V 1.5 10 m 
(b) Graphing B versus VH we find a slope of approximately 4.56 T/V. Algebraically we
nez
ne z
. So
VH . Thus the slope, m is equal to
I
I
 4.56 T/V  0.10 A 
mI
n

 1.90 1022 m3
ez
1.602 1019 C 1.5 104 m
see that B 


5. E  V / L  QdT / dx so Q 

V
L
dT
dx

12.8 106 V
 0.10 m 
 
10 K
 1.28 106 V/K.
0.1 m

6. From the table 3.03 mV corresponds to 58.4 C .
7. Over most of the table, 0.05 mV corresponds to a temperature change of 1 C. Therefore
for 0.01 C there is a corresponding voltage difference of  0.01 0.05 mV   5 107 V.
8. (a) P is in Group V; Ge is in Group IV; so adding phosphorous will create an n-type
semiconductor.
(b) Ga is in Group III; Ge is in Group IV; so adding gallium will create a p-type
semiconductor.
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Chapter 11
Semiconductor Theory and Devices
161
9. The relation between energy and photon wavelength is E 
hc

. The photon must
provide enough energy to excite the electron across the gaps, so  
(a)  
1240 eV  nm
 1850 nm
0.67 eV
(b)  
1240 eV  nm
 1130 nm
1.10 eV
(c)  
1240 eV  nm
 3440 nm
0.36 eV
(d)  
1240 eV  nm
 344 nm
3.6 eV
10. The relation between energy and photon wavelength is E 
wavelength 574 nm corresponds to E 
hc

hc
.
E
. A photon with
1240 eV  nm
 2.16 eV.
574 nm
11. Using Equation (11.9), the ratio of currents in forward and reverse is
19
23
eV / kT
e f  1 exp 1.60 10 C  1.5 V  / 1.38 10 J/K   293 K    1


I r eeVr / kT  1 exp 1.60 1019 C   1.5 V  / 1.38 1023 J/K   293 K    1


If
which simplifies to
If
Ir
 6.0 1025 .
12. From Equation (11.9), the ratio of the two currents is
If
Ir
e f 1
.
eeVr / kT  1
eV / kT

When the ratio of forward-bias to reverse-bias current is this high, the first term in the
denominator is nearly zero, and the first term in the numerator is much greater than one,
I
eV / kT
so to a good approximation f  e f . Solving for Vf,
Ir
23
kT  I f  1.38 10 J/K   293 K 
Vf 
ln   
ln 106   0.35 V.
e  Ir 
1.6 1019 C
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162
Chapter 11
Semiconductor Theory and Devices
13. The angle between the solar rays
and a vector normal to the surface
A is λ. Therefore   0 cos  .
(a) On the equinox   26 so   0 cos  26   0.900 0 ;
On the winter solstice   26  23  49 and   0 cos  49   0.656 0 ;
On the summer solstice   26  23  3 and   0 cos  3   0.999 0
(b) On the equinox   50 so   0 cos  50   0.6430 ;
On the winter solstice   50  23  73 and   0 cos  73   0.2920 ;
On the summer solstice   50  23  27 and   0 cos  27   0.8910
(c) On the equinox   60 so   0 cos  60   0.5000 ;
On the winter solstice   60  23  83 and   0 cos 83   0.122 0 ;
On the summer solstice   60  23  37 and   0 cos  37   0.799 0
14. We require 109 W  0.3  200W/m2  A, so rearranging
109 W
A
 1.67 107 m2
2
0.3  200 W/m 
which corresponds to a square array 4.1 km on a side.
15. In general I  I 0  exp(eV / kT )  1 and in Example 11.4
I0 
I
 18.16 μA.
exp(eV / kT )  1


 
0.250 eV
  1  1.99A
(a) I  18.16 106 A   exp 
 8.617 105 eV/K   250 K   


 



 
0.250 eV
  1  0.288 A  288 mA
(b) I  18.16 106 A   exp 
  8.617 105 eV/K   300 K   


 



 
0.250 eV
  1  0.006 A  6.00 mA
(c) I  18.16 106 A   exp 
  8.617 105 eV/K   500 K   


 

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Chapter 11
Semiconductor Theory and Devices
163
16. Because the tube is single-walled, we can find the surface area density, σ in SI units.
Each atom has mass 12 u, with u  1.6605 1027 kg .
   2.3 1019 atoms/m2  12 u  1.6605 1027 kg/u   4.58 107 kg/m2
(a) To determine the density of the material, we need the mass per unit volume. The
mass will equal the mass density, σ from above, times the area A of the cylindrical
wall. If the tube’s length is L with radius R, them m  2  RL . Therefore the
7
2
m 2 RL 2 2  4.58 10 kg/m 


 1300 kg/m3 .
density will be   
V
 R2 L
R
0.7 109 m
(b) The material is less dense than steel and has a greater tensile strength.
17. We know the density of the single-walled nanotube from the previous problem.
Consider a tube that is 1 nm long. The tube will contain one buckyball. The total mass
will be the mass of the tube plus the mass of the buckyball which is
m  V  (60)(12u) 1.6605 1027 kg/u     R2 L   1.20 1024 kg. We find the term
 V equal to 2.0 1024 kg , so the total mass equals 3.2 1024 kg. Therefore the density
of the nanotube peapod is
m 3.2 1024 kg
3.2 1024 kg
 

 2080 kg/m3 .
2
2

9

9
V
R L
  0.7 10 m  110 m 
18. (a) Using E  Eg for the conduction band we have E  EF  Eg  Eg / 2  Eg / 2. Then
exp  E  EF  / kT   exp  Eg / 2kT  . With Eg  1 eV for a semiconductor (and larger for
1
eV at room temperature, we can see that Eg 2kT and the
40
exponential term will be much greater than 1, so we can neglect the +1 term in the Fermi1
Dirac factor. This leaves FFD 
 exp   Eg / 2kT  .
exp  Eg / 2kT 
an insulator) and kT 


8.0 eV
  6.3 1068
(b) FFD  exp   Eg / 2kT   exp  
 2 8.617 105 eV/K   300 K  




1.11 eV
  2.8 1010
(c) FFD  exp   Eg / 2kT   exp  
5
 2 8.617 10 eV/K   293 K  


23
(d) For one mole of silicon there are 6 10 atoms, so there are still many conduction
electrons available.
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164
Chapter 11
19. (a) Replacing
0
Semiconductor Theory and Devices
with  0 we have for the new Bohr radius
4 0 2
 11.7a0  11.7  5.29 102 nm   0.619 nm.
2
me
(b) This value is about 2.6 times the lattice spacing. This is consistent with the fact that
the electron is very weakly bound, and hence the doped silicon should have a higher
electrical conductivity than pure silicon.
a0 
20. From the Bohr theory E0 
energy E0 
me4
2
2
 4 
2 2
0

2
me4
2
2

 4 
2 2
0
E0

2

. Replacing
13.6 eV
11.7 
2
0
with  0 we have a new Rydberg
 0.099 eV .
This is more than a factor of ten less than the band gap for pure silicon which is
consistent with the idea that the doped version has a higher electrical conductivity.
21. Answers will vary depending on the algorithm used, but one should find that using a
second-order method results in some improvement.
22. (a) I  I 0  exp  eV / kT   1 . To find the value of V for the diode, use the loop rule:
V  IR  6V so V  6V  IR . We are given that I 0  1.05μA and I = 140 mA with
T  293 K so
I 140 103

 1.33 105  exp  eV / kT  so
I 0 1.05 106
ln 1.33 105  
R

eV e  6V  IR 
. Solving for R we find

kT
kT

6 V  kT ln 1.33 105  / e
I


6 V  8.617 10-5 eV/K  293K ln 1.33 105   /e

  40.7 

0.140A
(b) VR  IR   0.140 A  40.7    5.70V
kT ln 8 8.617 10

23. (a) 7  exp  eV / kT   1 so V 
e
5
eV/K   293 K  ln 8
e
 52.5 mV
(b) 0.7  exp  eV / kT   1 so
5
kT ln 0.3 8.617 10 eV/K   293 K  ln 0.3

 30.4 mV
e
e
hc 1240 eV  nm
24. Eg 

 1.91 eV

650 nm
V
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Chapter 11
Semiconductor Theory and Devices
165
25. The GaN laser’s wavelength is 405 nm, so Eg 
26. (a) The total area is  580 10
6
 45 10
9
hc


1240 eV  nm
 3.06 eV.
405 nm
m   1.12 106 m2 so each side is the square
2
root of this which equals 1.08 mm.
(b) The area of each transistor is now  32 109 m   1.024 1015 m2 so the number is
2
1.12 106 m2
 1.09 109 or nearly a factor of 2 improvement.
15
2
1.024 10 m
27. From Problem 18(a) we have FFD  exp  Eg / 2kT  .
N
Si at T  0 C :


1.11 eV

FFD  exp  
 5.67 1011
 2  8.617 105 eV/K   273 K  


Si at T  75 C :


1.11 eV
 =9.16 109
FFD  exp  
 2 8.617 105 eV/K   348 K  


Ge at T  0 C :


0.67 eV
 =6.54 107
FFD  exp  
 2 8.617 105 eV/K   273 K  


Ge at T  75 C :


0.67 eV
 =1.41105
FFD  exp  
 2 8.617 105 eV/K   348 K  


The Fermi-Dirac factor is orders of magnitude higher in germanium, making the
conduction electron density too high. The result is a reverse-bias current that is too large.
See Physics Today December 1997 p. 38.
28. (a) Energy is power multiplied by time:
E  Pt   0.15 200 W   3.156 107 s   9.47 108 J.
(b) Converting we find 2.8 1012 kW  h  1.011019 J . Then the area is
1.011019 J
 1.06 1010 m2
8
2
9.47 10 J/m
c) Using 2.5 times the area found in (b) the fraction of the U.S. covered is
2.5 1.06 1010 m2 
 0.0028 or about one-fourth of one percent. See American Scientist
9.6 1012 m2
July-August 1993, p. 368.
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166
Chapter 11
Semiconductor Theory and Devices
29. From the diode equation the current is I  I 0 exp  eV / kT   1 . Then the ratio is
If
Ir

I 0 exp  eV f / kT   1
I 0 exp  eVr / kT   1
(a) At T = 77 K:
.


1.50 eV
exp 
 1
5
If
8.617  10 eV/K   77 K  



 1.5 1098
Ir


1.50 eV
exp 
 1
5
  8.617  10 eV/K   77 K  
At T = 273 K:


1.50 eV
exp 
 1
5
If
8.617  10 eV/K   273 K  



 4.9 1027
Ir


1.50 eV
exp 
 1
5
  8.617  10 eV/K   273 K  
At T = 350 K:


1.50 eV
exp 
 1
5
If
8.617  10 eV/K   350 K  



 4.0 1021
Ir


1.50 eV
exp 
 1
5
  8.617  10 eV/K   350 K  


1.50 eV
exp 
 1
5
If
8.617  10 eV/K   600 K  



 4.0 1012
At T = 600 K:
Ir


1.50 eV
exp 
 1
5
8.617

10
eV/K
600
K






(b) The ratio changes significantly as a function of temperature which is something diode
designers must keep in mind.
30.
(a) An electron can be produced if the 1.11 eV band gap can be overcome. If we divide
the total energy available by the band gap energy, the maximum number of electrons
1.04 106 eV
 9.37 105 .
1.11 eV
(b) If the silicon is cooled well below room temperature, very few electrons will be in
the conduction band. At room temperature, however, enough electrons are in the
conduction band so that additional current will be measured, tending to mask the
gamma-ray signal.
that can be produced is N 
31. From the diode equation the current is I  I 0 exp  eV / kT   1 . Then the ratio is
If
Ir

I 0 exp  eV f / kT   1
I 0 exp  eVr / kT   1

exp  eV f / kT   1
exp  eVr / kT   1
. Substituting the given values we find
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Chapter 11
Semiconductor Theory and Devices
167


0.25 eV
exp 
 1
5
If
8.617  10 eV/K   293 K  



 20000 . So the ratio of the currents in the
Ir


0.25 eV
exp 
 1
5
  8.617  10 eV/K   293 K  
forward- and reverse-bias is 20,000.
32. Young’s Modulus relates the elongation of an object due to an applied force. The formula
F L 
is Y  
 . Rearranging this formula to solve for F and substituting, we find
A  L 
  (1.6 109 m)2 
 L 
9
8
F  Y A

1050

10
Pa

  0.01  2.110 N .

L
4




33. The number of bits stored is (4.7 109 )(8)  3.76 1010 bits . To find the area we use
A    r22  r12     0.0582  0.0232   8.91103 m2 . The number of bits stored per
3.76 1010 bits
square meter is then
 4.22 1012 bits/m2 .
3
2
8.9110 m
34. 25 GB × 8 bits/byte = 200 Gbits With an inner radius of 2.3 cm and outer radius 5.8 cm,
the effective useful area is   0.058 m     0.023 m   8.91103 m2 . The average area
2
2
8.91103 m2
 4.46 1014 m2 . For an approximately rectangular bit, the area is
200 109
length times width, so the length is area divided by width, or
per bit is
4.46 1014 m2
 1.4 107 m = 0.14 μm.
7
3.2 10 m
35. 1 hectare = 104 m2, so 140 hectares = 1.40 × 106 m2.
The facility’s power output per unit area is
64 106 W
 45.7 W/m2 .
6
2
1.40 10 m
45.7 W/m2
 0.073  7.3% .
630 W/m2
This seems low compared with solar cells. However, the NSO facility does not use every
square meter of land surface, so the comparison is not an exact one.
The efficiency is this output divided by the sun’s input:
36.
3
4
(a) The mass of the dot is m  V   4820 kg/m3    2.50 109 m   3.15 1022 kg.
3
From the periodic table, the average mass of a CdS molecule is
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publicly accessible website, in whole or in part.
168
Chapter 11
Semiconductor Theory and Devices
112.4 u + 32.1 u = 144.5 u  2.40 1025 kg. The number of molecules is
3.15 1022 kg
 1312.5. The number of atoms is twice this or 2625.
2.40 1025 kg
(b) From Chapter 6, the energy levels are given by E 
n2 h2
. For n = 1, we find
8mL2
6.626 1034 J  s 

h2
E

 1.511020 J = 0.094 eV.
8mL2 8  9.111031 kg  2.0 109 m 2
2
(c) The difference in energy levels is
nu2 h 2 nl2 h 2
. With nu = 6 and nl = 5,

8mL2 8mL2
E   62  52  0.094 eV = 1.0 eV.
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Chapter 12
The Atomic Nucleus
169
Chapter 12
h
2h
1  cos   , so max  . From Chapter 3 Problem
mc
mc
51 we know that the kinetic energy of the recoiling particle is
2h
  /   hc
from above and for convenience letting
K 
 5.7 MeV. Using  

mc
1   /   
1. From Compton scattering  
x
2h
2h / (mc ) hc
x mc 2
mc 2
, we have K 
;

x ; K 1  x   x 2
mc
2
1  2h / (mc )  1  x 2
mc 2 2
x  Kx  K  0. This is a quadratic equation that can be solved numerically to find
2
2 1240 eV  nm 
2h

 2.26 105 nm. The
x  0.117  2h / mc , so  
mcx  938.27 106 eV   0.117 
photon's energy is E 
2. (a) integral: 6 Li ,
18
hc


1240 eV  nm
 54.9 MeV.
2.26 105 nm
F ; (b) half-integral:
3
7
He ,
Li ,
19
F
3. In each case the atomic number equals the number of protons (Z), and the atomic charge
is Ze.
4.
3
2
He : Z = 2, N = 1, A = 3, m = 3.02 u
4
2
He : Z = 2, N = 2, A = 4, m = 4.00 u
18
8
O : Z = 8, N =10, A = 18, m = 18.00 u
44
20
Ca : Z = 20, N = 24, A = 44, m = 43.96 u
209
83
Bi : Z = 83, N = 126, A = 209, m = 208.98 u
235
92
U : Z = 92, N = 143, A = 235, m = 235.04 u
6
Li : Z = 3, N = 3
C : Z = 6, N = 7
40
K : Z = 19, N = 21
102
Pd : Z = 46, N = 56
13
5.
Ca through 52 Ca ; Isobars
35
P , 36 S , 37 Cl , 38 Ar , 39 K , 40 Ca
6.
14
Isotopes
38
40
S,
40
Cl ,
40
Ar ,
40
K,
40
Ca ; Isotones
32
Mg , 34 Si
50
N (99.63%), 15 N (0.37%);
V (0.250%), 51 V (99.75%);
84
Sr (0.56%), 86 Sr (9.86%), 87 Sr (7.00%), 88 Sr (82.58%)
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170
Chapter 12
7.
39
Ar (269 y),
46
Ar (8.4 s).
41
Ar (1.822h),
250
42
No (0.25 ms), 251 No (0.8 s),
258
No (1.2 ms), 259 No (58 m),
Ar (32.9 y),
43
Ar (5.37 m),
252
44
No (2.30 s), 254 No (55 s),
260
No (106 ms).
The Atomic Nucleus
Ar (11.87 m),
256
45
Ar (21.48 s),
No (2.91 s),
8. From the text nuclear density is 2.3 1017 kg / m3 , which is 2.3 1014 times the density of
water.
9.
p
2.79N

m
0.51100

 2.786 N  2.787 e  2.786
 1.52 103
e 1.00116B
B
mp
938.27
10. From Appendix 8 the mass of the nuclide is 55.935 u or 9.29 1026 kg.
m
m
9.29 1026 kg



 2.29 1017 kg/m3
3
4 3 4 3
4
r
 r0 A
 1.2 1015 m   56 
3
3
3
11. The electron binding energy is virtually the same as in hydrogen, or 13.6 eV. The
deuteron rest energy is mc2  1876 MeV. The ratio of these is
13.6 eV
 7.25 109.
6
1876 10 eV
“Nuclear calculations”' might also refer to nuclear binding energies. For the deuteron the
binding energy is 2.2 MeV, so the ratio of the electronic to nuclear binding energy is on
the order of 105. Therefore it is generally safe to ignore electronic binding energies.
12. Using Equation (12.14), the values of Bd from Equation (12.9) and the atomic deuterium
mass in order to evaluate the second term inside the parenthesis, we find


Bd
  1.00059 Bd  2.22 MeV.
Emin  Bd 1 
 2M  2 H  c 2 


From this calculation we can see that the error is 0.00059 or 0.059%.
13. The incoming photon’s momentum is p. Its energy is hf = pc. By conservation of
momentum, the total momentum of the two particles after the reaction is also p, and the
total mass is mp + mn. (We will ignore electron masses here.) Conservation of relativistic
energy gives
pc  M d c 2 

p 2c 2   m p  mn  c 2

2
.
Squaring both sides of the equation:
 pc 
2

 2 pM d c3  M d2c 4   pc    mp  mn  c 2
2

2
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Chapter 12
The Atomic Nucleus
171
The first term on each side cancels. The resulting equation can be solved for 2pc :
 m
2 pc 
p
 mn  c 2

2
 M d2c 4
M d c2
  m  m 2  M 2 c 4 
p
n
d


2


Mdc


Now Bd  mp  mn  M d  c 2 , so we can substitute  mp  mn  c 2  Bd  M d c 2 , giving
  B  M c 2 2  M 2 c 4   2
Bd  M d2c 4  2 Bd M d c 2  M d c 4 
d
d
d


2 pc 



 
M d c2
M d c2





Bd2 
B 
2 pc   2 Bd 
 Bd  2  d 2 
2
Mdc 
Mdc 


Now for the photon pc = hf, and so

Bd2 
hf  Bd 1 
2
 2M d c 
which is equivalent to Equation (12.14).
14. (a) Think of the nucleus as the composite of deuteron and 1 n , so that
B   M  21 H   mn  M  31 H   c 2 . Use the atomic masses from Appendix 8:


B   2.014102 u  1.008665 u  3.016049 u  c 2 931.49 MeV/  u  c 2   6.26 MeV
(b) With the nucleus as two neutrons and a proton, we have:
B  2mn  M  11 H   M  31 H   c 2


B   2 1.008665 u + 1.00728u  3.016049 u  c 2 931.49 MeV/  u·c 2   8.48 MeV
(c) The difference is just the binding energy of the deuteron, 2.22 MeV. This makes
sense, because the first step (a) separates the triton into a neutron and deuteron, and a
second step of separating the deuteron into a proton and neutron completes the
disintegration of the original nucleus.
15. The distance equals the nuclear radius: r  r0 A1/3  1.2 fm   31/3   1.73fm
GMm  6.673 10
Fg  2 
r
11
N  m2 / kg 2 1.673 1027 kg 
1.73 10
15
m
2
2
 6.24 1035 N
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publicly accessible website, in whole or in part.
172
Chapter 12
The Atomic Nucleus
9
2
2
19
ke2  8.988 10 N  m / C 1.602 10 C 
Fe  2 
 77.1 N
2
15
r
1.73

10
m


2
To compare with the strong force, we need the potential energy:
11
2
2
27
GMm  6.673 10 N·m / kg 1.673 10 kg 
Vg 

 6.7 1037 MeV
15
13
r
1.73 10 m 1.602 10 J/MeV 
2
9
2
2
19
ke2  8.988 10 N  m / C 1.602 10 C 
Ve 

 0.83MeV
r
1.73 1015 m 1.602 1013 J/MeV 
2
The electrostatic force is about 50 times weaker than the strong force. The gravitational
force is almost 1038 times weaker than the strong force.
16. The required nuclear force must be at least of this magnitude:
9
2
2
19
ke2  8.988 10 N  m / C 1.602 10 C 
Fe  2 
 40.0 N. We can also compare the
2
15
r
 2.4 10 m 
2
9
2
2
19
ke2 8.988 10 N  m / C 1.602 10 C 

 0.60 MeV
potential energy: V 
r
 2.4 1015 m 1.602 1013 J/MeV 
2
17. (a) Think of the nucleus as the composite of
A1
Z
X and 1 n , so that
B   M  AZ1 X   mn  M  ZA X   c 2
(b) Details for 6 Li are shown. The other examples are similar. Use the atomic masses
from Appendix 8:
B   M  5 Li   mn  M  6 Li  c 2


  5.012540 u  1.008665 u  6.015122 u  c 2 931.49 MeV/  u·c 2   5.67 MeV
16
O : B   M  15 O   mn  M  16 O  c 2  15.7 MeV
207
Pb : B   M  206 Pb   mn  M  207 Pb  c 2  6.74 MeV
18. (a) As in the previous problem, consider the nucleus as the composite of
that B   M  ZZ 11Y   M  1 H   M  ZA X   c 2 .
(b) 8 Be :
B   M  7 Li   M  1 H   M  8 Be  c 2

A1
Z 1
Y and 1 H , so

  7.016004 u  1.007825 u  8.005305 u  c 2 931.49 MeV/  u·c 2   17.3 MeV
15
O : B   M  14 N   M  1 H   M  15 O  c 2  7.3 MeV
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publicly accessible website, in whole or in part.
Chapter 12
32
The Atomic Nucleus
173
S : B   M  31 P   M  1 H   M  32 S c 2  8.9 MeV
19. The energy release comes from the mass difference:
E  mc 2  3 M  4 He   M  12 C  c 2


 3  4.002603 u   12.000 u  c 2 931.49 MeV/  u  c 2   7.27 MeV
20. With an even number of nucleons the 4 He nucleus has integer spin. Because alignment is
unlikely, the net spin should be zero. The 3 He nucleus has three nucleons, so its spin is
half-integer. Because alignment is unlikely, the net spin should be one-half.
21. As in Problem 18, B   M  106 Pd   M  1 H   M  107 Ag  c2  5.8 MeV. The electron
binding energy is less be a factor of 5.8 MeV / 25.6 keV  227.
22. For 4 He the radius is r  r0 A1/ 3  1.2 fm   41/ 3   1.90fm.
V
For
e2 1 1.44 109 eV  m

 0.76 MeV
4 0 r
1.90 1015 m
40
Ca:
ECoul
2
3 Z  Z  1 e

 0.72 Z  Z  1 A1/3 MeV
5 4 0 R
 0.72  20 19  401/3 MeV  80 MeV
For 208 Pb: ECoul  0.72 82 81 2081/3 MeV  807 MeV
There is roughly a factor of ten between each of these three nuclides.
23. Refer to Equation (12.20), the vonWeizsäcker semi-empirical mass formula. Adding
another neutron to 18 O reduces  from  to 0, makes the symmetry term more
negative, and makes the surface term more negative. The net effect of these three terms
more than offsets the increase in the volume term.
24. Refer to Equation (12.20), the von Weizsäcker semi-empirical mass formula.  drops
from  in 42 Ca to 0 in 41 Ca . The volume term is substantially higher in 42 Ca , more
than offsetting the difference in the surface effect. The symmetry term is only slightly
higher for 42 Ca . Comparing 42 Ca and 42 Ti , we see that the coulomb term is higher in
42
Ti , because it has two more protons.
25. We begin with Equation (12.20) and substitute Equation (12.19) for the Coulomb term.
Therefore, we have: B  X   aV A  a A A
A
Z
2/3
 0.72  Z  Z  1  A
1/3
 aS
N  Z
A
2

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174
Chapter 12
The Atomic Nucleus
Evaluating this expression for 48 Ca , we have:
B  Ca   14MeV  48  13MeV  48
48
20
This gives B 
48
20
2/ 3
 20·19 
 0.72
1/ 3
48
Ca   401.492 MeV which equals
19MeV 
 28  20 
48
2

33MeV
.
483/ 4
 c2  u 
2
401.492 MeV 
  0.43102 u  c . From Equation (12.10), we can calculate the
 931.5MeV 
48
48
Ca   28  mn  20  M  1 H    B  20
Ca  / c 2  . Substituting, we find
mass: M  20
48
M  20
Ca   28  1.008665u   20  1.007825u   0.43102 u  c 2 / c 2   47.9681u, or
 931.5MeV 
4
2
47.9681u  
  4.468 10 MeV / c . The atomic mass given in Appendix 8 is
2
c

u


47.952534 u. Our calculation based on a semi-empirical formula is different by about
0.03%.
26. (a) We use Equation (12.20) to calculate the binding energy with Equation (12.19) for
the Coulomb term. We find that   (33MeV) 183/4  3.776 MeV for the pairing term,
so   3.776MeV for 18 N but 3.776 MeV for the others, which are even-even nuclei.
Once we determine the binding energy, we must divide by 18 to find the binding energy
per nucleon. Evaluating this expression for 186 C , we have:
B  C   14 MeV 18  13MeV 18
18
6
2/3
6  5 
 0.72
181/ 3
19 MeV 
12  6 
18
2
 3.776 MeV
 120.2466 MeV
which gives a binding energy per nucleon for 186 C of 6.68 MeV. The values for the other
elements are 7.25 MeV, 8.16 MeV, and 7.64 MeV in the order given.
(b) The nuclide 18 O is most stable, which is expected because it is the closest to having
the same number of neutrons and protons and it is even-even.
(c) Using the information in Appendix 8 and Equation (12.10), the binding energy of 18 O
is: B  188 O   10  mn  8M  1 H   M  188 O  c 2  0.1501u  c 2  139.8 MeV which gives a
binding energy of 7.77 MeV per nucleon.
27.  
ln 2
ln 2

 4.167 109 s 1
7
t1/2  5.271 y   3.156 10 s/y 
4.4 107 s 1
 1.06 1016 ;
9 1
 4.167 10 s
1 mol  60 g 
m  1.06 1016 

  1.05 μg
6.022 1023  mol 
N
R

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publicly accessible website, in whole or in part.
Chapter 12
The Atomic Nucleus
175
R0
ln 5
at t  T  3600 s ;  
5
T
ln 2 ln 2
ln 2


T
 3600 s   1550 s  26 minutes

ln 5
ln 5
28. R  R0e t 
t1/2
29. In general (using the definition of the mean value of a function)




t R(t ) dt
0

0
R(t ) dt

1
N0


0
t R(t ) dt
because all nuclei must decay between t = 0 and t   . Using R  R0e t we have
R 
R
N
1
1 t
  0  te t dt  0 % 2  2 0   1/2 .
0
N0
N0 
 N0  ln 2
U and 4% 235 U the number N of 238 U atoms is
1 mol 6.022 1023  0.96 
26
N  10 kg 

  6.15 10 . [A careful reading of example
0.235 kg
mol
 0.04 
12.11 indicates that the fuel rod has 10-kg of 235 U . The final ratio in the equation is to
determine the amount of 238 U present in the sample.] The half-life for 238 U alpha
emission is found in Appendix 8 as 4.47 109 y which equals 1.411017 s . Therefore,
ln 2
ln 2
R  N 
N
6.15 1026   3.02 109 Bq .

17
t1/2
1.4110 s
30. With 96%
31.  
238
ln 2
ln 2

 1.720 1017 s 1 using the t1/2 for
9
7
t1/2 1.277 10 y  3.156 10 s/y 
40
K from
Appendix 8.
N   60 kg 
1 mol 6.022 1023
 0.00012 0.003  3.252 1020
0.040 kg
mol
R   N  1.720 1017 s1  3.252 1020   5.6 103 Bq
32.  
ln 2
ln 2

 1.052 104 s 1
t1/2 109.8 min  60 s/min 


R  R0et  1.2 107 Bq  exp  1.052 104 s1   48 3600s   0.153 Bq
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publicly accessible website, in whole or in part.
176
Chapter 12
33.  
The Atomic Nucleus
ln 2
ln 2

 5.622 102 y 1. The mass decreases by the same exponential
t1/ 2 12.33 y
factor as N, so m  m0et   75 kg  exp   5.622 102 y1   7 y   51 kg
34. For convenience assume a mass of 1 kg of each material.
1 mol 6.022 1023
For H: N  1 kg 
 2.011026
0.003 kg
mol
ln 2
ln 2


 5.622 102 y1
t1/2 12.33 y 
3
R   N   5.622 102 y1  2.011026   1.13 1025 y1
1 mol 6.022 1023
 2.711024
For Rn: N  1 kg 
0.222 kg
mol
ln 2
ln 2


 66.28 y1
t1/2  3.82 d 1 y/365.25 d 
222
R   N   66.28 y1  2.711024   1.80 1026 y1
Pu: N  1 kg 
1 mol 6.022 1023
 2.52 1024
0.239 kg
mol
For
239

ln 2
ln 2

 2.875 105 y1
t1/2  24110 y 
R   N   2.875 105 y1  2.52 1024   7.245 1019 y1
The order of activity is
222
Rn  3 H 
35. The decays in question are
52
239
Pu .
Fe  n  51 Fe and
52
Fe  p  51 Mn.
neutron decay: Q   51.948117  50.956825  1.008665 u  c2  16.2MeV
which is not allowed because Q  0 .
proton decay: Q   51.948117  50.948216  1.007825 u  c2  7.4MeV
which is not allowed because Q  0 . Because
decay.
40
Fe has an excess of protons, it should
36. The separation energy is the energy required to “extract” a nucleon from the nucleus and
is equal to the absolute value of the Q values found: 7.7 MeV for the neutron and 5.6
MeV for the proton.
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Chapter 12
The Atomic Nucleus
177
37. We need not consider   decay, because there is no physical object with mass 1 and
charge 2. The   decay is p  n    , with
Q   mp  mn  me  c 2  1.0072876  1.008665  0.00054858 u·c 2  1.8MeV, which is
not allowed with Q  0 .
For electron capture p     n , with
Q   mp  me  mn  c 2  1.0072876  1.008665  0.00054858 u·c 2  0.79 MeV,
which is not allowed with Q  0 .
38. For 144 Sm:
Q   M  144 Sm   M  140 Nd   M  4 He   c 2
 143.911995  139.909310  4.002603 u  c 2  0.076 MeV
For 147 Sm:
Q   M  147 Sm   M  143 Nd   M  4 He   c 2
 146.914893  142.909810  4.002603 u·c 2  2.31 MeV
With Q  0 both isotopes may decay, though the lighter one is barely stable. The
147
Sm
11
abundance is greater because its decay has an extremely long half-life of 10 years.
39. The decay is 241 Am 237 Np  4He .
Q   M  241 Am   M  237 Np   M  4 He   c 2
  241.056823  237.048167  4.002603 u·c 2  5.64 MeV
By Equation (12.32) the emitted alpha particle's energy is
A4
237
K 
Q
 5.64 MeV   5.55 MeV . Then by conservation of energy the
A
241
daughter's kinetic energy is 0.09 MeV.
40. Let M be the mass of the decaying nucleus, E1 and p1 the energy and momentum of the
recoiling nucleus, and let E2 and p2 be the energy and momentum of the   . Then by
conservation of momentum p1  p2  0 , and by conservation of energy Mc 2  E1  E2 .
Then using the energy-momentum invariant E 2  p 2c 2  E02 one can rewrite the energy
conservation equation in terms of the two momenta. Then we would have two equations
in two unknowns ( p1 and p2 ), and by the laws of algebra there are unique solutions for
p1 and p2 . Hence there are unique solutions for E1 and E2 .
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178
Chapter 12
41.
The Atomic Nucleus
Br  76 As 4 He :
Q  79.918530  75.922394  4.002603 u·c2  6.0 MeV (not allowed)
80
80
Br 80 Kr    : Q   79.918530  79.916378 u·c2  2.0MeV (allowed) ;
80
Br 80 Se    :
Q   79.918530  79.916522  2  0.000549  u·c2  0.85 MeV (allowed)
80
Br    80 Se :
Q   79.918530  79.916522 u·c2  1.9MeV(allowed)
42.
Ac 223 Fr 4 He :
Q   227.027747  223.019731  4.002603 u  c2  5.0 MeV (allowed)
227
227
Ac 227 Th    :
Q   227.027747  227.027699 u  c2  0.045MeV (allowed, barely)
227
Ac 227 Ra    :
Q   227.027747  227.029171  2  0.000549  u  c2  2.3 MeV (not allowed)
227
Ac    227 Ra :
Q   227.027747  227.029171 u  c2  1.3MeV (not allowed)
43.
Pa 226 Ac 4 He :
Q   230.034533  226.026090  4.002603 u  c2  5.4 MeV (allowed)
230
230
Pa 230 U    :
Q   230.034533  230.033927  u  c2  0.56MeV (allowed)
230
Pa 230 Th    :
Q   230.034533  230.033127  2  0.000549  u  c2  0.29 MeV (allowed)
230
Pa    230 Th :
Q   230.034533  227.033127  u  c2  1.3MeV (allowed)
44. Looking at Figure 12.16 it appears the gamma ray energy can be 0.226 MeV or 0.230
MeV in a transition from Ex to the ground state. Clearly a 0.072 MeV transition is
possible. There are also two possible transitions from Ex to the 0.072 MeV level. The
energies of those gamma rays are 0.226 MeV  0.072 MeV  0.154 MeV and
0.230 MeV  0.072 MeV  0.158 MeV .
45. After six days the number of radon atoms remaining is
  ln 2  6 d  
N  N 0 exp  
  0.3367 N 0
3.82 d 

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Chapter 12
The Atomic Nucleus
179
Therefore the partial pressure of radon is 0.337 atm after 6 days. At first glance one
might assume that the partial pressure of the helium is therefore 0.663 atm. But the
problem leads to an interesting example.
Note that 222 Rn decays entirely by alpha decay. If we continue to follow the decay chain,
however, the daughter 218 Po , can decay either by alpha or   with a half-life of 3.10
minutes; however, 99.98% of the time, 218 Po decays by emitting an alpha to 214 Pb .
214
Pb is also unstable and decays via   to 214 Bi with a half-life of 26.8 minutes. 214 Bi
decays 99.9% of the time via   decay to
214
214
Po with a half-life of 19.9 minutes. Finally
210
Po decays by alpha decay to Pb which has a half-life of 22.3 year which is quite
long with respect to the 6 days of the problem. Notice that all of the other half-lives are
short with respect to 6 days. Therefore the decay chain has produces 3alphas, not 1.
Using standard chemistry, the ratio of the partial pressures is the same as the ratio of the
molecules. The partial pressure of the helium is therefore 3 0.663 atm  1.99 atm.
46.
Co decays by   to
gamma ray.
60
60
Ni . The daughter nucleus is in an excited state and emits a
47. There are three explanations. Looking at the Q formulas, we see that two extra electron
masses are required for   decay, and this energy may not be available in “close calls.”
Second, the product of   decay should be more stable with a greater N/A ratio. Finally,
successive alpha decays place daughters farther and farther below the line of stability.
Only   decays can move back in the direction of the stability line.
48. For the reaction
14
O 14 N    we have
Q  14.008595  14.003074  2  0.000549  u  c2  4.1 MeV which is allowed.
For the other reaction p  n    we have
Q  1.007825  1.008665  2  0.000549  u  c2  1.8 MeV which is not allowed.
49. From Equation (12.50) R 
N  206 Pb 
N  238 U 
 et  1 . Rearranging we have t 
t1/2
ln  R  1 .
ln 2
4.47 109 y
ln 1.76   3.65 109 y
ln 2
4.47 109 y
For R  3.1: t 
ln  4.1  9.10 109 y
ln 2
The large time difference implies different origins.
For R  0.76 : t 
50.
K , 87 Rb , 138 La , and 147 Sm have half-lives that are close to the age of the earth.
However, of these only 87 Rb and 147 Sm have reasonable abundances.
40
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publicly accessible website, in whole or in part.
180
Chapter 12
51. As in Problem 40 t 
52. R 
N  206 Pb 
N
238
U
The Atomic Nucleus
4.47 109 y
ln 1.01  6.4 107 y.
ln 2
 et  1  eln 2  1  1 where the substitution for  occurs since the time
given in the problem almost exactly matches the half-life of U-238. Thus
 t   ln(2) / t1/2  t  ln(2) . A more exact answer would be
 t   ln(2) / 4.47 109  4.6 109  1.03ln(2) and thus the ratio would be
R 
N  206 Pb 
N
238
U
 e  t  1  e
1.03 ln 2
 1  1.04 revealing a slightly higher amount of lead.
53. From the A values it is clear that there are 28 / 4  7 alpha decays. Seven alpha decays
reduces Z from 92 to 78, so there must be four   decays in order to bring Z up to 82.
There are other possible combinations of beta decays (including   and electron capture),
but the net result must be a change of four charge units. We would have to look at a table
of nuclides to determine the exact chain(s).
54. 0.0072 
N 0  235 U 
N0 
238
U
N  235 U 
N  238 U 

  0.0072 
N 0  235 U  exp  235t 
N 0  238 U  exp  238t 
so
exp  238t 
exp  235t 



ln 2
ln 2
9
  0.0072  exp  

4.5

10
y


  0.301

9
9
  0.7038 10 y 4.468 10 y 

or 30.1%.
  ln 2  24 h  
55. (a) After one day N  N 0 exp  
  0.519 N 0 so 51.9 g is still
25.39 h 

remainder (41.1 g) is almost all 248 Cf , because its half-life is 334 days.
252
Fm . The
(b) After one month the amount of fermium is
  ln 2  30  24 h  
9
N  N 0 exp  
  N 0  2.9 10   0 . Almost none of the fermium is
25.39 h


left. It is more difficult to determine the remaining amount of 248 Cf. Of the amount
created on the first day of the month the remaining amount is about
  ln 2  29 d  
N0 exp  
  0.942 N0
333.5
d


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Chapter 12
The Atomic Nucleus
181
Of the amount created on the fourth day of the month the remaining amount is about
  ln 2  25 d  
N0 exp  
  0.949 N0 . Because most of the fermium decayed within the
333.5
d


first four days, we can see that the sample should be between 94% and 95% remaining as
248
Cf by the end of the month.
(c) Five years is more than five half-lives of californium, so there is very little of it left.
However, the half-life of curium is 18.1years, so there is at least
  ln 2  5 y  
244
N0 exp  
  0.826 N0 of Cm after five years.
18.1 y 

(d) After 100 years more than five half-lives of 244 Cm have elapsed. The half life of its
decay product 240 Pu is quite long, 6593 years, so most of the sample is 240 Pu after 100
years.
240
236
U with a half-life of 6563 years, the next decay is to 232 Th,
with a half-life of 2.34 107 y. It takes this amount of time for “most”' of the sample to
be thorium.
(e) After
Pu decays to
56. (a) t1/2  7.7 1024 y  2.4 1032 s;
R  N 
(b)
ln 2 N A
ln 2
6.022 1023

 1.36 108 s1  kg 1
32
t1/ 2 0.128 kg 2.4 10 s 0.128 kg
10 s 1
 7.36 108 kg. This is not a realistic sample size.
8 1
1
1.36 10 s  kg
57. (a) The magnesium isotope has Z > N, which is generally not favorable to stability. On
the other hand, the sodium isotope has N − Z = 1, which is favorable in this range of A.
Therefore we expect 23Mg to be less stable.
(b) Emitting an alpha particle would still leave Z > N, and emitting an electron would
raise the Z value even further. Thus the only candidates are positron emission 23Mg→
23
Na + β+ and electron capture 23Mg + β− → 23Na. We find the Q values for both
reactions:
Positron emission: Q   M  23 Mg   M  23 Na   2me  c 2
Q   22.994125 u  22.989770 u  2(0.000549 u)  c2  0.003257 u  c2  3.03 MeV
Electron capture: Q   M  23 Mg   M  23 Na  c 2
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182
Chapter 12
The Atomic Nucleus
Q   22.994125 u  22.989770 u  c2  0.004355 u  c2  4.06 MeV
As expected, both processes are allowed.
58. For 36Ar: B  18mn  18M  1 H   M  36 Ar  c 2 ; B = 0.329274 u∙c2 = 306.7 MeV so
B/A = 8.52 MeV/nucleon.
For 76Se: B  42mn  34M  1 H   M  76 Se  c 2 ; B = 0.710766 u∙c2 = 662.1 MeV so
B/A = 8.71 MeV/nucleon.
This is the expected result, and matches the asymmetric shape of the curve in Figure 12.6.
59. The resulting nucleus is 16O. If it experienced alpha decay, the resulting nucleus would be
12
C. The Q for this reaction is given by Equation (12.31):
Q   M  16 O   M  12 C   M  4 He  c 2
Q  15.994914 u  12.0 u  4.002603 u  c2  0.007689 u  c2  7.16 MeV
For β− decay, 16O → 16F + β−, and Q is given by
Q   M  16 O   M  16 F c 2
Q  15.994914 u  16.011466 u  c2  0.016552 u  c2  15.4 MeV
For β+ decay, 16O → 16N + β+, and Q is given by
Q   M  16 O   M  16 N   2me  c 2
Q  15.994914 u 16.006101 u  2(0.000549 u)  c2  0.012285 u  c2  11.4 MeV
For electron capture, 16O + β− → 16N, and Q is given by
Q   M  16 O   M  16 N  c 2
Q  15.994914 u  16.006101 u  c2  0.011187 u  c2  10.4 MeV
Q for each reaction is negative, proving that 16O is stable.
60. The resulting nucleus is 13N. If it experienced alpha decay, the resulting nucleus would be
9
B. The Q for this reaction is given by Equation (12.31):
Q   M  13 N   M  9 B  M  4 He  c 2
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Chapter 12
The Atomic Nucleus
183
Q  13.005739 u  9.013329 u  4.002603 u  c2  0.010193 u  c2  9.49 MeV
For β− decay, 13N → 13O + β−, and Q is given by
Q   M  13 N   M  13 O  c 2
Q  13.005739 u  13.024810 u  c2  0.019071 u  c2  17.8 MeV
For β+ decay, 13N→ 13C + β+, and Q is given by
Q   M  13 N   M  13 C   2me  c 2
Q  13.005739 u 13.003355 u  2(0.000549 u)  c2  0.001286 u  c2  1.20 MeV
For electron capture, 13N + β− → 13C, and Q is given by
Q   M  13 N   M  13 C  c 2
Q  13.005739 u  13.003355 u  c2  0.002484 u  c2  2.22 MeV
Positron decay and electron capture are allowed, and this nucleus is unstable.
61. Adding charge dq to a solid sphere of radius r we have an energy change dE 
Q dq
4 0 r
4
 r 3 is the charge already there and  is the charge density. Also we know
3
16 2  2 r 4
dr .
dq   dV  4 r 2 dr . Making those substitutions gives dE 
3  4 0 
where Q 
Integrating from 0 to R we find E  
R
0
16 2  2 r 4
16 2  2 R5
. The charge density is
dr 
3  4 0 
15  4 0 
3  Ze 
16 2 R5  3Q 
3Q 2
.



3 
15  4 0   4 R  5  4 0  R 5  4 0  R
2
  Q / V  3Q / 4 R3 , so E 
2
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184
Chapter 12
The Atomic Nucleus
62. For x use the diameter 2r  2r0 A1/3  2 1.2 fm  21/3  3.02 fm. Then pmin  / 2x and
2
2
197.3 109 eV  m 

pmin



 569 keV
2m 8m  x 2 8  938 106 eV  3.02 1015 m 2
2
K min
h 2hx

 4 x  4  3.02 fm   38 fm.
p
These results are reasonable, because the energy is smaller than the deuteron's binding
energy, and the wavelength computed is the maximum wavelength, so the actual
wavelength may be less. Shorter wavelengths, corresponding to the size of the nucleus,
are allowed by the uncertainty principle.

63. The parity term in the semi-empirical formula, along with the more negative coulomb and
symmetry terms, keep larger odd-odd nuclei from being stable. The spins are significant
in this case, because transitions with very large J are very forbidden.
64. With an even number of neutrons this isotope is apparently the most stable judging from
the semi-empirical formula. The alpha decay would be 165 Ho 161 Tb 4 He with
Q  164.930319  160.927566  4.002603 u  c2  0.14 MeV . With a very low Q value,
this process is allowed, but barely. With such a small Q value, it is difficult to overcome
the nuclear attraction.
65. (a) Radon is created at a rate 1 N1 and lost at a rate 2 N 2 , so the net change is
dN2 / dt  1 N1  2 N2 .
dN 2
(b) Using separation of variables
 dt. Let u  1 N1  2 N2 so du  2 dN2
1 N1  2 N 2
du
 2 dt and ln u  2t  ln(const.).
(because we assume N1 is constant). Then
u
u  (const.)e2t  1 N1  2 N2 . To determine the constant, set N 2  0 at t = 0, so the
N
constant is 1 N1. 1 N1e2t  1 N1  2 N2 or rearranging N 2  1 1 1  e 2t  .
2
(c) For large t we have e 2t  0 , and so N2  1 N1 / 2 (which is constant).
66. (a) From the uncertainty principle Et  / 2 , so
6.582 1016 eV  s
E 

 1.73 106 eV.
10
2t
2 1.9 10 s 
(b) We have 5E / E  v / c , which we can solve for v:
5 1.73 106 eV 
5E
v
c
 2.998 108 m/s   2.01 cm/s.
E
129 103 eV
67. (a) Examining the mass numbers it must come from four alpha decays of
238
U.
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Chapter 12
The Atomic Nucleus
185
(b) From the table of nuclides the decays are:
218
Po  214Pb + 4He ; 214 Pb 214 Bi +   ;
214
Po 
210
Rn  218Po  4He ;
214
Bi  214 Po    ;
222
Pb + 4He
(c) The only half-life longer than an hour is the first decay, with t1/2  3.8
days. Therefore more than half the decays will occur in four days.
68. (a) Both 4 He and 16 O are “doubly magic”; that is both Z and N are magic numbers.
These elements are more tightly bound than nuclei around them.
(b) Because 208 Pb has Z = 82 and N = 126 it is doubly magic. It is particularly stable and
its binding energy is high. Other nuclei around it are unstable because of the large
Coulomb energy.
(c) 40 Ca is doubly magic and is the most stable of the calcium isotopes. It is the heaviest
nuclide with Z = N. 48 Ca is also stable and has Z = 20 and N = 28 which are both magic
numbers. Six isotopes of calcium are stable with Z = 20 and various values of N.
69.
4
2
He ;
16
8
O;
40
20
Ca ;
48
20
Ca ;
208
82
Pb
70. (a) We use the atomic masses in Appendix 8 to find Q and then Equation (12.32) to find
Kα for each isotope. For the decay of 218Ra, the reaction is 218Ra → 214Rn + 4He and
Q   M  218 Ra   M  214 Rn   M  4 He  c2
Q   218.007124 u  213.995346 u  4.002603 u  c2  0.009175 u  c2  8.55 MeV
 A4 
K  
 Q  8.39 MeV
 A 
Similarly we find K  7.46 MeV for 220Ra and K  6.56 MeV for 222Ra.
(b) Notice that the graph in Figure 12.12 is logarithmic, suggesting that we try an
exponential function of the form t1/ 2  Ae BK .
(The model of alpha decay in Section 6.7 also suggests an exponential function.) To fit
the straight line between the 218 and 222 isotopes as suggested, take the natural
logarithm:
ln t1/ 2  ln A  BK
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186
Chapter 12
The Atomic Nucleus
Thus we can find the slope B:
B
 ln t1/ 2
11.068  3.638

 8.04
K
8.39  6.56
where we used the half-lives given in Appendix 8. Now using the 222Ra data point in the
original equation we find A  3.06 1024. The parameters A and B complete the model.
(c) Using K  7.46 MeV for 220Ra, our function in part (b) estimates t1/2 = 27 ms. This is
within a factor of two of the actual value 17 ms. It is to be expected that the estimate will
be high, given the concave nature of the graph.
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Chapter 13
Nuclear Interactions and Applications
187
Chapter 13
1. (a)
4
2
He
(b) 11 H
(c)
18
9
F
76
34
(d)
Se
(e)
108
48
(b) 7 Li( p,  ) 4 He
(c) 15 N( , p)18 O
O(d , n)17 F
(e) 107 Ag(3 He,4 He) 106 Ag
(f) 162 Dy(3 He, p) 164Ho
2. (a)
16
(f) 23 He
Cd
(d)
76
Se(d , n)77 Br
6.022 1023 19 g
3. Probability is equal to nt , or nt 
3.2 cm   0.68 1024 cm2   0.10
3 
238 g cm
21
3
19
6
16
4. p  Ne , He  F , and Li  O
6.022 1023 10.5 g
 0.2 cm  17 1024 cm2   0.20
108 g
cm3
6. We can find the probability of landing in a solid angle  :
2

 6.022 1023  100 106 g 
d
27 cm 
3

nt

0.25

10
3

10
sr
  
  12 g  cm2 

d
sr 




5. The probability is nt : nt 
 3.76 1012


1
  6.24 1011 s 1
The rate of incident particles is  0.20 106 C/s  
 2 1.602 1019 C  


11 1
12
so the rate of detected particles is  6.24 10 s  3.76 10   2.35 s1 and the number
detected in one hour is 2.35 s1  3600 s   8460 .
O(n,  ) 13C
(d) 16 O( , p) 19 F
7. (a)
16
O(d , n) 17 F
(e) 16 O(d ,3 He) 15 N
(b)
16
(c)
16
(f)
16
O( , p) 15 N
O(7 Li, p) 22 Ne
All the products listed above are stable except the one in part (b).
8. (a) Q   M (16 O)  M (2 H)  M (4 He)  M (14 N)  u  c 2  3.11 MeV (exothermic)
(b) Q   M (12 C)  M (12 C)  M (2 H)  M (22 Na)  u  c 2  7.95 MeV (endothermic)
(c) Q   M (23 Na)  M (1 H)  M (12 C)  M (12 C)  u  c2  2.24 MeV (endothermic)
 24 
9. (b) K th   7.95 MeV     15.9 MeV
 12 
 24 
(c) K th   2.24 MeV     2.34 MeV
 23 
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188
Chapter 13
Nuclear Interactions and Applications
10. In the center of mass system we require
9.63 MeV  M  16 O  c2  9.63 MeV  14899.10 MeV  14908.73 MeV
We also have
M  4 He  c2  M  12 C c 2  3728.38 MeV  11177.93 MeV  14906.31 MeV
Therefore we need 14908.73 MeV 14906.31 MeV  2.42 MeV in bombarding energy.
16
In the lab frame
 2.42 MeV   3.23 MeV. The lifetime of the state can be
12
computed using the uncertainty principle:   / 2 so
6.582 1016 eV  s


 6.45 1022 s.
3
2
2  510 10 eV 
11. (a) Q  K y  KY  K x  1.1 MeV  8.4 MeV  5.5 MeV  4.0 MeV
(b) The Q value does not change for a particular reaction.
12. Q   M (20 Ne)  M (4 He)  M (12 C)  M (12 C)  u  c2  4.62 MeV
24
 4.62 MeV   5.54 MeV
20
The sum of the carbon kinetic energies is
K y  KY  Q  K x  4.62 MeV  45 MeV  40.4 MeV.
K th 
13. Letting M  mb  mB ,   1  v 2 / c 2 
1/2
2
, and    1  vcm
/ c2 
1/ 2
, we have from
conservation of energy and conservation of momentum:
mAc 2   ma c 2   Mc 2
 ma v   Mvcm
2
Squaring the second expression and dividing by c , we have:
 2 ma2 1   2    2 M 2 1   2  ;  2 ma2  ma2   2 M 2  M 2
 2 ma2v 2
c2

2
 2 M 2vcm
c2
From the energy equation  M  mA   ma , so  2 ma2  ma2   mA   ma   M 2 .
2
After multiplying the binomial and rearranging we have 2 ma mA  M 2  ma2  mA2 .
2   1 ma mA  M 2   ma2  2mAma  mA2   M 2   ma  mA 
K th     1 ma c
2
M

2

  ma  mA  c 2
2
2mA
2
 M   ma  mA    M   ma  mA    c 2

2mA
mA  ma  mB  mb
2mA
14. The energy available for the gamma ray is
Q   M (10 B)  M (n)  M (11 B)  u  c 2  11.5 MeV.
 Q
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Chapter 13
Nuclear Interactions and Applications
189
15. The kinetic energy is Q  5.0 MeV , or
K   M (9 Be)  M (4 He)  M (n)  M (12 C)  u  c 2  4.61 MeV  5.70 MeV  4.61 MeV
 10.3 MeV
16. (a) Q   M (16 O)  M (4 He)  M (1 H)  M (19 F)  u  c 2  8.11 MeV
20
8.11 MeV   10.14 MeV
16
(b) Q   M (12 C)  M (2 H)  M (3 He)  M (11 B)  u  c 2  10.46 MeV
K th 
14
10.46 MeV   12.21 MeV
12
6.022 1023
17. (a) N 
 40 103 g   4.0831020
59 g
K th 
Activation rate  n   4.083 1020  20 1028 m2 1.0 1018 m2  s 1   8.166 1011 s1
Then in one week the number of 60 Co produced is
8.166 1011 s1  7 86400 s    4.94 1017.
(b) Because the half-life of 60 Co is much longer than one week, we can assume almost
all the nuclei produced are still present. The activity is
ln 2
ln 2
R  N 
N
4.94 1017   2.06 109 Bq

8
t1/2
1.66 10 s
40 103 g
 1942 g
2.06 109 Bq
We would put 1.94 kg of 59 Co into the reactor for one week.
(c) 1.0 1014 Bq
18. The equation for K cm is correct because we know from classical mechanics that the
system is equivalent to a mass M x  M X moving with a speed vcm , so the center
1
2
of mass kinetic energy is Kcm   M x  M X  vcm
. Letting M  M x  M X we have by
2
conservation of momentum Mvcm  M x vx , or vcm  M x vx / M . Therefore
1
1 M2
1 M x2 2
2
Mvcm
 M x2 vx2 
vx
2
2 M
2 M
M v 2 1 M x2 2 M x vx2  M x  M x vx2  M  M x 
  Klab  K cm  x x 
Kcm
vx 
1 



2
2 M
2 
M 
2  M 
 MX

  K lab 
K cm

 Mx  MX 
Kcm 
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190
Chapter 13
 
19. Kcm
Nuclear Interactions and Applications
MX
14
Klab   6.7 MeV   5.21 MeV
Mx  MX
18
E*   M (14 N)  M (4 He)  M (18 F)  u  c2  4.41 MeV
  9.62 MeV
Ex  E*  Kcm
20. A small amount of the neutron’s energy must be used in recoil, so the amount of energy
that can go into excitation is
208
14 MeV 
 13.93 MeV
209
The net excitation energy is this plus the available mass-energy, so
E*  13.93 MeV +  M (208 Pb)  M (n)  M ( 209 Pb)  u  c 2  13.93 MeV +3.94 MeV
or E*  17.9 MeV. Because of the excited nucleus, we expect gamma decay.
21. From Chapter 12 the radius of the nucleus is r  14 N   1.2 fm 141/ 3  2.89 fm.
At that radius the potential energy is V 
 2  7  1.44 109 eV  m 
q1q2

 6.98 MeV.
4 0 r
2.89 1015 m
Q   M (14 N)  M (4 He)  M (1 H)  M (17 O)  u  c2  1.19 MeV
Kexit  Q  K  1.19 MeV  7.7 MeV  6.51 MeV  K ( p)  K (17 O).
Most of the energy at the forward angle will be in the proton, so there could be sufficient
energy to overcome the 6.98 MeV barrier in a tunneling process (though it appears to be
forbidden classically).
22. From Chapter 12 Problem 29 we see that the mean lifetime is
t
109 ms
  1/2 
 157 ms
ln 2
ln 2
6.582 1016 eV  s

 2.10 1015 eV.
Then from Equation (13.13)  
2
2  0.157 s 
17
Ne can decay by positron decay or electron capture.
23. E*   M (1 H)  M (16 O)  M (17 F)  u  c 2  0.60 MeV. No, this energy is greater than
0.495 MeV.
24. (a) E*   M (239 Pu)  M (n)  M (240 Pu)  u  c 2  6.53 MeV
239
1.00 MeV   1.00 MeV ; E  6.53 MeV  1.00 MeV  7.53 MeV
240
d  22 Ne  22 Ne  d
d  23 Na  22 Ne  3He
25. d  21Ne  22 Ne  p
(b) Kcm 
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Chapter 13
Nuclear Interactions and Applications
  18O  22 Ne  
  22 Ne  22 Ne + 4He
191
  19 F 
22
d  25Mg 22 Ne 5 Li
d  26 Mg 
d  27 Al  22 Ne  7 Be (and so on)


Ne  p
22
  21 Ne 
22
Ne  3He
Ne  6 Li
  24 Mg 22 Ne + 6Be
23
Na 
22
Ne  5Li
27
Al 
22
Ne 9 B (and so on)
26. E   M (239 Pu)  M (n)  M (95 Zr)  M (142 Xe)  3M (n)  u  c 2  183.6 MeV
27. (a) m   5.5 104 106 kg   550 kg
(b)
 550 kg 
6.022 1023
 1.4 1027 atoms
0.238 kg
(c) R   6.7 kg 1  s1   550 kg   3700 Bq
s
 3.2 10
3700 s  86400
d
N  U  7.2 N  U  e


N  U  993 N  U  e
1
(d)
8
235
28.
235
 1t
238
 2t
d 1
0
238
0
where the subscripts 1 and 2 refer to the 235 and 238 isotopes, respectively.
N 0  235 U  7.2

exp   1  2  t 
N 0  238 U  993


1
1

t
 t1/2 (235) t1/2 (238) 
 1  2  t  ln 2 


1
1
9
 ln 2 

  2.0 10 y   1.659
8
9
 7.04 10 y 4.47 10 y 
N0  235 U  7.2
7.2

exp   1  2  t  
exp 1.659   0.0381
Then
238
993
N0  U  993
which is more than five times higher than today. Note that with the ratio of abundances
equal to 0.0381, the percentage abundances needed to give that ratio are about 3.7% and
96.3%. Natural fission reactors cannot operate because of the relatively low abundance of
235
U today. The exception is a type of reactor known as the Canadian Deuterium Reactor
(CANDU) which uses heavy water rather than regular or light water as the moderator.
29. Here are three possibilities, out of many:
236
U 
95
Y 
138
I  3n ;
236
U 
94
Y 
140
I  2n ;
236
U 
97
Y 
136
I  3n
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192
Chapter 13
Nuclear Interactions and Applications
Note that the fragments (two nuclei plus some free neutrons) have to add up to mass
number A = 236 and atomic number Z = 92.
30. 1250 MWe  1.25 109 J/s 
86400 s/d
 6.74 1026 MeV/d
13
1.602 10 J/MeV
1
(fission)
0.235 kg
 1.32 kg/d
 6.74 1026 MeV/d   200 MeV  6.022
1023
31. For uranium we assume as in the preceding problem that each fission produces 200 MeV
of energy.
 6.022 1023 atoms  200 MeV
 5.13 1026 MeV
1.0 kg  

0.235
kg
atom


Converting to kWh, we find 2.30 107 kWh . For 1.0 kg of coal, Table 13.1 gives energy
output 3 × 107 J, which is equivalent to 8.33 kWh. Therefore we see that fission produces
over one million times more energy per kilogram of fuel.
32. The USA consumes about 1 × 1020 J per year, according to most sources. With a
population of 310 million, that’s 3.3 × 1011 J per year per person, or roughly 9 × 108 J per
day per person. The Rules of Golf specify the golf ball’s maximum mass of 45.93 g
(0.04593 kg). Multiplying this by c2 gives 4.13 × 1015 J of energy. If each person
consumes 9 × 108 J, that’s enough energy for 4.6 million people, a large city. The
estimate is reasonable.
6 109 J
 7.5 1021 J
33. (a) 2.0 1011 m3 
0.16 m3
1y
(b) 7.5 1021 J 
 15 y
5 1020 J
At this rate, it would not last long! Fortunately, other sources of energy are available,
although oil is still the primary source of energy for transportation.
(c) According to Table 13.1, 1 kg of uranium fuel produces 1014 J of energy, so the
amount of uranium required is
1 kg
7.5 1021 J  14  7.5 107 kg
10 J
3
3
34. (a) kT  8.617 105 eV/K   300 K   3.88 102 eV
2
2
3
3
kT  8.617 105 eV/K 15 106 K   1.94 keV
2
2
1
12
13
35. H  C  N    13 C     ;
(b)
Q   M (1 H)  M (12 C)  M (13 C)  u  c2  4.16 MeV
1
H  13C  14 N   ;
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Chapter 13
Nuclear Interactions and Applications
193
Q   M (1 H)  M (13 C)  M (14 N)  u  c2  7.55 MeV
1
H  14 N  15 O    15 N     ;
Q   M (1 H)  M (14 N)  M (15 N)  u  c2  10.05 MeV
1
H  15 N  12 C  4 He ;
Q   M (1 H)  M (15 N)  M (12 C)  M (4 He)  u  c 2  4.97 MeV
The total Q is Q  4.16 MeV  7.55 MeV  10.05 MeV  4.97 MeV  26.73 MeV.
36. (a) The temperature must be very high to overcome the Coulomb barrier so the two alpha
particles can interact through the nuclear force. A high density is necessary so that the
probability is high enough for a third alpha particle to interact with 8 Be before it decays.
(b) We use Equation (13.7) to find the Q value for the reaction:
Q  3M (4 He)  M (12 C)  u  c2  3  4.002603  12.0 u  c2  7.27 MeV.
37. (a) The temperature must be very high to overcome the Coulomb barrier so that the two
oxygen nuclei can interact through the nuclear force. As the value of Z increases, higher
and higher temperatures are required.
(b) We use Equation (13.7) to find the Q value for the reaction:
Q  2M (16 O)  M (32 S)  u  c2  2 15.994915  31.972071 u  c2  16.5MeV.
This includes the energy of the  ray in the released energy.
38. (a) This problem is similar to Example 13.9. First we will determine the Coulomb
potential energy that must be overcome, using Equation (12.2) to determine the radius.
r  r0 A1/3  1.2 1015 m  28  3.6 1015 m
Therefore the Coulomb barrier is:
1/3
9 109 N  m2 / C2  14  1.6 1019 C 

q1q2
V

 1.3 1011 J
15
4 0 r
3.6 10 m
We need at least this much kinetic energy to overcome the Coulomb barrier. We set this
3
value equal to the thermal energy, namely kT .
2
11
2 1.3 10 J 
2V
T

 6.3 1011 K
23
3k 3 1.38 10 J/K 
2
2
(b) We use Equation (13.7) to find the Q value for the reaction:
Q  2M (28 Si)  M (56 Ni)  u  c 2  2  27.976927   55.942136 u  c 2  10.9 MeV
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194
Chapter 13
Nuclear Interactions and Applications
This includes the energy of the  ray in the released energy.
39. (a) This problem is similar to Example 13.9. First we will determine the Coulomb
potential energy that must be overcome, using Equation (12.2) to determine the radius.
r  r0 A1/3  1.2 1015 m 12 
1/3
 2.7 1015 m
Therefore the Coulomb barrier is:
9 109 N  m2 / C2   6  1.6 1019 C 

q1q2
V

 3.11012 J
15
4 0 r
2.7 10 m
We need at least this much kinetic energy to overcome the Coulomb barrier. We set this
3
value equal to the thermal energy, namely kT .
2
12
2  3.110 J 
2V
T

 1.5 1011 K
23
3k 3 1.38 10 J/K 
2
2
(b) We use Equation (13.7) to find the Q value for the reaction:
Q  2M (12 C)  M (24 Mg)  u  c2  2 12.0   23.985042 u  c2  13.9 MeV
This includes the energy of the  ray in the released energy.
40. If the depth is d and the radius of the earth is r and 2/3 of the surface is covered with
water, then the volume of the oceans is
2
2
8
V   4 r 2  d    6.37 106 m   3000 m   1.02 1018 m3
3
3
With a density of 1000 kg / m3 , the mass is 1.02 1021 kg , so the number of water
6.022 1023
 3.411046.
0.018 kg
With a 0.015% abundance, the number of deuterium atoms (in D2O molecules) is
molecules is N  1.02 1021 kg 
 2 3.411046  0.00015  1.02 1043. There will be at most half this number of fusions,
or 5.11042. The energy released is
5.11042   4.0 MeV  1.602 1013 J/MeV   3.27 1030 J.
41. (a) We use temperatures because that is what we strive for experimentally. Kinetic energy
is useful for comparison with the Coulomb barrier or Q values.
3
kT
2
2  6000 eV 
2K
(c) T 

 4.64 107 K
5
3k 3 8.617 10 eV/K 
(b) K 
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Chapter 13
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195
42. 4 He  3He 7 Be   ; Q   M (4 He)  M (3 He)  M (7 Be) u  c2  1.59 MeV
2
H  2 H  3 H  1H ; Q   M (2 H)  M (2 H)  M (3 H)  M (1 H)  u  c2  4.03 MeV
1
H  2 H  3 He   ; Q   M (1 H)  M (2 H)  M (3 He)  u  c2  5.49 MeV
1
H  12C  13 N   ; Q   M (1 H)  M (12 C)  M (13 N)  u  c2  1.94 MeV
3
He  3 He  4 He  2 p ;
Q   M (3 He)  M (3 He)  M (4 He)  2M (1 H)  u  c 2  12.9 MeV
7
Li  1H 4 He  4 He ; Q   M (1 H)  M (7 Li)  2M (4 He)  u  c 2  17.3 MeV
3
H  2 H  4 He  n ; Q   M (3 H)  M (2 H)  M (4 He)  M (n) u  c2 =17.6 MeV
3
He  2 H  4 He  1 H ; Q   M (3 He)  M (2 H)  M (4 He)  M (1 H)  u  c 2  18.4 MeV
43. The reaction is 1 H  12 C  13 N   .
Q   M (12 C)  M (1 H)  M (13 N)  u  c2  1.94 MeV
 M (1 H)  M (12 C) 
Then K th  Q 

M (12 C)


which will be negative, so the threshold energy is not a concern.
However, the Coulomb barrier is
6 1.44 109 eV  m 
q1q2
6e2
V


 2.19 MeV.
4 0 r 4 0  rp  r0 A1/ 3  1.2 1015 m  1.2 1015 m  12 1/ 3

Setting this coulomb barrier energy equal to the thermal energy
2  2.19 106 eV 
2K
T

 1.69 1010 K.
3k 3  8.617 105 eV/K 

3
kT , we have
2
44. Q  3M (4 He)  M (12 C)  u  c 2  7.27 MeV
3
3
45. (a) K  kT  8.617 105 eV/K   300 K   3.88 102 eV
2
2
2  3.88 102 eV 1.602 1019 J/eV 
2K

 2720 m/s
(b) v 
m
1.675 1027 kg
(c)  
h
6.626 1034 J  s

 1.45 1010 m
27
p 1.675 10 kg   2720 m/s 
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196
Chapter 13
Nuclear Interactions and Applications
512
Bq  8.533 Bq. The initial rate was R0  4 105 Bq . Over 60 days the activity
60
 ln 2
has decreased to R0  R0e t   4 105 Bq  exp  
 60 d    1.571105 Bq.
 44.5 d

Therefore the fraction f of the ring that has worn off into the oil is
R
8.533 Bq
f 

 5.43 105
5
R0 1.57110 Bq
46. R 
47. (a) With a half-life of 6.01 h, not much of the original material remains after one week.
 ln 2

(b) R  R0e t  1011 Bq  exp  
 216 h    1.46 Bq
 6h

 ln 2

(c) R  R0e t   0.9 1011 Bq  exp  
 96 h    1.37 106 Bq
 6h

48. (a) We need 1000 Bq at t = 30 minutes with t1/2  83.1 minutes.
ln 2

R  R0e t   R0  exp  
 30 min    1000 Bq
 83.1 min

 ln 2
R0  1000 Bq  exp 
 30 min    1284 Bq
 83.1 min

Because only 27% go to the first excited state, we need activity R0 
R0
 4756 Bq.
0.27
ln 2
, so solving for N:
t1/2
Then R0   N  N
 4756 s
Rt
N  0 1/2 
ln 2
1
 83.1 min  60 s/min   3.42 10
7
ln 2
139 g
 7.89 1015 g. The fraction f activated is
(b) m   3.42 107 
23
6.022 10
15
7.89 10 g
f 
 1.10 106.
8
 0.717  10 g 
49. (a) I   0.15  5.0 105 s1  2 1.602 1019 C   2.40 1014 A
(b) One-tenth of the above value is 2.40 1015 A .
1 MeV

 1 particle 
15
50. (a) R   5000 J/s  

  5.89 10 Bq
13
1.602

10
J
5.3
MeV



N
R

5.89 10

15
s 1  138 d 86400 s/d 
ln 2
 1.013 1023
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Chapter 13
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197
 0.210 kg 
m  1.013 1023  
 3.53 102 kg
23 
 6.022 10 
(b) For 7 kW we need a mass of (7 / 5)  3.53 102 kg   0.0494 kg. Letting the initial
mass be m0 , we have the usual exponential decay and so m  m0 exp(t ) , or
  ln 2  730 d  
m0  m exp(t )   0.0494 kg  exp 
  1.93 kg.
138 d


51. The fraction absorbed is the patient’s cross-sectional area divided by 4  distance  .
2
Note that there are two gammas per decay: R  2  3 1014 Bq 
0.3 m2
4  4 m 
2
 8.95 1011 Bq
52. From Chapter 12 the diameter of the uranium nucleus is
d  2r  238 U   2 1.2 fm  2381/ 3  14.87 fm. The momentum of the neutron is
p
h


6.626 1034 J  s
 4.456 1020 kg  m/s and the kinetic energy is
14.87 1015 m
20
p 2  4.456 10 kg  m/s 
K

 5.927 1013 J  3.70 MeV.
27
2m
2 1.675 10 kg 
2
This is still much less than the neutron's rest energy, which justifies the non-relativistic
2  5.927 1013 J 
2K

 2.86 1010 K.
treatment. Then in thermal equilibrium T 
23
3k 3 1.38110 J/K 
Clearly this is not a realistic temperature in a reactor, so the proposed experiment is not
possible.
53. The alpha decay reaction is
239
Pu  235 U  4 He , and the energy released in each decay is
Q   M (239 Pu)  M (235 U)  M (4 He)  u  c2  5.25 MeV.
The activity of a 10 kg sample is
 6.022 1023 
ln 2
13
R  N 
10
kg



  2.30 10 Bq. Then
7
0.239
kg
24110
y
3.156

10
s/y





the power is P   0.6 5.25 MeV  1.602 1013 J/MeV  2.30 1013 s1   11.6 W .
54. Perhaps the best source of information for this question is the Internet site for the
International Union of Pure and Applied Chemistry (IUPAC) at www.iupac.org. As of
early summer 2011, elements with Z = 113, 114, 115, 116, and 118 had been observed
experimentally but not officially named by IUPAC. The current names for 114 and 116
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198
Chapter 13
Nuclear Interactions and Applications
are ununquadium and ununhexium, for example. Element 112 was recently named
Copernicum.
55. An Internet search will yield current information. In addition the April 2000 issue of
Physics Today magazine describes the relevance of radioisotopes for this procedure.
There is still considerable debate about the efficacy of beta as compared to gamma
emitters. Although the first study by Bottcher (1994) et al. used a gamma emitter, 192 Ir ,
recent studies have included beta emitters such as
32
P, 90 Sr ,
90
Y,
188
Re ,
99
Tc , and
137
Xe . Almost all of the trials have shown low-dose radiation to be effective in
preventing keloids and benign vascular malformations.
56. (a) The weighted average is 580 barns × 0.0072 + 0 barns × 0.993 = 4.2 barns.
(b) At the higher concentration, the weighted average becomes 580 barns × 0.020 + 0
barns × 0.98 = 11.6 barns.
101
132
57. The reaction is n  235
92 U  42 Mo  50 Sn  3n . Looking up the masses,
mass  n  235
92 U   1.008665 u  235.043924 u  236.052589 u
132
mass  101
42 Mo  50 Sn  3n   100.910347 u  131.917744 u  3 1.008665 u 
 235.854086 u
The mass difference is Δm = 0.198503 u.
 931.49 Mev/c 2  2
Energy released =  m  c  0. 198503 u 
 c  185 MeV.
u


2
58. (a) 1 kg 
6.022 1023 1mol 13.6 eV
 8.19 1027 eV
mol
0.001 kg 1 atom
(b) 1 kg 
6.022 1023 1 mol 2.224 106 eV
 6.70 1032 eV
mol
0.002 kg
1 atom
(c) 1 kg 
1u
931.49 106 MeV
 5.611035 eV
27
1.66 10 kg
u
(d) The results are as expected. The annihilation process in part (c) is the complete
conversion of matter into energy, which maximizes energy yield. The energy in (b) is
much greater than the energy in (a), because nuclear binding energies are so much larger
than atomic binding energies.
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Chapter 13
Nuclear Interactions and Applications
199
59. (a) Q  Kout  Kin  86.63 MeV  100 MeV  13.37 MeV
This agrees very nearly with the Q computed using atomic masses (assuming all the
masses are known).
Q   M (18 O)  M (30 Si)  M (14 O)  M (34 Si)  u  c2  13.27 MeV
Q
 M (18 O)  M (30 Si)  M (14 O)  33.979 u
2
c
which agrees very nearly with the value for the mass of 34 Si in Appendix 8.
1 mol 6.022 1023
60. (a) The number of atoms is 1000 100 kg 
 0.04  1.03 1028
0.235 kg
mol
(b) M  
(b)
1.03 1028
4  6 10 m 
6
2
 2.28 1013 m2
Then the activity for each square meter is
ln 2
R  N 
2.28 1013   1.74 104 Bq.

7
28.8 y  3.156 10 s/y 
61. (a) Starting with F ( E )  (const.)e E / kT E1/2 and setting dF / dE  0 , we find that the most
probable energy is E * 
1
1
1
kT . E*  kT  8.621105 eV/K  2 108 K   8620 eV
2
2
2
n(2 E )
 1
 exp     2 E  E    exp    E   exp     0.607
n( E )
 2
n(5E )
 exp     5E  E    exp  4 E   exp  2   0.135
n( E )
n(10 E )
 9
 exp    10 E  E    exp  9 E   exp     0.0111 .
n( E )
 2
(b)
 1 mol 
62. (a) P   nt   90 1027 cm2 1.85g/cm3   3.2 cm   6.022 1023 mol -1  
  0.0357
 9g 
(b) With this probability the number of incident alpha particles is
1.6 105 s 1
 4.48 106 s 1.
0.0357
 2.41104 y  3.156 107 s 
R
18
(c) N    4.48 106 s 1  

  4.92 10 and

ln 2
y



m   4.92 1018 
0.239 kg
 1.95 106 kg, which is reasonable.
23
6.022 10
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200
Chapter 13
Nuclear Interactions and Applications
63. m(40 K)   65 kg  0.0035 0.00012   2.73 105 kg
N   2.73 105 kg 
R  N 
6.022 1023
 4.111020
0.040 kg
ln 2
1y
4.111020   7050 Bq

9
7
1.28 10 y 3.156 10 s
The beta activity is  7050 Bq  0.893  6300 Bq.
64. We can determine the minimum mass as follows:
 1 MeV 
E  15 kilotons   4.2 1012 J/kiloton   6.3 1013 J 
 3.94 10 26 MeV.
13 
 1.6 10 J 
If each fission yields 200 MeV, then we have:
3.94 1026 MeV
Number of fissions =
= 1.97 1024 fissions which is the minimum
200 MeV/fission
number of fissions. This requires a mass of:
27
 235 u   1.66 10 kg 
M  1.97 1024  
  0.77 kg.

u
 atom  

The actual mass was larger since not every nucleus fissioned.
65. Using the mass of the 238 isotope, the rate of spontaneous fission per atom is
6.7
decays 238 u 1.66 1027 kg
decays


 2.65 1024
kg  s atom
1u
atom  s
The overall decay rate for the 238 isotope is computed using the half-life of
4.47  109 y :

R ln 2
ln 2
decays


 4.92 1018
9
7
N t1/2  4.47 10 y  3.15 10 s/y 
atom  s
Therefore the probability of spontaneous fission is
2.65 1024
4.92 1018
decays
atom  s  5.4 107. This result agrees with published values.
decays
atom  s
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Chapter 14
Particle Physics
201
Chapter 14
1. By conservation of momentum the photons must have the same energy. For each one
mc 2
938.27 106 eV

 2.27 1023 Hz.
E  hf  mc 2 and f 
15
h
4.136 10 eV  s
2. As in the text (see Example 14.1), let mc2  c / R , so
c 197.3 eV  nm
R 2 
 1.41106 nm  1.41 fm.
6
mc
140 10 eV
3. We will use a small nucleus (helium, diameter 3.8 fm) and a large nucleus (uranium,
diameter 14.9 fm) to obtain a range of values.
3.8 1015 m
For helium: t 
 1.27 1023 s
8
2.998 10 m/s
14.9 1015 m
For uranium: t 
 4.97 1023 s
2.998 108 m/s
4. We know that   D and the problem says specifically to choose   0.10D with the
diameter D  0.15 fm . Therefore   0.1(1.5 fm)  0.15 fm . We know from de Broglie's
relationship that p  h /  and we can determine the kinetic energy from this momentum
as follows: E 2   K  mc 2    pc    mc 2 
2
K
 pc 
2
  mc

2 2
2
2
2
 hc 
2 2
2
 mc  
   mc   mc
 0.15fm 
2
Electron:
2


2
1239.8 eV  nm
   0.511106 eV   0.511106 eV  8.26 GeV
K 

6
 0.15 fm 10 nm/1 fm  


Proton:
2


2
1239.8 eV  nm
   938.27 106 eV   938.27 106 eV  7.38 GeV
K 

6
 0.15 fm 10 nm/1 fm  


5. As in the preceding problem, we know that p  h /  and
K
 pc 
2
  mc

2 2
2


hc
2 2
2
 mc  
   mc   mc
18
1

10
m


2
2
2
 1239.8 eV  nm 
6
6
Electron: K  
   0.51110 eV   0.51110 eV  1.24 TeV
9
1

10
nm


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202
Chapter 14
Elementary Particles
2
2
 1239.8 eV  nm 
6
6
Proton: K  
   938.27 10 eV   938.27 10 eV  1.24 TeV
9
 110 nm 
The kinetic energies are so large that the rest energies do not affect our answer.
6. We use Equation (14.4) to estimate the range of the force.
c
197.33 eV  nm
R

 1.26 107 nm  1.26 1016 m
2
6
2mc
2  782 10 eV 
7. We use Equation (14.4) to estimate the range of the force.
c
197.33 eV  nm
R

 6.58 1010 nm  6.58 1019 m
2
9
2mc
2 150 10 eV 
8. (a)   and  e
(b)  e
(c)  
(d)  
(e)  
9. Assuming roughly an equal number of protons and neutrons, we can use the average of
their masses per baryon, or 1.674 1027 kg. The mass of Earth is 5.98 1024 kg, so the
5.98 1024 kg
baryon number is
 3.57 1051 .
27
1.674 10 kg
10. (a) We use Equation (14.6) to determine the width  from the mean lifetime.
6.58211016 eV  s
 
 8.02 106 eV

0.8211010 s
(b) This value of  is much smaller than the experimental uncertainty of the
measurement.
11. We use Equation (14.6) to determine the width  from the mean lifetime.
6.58211016 eV  s
 9.3 104 eV ;
J/Psi:   

7.11021 s
Upsilon:  


6.58211016 eV  s
 5.5 104 eV
1.2 1020 s
The full-width of the charged pion is  


6.58211016 eV  s
 2.5 108 eV . This
2.6 108 s
value is more than 1012 times smaller than either of the previous two.
12. In the first reaction strangeness is violated ( 1  1 ). The second reaction is allowed.
0
The fact that the K 0 has strangeness +1 and K has strangeness 1 is consistent with the
idea that it is not its own antiparticle. However, M. Gell-Mann and A. Pais found that
0
0
K0        K (and the reverse operation K        K0 ) can occur by the
laws of quantum mechanics (see Feynman Lectures in Physics vol. 3 p. 11-16). So in
this case the K 0 does act as its own antiparticle.
13. In both (a) and (b) the baryon number is not conserved.
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Chapter 14
Particle Physics
203
14. (a)  and e lepton numbers are not conserved
(b) charge is not conserved
(c) momentum-energy is not conserved
15. Let subscript 1 refer to the  , subscript 2 to the  , and no subscript to the photon. From
conservation of momentum ∣ p∣∣ p2 ∣ E / c . From conservation of energy
p22c 2   m2c 2   E  E 2   m2c 2   E , so
2
m1c 2 
2
 m c    m c   1193 MeV   1116 MeV 
E
2 2
1
2 2
2
2m1c 2
2
2 1193 MeV 
2
 74.5 MeV.
16. The  0 begins with energy E  K  E0  855 MeV. Because the photon energies are
equal, we know by conservation of momentum that the two photons make equal angles 
above and below the original line of motion for the  0 . Let each photon energy be E  .
E 855
MeV  427.5 MeV.
The conservation of energy and momentum give: E  
2
2
855 MeV   135 MeV 
E 2  E02
2E
p
cos  

 844.3 MeV/c.
c
c
c
pc
844.3 MeV
cos  

 0.9875 from which we find   9.1.
2 E 2  427.5 MeV 
2
2
17.
(a)   is needed to conserve lepton number.
(b) A kaon is needed to conserve strangeness, and to conserve charge too the kaon must
be a K  .
2e e e
1
S  3  0  0
q     0;
18. n (udd):
B  3   1;
3 3 3
3
2e 2e e
1
q     e ; B  3   1;
S  0  0 1  1
  (uus):
3 3 3
3
2e e 2e
1
q     e ; B  3   1;
 C (udc):
S  000  0
3 3 3
3
2e e
1 1
q    e;
B    0;
19.   ( ud ):
S  00  0
3 3
3 3
2e e
1 1
q    e;
B    0;
S  0 1  1
K  ( us ):
3 3
3 3
2e 2e
1 1
q    e;
B    0;
S  00  0
D0 ( cu ):
3 3
3 3
D : cd
D : cd
20. D0 : cu
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204
Chapter 14
Elementary Particles
21. The B and B have zero strangeness and zero charm, but they have masses greater than
5000 MeV / c2 and charge +1 and 1 respectively. This leads one to conclude that the
quark configuration should be bu for the B and bu for the B . The B0 has zero charge,
so it should be either bd or bd . The only way to distinguish between these two
possibilities is by looking at the conservation laws in the appropriate decay reactions. it
0
turns out that the B0 is bd and the antiparticle B is bd .
22. Charge, baryon number, strangeness, topness, and bottomness are zero. Charm  1 .
0
From table 14.6 D0 in the text has configuration cu , so cu must be D .
23.
(a) The   ,  0 , and the K  have mean lifetimes on the order of 1010 s, which indicates
the decays are due to the weak interaction. Remember that there can be quark
transformations in weak interactions.
(b)   0  K  , sss  uds  us ; an s quark transformed into a d quark and a uu
quark-antiquark pair was created. 0  p    , uds  uud  ud ; the s quark
transformed into a d quark and a uu quark-antiquark pair was created. K       ,
us  no quarks; the s quark transformed into a u quark and the uu quark-antiquark
pair annihilates. There are no quarks in the final decay products.
24. We use Table 14.5 for quark properties and Table 14.4 or Table 14.6 to identify the
hadrons. The spin is 1/2 for all quarks and antiquarks.
(a) cd ; spin is 0 or 1; charge is 1; baryon number is 0; C = 1; S, B, T = 0. This is a D
meson.
(b) uds; spin is 1/2 or 3/2; charge is 0; baryon number is 1; S  1 ; C, B, T  0 . This
could be a  0 or a  baryon.
(c) sss ; spin is 1/2 or 3/2; baryon number is 1 ; charge is 1; S = 3; C, B, T  0 . This is a
 baryon.
(d) cd ; spin is 0 or 1; charge is 1 ; baryon number is 0; C  1 ; S , B, T  0 . This is a
D meson.
25. Assuming an average nucleon mass of 1.674 1027 kg and noting that 10 out of 18
nucleons in water are protons, we have
3
3
kg
1
10
 1.16 1032 protons
3.5 105 L  10 Lm 1000
3
27
m
1.673 10 kg 18
Then by the laws of radioactive decay
ln 2 decays
1.16 1032 protons
 0.080 decays/y
1033 y
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Chapter 14
Particle Physics
205
26. Baryon number is not conserved in any of the three. This is a common problem in proton
decay schemes, because there are no lighter baryons. In addition, (a) violates electron
lepton number, (b) violates muon lepton number, and (c) violates strangeness.
27.
31 kg
ln 2 decays 1 y
 3.5 108 decays/d
27
1.673 10 kg 1033 y 365.25 d
28. The lifetime of the   is 8.0 1011 s. Due to relativity it travels farther than one might
expect.
K  E0 3.6 GeV  1.189 GeV
1


 4.028 
E0
1.189 GeV
1  v2 / c2
v  c 1  1/  2  c 1  1/ (4.028) 2  0.969 c
d  vt    vt   4.028 0.969  2.998 108 m/s 8.0 1011 s   9.4cm.
29. We begin with Equation (14.10). Since K
mc 2 , this simplifies to Ecm  2mc 2 Klab .
From the problem statement, we know that Ecm  2K so Ecm  2K  2mc 2 Klab . This can
be rearranged to find 4K 2  2mc 2 Klab or K lab 
2K 2
.
mc 2
2 K 2 2  31 GeV 


 2050 GeV.
mc 2
0.938 GeV
2
30. From the preceding problem, we know that Klab
31.  
K  E0 7000 GeV  0.938 GeV
1
;

 7464 
E0
0.938 GeV
1  v2 / c2
v  c 1  1/  2  c 1  1/ 74642  0.999999991c
32. The maximum energies result from a head-on collision. Before the collision the proton
has energy E  K  E0  1008.27 MeV. After the collision the proton has energy and
momentum E1 , p1 and the recoiling particle has E2 , p2 . From conservation of momentum
and energy
1008.27 MeV    938.27 MeV 
E 2  E02
p

 369.13 MeV/c  p1  p2 ; with
c
c
E  E0  d   1008.27 MeV  1875.61 MeV  2883.9 MeV  E1  E2 .
2
2
Using the energy-momentum invariant E02  E 2  p 2c 2 for both the proton and deuteron,
these four equations can be solved to yield p1  119 MeV / c , p2  488MeV / c ,
E1  946 MeV , E2  1938MeV , so the deuteron's kinetic energy is
K  1938 MeV 1876 MeV  62 MeV .
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206
Chapter 14
Elementary Particles
Similarly for the triton (t) with E0  2809 MeV , we can follow the same procedure:
1008.27 MeV    938.27 MeV 
E 2  E02
p

 369.13 MeV/c  p1  p2 ; with
c
c
E  E0  t   1008.27 MeV  2809 MeV  3817 MeV  E1  E2 .
2
2
Using the energy-momentum invariant E02  E 2  p 2c 2 for both the proton and triton,
these two equations can be solved to yield p1  177MeV / c , p2  546 MeV / c ,
E1  955 MeV , E2  2862 MeV , so the triton’s kinetic energy is
K  2862 MeV  2809 MeV  53 MeV .
Note: At this kinetic energy (70 MeV) you will get the same results to two significant
figures using a non-relativistic calculation.
33. To conserve baryon number we must produce both a proton and antiproton, so in the cm
system we need Ecm  4E0 . We can use Equation (14.10) and solve for K to find
K
2
Ecm
  m1c 2  m2c 2 
2m2c 2
2

2
Ecm
 4  mc 2 
2mc 2
 4mc 
we have K 
2 2
Ecm  4mc
2
where we have used m1  m2  m . With
 4  mc 2 
2mc
E 2  E02

34. (a) p 
c
2
2
2
 6mc 2  6  938.27 MeV   5630 MeV.
 948.27 MeV    938.27 MeV 
2
2
c
 137.35 MeV/c
J 

137.35 106 eV 1.602 1019

p
eV 

R

 0.327 m .
qB
 2.998 108 m/s   e 1.4 T 
(b) From Equation (14.9) we have f 

f 
E 948.27 MeV

 1.0107 so
E0 938.27 MeV
 e 1.4 T 
2  938.27 10 eV  1.0107 
6
35. With p   mv we have f 
f 
v
2 R

eB
eB
with
1  v2 / c2 
2 m
2 m
 2.998 10
8
m/s   2.11107 Hz .
2
p
. With p  qBR we find
2 mR
qBR
qB
qB


1  v2 / c2 .
2 mR 2 m 2 m
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Chapter 14
Particle Physics
207
36. Elab  K  m1c 2  m2c 2 ; plab c 
 K  m c   m c 
2 2
1
2 2
1
Using the fact that E 2  p 2c 2 is invariant, so
2
2
Ecm
 Elab
  plab c    K  m1c 2  m2c 2    K  m1c 2    m1c 2 
2
2
2
2
 K 2   m1c 2  m2c 2   2 K  m1c 2  m2c 2   K 2  2Km1c 2   m1c 2    m1c 2 
2
2
2
  m1c 2  m2c 2   2 Km2c 2
2
For the nonrelativistic limit we have
Ecm 
 m1c2  m2c2   2Km2c2   m1c2  m2c 2  1 
2
2 Km2c 2
 m1c2  m2c2 
2
.
With K  mc 2 we can look at the binomial expansion of the square root:


Km2c 2
2
2 
;
Ecm   m1c  m2c  1 
  m c 2  m c 2 2 
1
2


Kcm
Km2c 2
m2
 Ecm   m1c  m2c  

K.
2
2
m1c  m2c
m1  m2
2
2
37. As in the preceding problem Ecm 
m c
1
2
 m2c 2   2Km2c 2 
2
 2mc 
2 2
 2Km2c 2 .
(a) If K  mc 2 we neglect K, so Ecm  2mc 2 .
(b) If K  mc 2 we neglect the first term and Ecm  2Km2c 2  2Kmc 2 .
In (a) we interpret the result to mean that at very low energies there is no extra energy
available (beyond the masses of the two original particles). In (b) we see that the
available center of mass energy increases only in proportion to K , thus illustrating the
great advantage of colliding beam experiments over fixed target experiments.
38.  
K  E0 50000 MeV  0.511 MeV
1
;

 97848 
E0
0.511 MeV
1  v2 / c2
v  c 1  1/  2  c 1  1/ 978482  c 1  5.2 1011   0.999999999948c
39. In each case the desired ratio is simply equal to the relativistic factor  .
(a)  
K  E0 12 MeV  938.27 MeV

 1.01
E0
938.27 MeV
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208
Chapter 14
Elementary Particles
K  E0 120 MeV  938.27 MeV

 1.13
E0
938.27 MeV
K  E0 1200 MeV  938.27 MeV
(c)  

 2.28
E0
938.27 MeV
(b)  
K  E0 33000 MeV  938.27 MeV

 36.17
E0
938.27 MeV
Therefore in the lab frame v  c and the time for 160,000 revolutions is
2 R
800 m
tN
 160000 
 0.427 s. The statement is accurate.
v
2.998 108 m/s
40.  
41. The spaceship of mass m = 15,000 kg has to escape the gravity of the sun, mass
M  2.0 1030 kg , starting from a distance of Earth’s orbit, r  1.5 1011 m . The
gravitational potential energy goes from U = −GMm/r to zero, a gain of GMm/r, so that is
the amount of energy required. That amount of energy equals the annihilation energy
mac2, where ma is the total amount of matter and antimatter involved. Thus GMm/r =
mac2, or
11
2
2
30
4
GMm  6.67 10 N  m /kg  2.0 10 kg 1.5 10 kg 
ma 

 1.48 104 kg
2
11
8
rc 2
1.5 10 m 3.0 10 m/s 
The amount (each) of matter and antimatter is half of this, or 7.4 105 kg. That means
only 74 mg of each! The matter-antimatter reaction is very efficient.
42.
(a) At that kinetic energy the protons are traveling at nearly the speed of light, so to the
stated precision the time required is the circumference divided by the speed of light:
2 r
6300 m
t

 2.1105 s or 21 μs.
8
c
3.0 10 m/s
(b) In a colliding-beam experiment, the full energy of both beams is available:
1.0 TeV  1.0 TeV  2.0 TeV.
(c) To obtain the desired energy, 2.0 TeV, Equation 14.10 gives the energy required for a
fixed-target experiment. We use 2.0 TeV = 2000 GeV, so that we can use 0.938 GeV
for the proton’s mass. Solving Equation 14.10 for the required energy K,
K
2
Ecm
  m1c 2  m2c 2 
2m2c 2
2
 2000 GeV    2  0.938 GeV  

2  0.938 GeV 
2
2
 2.1106 GeV
or 2.11015 eV . That is 300 times the beam energy and 150 times the cm energy of
the LHC. It is unlikely that an accelerator this large could ever be built.
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Chapter 14
Particle Physics
43. (a)
209
(b)
44. Here is the diagram:
The neutrino is a muon neutrino. The diagram for the first part has the pion as a single
solid horizontal line, and the second part is as shown, with the quark structure included.
45.
(a) With 100 GeV per nucleon, the total energy is 197 100 GeV  19.7 TeV per gold ion
with A = 197. For a colliding-beam experiment, the center of mass energy is twice
this amount, or 39.4 TeV.
(b) We can consider each nucleon or the entire nucleus and get the same result. For each
nucleon K = 100 GeV and mc2 = 938 MeV, so
K  mc 2 100.938 GeV


 107.6
mc 2
0.938 GeV
Then from relativity   1 
1
2
 0.99996 or v = 0.99996c.
46.
(a) The quark configuration changes from uud to udd, which means that an up quark has
changed into a down quark.
(b) In this case uds has changed to uud  ud . Note that strangeness is not conserved. The
strange quark transforms via the weak interaction into an up quark and the anti-up
down pair (the pion).
(c) In this case ds has changed to du  ud . Note that as in part (b) strangeness is not
conserved. The strange quark transforms via the weak interaction into an anti-up
quark and the up anti-down pair (the positive pion).
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210
Chapter 14
Elementary Particles
47. (a) Charge = 0; baryon number = 0; charm = 0; strangeness = 0; Le =  1 ; spin = 1/2
(b) The unknown can be a single particle. The  e satisfies the requirements.
48. None of the decays are allowed.
(a) There is not enough mass-energy in  0 to produce   and   and strangeness is not
conserved.
(b) Strangeness is not conserved.
(c) There is not enough mass-energy in  0 to produce K  and p.
49.
(a) For a stationary target the sum of the rest energies of the products equals the total
center of mass energy, so
Ecm  2 E0  2 E0  K    m p  m  mK  c 2
 938 MeV  1116 MeV  494 MeV  2548 MeV
E2
Rearranging we have cm  2 E0  K ;
2 E0
 2548 MeV   2 938 MeV  1585 MeV.
E2
K  cm  2 E0 


2 E0
2  938 MeV 
(b) In a colliding beam experiment the total momentum is zero, and we have by
conservation of energy 2E0  2K  E0   m  mK  c 2 ;
2
 E0   m  mK  c 2 938 MeV+1116 MeV + 494 MeV
K

 336 MeV.
2
2
50.
(a)
(b)
(c)
(d)
baryon number and  lepton number are not conserved, as well as mass-energy
allowed if neutrinos are added to conserve lepton number
strangeness is not conserved
baryon number is not conserved
(a)
(b)
(c)
(d)
(e)
allowed
 e should be  e to conserve electron lepton number
strangeness is not conserved
strangeness is not conserved in a strong interaction
allowed
51.
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Chapter 14
Particle Physics
211
52.
(a)
(b)
(c)
(d)
baryon number and electron lepton number not conserved
not allowed; charge is not conserved
allowed
allowed
(a)
(b)
(c)
(d)
strangeness is not conserved
charge is not conserved
baryon number is not conserved
strangeness is not conserved
53.
54.
(a)   h / p and p  2mK , so  
(b) p  E / c  K / c , so  
h
2mK
hc
K
55.
(a) From Equation (14.10) we have Ecm 
m c
1
2
 m2c 2   2Km2c 2 and with
2
m1  m2  m and K  mc 2 , we have
Ecm  2Kmc 2  2  0.938GeV  7000 GeV   114.5 GeV.
(b) For colliding beams the available energy is the sum of the two beam energies, or
14.0 TeV . This is an improvement over the fixed-target result by a factor of
14000 GeV
 122.
114.5 GeV
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212
Chapter 15
General Relativity
Chapter 15
1. From Newton's second law we have for a pendulum of length L:
m g
d 2 d 2 m g
F  mG g sin   mI a  mI L 2 ; 2  G sin   G  , where we have made the
dt
mI L
mI L
dt
small angle approximation sin    . This is a simple harmonic oscillator equation with
solution   0 cos(t ) where  0 is the amplitude and the angular frequency is

mG g
mI L
2
. The period of oscillation is T 
 2
. Therefore two masses
mI L

mG g
with different ratios mI / mG will have different small-amplitude periods.
2. Beginning with Equation (15.3)
2
5
8 1
gHf  9.80 m/s  4.80 10 m 10 s 
f  2 
 5.2 103 Hz.
2
8
c
 2.998 10 m/s 
3. Beginning with Equation (15.4)
f
GM  1 1 
GM
GM
 2   
r  r1  
 r1  r2  . Use r1  r2  H and let r1  r2  r .
2  2
f
c  r1 r2 
r1r2c
r1r2c 2
f gH
From classical mechanics g  GM / r 2 , so
 2 .
f
c
4.
1 1
   We use r2  6378 km and r1   6378  10  km .
 r1 r2 
6.673 1011 m3  kg 1  s 2  5.98 1024 kg  

T
1
1





2
3
3
T
 6388 10 m 6378 10 m 
 2.998 108 m/s 
T
GM
 2
T
c
 1.09 1012
which is the same as in the example, to three significant digits.
5. The distance d is the sum of the radii of the earth's orbit and Venus's orbit (assuming
circular orbits).
d  149.6 109 m  108.2 109 m  257.8 109 m
9
2d 2  257.8 10 m 

 1719.8 s.
The round-trip time is t 
c
2.998 108 m/s
Using Shapiro's measurement of 200μs, the percent change is therefore
200 106 s
100%   1.16 105 % .
1719.8 s
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Chapter 15
General Relativity
213
6. Using the mass and the radius of the neutron star and given that the detection is at a very
large distance,
11
3
1
2
30
f
GM  1 1  GM  6.673 10 m  kg  s  5.8 10 kg 
 2    2 
 4.31101
2
4
8
f
c  r1 r2  rc
1.0 10 m  2.998 10 m/s 
The wavelength is affected by the same factor, so the redshift at the given wavelength is
   4.31101   550 nm   240 nm, where we have rounded to the two significant
figures given in the problem. Thus the new wavelength is 550 nm + 240 nm = 790 nm,
which is above the visible range in the infrared.
7. Using the mass and radius of the sun
11
3
1
2
30
f GM  6.673 10 m  kg  s 1.99 10 kg 
 2 
 2.123 106
2
f
rc
 6.96 108 m  2.998 108 m/s 
The redshift of the wavelength is affected by the same factor, so the redshift at the two
wavelengths equals:    2.123 106   400 nm   8.49 104 nm and
   2.123 106   700 nm   1.49 103 nm.
8. We can assume that g is constant over this distance. We find the frequency of the gamma
E
14.4 103 eV
 3.48 1018 Hz
ray: f  
15
h 4.136 10 eV  s
We can use Equation (15.3) to find:
2
f gH  9.80 m/s   381 m 
 2 
 4.15 1014 which equals a 4.15 10-12 %
2
f
c
 2.998 108 m/s 
change. The change in frequency is f  4.15 1014  3.48 1018 Hz   145 kHz .
9. We assume that g is constant over this distance. Using E  hf we find:
9.80 m/s 2   22.5 m  14.4 103 eV 

gHf gHE
f  2  2 
 8541 Hz
c
c h  2.998 108 m/s 2  4.136 1015 eV  s 
From the previous problem we know the frequency of the gamma ray is 3.48 1018 Hz so
8541 Hz
the percentage change is
100%   2.45 1013% .
18
3.48 10 Hz
10. We can ignore any blueshift due to each planet’s gravity, which is small compared with
the sun’s gravity.
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214
Chapter 15
General Relativity
(a) Mercury’s orbit is highly eccentric, but for this problem it will be good enough to use
the mean orbital radius (semi-major axis) 5.79 × 1010 m. Then by Equation 15.4,
f
GM
 2
f
c
1 1
  
 r1 r2 
 6.67 10

N  m 2 /kg 2 1.99 1030 kg  
1
1




2
8
10
 6.96 10 m 5.79 10 m 
3.00 108 m/s 
11
f
 2.09 106
f
(b) Now the distance at which the signal is received is Earth’s mean orbital radius
1.50 1011 m.
f

f
6.67 1011 N  m2/kg2 1.99 1030 kg   1  1 
2
 6.96 108 m 1.50 1011 m 
3.00 108 m/s
f
 2.11106
f
As we might expect, the added distance from Mercury to Earth contributes little to the
redshift.
11. We need to rewrite the frequency shift in Equation 15.4 into a wavelength shift.
f = c/λ, so df / d   c /  2 . For small changes df  f and d    , so
f  




c 
 
f


or



f
. Thus the laser’s wavelength shift is
f
1 1
   . In this case r1 is large, so we may use 1/ r1 = 0, and so
 r1 r2 

f GM
 2
f
c

 6.67 1011 N  m2 /kg2  4.5 1030 kg   0.278
GM


2
c 2 r2
3.00 108 m/s  1.2 104 m 
Therefore   0.278  632.8 nm   175.9 nm and the new wavelength is
    632.8 nm 175.9 nm = 456.9 nm or 460 nm to two significant figures.
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Chapter 15
General Relativity
215
11
3
1
2
22
2GM 2  6.673 10 m  kg  s  7.35 10 kg 
12. rs  2 
 1.09 104 m
2
8
c
 2.998 10 m/s 
11
3
1
2
27
2GM 2  6.673 10 m  kg  s 1.90 10 kg 
13. rs  2 
 2.82 m
2
8
c
2.998

10
m/s


14.
(a) From Equation (15.7) we have
1.0546 1034 J  s  2.998 108 m/s 
c3
T

8 kGM 8 1.3811023 J/K  6.673 1011 m3  kg 1  s 2 1.99 1030 kg 
3
 6.17 108 K
(b) Using Equation (15.7)
1.05 1034 J  s  3.0 108 m/s 

c3
T

8 kGM 8 1.38 1023 J/K  6.67 1011 N  m 2 /kg 2  6 109  2.0 1030 kg 
3
 1.0 1017 K
Such low temperatures, especially for part (b) would make the object extremely difficult
to observe.
15. Rearranging Equation (15.7), we have
1.0546 1034 J  s  2.998 108 m/s 
c3
M

 4.19 1020 kg
23
11
3
1
2
8 kGT 8 1.38110 J/K  6.673 10 m  kg  s   293 K 
3
which is about
4.19 1020 kg
 2.111010 solar masses.
30
1.99 10 kg
11
3
1
2
20
2GM 2  6.673 10 m  kg  s  4.19 10 kg 
rs  2 
 6.22 107 m
2
8
c
 2.998 10 m/s 
16.
(a) The mass is 1109  M Sun   1109 1.99 1030 kg   1.99 1039 kg . We use Equation
(15.5) to determine the Schwarzschild radius.
11
3
-1
-2
30
2GM 2  6.673 10 m  kg  s 1.99 10 kg 
rs  2 
 2.95 1012 m
2
c
 2.998 108 m/s 
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216
Chapter 15
General Relativity
Neptune’s orbit has a semi-major axis of 4.50 1012 m , so this black hole would be
smaller than our solar system.
(b) We use the result of Example 15.3, with the value of  from the example (or from
problem 16 below).
1.99 1039 

M 03
t

 6.63 10101 s
15
3
3 3  3.965 10 kg /s 
3
This is much, much greater than the age of our universe!
17. (a) We use the results of Example 15.3 that give the time in terms of the mass.
Rearranging the equation, using the age of the universe as 13.7 billion years, and using
the value of  from Problem 16, we have
1/ 3
7



15
3
9 3.156 10 s 
11
M 0   3 t   3  3.965 10 kg /s  13.7 10 y
   1.73 10 kg.
1y



(b) Current evidence for the smallest black holes require a mass of about 5 to 20 times the
mass of the Sun. The mass from part (a) is only M 0  1019 M Sun which is too small to
form a black hole.
1/ 3
18. From Equation (15.10) we see

8 
3
2  5.6705 108 W  m 2  K 4 1.0546 10 34 J  s   2.9979 108 m/s 
4
2 4c 6

k 4G 2
8 
3
1.3807 10
23
J/K   6.6726 10 11 N  m 2  kg 2 
4
6
2
 3.965 1015 kg 3 / s
19. We could use Equation 15.9, but it is simpler to start at Equation 15.10 and use the fact
the energy E released by the loss of mass M is E = Mc2. Then by Equation 15.10
dE
dM
c 2
 c2
 2
dt
dt
M
where α = 3.96 × 1015 kg3/s was given in Example 15.3. With energy being lost at a rate
(negative, for a loss) 4.2 1015 J / s , we solve for M to find
1/ 2
 c 
M  

 dE / dt 
2
  3.0 108 m/s 2  3.96 1015 kg3 /s  

 


4.2 1015 J/s


1/ 2
 2.9 108 kg.
This is a fairly modest mass—much smaller than a planet.
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publicly accessible website, in whole or in part.
Chapter 15
General Relativity
20.
(a) rS 
217
11
2
2
22
2GM 2  6.67 10 N  m /kg  7.3 10 kg 

 1.1104 m  0.11 mm
2
2
8
c
3.0 10 m/s 
1.05 1034 J  s  3.0 108 m/s 

c3
(b) T 

 1.7 K
8 kGM 8 1.38 1023 J/K  6.67 1011 N  m 2 /kg 2  7.3 10 22 kg 
3
GMm   6.67 10

(c) U  
R
11
N  m2 /kg 2  7.3 1022 kg  6.0 1024 kg 
6.4 106 m
 4.6 1030 J
This is a huge amount of gravitational energy.
21. We use Equation (15.4) and find
f
GM  1 1 
 2   
f
c  r1 r2 
 6.6726 10

N  m 2  kg 2  5.976  1024 kg   1 1 
  
2
 r1 r2 
 2.998 108 m/s 
11


1
1

   4.439 103 m  

6
7
6
  2.02 10  6.37 10  m 6.37 10 m 
 5.30 1010.
The frequency change is then f   5.30 1010  1575.42 MHz   0.834 Hz. This is a
small, but significant change with the precision required of the GPS system.
22. Set the change in the photon's energy equal to the change in gravitational potential
1 1
GMm  GMm 

energy: E  h f  
  GMm    where M is the mass of the
r1
r2 

 r1 r2 
earth and m is the equivalent mass of the photon. Now m  E / c 2  hf / c 2 , so
h f  
GMhf
c2
1 1
f
GM
 2
   and
f
c
 r1 r2 
1 1
  .
 r1 r2 
23. The problem statement advises us to assume that g is constant. Therefore we can use
f gH
Equation (15.3) to determine the change in frequency. We have
 2 so
f
c
  9.80 m/s   3.52 105 m  
 gH 
  259.7 106 Hz   9.97 103 Hz .
f   2  f  
2
  2.998 108 m/s 

 c 


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218
Chapter 15
General Relativity
We can obtain a more precise answer using the formula from the previous problem and
recalling that the earth's radius is 6378 km,
f
GM  1 1 
 2   
f
c  r1 r2 
 6.673 10

m3  kg 1  s 2  5.976 1024 kg   1 1 
  
2
8
 r1 r2 
2.998

10
m/s


11
1
1


   4.437 103 m  


6
6
 6.730 10 m 6.378 10 m 
 3.639 1011.
Therefore f   3.639 1011  259.7 106 Hz   9.45 103 Hz, which is about a 5%
difference.
2GM
dr. For a small change let
r3
dg  dg and dr  r  3 m. Also notice that the distance from the center of the earth is
24. g  GM / r 2 which can be differentiated to give dg  
6.378 106 m  3.50 105 m  6.728 106 m. Then
2  6.673 1011 m3  kg 1  s 2  5.98 1024 kg 
g 
 3 m   8 106 m/s 2 .
3
6
 6.728 10 m 
This is about 106 g , so it is a very small effect.
25. If we use   h / mc for a relativistic particle of mass m, we have  
h
2 Gm
  rs 
.
mc
c2
Solving for m we have
hc
m

2 G
 6.626 10 J  s  2.998 10 m/s   2.18 10
2  6.673 10 m  kg  s 
34
8
11
3
1
8
2
kg.
The Planck energy is
EPl  mc 2   2.18 108 kg  2.998 108 m/s   1.96 109 J  1.22 1028 eV.
2
26.
(a) The combination of G, h, and c that has the right units is
Gh
Pl 

c3
 6.673 10
11
m3  kg 1  s 2  6.626 1034 J  s 
 2.998 10
8
m/s 
3
 4.05 1035 m.
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publicly accessible website, in whole or in part.
Chapter 15
General Relativity
219
h
hc 1240 109 eV  m


 1.02 1034 m which is the same order of
2
28
mc mc
1.22 10 eV
magnitude as (a) and close to the known value 1.62 1035 m.
(b)  
27. The combination of constants that gives units of time is
tPl 
Gh

c5
 6.673 10
11
m3  kg 1  s 2  6.626 1034 J  s 
 2.998 10
8
m/s 
5
The time for light to travel the Planck length is t 
Pl
c

 1.35 1043 s.
Gh
 1.35 1043 s, as we
5
c
found above.
28. As in Problem 23,
f
GM  1 1 
 2   
f
c  r1 r2 
 6.673 10

11
m3  kg 1  s 2  5.976 1024 kg   1 1 
  
2
 r1 r2 
 2.998 108 m/s 
1
1


   4.439 103 m  


7
6
6
 3.587 10 m  6.378 10 m 6.378 10 m 
 5.911010.
Therefore f   5.911010  2.0 109 Hz   1.18 Hz.
29. Rewrite the entropy in terms of U using M  U / c 2 : S 
Then
1 S 16 2GUk
; solving for T and using


T U
hc5
8 2GU 2 k
hc5
 h / 2 , we find
hc5
c5
c3
, where in the last step we again used U = Mc2. This
T


2
16 GUk 8 GUk 8 GMk
is the desired result shown in Equation (15.7).
30. We assume a constant downward acceleration g. From classical mechanics, the
magnitude of the fall assuming only an initial horizontal velocity is Δy = ½ gt2, where the
time of flight is t = L/c. Making this substitution:
2
1 2 1 L
gL2
y  gt  g    2 .
2
2 c
2c
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220
Chapter 15
General Relativity
The distance L between New York and Los Angeles is roughly 4000 km = 4 × 106 m.
9.8 m/s  4.0 10 m 
Inserting numerical values: y 
2  3 10 m/s 
2
6
8
2
2
 8.7 104 m = 0.87 mm.
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publicly accessible website, in whole or in part.
Chapter 16
Cosmology and Modern Astrophysics – The Beginning and the End
221
Chapter 16
1 au
1.496 1011 m/au   3.086 1016 m
tan1
1 ly   2.9979 108 m/s   365.25d/y 86400 s/d   9.4611015 m
1. 1 pc 
1 pc 
3.086 1016 m
 3.26 ly
9.4611015 m/ly
2. As in Example 16.3 we have 7.0  exp  mc 2 / kT  so ln 7.0  mc2 / kT . Then
T
mc 2
939.56563 MeV  938.27231 MeV

 7.71109 K.
11
k ln 7.0
8.617 10 MeV/K   ln 7.0
135 MeV
 1.57 1012 K.
11
8.617 10 MeV/K
From Figure 16.6 the time associated with this temperature is about 104 s.


number of d
10 MeV  6 MeV
  1.00 ;
 exp 
4.
  8.617 1011 MeV/K 1014 K  
number of u




number of u
1300 MeV  10 MeV
  1.16 ;
 exp 
  8.617  1011 MeV/K 1014 K  
number of c




number of u
115 MeV  10 MeV
  1.01 ;
 exp 
 8.617 1011 MeV/K 1014 K  
number of s


3. Using mc2  135 MeV  kT : T 


number of d
115 MeV  6 MeV
  1.01 ;
 exp 
  8.617 1011 MeV/K 1014 K  
number of s




number of u or d
1300 MeV  6 MeV
  1.16 ;
 exp 
  8.617 1011 MeV/K 1014 K  
number of c




number of u or d
174000 MeV  6 MeV
  5.9 108 ;
 exp 
 8.617 1011 MeV/K 1014 K  
number of t




number of u or d
4250 MeV  6 MeV
  1.64 ;
 exp 
  8.617 1011 MeV/K 1014 K  
number of b




number of s
1300 MeV  115 MeV
  1.15 ;
 exp 
 8.617 1011 MeV/K 1014 K  
number of c


Similarly,
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222
Chapter 16
Cosmology – The Beginning and the End
 174000 MeV  115 MeV 
number of s
  5.8 108 ;
 exp 
11
14
  8.617 10 MeV/K 10 K  
number of t




number of s
4250 MeV  115 MeV
  1.62 ;
 exp 
 8.617 1011 MeV/K 1014 K  
number of b


 174000 MeV  1300 MeV 
number of c
  5.1108 ;
 exp 
11
14
 8.617 10 MeV/K 10 K  
number of t




number of c
4250 MeV  1300 MeV
  1.41 ;
 exp 
 8.617 1011 MeV/K 1014 K  
number of b


 174000 MeV  4250 MeV 
number of b
  3.6 108 .
 exp 
11
14
  8.617 10 MeV/K 10 K  
number of t


5. mc2  kT so we have:
mc 2
0.5110 MeV
electron: T 

 5.93 109 K ;
11
k
8.617 10 MeV/K
2
mc
105.66 MeV
muon: T 

 1.23 1012 K
11
k
8.617 10 MeV/K
6. As in the preceding problem we have for the 2.2 eV / c2 mass,
mc 2
2.2 eV
T

 2.55 104 K.
5
k
8.617 10 eV/K
Using Figure 16.6, this temperature corresponds to a time of approximately 108 seconds.
If the mass of the neutrino is 1.0 104 eV / c2 then
mc 2
104 eV
T

 1.16 K.
k
8.617 105 eV/K
Note this answer is lower than the present cosmic background temperature!
7. The   ( E0  140 MeV ) is more massive than the  0 ( E0  135 MeV ), so the   would
have been formed for a longer time. With mc2  k T we have
mc 2
5 MeV
T 

 5.80 1010 K.
11
k
8.617 10 MeV/K
8. Set the deuteron binding energy 2.22 MeV equal to kT :
2.22 MeV
2.22 MeV
T

 2.58 1010 K.
11
k
8.617 10 MeV/K
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publicly accessible website, in whole or in part.
Chapter 16
Cosmology and Modern Astrophysics – The Beginning and the End
223
9. Set the hydrogen binding energy 13.6 eV equal to kT :
13.6 eV
13.6 eV
T

 1.58 105 K.
5
k
8.617 10 eV/K
0.3GN 2 m2 3.9 2 N 5/3
10. Beginning with Equation (16.15),
, one can solve for V and find:

V 4/3
2mV 5/3
3.9 2
6.5 2
V 1/ 3 

.
0.6m3 N 1/ 3G N 1/ 3m3G
11. Notice that Example 16.5 indicates a neutron star with a mass of two solar masses.
mass
2M
3M
where M  1.99 1030 kg (mass of sun) and R  11 km .



3
4
volume
 R3 2 R
3
3 1.99 1030 kg 
3M


 7.14 1017 kg/m3
3
3
3
2 R
2 1110 m 
3 1.67 1027 kg 
mp

 2.311017 kg/m3 and the
4 3 4 1.2 1015 m 3


 r0
3
neutron star is about three times as dense.
For the nucleon we have  
12. We see from Equation (16.16) that V 1/3 is proportional to N 1/ 3 . However, V 1/3 is
proportional to R, so R is proportional to N 1/ 3 . The reason this makes sense is that the
gravitational pressure increases in proportion to N 2 [see Equation (16.12)] while the
neutron pressure increase in proportion to N 5/3 [see Equation (16.14)]. The gravitational
pressure increases more rapidly than the neutron pressure so that the radius of the neutron
star decreases as the number of neutrons increases.
13. (a) Beginning with Equation (16.12):
2
Nm 

M2
P  0.3G 4 / 3  0.3G
4/3
V
4
3

R


3

 0.3  6.67 10
11
1
m  kg ·s
3
2

1.99 10
30
kg 
2
3
4
8
   6.96 10 m  
3

4/3
 5.00 1013 N/m 2 .
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224
Chapter 16
(b)
P  0.3G
M2
4/3
4
3
 R 
3

33
 3.2110 N/m 2 .
 0.3  6.67 1011 m3  kg 1  s 2 
Cosmology – The Beginning and the End
4 1.99 1030 kg 
2
3
4
4
  1.110 m  
3

4/3
14. We begin with Equation (16.1).
 1 Mpc 
71 km/s
4
v  HR 
4.0 109 ly  

  8.7110 km/s  0.29c.
6
Mpc
 3.26 10 ly 
v
15000 km/s
15. R  
 211 Mpc or about 689 Mly .
H 71 km/s/Mpc
16.
(a) From Equation (16.18) we see that for a redshift of 3.8 we have  / 0  3.8 , so
3.8 
R
1 
 1 which can be solved to find   0.917. Then
1 
v 0.917  299790 km/s 

 3872 Mpc.
H
71 km/s/Mpc
 3.26 Mly 
(b) 3872 Mpc 
  12.6Gly.
 1 Mpc 
17. The redshift is z = Δλ/λ0 = 3.8, so   z0  3.8  21 cm   80 cm and the new
wavelength is 21 cm + 80 cm = 101 cm. There could be another spectral line at this
wavelength, but it is unlikely to lead to confusion. As Figure 16.1 shows, the pattern of
spectral lines shifts as a group, with each line experiencing the same redshift factor z.
This makes it possible to identify the 21-cm line within the pattern of the entire spectrum.
1 
18. Beginning with Equation (16.18), the speed is given by z  6.42 
 1 which can
1 
be solved to find   0.964 or v  0.964c. The distance is then given by Hubble’s law
v 0.964  299790 km/s 
R 
 4070 Mpc . Expressed in ly,
H
71 km/s/Mpc
3.26 ly
R  4070 Mpc 
 13.3 Gly.
Mpc
19.
(a) From Equation (16.18), for a redshift of 8.6 we have  / 0  8.6, so
8.6 
1 
 1 which can be solved to find   0.979 or v  0.979c.
1 
© 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 16
Cosmology and Modern Astrophysics – The Beginning and the End
(b) R 
225
v 0.979  299790 km/s 

 4130 Mpc ;
H
71 km/s/Mpc
 3.26 Mly 
4130 Mpc 
  13.5 Gly.
 1 Mpc 
20.
(a) Type I supernova have no hydrogen spectral lines. Type Ia has strong silicon spectral
lines but no helium lines. Type Ib has very weak or no silicon lines and has neutral
helium lines. Type Ic has weak or no helium lines and weak or no silicon lines.
(b) All supernovas other than Type Ia are caused by gravitational contraction of stars
much more massive than our Sun.
21.
 1 nucleon 
3
(a)  3 1028 kg/m3  
  0.18 nucleons/m
27
1.67

10
kg


(b) R  16.6 ly   9.46 1015 m/ly   1.57 1017 m
60 1.99 1030 kg 
1 nucleon
 7.15 1058 nucleons
27
1.67 10 kg
The nucleon density is the number of nucleons divided by the volume, or
7.15 1058
 4.41106 nucleons/m3 . The nucleon density in our neighborhood
3
4
 1.57 1017 m 
3
is much larger than it is for the universe as a whole.
22.
(a) The solid blue curve represents the current, best idea of the expansion of the universe.
The expansion of the universe is accelerating. Current models suggest that about 5%
of the mass-energy is ordinary matter, about 23% is dark matter, and about 72% is
dark energy. Most of the mass is represented by dark energy. The solid black curve
represents the known mass of the universe. It is an open, low density universe, and
the expansion of the universe is slowing down. The blue dashed curve represents a
flat, critical density universe. The expansion rate slows down until the curve
becomes more horizontal. The black dashed curve represents a closed, high density
universe which will expand for a few more billion years, but eventually will turn
around and collapse. There is not enough mass in the universe for this to happen.
(b) Yes, the black dashed curve represents a closed universe. See the explanation in part
(a).
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publicly accessible website, in whole or in part.
226
Chapter 16
23. H   22 km/s/Mly 
Cosmology – The Beginning and the End
1 Mly
 2.32 1018 s 1
12
10  9.46110 km 
6
3  2.32 1018 s 1 
3H 2
c 

 9.63 1027 kg/m3  9.63 1030 g/cm3 .
11
3
1
2
8 G 8  6.67 10 m  kg  s 
2
24. From Wien’s law maxT  2.898 10 m  K. max
3
2.898 103 m  K

 1.06 mm.
2.725 K
25. For H  64 km / s / Mpc we have
1 Mpc


18 1
H   64000 m/s/Mpc  
  2.07 10 s ;
22
 3.086 10 m 
3  2.07 1018 s 1 
3H 2
c 

 7.7 1027 kg/m3 .
11
3
1
2
8 G 8  6.67 10 m  kg  s 
2
For H  78 km / s / Mpc we have
1 Mpc


18 1
H   78000 m/s/Mpc  
  2.53 10 s ;
22
 3.086 10 m 
3  2.53 1018 s 1 
3H 2
c 

 1.2 1026 kg/m3 .
11
3
1
2
8 G 8  6.67 10 m  kg  s 
2
26.
(a) The only combination of the constants that gives time is m 
(b) t p 
Gh

c5
 6.67 10
11
1
m  kg  s
3
2
 2.998 10
8
 6.626 10
m/s 
5
34
J  s
1
2
1
5
2
2
, n  , and l   .
 1.35 1043 s
27. A complete derivation can be found in Section 2.10.
28. Let dm be the mass of a spherical shell of thickness dr, so dm   d (vol)  4 r 2 dr .
GM dm
4
where M    r 3 is the mass inside radius r. Thus
dV  
r
3
2
R
16 4
16 2  2 R5G
dV  G  2
r dr and therefore V   dV  
.
0
3
15
16 2  2 R5G  9M 2 
3GM 2


.
Now   3M / 4 R3 , so V  

2 6 
15
5R
 16 R 
29. From the slopes it appears that t0   / 2 .
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Chapter 16
Cosmology and Modern Astrophysics – The Beginning and the End
227
2
2
 d   mc 
30. In general, as in Example 16.7, we have t    
 . For a distance of 50 kpc ,
 2c   E 
 50 kpc  3.086 1019 m  2.57 1012 s. Note that
d

2c 2  3.00 108 m/s 
kpc
2.57 1012 s   2.57 s 10 MeV/10 eV  , so for any distance (in kpc)
2
2
 distance   m c 2   10 MeV 
t   2.57 s  
 

 .
E

 50 kpc  10 eV  
2
31. We know from Section 9.7 [see the comment after Equation (9.56)] that the energy flux
rate    T 4 is related to the energy density  e by a factor of c / 4 , such that
c
c
  (c / 4) e . But using E  mc 2 we have e  rad c 2 , so  T 4  e  rad c 2 or,
4
4
4
4 T
rearranging rad  3 .
c
8
2
4
4 T 4 4  5.67 10 W  m  K   2.725 K 
 3 
 4.63 1031 kg/m3
3
8
c
3.00 10 m/s 
4
32. rad
This is substantially less than the matter density of about 3 1028 kg/m3 , so the universe
is matter dominated.
33. At 380,000 years the temperature is about 102 K. Then
8
2
4
2
4 T 4 4  5.67 10 W  m  K 10 K 
 3 
 8.4 1025 kg/m3 .
3
8
c
3.00 10 m/s 
4
rad
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publicly accessible website, in whole or in part.
228
Chapter 16
Cosmology – The Beginning and the End
34. The 300 day mark is almost exactly 200 days after the peak, so use t  200 d in the
exponential decay formula exp  (ln 2)t / t1/2  .
56
Ni: exp  (ln 2)t / t1/2   exp  (ln 2)  200 d  /  6.1 d    1.35 1010
Co: exp  (ln 2)t / t1/2   exp  (ln 2)  200 d  /  77.1 d    0.166
The cobalt must be primarily responsible, with almost no contribution from the
56
35. Redshift 

0

nickel.
1 
580.0 nm  121.6 nm
1
 3.77 ; 3.77 
1 
121.6 nm
Therefore   0.916 and v  0.916c.
36. By definition
equivalent to

0

0
 z . From Equation (16.18) we know
1  z 1 

0

1 
 1 ; this is
1 
1 
.
1 
37. Using the binomial expansion with the result of the preceding problem
 
 

1  z  1   ... 1   ...   1   so we see that to first order z   .
2
2



1 
38. 5.34 
 1 so   0.951 and v  0.951c ;
1 
R
v 0.951 299790 km/s 

 4016 Mpc.
H
71 km/s/Mpc
39. Q   2mp  md  me  u·c 2  0.43 MeV.
There are three particles in the final state, but it is possible for the deuteron and positron
to have negligible energy, in which case the neutrino energy is 0.43 MeV.
We begin with Equation (16.21):
3H 2  H in (km/s)/Mpc 
3 104
c 


11
3
1
2
8 G 
100
 8  6.6726 10 m  kg  s 
2
40.

Mpc
 H 
13
2
2
2
3
1
2 


 1.789 10 km  s  Mpc  m  kg  s  
22
 100 
 3.086 10 m 
2
2
2
 3 m 
10

km 

2
2
 H 
 H 
26
3
29
3

 1.88 10 kg/m   
 1.88 10 g/m 
 100 
 100 
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publicly accessible website, in whole or in part.
Chapter 16
Cosmology and Modern Astrophysics – The Beginning and the End
229
41.
(a) a  Ct n ;
q  a
da / dt  nCt n1 ;
d 2 a / dt 2
2
d 2 a / dt 2  n(n  1)Ct n2 ;
an(n  1)Ct n 2
an(n  1)
n(n  1)

 2 n 
2 2 2 n2
nCt
n Ct
n2
 da 
 
 dt 
where in the last step we used the definition a  Ct n . From this we see that there is
deceleration  q  0  only if 0  n  1 .
(b) H 
42.
(a)
1 da nCt n 1 n

 with 0  n  1 . There is an inverse dependence on time.
a dt
Ct n
t
X n exp   En / kT 

 exp    En  E p  / kT   exp    mn  m p  c 2 / kT  where is
X p exp   E p / kT 
last step uses the fact that since E  K  mc 2 , the kinetic energy term will cancel.
X
The difference in rest energy equals 1.3 MeV so we have n  exp  1.3 MeV / kT  .
Xp
X
1
 exp  1.3 MeV / kT 
(b) n 
X p 6.7
Take the natural logarithm of both sides and solve for kT. We find kT  0.68 MeV
and thus T  7.9 109 K .
(c) Using the equipartition theorem, this temperature corresponds to an energy of
K  32 kT  1.02 MeV . According to Figure 16.6, this temperature corresponds to a
time of perhaps 0.1 s or 1 s, which is before the time at which nuclei could form.
43. For a uniform distribution assume a cubical lattice of stars, each separated by distance d.
Then each star occupies a cube of volume d 3 , and the density of that cube and the
universe as a whole is   M / d 3 . Solving for d,
1/ 3
M 
d  
  
1/ 3
 2.0 1030 kg 

27
3 
 9 10 kg/m 
 6.0 1018 m. Converting to lightyears,
1 ly


d  6.0 1018 m 
  640 ly.
15
 9.45 10 m 
Our local region is much denser than this, because we know there are many stars closer
than 640 ly to the sun. (Other than the sun, the closest star system, Alpha Centauri, is
just over 4 ly away.) This density is to be expected, because galaxies are relatively
dense collections of stars.
© 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.