INSTRUCTOR
SOLUTIONS
MANUAL
Instructor Solutions Manual
for
Modern Physics
Sixth Edition
Paul A. Tipler
Ralph A. Llewellyn
Prepared by
Mark J. Llewellyn
Department of Electrical Engineering and Computer Science
Computer Science Division
University of Central Florida
W. H. Freeman and Company
New York
Instructor Solutions Manual to Accompany Tipler & Llewellyn Modern Physics, Sixth
Edition
© 2012, 2008, 2003 by W.H. Freeman and Company
All rights reserved.
Published under license, in the United States by
W. H. Freeman and Company
41 Madison Avenue
New York, NY 10010
www.whfreeman.com
Preface
This book is an Instructor Solutions Manual for the problems which appear
in Modern Physics, Sixth Edition by Paul A. Tipler and Ralph A. Llewellyn.
This book contains solutions to every problem in the text and is not intended
for class distribution to students. A separate Student Solutions Manual for
Modern Physics, Sixth Edition is available from W. H. Freeman and
Company. The Student Solutions Manual contains solutions to selected
problems from each chapter, approximately one-fourth of the problems in
the book.
Figure numbers, equations, and table numbers refer to those in the text.
Figures in this solutions manual are not numbered and correspond only to
the problem in which they appear. Notation and units parallel those in the
text.
Please visit W. H. Freeman and Company’s website for Modern Physics,
Sixth Edition at www.whfreeman.com/tiplermodernphysics6e. There you
will find 30 More sections that expand on high interest topics covered in the
textbook, the Classical Concept Reviews that provide refreshers for many
classical physics topics that are background for modern physics topics in the
text, and an image gallery for Chapter 13. Some problems in the text are
drawn from the More sections.
Every effort has been made to ensure that the solutions in this manual are
accurate and free from errors. If you have found an error or a better solution
to any of these problems, please feel free to contact me at the address below
with a specific citation. I appreciate any correspondence from users of this
manual who have ideas and suggestions for improving it.
Sincerely,
Mark J. Llewellyn
Department of Electrical Engineering and Computer Science
Computer Science Division
University of Central Florida
Orlando, Florida 32816-2362
Email: markl@cs.ucf.edu
Table of Contents
Chapter 1 – Relativity I
1
Chapter 2 – Relativity II
31
Chapter 3 – Quantization of Charge, Light, and Energy
53
Chapter 4 – The Nuclear Atom
79
Chapter 5 – The Wavelike Properties of Particles
109
Chapter 6 – The Schrödinger Equation
127
Chapter 7 – Atomic Physics
157
Chapter 8 – Statistical Physics
187
Chapter 9 – Molecular Structure and Spectra
209
Chapter 10 – Solid State Physics
235
Chapter 11 – Nuclear Physics
259
Chapter 12 – Particle Physics
309
Chapter 13 – Astrophysics and Cosmology
331
Chapter 1 – Relativity I
1-1.
(a) Speed of the droid relative to Hoth, according to Galilean relativity, uHoth , is
uHoth  uspaceship  udroid
 2.3 108 m / s  2.1108 m / s
 4.4 108 m / s
(b) No, since the droid is moving faster than light speed relative to Hoth.
1-2.
4
2 L 2  2.74 10 m 
(a) t 

 1.83 104 s
8
c
3.00 10 m / s
(b) From Equation 1-6 the correction  t 
2L v 2

c c2
 t  1.83 104 s 104   1.83 1012 s
2
c
4 km / s
 1.3 105
c 299, 796 km / s
No, the relativistic correction of order 10-8 is three orders of magnitude smaller than
(c) From experimental measurements

the experimental uncertainty.
0.4 fringe

1.0 fringe
 v2 
1.0
2
 29.9 km / s   2.22 103  v  47.1 km / s
0.4
1-3.
 29.8km / s 
1-4.
(a) This is an exact analog of Example 1-1 with L = 12.5 m, c = 130 mph, and v = 20
2
 v km / s 
2
mph. Calling the plane flying perpendicular to the wind plane #1 and the one flying
parallel to the wind plane #2, plane #1 win will by Δt where
Lv 2 12.5mi  20mi / h 
t  3 
 0.0023 h  8.2s
3
c
130mi / h 
2
(b) Pilot #1 must use a heading   sin 1  20 /130   8.8 relative to his course on both
legs. Pilot #2 must use a heading of 0 relative to the course on both legs.
Chapter 1 – Relativity I
1-5.
(a) In this case, the situation is analogous to Example 1-1 with L  3 108 m,
v  3 104 m / s, and c  3 108 m / s If the flash occurs at t = 0, the interior is dark
until t =2s at which time a bright circle of light reflected from the circumference of
the great circle plane perpendicular to the direction of motion reaches the center, the
circle splits in two, one moving toward the front and the other moving toward the rear,
their radii decreasing to just a point when they reach the axis 10-8 s after arrival of the
first reflected light ring. Then the interior is dark again.
(b) In the frame of the seated observer, the spherical wave expands outward at c in all
directions. The interior is dark until t = 2s at which time the spherical wave (that
reflected from the inner surface at t = 1s) returns to the center showing the entire inner
surface of the sphere in reflected light, following which the interior is dark again.
1-6.
Yes, you will see your image and it will look as it does now. The reason is the second
postulate: All observers have the same light speed. In particular, you and the mirror are
in the same frame. Light reflects from you to the mirror at speed c relative to you and the
mirror and reflects from the mirror back to you also at speed c, independent of your
motion.
1-7.
N 
2 Lv 2
(Equation 1-10) Where λ = 590 nm, L = 11 m, and ΔN = 0.01 fringe
c2
v2 
2
N  c 2
  0.01 fringe   590 109 m  3.00 108 m / s  2 11m 
2L
v  4.91103 m / s  5 km / s
1-8.
(a) No. Results depends on the relative motion of the frames.
(b) No. Results will depend on the speed of the proton relative to the frames. (This
answer anticipates a discussion in Chapter 2. If by “mass”, the “rest mass” is implied,
then the answer is “yes”, because that is a fundamental property of protons.)
2
Chapter 1 – Relativity I
(Problem 1-8 continued)
(c) Yes. This is guaranteed by the 2nd postulate.
(d) No. The result depends on the relative motion of the frames.
(e) No. The result depends on the speeds involved.
(f) Yes. Result is independent of motion.
(g) Yes. The charge is an intrinsic property of the electron, a fundamental constant.
1-9.
The wave from the front travels 500 m at speed c + (150/3.6) m/s and the wave from the
rear travels at c – (150/3.6) m/s. As seen in Figure 1-14, the travel time is longer for the
wave from the rear.
t  tr  t f 
500m
500m

8
3.00 10 m / s  150 / 3.6  m / s 3.00  10 m / s  150 / 3.6  m / s
8
 3 108  150 / 3.6  3 108  150 / 3.6







 500
  3 108   2 150 / 3.6   3 108   150 / 3.6 2 


 500
2 150 / 3.6 
 3 10 
8 2
 4.63  1013 s
*
1-10.
A
*
B
*
v
C
While the wavefront is expanding to the position shown, the original positions of
A, B, and C have moved to the * marks, according to the observer in S.
(a) According to an S  observer, the wavefronts arrive simultaneously at A and B .
(b) According to an S observer, the wavefronts do not arrive at A and C simultaneously.
(c) The wavefront arrives at A first, according to the S observer, an amount Δt before
arrival at C  , where
3
Chapter 1 – Relativity I
(Problem 1-10 continued)
t 
BC  BA

cv cv
since BC   BA  L, Thus
c  v  c  v 
 2v 
t  L 
 L  2 2 
2
2

 c v 
c  v 
1-11.
β
  1/ 1   2 
1/ 2
0
0.2
0.4
0.6
0.8
0.85
0.90
0.925
0.950
0.975
0.985
0.990
0.995
1-12.
vx 

t1    t1  20 
c 

1
1.0206
1.0911
1.2500
1.6667
1.8983
2.2942
2.6318
3.2026
4.5004
5.7953
7.0888
10.0125
vx 

t2    t2  20 
c 

( from Equation 1-19)
vx
vx 

(a) t2  t1    t2  20  t1  20     t2  t1 
c
c 

(b) The quantities x1 and x2 in Equation 1-19 are each equal to x0 , but x1 and x2 in
Equation 1-18 are different and unknown.
1-13. (a)
  1/ 1  v 2 / c 2 
1/ 2
 1/ 1   0.85c  / c 2 


2
1/ 2
 1.898
x    x  vt   1.898 75m   0.85c   2.0  105 s    9.537  103 m
y  y  18m
z  z  4.0m
t     t  vx / c 2   1.898  2.0 105 s   0.85c  75m  / c 2   3.756  105 s
4
Chapter 1 – Relativity I
(Problem 1-13 continued)
(b)
x    x  vt    1.898  9.537 103 m   0.85c   3.756 10 5 s    75.8m
difference is due to rounding of  , x, and t .
y  y  18m
z  z  4.0m
t    t   vx / c 2   1.898 3.756 105 s   0.85c   9.537 103 m  / c 2   2.0 10 5 s
1-14. To show that Δt = 0 (refer to Figure 1-8 and Example 1-1).
t1 
L
c v
2
2

L
c v
2
2

2L
c v
2
2

2L
1
c 1  v2 / c2
t2 , because length parallel to motion is shortened, is given by:
t2 
L 1  v 2 / c 2 L 1  v 2 / c 2 2 Lc 1  v 2 / c 2

 2
cv
cv
c 1  v 2 / c 2 
t2 
2L
c

1  v2 / c2
1  v2 / c2

2

2L
1
 t1
c 1  v2 / c2
Therefore, t2 – t1 = 0 and no fringe shift is expected.
1-15. (a) Let frame S be the rest frame of Earth and frame S  be the spaceship moving at speed
v to the right relative to Earth. The other spaceship moving to the left relative to Earth
at speed u is the “particle”. Then v = 0.9c and ux = −0.9c.
u x 

ux  v
1  ux v / c 2
(Equation 1-22)
0.9c  0.9c
1.8c

 0.9945c
2
1.81
1   0.9c  0.9c  / c
(b) Calculating as above with v  3.0 104 m / s  ux
ux 
1
3.0  104 m / s  3.0  104 m / s
6.0  104 m / s

 6.0  104 m / s
8
4
4
1  10
 3.0  10 m / s  3.0  10 m / s 
 3.0 10 m / s 
8
2
5
Chapter 1 – Relativity I
1-16.
ax 
dux
dt 
where u x 
And t     t  vx / c 2 
ux  v
1  ux v / c 2
(Equation 1-22)
(Equation 1-18)
dux   ux  v   vdu x / c 2 1  u x v / c 2   1  u x v / c 2  du x
2
1
 v 
2
 2   ux  v  du x  1  u x v / c  du x
c
 
2
1  ux v / c2 
dt     dt  vdx / c 2 
 v 
u  v  du x / dt   1  u x v / c 2   du x / dt 

2  x

du
c
ax  x   
3
dt 
 1  u v / c 2 
x
 dux / dt  1  v 2 / c 2 

 1  u x v / c
ay 
du y
dt 


2 3
where u y 
ax
 3 1  u x v / c 2 
uy
3
(Equation 1-22)
 1  ux v / c 2 
du y   du y /   1  u x v / c 2    u y /   1  u x v 2 / c 2  du x
1
2
 du  1  u v / c    u v / c  du

 1  u v / c 
2
y
2
x
y
x
2 2
x
2
2
duy  du y / dt  1  u x v / c    u y v / c   du x / dt 
ay 

2
dt 
 1  u v / c 2   1  u v / c 2 
x

x
a y 1  u x v / c 2   ax  u y v / c 2 
 2 1  u x v / c 2 
3
az is found in the same manner and is given by: az 
6
az 1  ux v / c 2   ax  u z v / c 2 
 2 1  ux v / c 2 
3
Chapter 1 – Relativity I
1-17. (a) As seen from the diagram, when the observer in the rocket ( S  ) system sees 1 c∙s
tick by on the rocket’s clock, only 0.6 c∙s have ticked by on the laboratory clock.
ct
ct 
4
x
3
2
1
1
0
0
1
2
3
4
x
(b) When 10 seconds have passed on the rocket’s clock, only 6 seconds have passed on
the laboratory clock.
1-18. (a)
y
u x  0
u y  0
c
y
x
v
x
ux 
uy 
(b)
u x  v
0v

v
2
1 0
1  vu x / c
u y
 1  vu x / c
2


(Equation 1-23)
c
c

 1  0  
u  ux2  u y2  v 2  c 2 /  2  v 2  c 2 1  v 2 / c 2   c
7
Chapter 1 – Relativity I
1-19. By analogy with Equation 1-23,
1-20
(a) ux 
ux  v
0.9c  0.9c
1.8c


 0.9945c
2
2
1  vux / c 1   0.9c  0.9c  / c
1.81
(b) ux 
1.8c /1.81  0.9c  1.8   0.9 1.81 c  3.429 c  0.9997c
ux  v

2
1  vux / c 1  1.8c /1.81 0.9c  / c 2 1.81  1.8  0.9 
3.430
1

(a)
(Equation 1-16)
1  v2 / c2
 1  v / c
2
2
 1  v
 1     2
 2  c

2 1/ 2
1 v2 3 v4
 1


2 c2 8 c2
1
(b)

2
  1  3  1  v 2 
         2  
  2  2  2!  c 
1 v2
 1
2 c2
 1  v 2 / c 2  1  v 2 / c 2 
1/ 2
2
2
2
 1   v   1  1  1  v 
 1      2         2  
 2   c   2  2  2!  c 
 1
(c)
1 v2 1 v4


2 c2 8 c2
 1 
t  t 
1 v2
2 c2
1 v2 3 v4


2 c2 8 c4
  1  1
1-21.
 1
1


1
1


1 v2 1 v4

2 c2 8 c4
1 v2
2 c2
(Equation 1-26)
t  t  t   t 
1 v2
1 v2

  1 



1

t 
t 
2 c2
2 c2
t  t 
v  2c
t 
2
2
t  t  

v  c 2

t  

1/ 2
8
 c  2  0.01
1/ 2
 0.14c
Chapter 1 – Relativity I
S
1-22.
Orion (#2)
v = 1000 km/h
Lyra (#1)
S
Earth
2.5 103 c y
2.5 103 c y
(a) Note that   1/ 1  v 2 / c 2  1 and 1 c y  c  3.15 107 s 
From Equation 1-27: t  t2  t1  0 , since the novas are simultaneous in system S
(Earth). Therefore, in S  (the aircraft)
v
t   t2  t1   2  x2  x1 
c
6
10 m / h

2.5 103  2.5 103   c   3.15 107 s 
2 
 3600s / h  c
 1.46 105 s  40.5h
(b) Since t  is positive, t2  t1 ; therefore, the nova in Lyra is detected on the aircraft
before the nova in Orion.
1-23. (a)
L  Lp / 
(Equation 1-28)
=Lp 1  v 2 / c 2  1.0m 1   0.6c  / c 2 


2
1/ 2
 0.80m
(b) t  L / v  0.80m / 0.6c  4.4 109 s
(c)
The projection OA on
the x axis is L. The
length OB on the ct axis
yields t.
ct 
ct
back of meterstick
passes x = 0
x
B
meterstick
AA
0
x
9
Chapter 1 – Relativity I
t 
t  t  
1-24. (a)

(Equation 1-26)
1  v2 / c2
2.6  108 s
1   0.9c  / c 2 


2
1/ 2

2.6 108 s
0.19
 5.96  108 s
(b) s  vt   0.9   3.0 108 m / s  6.0 108 s   16.1m
(c) s  vt   0.9   3.0 108 m / s  2.6 108 s   7.0m
(d)
 s 
2
  ct    x 
2
2
(Equation 1-31)
2
 c  6.0  108    16.1m   324  259  65  s  7.8m
2
1-25. From Equation 1-28, L  Lp /   Lp 1  v 2 / c 2
where L  85m and Lp  100m
1  v 2 / c 2  L / Lp  85 /100
Squaring 1  v 2 / c 2  85 /100 
2
2
 v 2  1   85 /100   c 2  0.2775c 2 and v  0.527c  1.58  108 m / s


1-26. (a) In the spaceship the length L = the proper length Lp; therefore, ts 
Lp
c

Lp
c

2 Lp
c
(b) In the laboratory frame the length is contracted to L  Lp /  and the round trip time is
tL 

L
L
L


c  v c  v c 1  v 2 / c 2 
2 Lp / 
c 1  v 2 / c 2 

2 Lp 1  v 2 / c 2
c

1  v2 / c2

2

2 Lp
c 1  v2 / c2
(c) Yes. The time ts measured in the spaceship is the proper time interval τ. From time
dilation (Equation 1-26) the time interval in the laboratory tL   ; therefore,
tL 
1
2 Lp
1  v2 / c2
c
which agrees with (b).
10
Chapter 1 – Relativity I
1-27. Using Equation 1-28, with LAp and LBp equal to the proper lengths of A and B and LA =
length of A measured by B and LB = length of B measured by A.
LA  LAp /   100m 1   0.92c   39.2 m
2
LBp   LB  36 / 1   0.92c  / c 2  91.9m
2
1-28.
In S  : x  1.0m cos30  0.866m
y   1.0m sin 30  0.500m where    30
In S : x  x 1   2  0.866m 1   0.8   0.520m
2
y  y   0.500m
0.500
where   tan 1
 43.9
0.520
L
1-29. (a)
 x 
2
  y  
2
 0.520m 
2
  0.500   0.721m
2
In S  : V   a  b  c   2m  2m  4m   16m3
In S: Both a and c have components in the x direction.
ax  a sin 25   2m  sin 25  0.84m and cx  c cos 25   4m  cos 25  3.63m
ax  ax 1   2  0.84 1   0.65   0.64m
2
cx  cx 1   2  3.634 1   0.65   2.76m
2
a y  ay  a cos 25  2 cos 25  1.81m and c y  cy  c sin 25  4sin 25  1.69m
a  ax2  a y2 
 0.64   1.81
c  cx2  c y2 
 2.76   1.69 
2
2
2
 1.92m
2
 3.24m
b (in z direction) is unchanged, so b  b  2m
 (between c and xy-plane)  tan 1 1.69 / 2.76   31.5
 (between a and yz-plane)  tan 1  0.64 /1.81  19.5
V  (area of ay face) b (see part[b])
V   c  a sin 78   b   3.24m 1.92m sin 78  2m   12.2m3
11
Chapter 1 – Relativity I
(Problem 1-29 continued)
(b)
y
a
b
19.5◦
c
x
31.5◦
z
1-30.
 
c

f
c
1 
1 
fo

2
   1 v / c
  
 o  1  v / c
1 v / c
o (Equation 1-36)
1 v / c
2
  
   1  v / c   1  v / c
 o 
2
2
2

  
v    
v 1     / o 
   1  1     
Solving for v / c,
c  o 
c 1     / o 2

 o 

2
v 1   590nm / 650nm 
o  650nm. For yellow    590nm.

 0.097
c 1   590nm / 650nm 2
Similarly, for green    525nm 
and for blue    460nm 
v
 0.210
c
v
 0.333
c
12
Chapter 1 – Relativity I
1-31.
 
c
c
1 v / c


o
f
1 v / c
1 
fo
1 
(Equation 1-37)
1  1.85  10 m / s  /  3.00 10 m / s  
   o  
1 v / c

 1 
1  
8
o
o
1 v / c
 1  1.85  10 m / s  3.00 10 m / s  
7

8
1/ 2
 1  0.064

1-32. Because the shift is a blue shift, the star is moving toward Earth.
f 
1 
f o where f  1.02 f o
1 
1.02 
2
1.02   1  0.0198

2
1.02   1
2
1 

1 

v  0.0198c  5.94 106 m / s
1-33.
f 
1 
fo
1 
 
1 
1 
o 
 656.3nm 
1 
1 
For   103 :    656.3nm 
1  103
 657.0nm
1  103
For   10 :    656.3nm 
1  102
 662.9nm
1  102
2
  101 :    656.3nm 
1  101
 725.6nm
1  101
1-34. Let S be the rest frame of Earth, S  be Heidi’s rest frame, and S  be Hans’ rest frame.
1
 S 
 1.1198
2
1   0.45c / c 
 S  
1
1   0.95c / c 
2
 3.206
When they meet, each will have traveled a distance d from Earth.
Heidi: d  0.45c tHeidi
Hans: d  0.95c tHans
and tHans  tHeidi  1 in years.
Therefore,  0.95c  0.45c  tHeidi  0.95c and tHeidi  1.90 y; tHans  0.90 y
(a) In her reference frame S  , Heidi has aged t  when she and Hans meet.
13
Chapter 1 – Relativity I
(Problem1-34 continued)
v
0.45c




t    S   t  2 x   1.1198 1.90  2  0.45c 1.90  
c
c




2
 1.198 1.90 y  1   0.45   1.697 y


In his reference frame S  , Hans has aged t  when he and Heidi meet.
v
2


t    S   t  2 x   3.3026 1.90 y  1   0.95   0.290 y


c


The difference in their ages will be 1.697  0.290  1.407 y  1.4 y
(b) Heidi will be the older.
1-35. Distance to moon = 3.85 108 m  R
Angular velocity ω needed for v = c:
  v / R  C / R   3.00 108 m / s  /  3.85 108 m   0.78rad / s
Information could only be transmitted by modulating the beam’s frequency or intensity,
but the modulation could move along the beam only at speed c, thus arriving at the moon
only at that rate.
1-36. (a) Using Equation 1-28 and Problem 1-20(b).
t   t /   t 1  v 2 / 2c 2   t  tv 2 / 2c 2
where t  3.15  107 s / y
v  2 RE / T   2   6.37 106 m  / 108min  60 s / min 
v  6.177  103 m / s  2.06  105 c
Time lost by satellite clock = tv 2 / c 2   3.15 10 7 s  2.06  105  / 2  0.00668s  6.68ms
2
(b)
1s  t  v 2 / 2c 2 
t  2 /  v 2 / c 2   2 /  2.06 105   4.71109 s  150 y
2
14
Chapter 1 – Relativity I
cos    
(Equation 1-41)
1   cos  
where    half-angle of the beam in S   30
cos 30  0.65
For   0.65, cos  
 0.97 or   14.1
1   0.65  cos 30
1-37.
cos  
A
0.75m
1.5m
lamp
14.1
beam
The train is A from you when the headlight disappears, where
A
0.75m
 3.0m
tan14.1
1-38. (a) t  t0 For the time difference to be 1s, t  t0  1s
 1
t  t /   1  t 1    1
 
1
1 v2
Substituting  1 
(From Problem 1-20)

2 c2
3.0 108 


1 v2 
2
2
t 1  1 
  1  t  2c / v  2
2
2 c2 

1.5 106 / 3.6 103 
2
 1.04 1012 s  32, 000 y
(b)
t  t0  273  109 s. Using the same substitution as in (a).
t 1  1/    273 109 and the circumference of Earth C  40, 000km, so
4.0 107 m  vt or t  4.0 107 / v, and
4.0 10 / v   2c / v
7
2
2
 273 10  , or v 
9
2c 2  273 109 
4.0 107
 1230 m / s
Where v is the relative speed of the planes flying opposite directions. The speed of
each plane was 1230m / s  / 2  615 m / s  2210 km / h  1380 mph.
15
Chapter 1 – Relativity I
1-39.
y
y
S (other)
c
S  (Earth)
v
θ
x
x
(a)
cx  c cos   v
c y  c sin 
tan   
c y
cx

c sin 
sin 

c cos   v cos   v / c
(b) If   90 , tan   
1
1
 1
v/c 
 90     45
e.g., if v  0.5c,    63
1-40. (a) Time t for information to reach front of rod is given by:
ct 
Lp

 vt  t 
Lp
 (c  v )
Distance information travels in time t:
c (c  v)(c  v) / c 2
c 1  v2 / c2
cv
ct 

Lp 
Lp 
Lp
2
 (c  v )
(c  v )
cv
(c  v )
cLp
Since
(c  v) (c  v)  1 for v > 1, the distance the information must travel to reach
the front of the rod is  Lp ; therefore, the rod has extended beyond its proper length.
(b)

1
1 v / c
(ct  Lp ) 
1
Lp
1 v / c
16
Chapter 1 – Relativity I
Δ
0
0.11
0.29
0.57
0.73
1.17
2.00
2.50
3.36
5.25
8.95
v/c
0
0.10
0.25
0.40
0.50
0.65
0.80
0.85
0.90
0.95
0.98
Coefficient of extension vs. v/c
10
9
8
7
Delta
6
5
4
3
2
1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
v/c
(c) As v  c,    , the maximum length of the rod   also.
1-41. (a) Alpha Centauri is 4 c∙y away, so the traveler went L  1   2 8c  y  in 6 y, or
8c  y 1  v 2 / c 2  v  6 y 
1  v 2 / c 2  v  6 / 8c    3 / 4  v / c 
 1   2  3 / 4  2
2
2
3 / 4  2   2  1
 2  1/ 1  0.5625 
v  0.8c
(b)
t  t0    6 y  and   1/ 1   2  1.667
t  1.667  6 y   10 y or 4y older than the other traveler.
17
Chapter 1 – Relativity I
(Problem 1-41 continued)
(c)
ct 
10
ct
(c∙y)
8
x
6
4
2
Alpha Centauri
0
Earth
0
2
4
6
8
10
x
1-42. Orbit circumference  4.0 107 m .
Satellite speed v  4.0 107 m /  90 min  60s / min   7.41103 m / s
t  t0  tdiff
1 
t  t /   tdiff  t 1  1 /    t   2  (Problem 1-20)
2 
tdiff   3.16  107 s  1 / 2   7.41  103 / 3.0  108 
 0.0096s  9.6ms
1-43. (a) t   t 
v
c2
x (Equation 1-20)
For events to be simultaneous in S , t   0.
v
v
1.5 103 m 
2 
c
c
6
8
v   2 10 s  3 10 m / s  s /1.5 103 m
t 
2
x  2 106 s 
 1.2 108 m / s  0.4c
(b) Yes.
18
2
Chapter 1 – Relativity I
(Problem 1-43 continued)
ct 
ct
(c)
1000
x
500
B
0
500
1000
1500
x
0
A
(d)
 s 
2
  x    ct 
2
2
(Equation 1-33)
 1.5  103    3  108 m / s  2  106 s  
 2.25  106  3.6  105  1.89  106 m2
s  1370m
L  s  1370m
2
2
1-44. (a)   1/ 1   2  1/ 1   0.92   2.55
2
(b)   2.6 108 s
t  lab     2.55  2.6 108 s   6.63 108 s
(c) N  t   N0 et /  (Equation 1-29) L  1   2 L0  1   0.92   50m   19.6m
2
Where L is the distance in the pion system. At 0.92c, the time to cover 19.6m is:
t  19.6m / 0.92c  7.0 108 s. So, for N0 = 50,000 pions initially, at the end of 50m
in the lab, N   5.0 104  e7.0/ 2.6  3,390
(d) 47
19
Chapter 1 – Relativity I
1-45.
 1 v2 
L  Lp  L  Lp  Lp 1/    Lp 1  1/    Lp  2  (See Problem 1-20)
2c 
For Lp  11m and v  3 104 m / s
L  11 0.5 108   5.5 108 m
"Shrinkage" 
5.5 108 m
 550 atomic diameter
1010 m / atomic diameter
1-46. (a)
ct 
ct
1500
B
1000
500
A
0
0
500
1000
1500
2000
x
(b) Slope of ct  axis  2.08  1/  , so   0.48 and v  1.44 108 m / s
(c)
ct    ct and   1/ 1   2 so ct  1   2  ct
ct  877m t   1.5  877  / c  4.39 s
For ct   1000m and   0.48
(d) t  t   1.14t   t   5 s /1.14  4.39 s
1-47. (a) L  Lp /   Lp 1  u 2 / c 2  100m 1   0.85  52.7m
2
(b) u 
u u
0.85c  0.85c 1.70c


 0.987c
2
2
1  uu / c
1.72
1   0.85
(c) L  Lp /    Lp 1  u2 / c2  100m 1  1.70 /1.72   16.1m
2
(d) As viewed from Earth, the ships pass in the time required for one ship to move its
L
52.7m
 2.1107 s
own contracted length. t  
8
u 0.85  3.00 10 m / s
20
Chapter 1 – Relativity I
(Problem 1-47 continued)
0.987c
(e)
100 m
16 m
1-48. In Doppler radar, the frequency received at the (approaching) aircraft is shifted by
approximately f / f0  v / c. Another frequency shift in the same direction occurs at the
receiver, so the total shift f / f0  2v / c.
v   c / 2  8 107   120m / s .
1-49.
#1
#2
For star #1:
v  32km / s  3.2 104 m / s
cm
period  115d
L
c  c  3.2 104
c+
c−
Earth
c  c  3.2  104
Simultaneous images of star#1 in opposition
will appear at Earth when L is at least as large
as:
21
Chapter 1 – Relativity I
(Problem 1-49 continued)
L
L
 57.4d 
4
c  3.2 10
c  3.2  104
c  3.2  104
L
L
c  3.2  104

75.5
d

24
h
/
d

3600
s
/
h


c  3.2  104 c  3.2  104
c  3.2  104 c  3.2  104
 c  3.2 10  L   57.5d  24  3600   c  3.2 10  L
c   3.2 10 
c   3.2  10 
4
4
4 2
2
4 2
2
2
 c  3.2 104    c  3.2  104   L  c 2   3.2  104    57.5  24  3600 




c 2   3.2  104 2   57.5  24  3600 


L 
6.4 104
L  6.99  1018 m  739c y
1-50.
t2  t1    t2  t1  
v
 xb  xa  (Equation 1-20)
c2
(a) t2  t1  0   t2  t1    v / c 2   xb  xa  
 0.5 1.0 y   v / c 2   2.0 1.0  c
Thus, 0.5   v / c   v  0.5c in the  x direction.
(b) t     t  vx / c 2 
Using the first event to calculate t  (because t  is the same for both events),
2
t   1 / 1   0.5  1 y   0.5c 1c y  / c 2   1.155 1.5  1.7 y


(c)
 s 
2
  x    ct   1c y    0.5c y   0.75  c y 
2
2
2
2
(d) The interval is spacelike.
(e) L  s  0.866c y
22
2
 s  0.866c y
y
Chapter 1 – Relativity I
1-51. (a)
ct (c∙y)
1.7y
1
2
1.0
x
x (c∙y)
-2.0
-1.0
1.0
2.0
Because events are simultaneous in S  , line between 1 and 2 is parallel to x axis.
Its slope is 0.5   .
v  0.5c.
(b) From diagram t   1.7 y.
1-52.
xb  xr    xb  xr   v  tb  tr 
(1)
tb  tr    tb  tr   v  xb  xr  / c 2 
(2)
Where xb  xr  2400m tb  tr  5 s xb  xr  2400m tb  tr  5 s
Dividing (1) by (2) and inserting the values,
2400  v  5 106 
400
2

 2400  v  2400  / 5 106 c 2  2400  5 10 6 v
6
6
2
5 10
5 10  v  2400  / c
  2400 2

6
v

5

10
  4800  v  2.69 108 m / s in  x direction.
6 2
 5 10 c

1-53.
ux  0.85c cos50
u y  0.85c sin 50
  0.72   1/ 1   2  1.441
v  0.72c
23
Chapter 1 – Relativity I
(Problem 1-53 continued)
ux 
ux  v
1  vux / c 2
ux 
0.85c cos 50  0.72c
0.1736c

 0.286c
2
1   0.72c   0.85c cos 50 / c  1  0.3934
uy 
uy 
uy
 1  vux / c 2 
0.85c sin 50
1.441 1   0.72c   0.85c cos50 / c 2  
 0.745c
u  ux2  uy2  0.798c
tan    uy / ux  0.745 /  0.286    111 with respect to the  x axis.
1-54. This is easier to do in the xy and xy planes. Let the center of the meterstick, which is
parallel
to
the
x-axis
and
x  y  x  y  0 at t  t   0.
moves
upward
with
speed
vy
in
S,
at
The right hand end of the stick, e.g., will not be at
t   0 in S  because the clocks in S  are not synchronized with those in S. In S  the
components of the sticks velocity are:
uy 
ux 
uy
 1  vu x / c 2 

vy

because u y  v y and u x  0
ux  v
 v because u x  0
1  vu x / c 2
When the center of the stick is located as noted above, the right end in S  will be at:
x    x  vt   0.5 because t  0. The S  clock there will read: t     t  vx / c 2   0.5 v / c 2
Because t = 0. Therefore, when t   0 at the center, the right end is at xy given by:
x  0.5
and    tan 1
For   0.65
v y  0.5 v 
  c 2 
v  0.5 v 
y
 tan 1 y  2  / 0.5  tan 1  v y v / c 2  1   2
x
  c 
y   uy t  
    0.494v y / c 
24
Chapter 1 – Relativity I
1-55.
0  x 2  y 2  z 2   ct 
2
   x  vt     y 2  z 2  c  t   vx   / c 2 
2
2
 x2   2  c 2 2 v 2 / c 4   y 2  z 2  t 2   2 v 2  c 2 2   x t   2 2 v  2vc 2 2 / c 2 
 x2  y 2  z 2   ct 
2
1-56. The solution to this problem is essentially the same as Problem 1-53, with the manhole
taking the place of the meterstick and with the addition of the meterstick moving to the
right along the x-axis. Following from Problem 1-53, the manhole is titled up on the right
and so the meterstick passes through it; there is no collision.
y
manhole
x
meterstick
1-57. (a)
t2    t2  vx2 / c 2  and t1    t1  vx1 / c 2 
t2  t1   t2  t1  v  x2  x1  / c 2    T  vD / c 2 
(b) For simultaneity in S , t2  t1, or T  vD / c2
 v / c  cT / D.
Because v / c  1, cT / D is also <1 or D  cT .
(c) If D < cT, then T  vD / c 2   T  vcT / c 2   T 1  v / c  . For T > 0 this is always
positive because v/c < 1. Thus, t2  t1   T  vD / c 2  is always positive.
(d) Assume T  D / c with c  c. Then
 c v
T  vD / c 2   D / c    vD / c 2    D / c    
 c c 
This changes sign at v / c  c / c which is still smaller than 1. For any larger v
still smaller than c)
t2  t1   T  vD / c 2   0 or t1  t2
25
Chapter 1 – Relativity I
1-58.

v  0.6c
1
1   0.6 
2
 1.25
(a) The clock in S reads   60min  75min when the S  clock reads 60 min and the first
signal
from
S  is
sent.
At
that
time,
the
S  observer
is
at
v  75min  0.6c  75min  45c min. The signal travels for 45 min to reach the S
observer and arrives at 75 min + 45 min = 120 min on the S clock.
(b) The observer in S sends his first signal at 60 min and its subsequent wavefront is
found at x  c  t  60 min  .
The S  observer is at x  vt  0.6ct and receives the
wavefront when these x positions coincide, i.e., when
c  t  60 min   0.6ct
0.4ct  60c min
t   60c min  / 0.4c  150 min
x  0.6c  0 min   90c min
The confirmation signal sent by the S  observer is sent at that time and place, taking
90 min to reach the observer in S. It arrives at 150 min + 90 min = 240 min.
(c) Observer in S:
Sends first signal
Receives first signal
Receives confirmation
60 min
120 min
240 min
The S  observer makes identical observations.
1-59. Clock at r moves with speed u  r , so time dilation at that clock’s location is:
t0  t  tr  t0 1  u 2 / c 2  t0 1  r 2 2 / c 2
 1

c, tr  t0 1  r 2 2 / c 2 
 2

 1

t0 1  r 2 2 / c 2   t0
t  t0
r 2 2
 2

And, r


t0
t0
2C 2
Or, for r
26
Chapter 1 – Relativity I
1-60.
vB
ux 
ux  v
1  ux v / c 2
uy 
(a) For vBA : v  vB , vAx  0, vAy  vA . So, vAx 
vAy
 B 1  vAx v / c
2


v A
where  B 
B
vBA  vAx2  vAy2  vB2   vA /  B 
 
tan  BA
1  vB2 / c 2
2
vAy vA /  B
v

 A
vAx
 vB
 B vB
vBy
 A 1  vBx v / c
vAB 
vAx  v
 vB
1  vAx v / c 2
1
 
(b) For vAB : v  vA , vBy  vB , vBx  0. So, vBx
 
vBy
 1  ux v / c 2 
VA
Space station
vAy 
uy
2


vB
A
 v A    vB /  A 
 
tan  AB
2
where  A 
vBx  v
 vA
1  vBx v / c 2
1
1  vA2 / c 2
2
vB
v
 B
 A  v A 
 A vA
(c) The situations are not symmetric. B viewed from A moves in the +y direction, and A
viewed from B moves in the –y direction, so tan  A   tan  B  45 only if vA  vB
and  A   B  1, which requires vA  vB  0.
27
Chapter 1 – Relativity I
1-61.
1-62. (a) Apparent time A  B  T / 2  t A  tB and apparent time B  A  T / 2  t A  tB where
tA = light travel time from point A to Earth and tB = light travel time from point B to
Earth.
A B 
T
L
L
T
2vL


  2 2
2 cv cv 2 c v
B A
T
L
L
T
2vL


  2 2
2 cv cv 2 c v
(b) Star will appear at A and B simultaneously when tB  T / 2  t A or when the period is:
L 
4vL
 L
T  2 t B  t A   2 

 2 2

c  v c  v  c  v
1-63. The angle of u  with the x axis is:
uy
uy
1  vux / c
tan   

ux
ux  v
 vu 
 1  2x 
c 

tan   
uy
  ux  v 

2

u sin 
sin 

  u cos   v    cos   v / u 
28
Chapter 1 – Relativity I
#1
1-64.
d  vt cos 
θ
d
xEarth  vt sin 
#2
xEarth
L
L
to Earth
(a) From position #2 light reaches Earth at time t2  L / c. From position #1 light reaches
L d
Earth at time t1    t.
c c
L L d

    t 
c c c

vt cos 

 t
c
 t 1   cos  
t Earth  t2  t1 
t Earth
t Earth
(b)
Bapp 
vapp
c

xEarth
vt sin 
 sin 


ctEarth ct 1   cos   1   cos 
(c) For   0.75 and vapp  c  app  1, the result in part (b) becomes
1
0.75sin 
 sin   cos   1/ 0.75
1  0.75cos 
Using the trigonometric identity
p sin   q cos   r sin   A where r 
sin   cos   2 sin   45   1/ 0.75
sin   45   1/ 0.75 2
  45  70.6
  25.6
29
p2  q2
sin A  p / r
cos A  q / r
Chapter 2 – Relativity II
2-1.
2
u yB
 uo2 1  v 2 / c 2 
2
uxB
 v2
2
1   u xB
 uB2  / c 2  1  v 2 / c 2   uo2 / c 2 1  v 2 / c 2 
1  v

2
/ c 2 1  uo2 / c 2 
 1  v 2 / c 2 
1/ 2
p yB 
mu yB
2
2
1   u xB
 u yB
 / c2
1  u
2
o
/ c2 
1/ 2
muo 1  v 2 / c 2

1  v 2 / c 2 1  uo2 / c 2
 muo / 1  uo2 / c 2   p yA
2-2.
d  mu   m  ud    du 
  1  2u   u 2 3/ 2  u 2  1/ 2 
 m u    2  1  2   1  2   du
  2  c   c 
 c  
 u 2  u 2  3/ 2  u 2  u 2  3/ 2 
 m  2 1  2   1  2 1  2   du
 c  c 
 c  c  
 u2 
 m 1  2 
 c 
2-3.
(a)  
1
1 u / c
2
2
3/ 2
du
1

1   0.6c  / c 2
2

1
 1.25
0.8
(b) p   mu    mc   u / c  / c  1.25  0.511 MeV  0.6  / c  0.383 MeV / c
2
(c) E   mc 2  1.25  0.511 MeV   0.639 MeV
(d) Ek    1 mc2  0.25  0.511 MeV   0.128 MeV
31
Chapter 2 – Relativity II
2-4.
The quantity required is the kinetic energy. Ek    1 mc 2  1  u 2 / c 2 





2
(a) Ek   1   0.5

2
(b) Ek   1   0.9 


1/ 2
2
(c) Ek   1   0.99 

2-5.
1/ 2

1/ 2
 1 mc 2

 1 mc 2  0.155mc 2


 1 mc 2  1.29mc 2

1/ 2

 1 mc 2  6.09mc 2

E  mc 2  m  E / c 2 
10 J
3.08 10 m / s 
8
2
 1.11016 kg
Because work is done on the system, the mass increases by this amount.
2-6.
m(u)  m / 1  u 2 / c 2
(a)
(Equation 2-5)
m(u )  m
 0.10 
m
1
1 u / c
2
2
m / 1  u 2 / c2  m
 0.10
m
1
 1  0.10 
1 u / c
2
2
 1.10  1  u 2 / c 2  1/ 1.10 
Thus, u 2 / c2  1  1/ 1.10   0.1736  u / c  0.416
2
(b) m(u)  5m
m
1 u / c
2
2
 5m  1  u 2 / c 2  1/ 25
Thus, u 2 / c2  1 1/ 25  0.960  u / c  0.980
(c) m(u)  20m
m
1 u / c
2
2
 20m  1  u 2 / c 2  1/ 400
Thus, u 2 / c2  1  1/ 400  0.9975  u / c  0.99870
32
2
Chapter 2 – Relativity II
2-7.
(a) v   Earth  moon distance  / time  3.8 108 m /1.5s  0.84c
(b)
Ek   mc 2  mc 2  mc 2   1
(Equation 2-9)
  1 / 1   0.84   1.84
2
mc 2 (proton)  938.3MeV
Ek  938.3MeV 1.87  1  813MeV
(c) m(u ) 
m
1 u / c
2
2

938.3MeV / c 2
1   0.84 
2
 1.730 103 MeV / c 2  1.730GeV / c 2
1 2 1
2
mv   938.3MeV / c 2   0.84c   331MeV
2
2
813Mev  331MeV
% error 
100  59%
813MeV
(d) Classically, Ek 
2-8.
Ek  mc 2   1
(Equation 2-9)
Ek  u2   Ek  u1   W21  mc 2   u2   1  mc 2   u1   1
Or, W21  mc 2   u2     u1 
1/ 2
1/ 2
(a) W21  938.3MeV 1  0.162 
 1  0.152    1.51MeV


1/ 2
1/ 2
 1  0.852    57.6MeV
(b) W21  938.3MeV 1  0.862 


(c) W21  938.3MeV 1  0.962 

2-9.
1/ 2
 1  0.952 
1/ 2
  3.35 103 MeV  3.35GeV

E   mc2 (Equation 2-10)
(a) 200GeV    0.938GeV  where mc2  proton   0.938GeV

1
1  v2 / c2
v
1
 1 2
c
2

200GeV
 213
0.938GeV
(Equation 2-40)
v
1
 1
 1  0.00001102 Thus, v  0.99998898c
2
c
2  213
33
Chapter 2 – Relativity II
(Problem 2-9 continued)
(b) E  pc for E
mc2 where E  200GeV 197  3.94  104 GeV (Equation 2-36)
p  E / c  3.94 104 GeV / c
(c) Assuming one Au nucleus (system S  ) to be moving in the +x direction of the lab
(system S), then u for the second Au nucleus is in the −x direction. The second Au’s
energy measured in the S  system is:
E     E  vp x    213  3.94  104 GeV  v 3.94  104 GeV / c 
  213  3.94  104 GeV  1  v / c 
  213  3.94  104 GeV   2 
 1.68  107 GeV
px    p x  vE / c 2    213  3.94  104 GeV  v 3.94  104 GeV / c 2 
   213  3.94  104 GeV   2 
 1.68  107 GeV / c
2-10. (a) E  mc 2  103 kg  c 2  9.0 1013 J
(b) 1kWh  1103 J  h / s  3600s / h   3.6 106 J
So, 9.0 1013 J / 3.6 106 J / kWh  2.5 107 kWh
@$0.10 / kWh would sell for $2.5 106 or $2.5 million.
(c) 100W  100 J / s, so 1g of dirt will light the bulb for:
9.0 1013 J
9.0 1011 s
 9.0 1011 s 
 2.82 104 y
7
100 J / s
3.16 10 s / y
2-11.
E   mc2
(Equation 2-10)
where   1/ 1  u 2 / c 2  1/ 1   0.2   1.0206
2
E  1.0206 0.511MeV   0.5215 MeV
Ek   mc 2  mc 2  mc 2    1
(Equation 2-9)
=  0.511MeV 1.0206  1  0.01054MeV
34
Chapter 2 – Relativity II
(Problem 2-11 continued)
E 2   pc    mc 2 
2

2
(Equation 2-32)

2
1 2
1
2
2
E   mc 2   2  0.5215MeV    0.511MeV  
2

c
c 
2
2
 0.01087 MeV / c  p  0.104 MeV / c
p2 
2-12.
E   mc 2
(Equation 2-10)
  E / mc  1400MeV / 938MeV  1.4925
2
(a)
  1/ 1  u 2 / c 2  1.4925  1  u 2 / c 2  1/ 1.4925
2
u 2 / c 2  1  1/ 1.4925  0.551  u  0.74c
2
(b)
E 2   pc    mc 2 
2
2
(Equation 2-32)
2
1
1
2
2
p   E 2   mc 2    1400MeV    938MeV    1040MeV / c



 c
c
2-13. (a)
pR   mS v
pN  mS v
  1/ 1  v 2 / c 2  1/ 1   2.5 105 / 3.0 108 
2
  1.00000035
3
3
pR  pN  mS  2.5 10   mS  2.5 10    1
3.5 107



 3.5  107
3
pR

1.00000035
 mS  2.5 10 
(b)
ER   mS c 2  mS c 2     1 mS c 2
1
mS v 2
2
2
2
ER  EN    1 c  0.5v

ER
  1 c 2
EN 
3.5  107 c 2  0.5v 2

3.5  107 c 2
 1
0.5  2.5 105 
3.5 107 c 2
2
 0.0079
35
Chapter 2 – Relativity II
2-14.
u  2.2 106 m / s and   1/ 1  u 2 / c 2
(a)
Ek  0.511MeV    1  0.511MeV 1/ 1  u 2 / c 2  1  0.5110  2.689  105 


 1.3741105 MeV
2
1
1
mu 2  mc 2  u 2 / c 2    0.5110MeV / 2   2.2 106 / c 2 
2
2
5
 1.374 10 MeV
Ek (classical) 
% difference =
1109
100  0.0073%
1.3741105
(b)
p
2
1
1
E 2   mc 2  
c
c
mc 2
 mc   1 
c
2
 mc    mc 
2 2
2 2
2


6
8 2 
1/ 1   2.2 10 / 3.0 10    1



1/ 2
 0.5110MeV / c  7.33 103   3.74745  103 MeV / c
p (classical)  mu 
mc 2  u 
6
8
   0.5110MeV / c   2.2  10 / 3.0  10 
2 
c c
 3.74733 10 3 MeV / c
% difference =
1.2 107
 100  0.0030%
2.74745 103
2-15. (a)
60W  60 J / s  3.16 107 s / y   1.896 109 J
m  E / c 2  1.896 109 J /  3.0 108 m / s   2.1108 kg  2.1105 g  21 g
2
(b) It would make no difference if the inner surface were a perfect reflector. The light
energy would remain in the enclosure, but light has no rest mass, so the balance
reading would still go down by 21 μg.
2-16.
4
He  3H  p  e
Q  m  3H   mp  me  m  4 He  c 2
 2809.450MeV  938.280MeV  0.511MeV  3728.424MeV  19.827MeV
36
Chapter 2 – Relativity II
2-17.
H  2H  n
Energy to remove the n = 22.014102u  2 H   1.008665u  n   3.016049u  3H 
3
 0.006718u  931.5MeV / u  6.26MeV
2-18. (a) m 
E Eu
4.2eV
 2 
u  4.5 109 u
2
6
c
uc
931.5 10 eV
(b) error =
2-19. (a)
m
4.5 109 u

 7.7 1011  7.7 109 %
m  Na   m  Cl  23u  35.5u
m  m  He   2m  H  
4
2
m  4 He  c 2  2m  2 H  c 2
u
uc 2
 3727.409MeV  2 1875.628MeV  u / 931.5MeV
 0.0256u
(b) E  m c 2   0.0256uc 2  931.5MeV / uc2   23.8MeV
(c)
dN
P
1W
1eV



 2.62 1011 / s
dt E 23.847MeV 1.602 1019 J
2-20.
v
hf
M
(a) before photon absorbed:
1.01M
after photon absorbed:
E  hf  Mc2
E f  Ekinetic  1.01Mc 2
Conservation of energy requires: hf  Mc 2  Ekinetic  1.01Mc 2
Rearranging, the photon energy is:
hf  1.01Mc 2  Mc 2  Ekinetic
hf  0.01Mc 2  Ekinetic  0.01Mc 2
(b) The photon’s energy is hf  0.01Mc2 because the particle recoils from the absorption
of the photon due to conservation of momentum. The recoil kinetic energy (which is
greater than 0) must be supplied by the photon.
37
Chapter 2 – Relativity II
2-21. Conservation of energy requires that Ei2  E 2f , or
 pi c 
2
  2m p c 2    p f c    2m p c 2  m c 2  and conservation of momentum requires that
2
2
2
pi  p f , so
4  m p c 2   4  m p c 2   2m p c 2  2m c 2   m c 2 
2
2
0  2m p c 2  2m c 2   m c 2 
2
2


m c2 
m
0  m c 2  2   2   m c 2  2  


2m p c 
2m p





Thus, m c 2  2  m / 2mp  is the minimum or threshold energy Ei that a beam proton
must have to produce a π0.


m c 2 
135 
2
E  m c  2 

135
MeV
2



  280MeV


2m p c 2 
2  938 


2-22.
E  mc 2  m  E / c 2 
2-23. (a)
200 106 eV
 3.57 1025 g
32
 5.6110 eV / g 
3
kT per atom
2
3
E  1.38 1023 J / K 1.50 107 K 
2
E  3.105 1016 J / atom
E
H atoms/kg  1kg /1.67 1027 kg / atom  6.0 1026 atoms
Thermal energy/kg   3.1 1016 J / atom  6.0  1026 atoms   1.86  1011 J
(b)
E  mc 2
 m  E / c2
m  1.86 1011 J / c 2  2.07 106 kg
38
Chapter 2 – Relativity II
p  E / c  1.0 J / s  / c
2-24. 1.0W  1.0 J / s 
(a) On being absorbed by your hand the momentum change is p  1.0 J / s  / c and,
from the impulse-momentum theorem,
F t  p where t  1s  F  1.0 J / s  / ct  1.0 / c  N  3.3 109 N
This magnitude force would be exerted by gravity on mass m given by:
m  F / g  3.3 109 N /  9.8m / s 2   3.4 1010 kg  0.34 g
(b) On being reflected from your hand the momentum change is twice the amount in part
(a) by conservation of momentum. Therefore, F  6.6 109 and m  0.68 g.
2-25. Positronium at rest:  2mc 2   Ei2   pi c 
2
2
Because pi  0, Ei  2mc2  2  0.511MeV   1.022MeV
After photon creation;  2mc 2   E 2f   p f c 
2
2
Because pf  0 and energy is conserved,  2mc 2   E 2f  1.022MeV  or
2
2
2mc2  1.022MeV for the photons.
2-26.
E 2   pc 2    mc 2 
2
E   pc    mc

2
2
(Equation 2-31)
 
1/ 2
2 2
 mc 1   p 2 / m2 c 2  
1/ 2
  pc 2 
 mc 1   2  
  mc  
2
1/ 2
2
p2 

2 
  mc 1  2m 2 c 2 


2
 1
 mc 1   p / mc  
 2
2
 mc 2  p 2 / 2m
2-27.
E 2   pc 2    mc 2 
2
2
(Equation 2-31)
(a)  pc   E 2   mc 2    5MeV    0.511MeV   24.74 MeV 2
2
2
2
2
or, p  24.74 MeV / c  4.97MeV / c
(b) E   mc 2
 1  u 2 / c 2   mc 2 / E 
   E / mc 2  1/ 1  u 2 / c 2
2
u / c  1   mc 2 / E  


1/ 2
1/ 2
2
 1   0.511/ 5.0  


39
 0.995
2
Chapter 2 – Relativity II
2-28.
4
3
E/mc2
2
1
0
0
2-29.
1
E 2   pc 2    mc 2 
2
1746MeV 
2
2
2
3
(Equation 2-31)
  500MeV    mc 2 
2
2
1/ 2
2
2
mc 2  1746MeV    500MeV  


E   mc 2
 1673MeV
 m  1673MeV / c 2
   1/ 1  u 2 / c 2  E / mc 2
2
u / c  1   mc 2 / E  


1/ 2
2-30. (a) BqR  m u  p
B
p/mc
4
1/ 2
2
 1  1673MeV /1746MeV  


 0.286  u  0.286c
(Equation 2-37)
m u
and E   mc 2 which we have written as (see Problem 2-29)
qR
2
u / c  1   mc 2 / E  


1/ 2
1/ 2
2
 1   0.511MeV / 4.0MeV  


And   1/ 1  u 2 / c 2  1/ 1   0.992   7.83
2
9.1110 kg   7.83 0.992c   0.316T
Then, B 
1.60 10 C  4.2 10 m 
31
19
2
(b)  m exceeds m by a factor of γ = 7.83.
40
 0.992
Chapter 2 – Relativity II
2-31. (a) p  qBR  e  0.5T  2.0  
3.0 108 m / s
 300MeV / c
c
2
2
Ek  E  mc 2   pc    mc 2  


1/ 2
(b)
 mc 2
1/ 2
2
2
  300MeV    938.28MeV  


 46.8MeV
 938.28MeV
2-32. The axis of the spinning disk, system S  , is the z-axis in cylindrical coordinates.
r  r, z  z,      t  dr  dr , dz  dz, d  d   dt
Therefore, d  d    dt and d 2  d 2  2 d  dt   2 dt 2 . Substituting for
d 2 in Equation 2-43 yields
ds 2  c2 dt 2  dr 2  r 2 (d 2  2 d  dt   2 dt 2 )  dz 2 
Simplifying, we obtain
ds 2  (c2  r 2 2 ) dt 2  (dr 2  r 2 d 2  2r 2 d  dt  dz 2 ) which is Equation 2-44.
2-33.   4GM / c2 R
(Equation 2-44)
Earth radius R  6.37 106 m and mass M  5.98 1024 kg

4  6.67 1011 N m2 / kg 2  5.98 1024 kg 
 3.00 10 m / s   6.37 10 m 
8
2
6
 2.78 109 radians
  2.87 104 arc seconds
2-34. Because the clock furthest from Earth (where Earth’s gravity is less) runs the faster,
answer (c) is correct.
41
Chapter 2 – Relativity II
2-35.
6  6.67 1011 N  m2 / kg 2 1.99 1030 kg 
6 GM
  2

2
c 1   2  R  3.00 108 m / s  1  0.0482  7.80 1011 m 
(Equation 2-51)
 3.64 108 radians/century  7.55 103 arc seconds/century
r 2 d 
2-36. From Equation 2-45, dt  dt   2 2 2
c r 
2
2r 2 d dt   r 2 d  
and dt  dt   2 2 2   2 2 2  .
c r 
c r  
2
2
Substituting dt and dt 2 into Equation 2-44 yields
2

2r 2 d dt   r 2 d   
2
ds  (c  r  )  dt   2 2 2   2 2 2  
c r 

 c  r   
 2 2


r 2 d  
2
2
  dr  r d   2r  d   dt   2 2 2   dz 2 
c r  



2
2
2
2
 r  d  
 2r  d  dt  
2
ds  (c  r  )dt 
2
2
2
2
2
2
2
c 2  r 2 2
2
2



2
r

d



2
2
2
2
2

 dr  r d   2r  d  dt   2 2 2  dz


c r 


Cancelling the 2r 2 d dt  terms, one of the 
 r  d  

2
r d 
2
2
2
c 2  r 2 2

2  r 2 d  
c 2  r 2 2
2
terms, and noting that
c 2 r 2 d  2
c 2  r 2 2
we have that
ds 2  (c 2  r 2 2 )dt 2  (dr 2 
c 2 r 2 d  2
 dz 2 ) which is Equation 2-46.
2
2 2
c r 
2-37. The transmission is redshifted on leaving Earth to frequency f , where f0  f  f0 gh / c 2 .
Synchronous satellite orbits are at 6.623RE where
g
GM E
 6.623RE 
2

9.9m / s 2
 6.623
2
 0.223m / s 2
42
Chapter 2 – Relativity II
(Problem 2-37 continued)
h  6.623RE  6.623  6.37 106 m   4.22 107 m
f0  f   9.375 109 Hz  0.223m / s 2  4.22 107  /  3.00 108   0.980Hz
2
f  f0  0.980Hz  9.374999999 109 Hz
2-38.
Earth
white dwarf
distant star
57 c∙y
35 c∙y
On passing “below” the white dwarf, light from the distant start is deflected through an
angle:
  4GM / c R 
2
4  6.67 1011 N  m2 / kg 2   3 1.99 1030 kg 
 3.00 10 m / s  10 m 
2
8
7
(Equation 2-44)
 1.77 103 radians  0.051 or the angle between the arcs is 2  0.102
2-39. The speed v of the satellite is:
v  2 R / T  2  6.37 106 m  /  90 min 60s / min   7.42 103 m / s
Special relativistic effect:
After one year the clock in orbit has recorded time t  t /  , and the clocks differ by:
t  t   t  t /   t 1  1/    t  v 2 / 2c 2  , because v
c. Thus,
t  t    3.16 107 s  7.412 103  /  2  3.00 108 m   0.00965s  9.65ms
2
2
Due to special relativity time dilation the orbiting clock is behind the Earth clock by
9.65ms.
43
Chapter 2 – Relativity II
(Problem 2-39 continued)
General relativistic effect:
2
5
f gh  9.8m / s  3.0 10 m 


 3.27 1011 s / s
2
8
f0 c2
 3.0 10 m / s 
In one year the orbiting clock gains  3.27 1011 s / s  3.16 107 s / y   1.03ms .
The net difference due to both effects is a slowing of the orbiting clock by 9.65−1.03 =
8.62 ms.
2-40. The rest energy of the mass m is invariant, so observers in S  will also measure m =
4.6kg, as in Example 2-9. The total energy E  is then given by:
 mc    E  pc 
2 2
2
2
Because, p  0, E  mc 2  4.6kg   3.0 108 m / s   4.14 1014 J
2
2-41. (a)
E   me c 2
   E / mec 2  50 10 MeV / 0.511MeV  9.78 104
L  L0 /   102 m
L0  9.78 104 102 m   978m (length of one bundle)
The width of one bundle is the same as in the lab.
(b) An observer on the bundle “sees” the accelerator shortened to 978m from its proper
length L0 , so L0    978  978 104  978  9.57 107 m.
(Note that this is about
2.5 times Earth’s 40,000km circumference at the equator.)
(c) The e+ bundle is 10−2 m long in the lab frame, so in the e− frame its length would be
measured to be: L  102 m  /   102 m / 9.78 104  1.02 107 m .
44
Chapter 2 – Relativity II
2-42.
Ek   mc2  mc2  mc2   1 If Ek  mc2  938MeV , then   2.
(a)
 mc 
2 2
 pc 
2
 E 2   pc 
2
(Equation 2-32) Where E   mc 2  2  938MeV 
 E 2   mc 2    2  938    938   2.46 106
2
p   2.64 106 
1/ 2
2
2
/ c  1.62 103 MeV / c
(b) p   mu  u  p /  m  1.62 103 MeV / c  /  2   938MeV / c 2   0.866c
2-43. (a) The momentum of the ejected fuel is:
p   mu  mu / 1  u 2 / c 2  103 kg  c / 2  / 1   0.5  1.73 1011 kg  m / s
2
Conservation of momentum requires that this also be the momentum ps of the
spaceship: ps  msus / 1  us2 / c 2  1.73 1011 kg  m / s
ms us / 1  us2 / c 2  1.73 1011 kg  m / s 
Or,
2
ms2 cs2  1  us2 / c 2 1.73 1011 kg  m / s   1.73 1011 kg  m / s 2    3.33 105 kg 2  us2
2
10 kg 
6
Or,
2
us2   3.33 105 kg 2  us2  1.73 1011 kg  m / s 
2
us  1.73 1011 kg  m / s  /106 kg  1.73 105 m / s  5.77 10 4 c
(b) In classical mechanics, the momentum of the ejected fuel is: mu  mc / 2  103 c / 2,
which must equal the magnitude of the spaceship’s momentum msus, so
103 kg  3.0 108 m / s 
3
us  10  c / 2  / ms 
 5.0 104 c  1.5 105 m / s
6
2 10 kg 
(c) The initial energy Ei before the fuel was ejected is Ei  ms c 2 in the ship’s rest
system. Following fuel ejection, the final energy Ef is:
E f  energy of fuel + energy of ship  mc2 / 1  u 2 / c 2   ms  m  c 2 / 1  us2 / c 2
Where u  0.5c and us c, so E f  1.155mc 2   ms  m  c 2  1.155  1 mc 2  ms c 2
The change in energy ΔE is:
E  E f  Ei   0.155 103 kg  c 2  106 kg  c 2   106 kg  c 2 
E  155kg  c 2 or 155 kg  E / c 2 of mass has been converted to energy.
45
Chapter 2 – Relativity II
2-44. The observer at the pole clock sees the light emission of the equatorial clock as transverse
Doppler effect, measuring frequency f, where
f0 / f   1   cos  
(Equation 1-35a)
   / 2 for the equatorial clock, so f / f 0  1  v 2 / c 2  1 
1 v2
2 c2
f / f0  1  1.193 1012 (red shift)
The observer at the equatorial clock sees a gravitational blue shift for the pole clock and
observes
f / f 0  1  gh / c 2
(Equation 2-45)
f / f 0  1  2.897 1012 (blue shift)
2-45. (a)
p  300 BR  q / e 
(Equation 2-38)
p  300 1.5T   6.37 106  1  2.87 109 MeV
mc 2 , E  Ek and E  pc (Equation 2-32)  Ek  pc  2.87 109 MeV
For E
(b)
For E  pc, u  c and
T  2 R / c  2  6.37 106 m  / c  0.133s
2-46.
f
 1  GM / c 2 R
f0
(Equation 2-47)
The fractional shift is:
f0  f
f
 1   GM / c 2 R  7 104
f0
f0
The dwarf’s radius is:
R  GM / c  7  10
2
4

6.67  1011 N  m2 / kg 2  2  1030 kg 
 3.00 10 m / s   7 10 
2
8
Assuming the dwarf to be spherical, the density is:

M
2 1030 kg

 5.0 1010 kg / m3
V 4  2.12 106 m 3 / 3
46
4
 2.12  106 m
Chapter 2 – Relativity II
2-47. The minimum energy photon needed to create an e− − e+ pair is Ep = 1.022 MeV (see
Example 2-13). At minimum energy, the pair is created at rest, i.e., with no momentum.
However, the photon’s momentum must be p  E / c  1.022MeV / c at minimum. Thus,
momentum conservation is violated unless there is an additional mass “nearby” to absorb
recoil momentum.
2-48.
  1  vux / c 2  


uy
  m 

py   muy  
2
 1  u 2 / c 2 
  1  vux / c  
Canceling γ and 1  vux / c 2  , gives: py 
In an exactly equivalent way, pz  pz .
mu y
1  u 2 / c2
 py
2-49. (a) u   u  v  / 1  uv / c 2  where v  u, so u  2u / 1  u 2 / c 2  . Thus,
the speed of the particle that is moving in S  is: u  2u / 1  u 2 / c 2  from which we
see that:
2
4u
 u 
1     1  2 / 1  u 2 / c 2 
c
c
2
2
 1  2u 2 / c 2  u 4 / c 4  4u 2 / c 2  / 1  u 2 / c 2 
 1  u 2 / c 2  / 1  u 2 / c 2 
2
  u 2 
And thus, 1    
  c  
1/ 2
2
1  u 2 / c2

1  u 2 / c2
(b) The initial momentum pi in S  is due to the moving particle,
pi  mu / 1   u / c  where u and 1   u / c  were given in (a).
2
pi  m
2
2u 1  u 2 / c 2 
1  u
2
/c
2
1  u
2
/c
2

 2mu / 1  u 2 / c 2 
(c) After the collision, conservation of momentum requires that:
p f  Mu / 1  u 2 / c 2 
1/ 2
 pi  2mu / 1  u 2 / c 2 
47
or M  2m / 1  u 2 / c 2 
1/ 2
Chapter 2 – Relativity II
(Problem 2-49 continued)
(d) In S: Ei  2mc 2 / 1  u 2 / c 2 and E f  Mc 2 (M is at rest.) Because we saw in (c) that
M  2m / 1  u 2 / c 2 
1/ 2
, then Ei  E f in S .
In S : Ei  mc 2  mc 2 / 1   u / c  and substituting for the square root from (a),
2
Ei  2mc2 / 1  u 2 / c2  and E f  Mc2 / 1  u 2 / c 2 . Again substituting for
M from (c), we have: Ei  E f .
2-50. (a) Each proton has Ek  mp c 2   1 , and because we want Ek  m p c 2 , then γ = 2 and
u = 0.866c. (See Problem 2-40.)
ux  v
where u = v and ux =−u yields:
1  ux v / c 2
2  0.866c 
2u
ux 

 0.990c
2
2
1  u / c 1   0.866c 
(b) In the lab frame S  : ux 
(c) For u  0.990c,   1/ 1   0.99   7.0 and the necessary kinetic energy in the
2
lab frame S is: Ek  mp c 2   1  mp c 2  7  1  6mp c 2
2-51. (a) pi  0  E / c  Mv or v  E / Mc
(b) The box moves a distance x  vt , where t  L / c,
so x   E / Mc  L / c   EL / Mc 2
(c) Let the center of the box be at x = 0. Radiation of mass m is emitted from the left
end of the box (e.g.) and the center of mass is at:
xCM 
M  0  m  L / 2
M m

mL
2  M  m
When the radiation is absorbed at the other end the center of mass is at:
xCM 
M  EL / Mc 2   m  L / 2  EL / Mc 2 
M m
48
Chapter 2 – Relativity II
(Problem 2-51 continued)
Equating the two values of xCM (if CM is not to move) yields:
m   E / c 2  / 1  E / Mc 2 
Mc2 , then m  E / c 2 and the radiation has this mass.
Because E
2-52. (a) If v mass is 0:
E2   p c    m c  and Ev2   pv c   0
2
2
2
Ek   Ev  139.56755MeV  105.65839MeV
m c 2    1  Ev  33.90916MeV

Ev  pv c  E2   m c 2 

2 1/ 2
 33.90916  m c 2    1
Squaring, we have
 m c    1   33.90916

2 2
2
2
 2  33.90916   m c 2     1   m c 2     1
2
2
Collecting terms, then solving for    1 ,
 33.90916 
 1 
2
2  m c 2   2  33.90916  m c 2
2
Substituting m c 2  105.65839MeV
  1  0.0390    1.0390 so,
1
2
2 1/ 2
Ek   4.12MeV and pu  109.78   105.66    29.8MeV / c

c
Ev  29.8MeV and pv  29.8MeV / c
(b)
If v mass = 190 keV , then Ev2   pv c  +  m v c 2  and
2
2
Ek   Ekv  139.56755MeV  105.65839MeV   0.190MeV  33.71916MeV
Solving as in (a) yields
E  109.78MeV ,
2-53.
f
 1  Gm / c 2 R
f0
p  29.8MeV / c, Ev  29.8MeV , and pv  29.8MeV / c
(Equation 2-47)
Since c  f  and c  f 0 0 ,
6.67 10 N  m / kg 1.99 10 kg 

  0  1  GM / c 2 R  1 
2
 c 
 3.00 108 m / s   6.96 106 m 
c
11
0
 1  0.000212  0.999788
49
2
2
30
Chapter 2 – Relativity II
(Problem 2-53 continued)
  0 / 0.999788  720.00nm / 0.999788  720.15nm
    0  0.15nm
u y   u y /   1  vu x / c 2 
2-54.
du y
ay 
du y
dt


1  vu
1
u y  vdx 
2 2
 2  1  vu x / c 
  c 
  dt  vdx / c 2 
/ c2  
1
x
2 1
2
2 2 

1   du y / dt  1  vu x / c    u y v / c   du x / dt  1  vu x / c  
ay  2

 
1  v  dx / dt  / c2 


2
ay
ax u y v / c
ay 

2
3
 2 1  vu x / c 2   2 1  vu x / c 2 
2-55. (a)
Fx 
dpx d  mv 

dt
dt
Fx  ma x because u x  0
Fx   m  dv / dt   mv d 1  v 2 / c 2 

Fx 
Fx 
ma x
1  v 2 / c2 
ma x

m  v2 / c2  ax
1  v / c 
1  v / c   m  v / c  a
1  v / c 
1/2
2
 / dt

2 3/2
2
2
2
1/2
2
2
x
2 3/2
Fx   3ma x
Because u x  0, note from Equation 2-1 (inverse form) that a x  a x /  3 .
Therefore, Fx   3ma x /  3  ma x  Fx
(b)
Fy 
dp y

d  mv y 
Fy  ma y because u y  u x  0
dt
dt
Fy   ma y because S  moves in +x direction and the instantaneous transverse
impluse (small) changes only the direction of v. From the result of Problem 2-52
(inverse form) with u y  ux  0, a y  a y /  2
Therefore, Fy   ma y   ma y /  2  ma y /   Fy / 
50
Chapter 2 – Relativity II
2-56. (a) Energy and momentum are conserved.
Initial system: E  Mc2 ,
p0
Invariant mass:  Mc 2   E 2   pc    Mc 2   0
2
2
2
Final system:
 2mc    Mc 
2 2
2 2
0
For 1 particle (from symmetry)
 mc    Mc / 2
2 2
2
2
 p 2c 2   Mc 2 / 2    muc 
2
2
 Mc 2 
2
  u / c 
Rearranging, 1  
2 
 2mc 
  2mc 2 2 

Solving for u, u  1  
2 
  Mc  
2
2
 Mc 2 
1
  

2 
2
2
 2mc  1  u / c
2
1/ 2
c
(b) Energy and momentum are conserverd.
Initial system: E  4mc 2
Invariant mass:
 Mc    4mc    pc 
2 2
2 2
2
Final system:
Invariant mass:  2mc 2    4mc 2    pc 
2
2
 4mc 2 2   Mc 2 2 
u pc 



2
c E
4mc
1/ 2
2
2
2
 Mc 2 
 u   4mc    Mc 


1


 
2
2 
c
 4mc 
 4mc2 
2
  Mc 2 2 

u  1  
2 
  4mc  
2
2
1/ 2
c
51
2
where  pc    4mc 2    Mc 2 
2
2
2
Chapter 3 – Quantization of Charge, Light, and Energy
3-1.
The radius of curvature is given by Equation 3-2.
R


mu
2.5 106 m / s
  m  3.911025 m / s  C  T 
 m
19
qB
 1.60 10 C   0.40T  
Substituting particle masses from Appendices A and D:
R (protron)  1.67 1027 kg  3.911025 m / s  C  T   6.5 102 m
R (electron)   9.111031 kg  3.911025 m / s  C  T   3.6 105 m
R (deuteron)   3.34 1027 kg  3.911025 m / s  C  T   0.13m
R (H 2 )   3.35 1027 kg  3.911025 m / s  C  T   0.13m
R (helium)   6.64 1027 kg  3.911025 m / s  C  T   0.26m
3-2.
u

s

r
u
For small values of  , s  ; therefore,  =
mu 2
Recalling that euB 
r
 r
mu
eB
53
s

r r
 
mu / eB

eB
mu
Chapter 3 – Quantization of Charge, Light, and Energy
3-3.
B
E
,
u
pc 
2
u pc

, and pc  E 2   mc 2 
c E
 0.561MeV    0.511MeV 
2
2
 0.2315MeV
u 0.2315MeV

 0.41
c 0.561MeV
2.0 105V / m
 B
 1.63 103 T  16.3G
0.41c
3-4.
F  quB and FG  mp g


19
6
5
FB quB 1.6 10 C  3.0 10 m / s   3.5 10 T 


 1.03 109
27
2
FG m p g
1.67 10 kg 9.80m / s 
3-5.
(a)
mu  2 Ek / e  e / m  
R

qB
 e / m  B 
1/ 2
4
1 2 Ek / e
1   2   4.5 10 eV / e  




B e/m
0.325T  1.76 1011 kg


(b)
frequency
f 
u
2 R

1/ 2
 2.2 103 m  2.2mm
 2 Ek / e  e / m 
2 R
 2   4.5 104 eV / e 1.76 1011 C / kg  


3
2  2.2 10 m 
1/ 2
 9.1109 Hz
period T  1/ f  1.11010 s
3-6.
(a) 1/ 2mu 2  Ek , so u 
 2Ek / e  e / m 
 u   2  2000eV / e  1.76 1011 C / kg 
(b)
t1 
1/ 2
x1
0.05m

 1.89 109 s  1.89ns
7
u 2.65 10 m / s
54
 2.65 107 m / s
Chapter 3 – Quantization of Charge, Light, and Energy
(Problem 3-6 continued)
(c) mu y  F t1  eEt1
 u y   e / m  Et1  1.76 1011 C / kg  3.33 103V / m 1.89 109 s   1.11106 m / s
3-7.
3-8.
NEk  W  CV T 
3
 4.186 J / cal   1.311014
CV T  5 10 cal / C   2C 
N


Ek
2000eV
1.60 1019 J / eV 
Q1  Q2   25.41  20.64 1019 C  4.47 1019 C   n1  n2  e
Q2  Q3   20.64  17.47  1019 C  3.17 1019 C   n2  n3  e
Q4  Q3  19.06  17.47  1019 C  1.59 1019 C   n4  n3  e
Q4  Q5  19.06  12.70 1019 C  6.36 1019 C   n4  n5  e
Q6  Q5  14.29  12.70 1019 C  1.59 1019 C   n6  n5  e
where the ni are integers. Assuming the smallest Δn = 1, then Δn12 = 3.0, Δn23 = 2.0,
Δn43 = 1.0, Δn45 = 4.0, and Δn65 = 1.0. The assumption is valid and the fundamental
charge implied is 1.59 1019 C.
3-9.
For the rise time to equal the field-free fall time, the net upward force must equal the
weight. qE  mg  mg  E  2mg / q.
3-10. (See Millikan’s Oil Drop Experiment on the home page at
www.whfreeman.com/tiplermodernphysics6e.) The net force in the y-direction is
mg  bv y  ma y . The net force in the x-direction is qE  bvx  max . At terminal speed
ax  ay  0 and vx / vt  sin  .
sin  
vx  qE / b  qE


vt
vt
bvt
55
Chapter 3 – Quantization of Charge, Light, and Energy
3-11. (See Millikan’s Oil Drop Experiment on the home page at
www.whfreeman.com/tiplermodernphysics6e.)
(a) At terminal speed mg  bvt where m  4 / 3 a3 oil and b  6 a. Substituting gives
 18  1.80 105 N  s / m 2  5.0 103 m / 20s  
18   vt 
2

a  
  a
4  oil g 
 4  0.75 1000kg / m3  9.8m / s 2  

 1.66 106 m  1.66  103 mm
1/ 2
m  4 1.66 106 m   750kg / m3  / 3  1.44 1014 kg
3
(b)
3-12.
19
5
FE  2  1.60 10 C  2.5 10 V / m 
F  qE and FG  mg 

 0.57
FG
1.44 1014 kg  9.8m / s2 
mT  2.898 103 m  K
(a) m 
2.898 103 m  K
 9.66 104 m  0.966mm
3K
(b) m 
2.898 103 m  K
 9.66 106 m  9.66 m
300 K
(c) m 
2.898 103 m  K
 9.66 107 m  966nm
3000 K
3-13. Equation 3-4: R   T 4 . Equation 3-6: R 
1
cU .
4
From Example 3-4: U  8 5 k 4T 4  / 15h3c 2 


R 1/ 4  cU 1

 c  8 5 k 4T 4  / 15h3c 2T 4 
4
4
T
T
4
2 5 1.38 1023 J / K 
15  6.63 10
34
4
J  s   3.00 10 m / s 
3
8
2
56
 5.67 108W / m2 K 4
Chapter 3 – Quantization of Charge, Light, and Energy
3-14. Equation 3-18: u    
8 hc 5
ehc /  kT  1
u  f  df  u    d   u  f   u  f 
d
Because c  f  ,
df
d
c/ f 2
df
8 hc  f / c   c  8 f 2
hf
u f  
 2  3
hf / kT
hf / kT
e
1  f 
c e
1
5
3-15.
2.898 103 m  K
(a) mT  2.898 10 m  K  m 
 1.07 103 m  1.07mm
2.7 K
3
(b) c  f   f 
c
m

3.00 108 m / s
 2.80 1011 Hz
3
1.07 10 m
(c) Equation 3-6:
R
1
c
cU   8 5 k 4T 4 /15h3c3 
4
4
 3.00 10 m / s 8 1.38 10 J / K   2.7 

 4 15  6.63 10 J  s   3.00 10 m / s 
8
4
23
5
3
34
Area of Earth: A  4 rE2  4  6.38 106 m 
8
3
4
 3.01106 W / m2
2
Total power = RA   3.01106W / m2   4   6.38 106 m   1.54 109W
2
3-16.
mT  2.898 103 m  K
2.898 103 m  K
 4140 K
(a) T 
700 109 m
(b) T 
2.898 103 m  K
 9.66 102 K
3 102 m
(c) T 
2.898 103 m  K
 9.66 104 K
3m
57
Chapter 3 – Quantization of Charge, Light, and Energy
3-17. Equation 3-4: R1   T14
3-18. (a) Equation 3-17: E 
(b) E 
R2   T24    2T1   16 T14  16R1
4
hc / 10hc / kT 
hc / 
0.1kT

 0.1
 0.951kT
hc /  kT
hc / kT  / 10 hc / kT 

e
1 e
1 e 1
hc /  0.1hc / kT  10kT
hc / 

 10
 4.59 104 kT
hc /  kT
hc / kT  /  0.1hc / kT 

e
1 e
1 e 1
Equipartition theorem predicts E  kT . The long wavelength value is very close to
kT, but the short wavelength value is much smaller than the classical prediction.
3-19. (a) mT  2.898 103 m  K  T1 
R1   T14
2.898 103 m  K
 107 K
27.0 106 m
and R2   T24  2R1  2 T14
 T24  2T14 or T2  21/ 4 T1   21/ 4  107 K   128K
2.898 103 m  K
(b) m 
 23 106 m
128K
3-20. (a) mT  2.898 103 m  K
(Equation 3-5)
2.898 103 m  K
m 
 1.45 107 m  145nm
4
2 10 K
(b) m is in the ultraviolet region of the electromagnetic spectrum.
58
Chapter 3 – Quantization of Charge, Light, and Energy
3-21. Equation 3-4: R   T 4
Pabs  1.36 103W / m2  RE2 m2  where RE = radius of Earth
Pemit   RW / m2  4 RE2   1.36 103W / m2  RE2 m2 
  RE2
R  1.36 103W / m2  
2
 4 RE
T4 
 1.36 103 W
 T 4

2
4
m

1.36 103W / m2
 T  278.3K  5.3C
4  5.67 108W / m2  K 4 
3-22. (a) mT  2.898 103 m  K  m 
f m  c / m 
2.898 103 m  K
 8.78 107 m  878nm
3300K
3.00 108 m / s
 3.42 1014 Hz
7
8.78 10 m
(b) Each photon has average energy E  hf and NE  40J / s.
N
40 J / s
40 J / s

 1.77 1020 photons / s
34
14
hf m
 6.63 10 J  s 3.42 10 Hz 
(c) At 5m from the lamp N photons are distributed uniformly over an area
A  4 r 2  100 m2 . The density of photons on that sphere is  N / A / s  m2 .
The area of the pupil of the eye is   2.5 103 m  , so the number of photons
2
entering the eye per second is:
n   N / A   2.5 10 m 
3
2
1.77 10

20
/ s     2.5 103 m 
100 m2
 1.77 1020 / s     2.5 103 m   1.10 1013 photons / s
2
59
2
Chapter 3 – Quantization of Charge, Light, and Energy
3-23. Equation 3-18: u    
8 hc 5
A 5
Letting
A


hc
,
B

hc
/
kT
,
and
U




ehc /  kT  1
eB /  1
  5  1 e B /    B 2 

du
d  A 5 
5 6 


 B/

  A
2
B/
d  d   eB /   1
e  1
e

1




6
6 B / 
A
 B B/
 A e
B


e  5  e B /   1  
 5 1  e  B /     0
2 
2 
B/
B
/

  e  1  

 e  1  
The maximum corresponds to the vanishing of the quantity in brackets.
Thus,
5 1  e B /    B . This equation is most efficiently solved by iteration; i.e., guess at a
value for B/λ in the expression 5 1  e B /   , solve for a better value of B/λ; substitute the
new value to get an even better value, and so on. Repeat the process until the calculated
value no longer changes. One succession of values is: 5, 4.966310, 4.965156, 4.965116,
4.965114, 4.965114. Further iterations repeat the same value (to seven digits), so we
have:
6.63 1034 J  s  3.00 108 m / s 

hc
hc
 4.965114 
 mT 

m
 m kT
 4.965114  k  4.965114  1.38 1023 J / K 
B
mT  2.898 103 m  K
(Equation 3-5)
3-24. Photon energy E  hf  hc / 
(a) For λ = 380nm: E  1240eV  nm / 380nm  3.26eV
For λ = 750nm: E  1240eV  nm / 750nm  1.65eV
(b) E  hf   4.14 1015 eV  s 100 106 s 1   4.14 107 eV
3-25. (a) hf  hc /   0.47eV .
max
 4.14 10
hc


4.87eV
15
eV  s  3.00 108 m / s 
4.87eV
60
 2.55 107 m  255nm
Chapter 3 – Quantization of Charge, Light, and Energy
(Problem 3-25 continued)
(b) It is the fraction of the total solar power with wavelengths less than 255nm, i.e., the
area under the Planck curve (Figure 3-6) up to 255nm divided by the total area. The
latter is: R   T 4   5.67 108W / m2  K 4   5800K   6.42 107W / m2 .
4
Approximating the former with u     with   127nm and   255nm :
 8 hc 127  109 m 5 
  255  109 m   1.23  104 J / m3
u 127nm   255nm   
9
 ehc / kT 12710   1 


R  0  255nm 
c
1.23  104 J / m 3  

4
R
8
4
3
 3.00  10 m / s 1.23  10 J / m  fraction = 1.4  104

 4   6.42  107W / m2 
R  0  255nm  
3-26. (a) t 
hc


1240eV  nm
 653nm,
1.9ev
ft 

h

1.9eV
 4.59 104 Hz
15
4.136 10 eV  s
1  hc
 1  1240eV  nm

(b) V0       
 1.9eV   2.23V
e 
 e  300nm

1  hc
 1  1240eV  nm

(c) V0       
 1.9eV   1.20V
e 
 e  400nm

3-27. (a) Choose λ =550nm for visible light. nhf  E 
dn
dE
hf 
P
dt
dt
 0.05  100W  550  109 m 
dn P P



 1.38  1019 / s
34
8
dt hf
hc  6.63  10 J  s  3.00  10 m / s 
(b) flux 
number radiated / unit time 1.38 1019 / s

 2.75 1017 / m2  s
2
area of the sphere
4  2m 
61
Chapter 3 – Quantization of Charge, Light, and Energy
3-28. (a) hf    ft 
(b) f  c /  

h

4.22eV
 1.02 1015 Hz
15
4.14 10 eV  s
3.00 108 m / s
 5.36 1014 Hz
9
560 10 m
No.
Available energy/photon hf   4.14 1015 eV  s  5.36 1014 Hz   2.22eV .
This is less than  .
3-29. (a) E  hf  hc /     hc / E
For E  4.26eV :   1240eV  nm /  4.26eV   291nm
and since f  c /  ,
f   3.00 108 m / s  /  291nm   1.03 1015 s 1
(b) This photon is in the ultraviolet region of the electromagnetic spectrum.
3-30. (a) First, add a row f 1014 Hz to the table in the problem, then plot a graph
Ek ,max versus f . The slope of the graph is h/e; the intercept on the  Ek ,max axis is
work function. The graph below is a least squares fit to the data.
λ nm
544
Ek ,max eV
0.360 0.199 0.156 0.117 0.062
f 1014 Hz
5.51
594
5.05
62
604
4.97
612
4.90
633
4.74
Chapter 3 – Quantization of Charge, Light, and Energy
(Problem 3-30 continued)
Slope  h / e  3.90 1015 eV  s
(b)
(3.90 1015  4.14 1015 ) eV  s
 5.6 percent
4.14 1015 eV  s
(c) The work function is the magnitude of the intercept on the Ek ,max axis, 1.78 eV.
(d) cesium
En
hc
3-32. (a)  
hc
3-31.



(b) Ek 

hc

 60   6.63 1034 J
s  3.00 108 m / s 
9
550 10 m
1240eV nm
 1.90eV
653nm
 
1240eV nm
 1.90eV  2.23eV
300nm
63
 2.17 1017 J
Chapter 3 – Quantization of Charge, Light, and Energy
3-33. Equation 3-25: 2  1   
 6.63 10
 
9.1110
34
31

1
100 
J s  1  cos135 
kg  3.00 10 m / s 
3-36.
p
h

8
 4.14 1012 m  4.14 103 nm
4.14 103 nm
100  5.8%
0.0711nm
3-34. Equation 3-24: m 
3-35.
h
1  cos 
mc
1.24 103
1.24 103
nm 
 0.016nm
V
80 103V
hc
c

(a) p 
1240eV nm
6.63 1034 J s
 3.10eV / c 
 1.66 1027 kg m / s
c  400nm 
400 109 m
(b) p 
1240eV nm
6.63 1034 J s
 1.24 104 eV / c 
 6.63 1024 kg m / s
c  0.1nm 
0.1109 m
(c) p 
1240eV nm
6.63 1034 J s
5

4.14

10
eV
/
c

 2.211032 kg m / s
3 102 m
c  3 107 nm 
(d) p 
1240eV nm
6.63  1034 J s
 620eV / c 
 3.32  1025 kg m / s
9
c  2nm 
2  10 m
2  1 
6.63 1034 J s  1  cos110 

h
1

cos


 3.26 1012 m


31
8
mc
 9.1110 kg  3.00 10 m / s 
6.63 1034 J s  3 108 m / s 

hc
1  
 2.43 1012 m
E1  0.511106 eV 1.60 1019 J / eV 
2  1  3.26 1012 m   2.43  3.26 1012 m  5.69 1012 m
64
Chapter 3 – Quantization of Charge, Light, and Energy
(Problem 3-36 continued)
E2 
hc
2

1240eV nm
 2.18 105 eV  0.218MeV
3
5.69 10 nm
Electron recoil energy Ee  E1  E2 (Conservation of energy)
Ee  0.511MeV  0.218MeV  0.293MeV . The recoil electron momentum makes an
angle θ with the direction of the initial photon.
PE
θ
h/λ1
110°
h/λ2
20°
h
2
cos 20  pe sin   1/ c  E 2   mc 2  sin 
2
 3.00 10 m / s  6.63 10
8
sin  
 5.69 10
12
(Conservation of momentum)
34
J s  cos 20
2
2
m   0.804MeV    0.511MeV  


1/ 2
1.60 10
13
J / MeV 
 0.330 or   19.3
3-37. (a) First, add a row ( 1  cos  ) to the table in the problem, then plot a graph of  versus
( 1  cos  ). The slope of the graph is the Compton wavelength of the electron.
 pm
 degrees
1  cos 
0.647 1.67 2.45 3.98 4.80
45
75
90
135
170
0.293 0.741 1.000 1.707 1.985
65
Chapter 3 – Quantization of Charge, Light, and Energy
h
(4.5  1.0) nm
 slope 
 2.43nm
mc
(1.88  0.44)
(b) The graph above is a least squares fit to the data. The percent difference is
(2.43  2.426) nm
0.004 nm
100 
100  0.15percent
2.426 nm
2.426 nm
3-38.
  2  1   
1  100 
3-39. (a) E1 
h
1  cos   100 0.00243nm 1  cos90  0.243nm
mc
hc
1

(b) 2  1 
(c) E2 
h
1  cos   0.011 Equation 3-25
mc
hc
2
1240eV nm
 1.747 104 eV
0.0711nm
h
1  cos   0.0711nm   0.00243nm 1  cos180   0.0760nm
mc

1240eV nm
 1.634 104 eV
0.0760nm
(d) Ee  E1  E2  1.128 103 eV
66
Chapter 3 – Quantization of Charge, Light, and Energy
3-40. (a)    h / mc 1  cos  
From protons:
   6.63 1034 J s  / 1.67 1027 kg  3.00 108 m / s  1  cos120 
  1.99 1015 m  1.99 106 nm
(b) Similarly, for electrons ( m  9.111031 kg )
  2.43 1012 m  2.43 103 nm
(c) Similarly, for N2 molecules ( m  4.68 1026 kg )
  4.72 1017 m  4.72 108 nm
3-41.
2  1 
h
1  cos   0.0711   0.00243nm 1  cos 
mc
λ2
0.0750
0.0730
0.0710
0
Slope =

1
1  cos 
θ
1  cos 
2 (nm)
0°
0
0.0711
45°
0.293
0.0718
90°
1
0.0735
135°
1.707
0.0752
2
 0.0745  0.0720  nm  2.381103
1.50  0.45
h
 h   2.381103 nm  9.111031 kg  3.00 108 m / s   6.5110 34 J s
mc
67
Chapter 3 – Quantization of Charge, Light, and Energy
3-42. (a) Compton wavelength =
electron:
h
mc
h
6.63 1034 J s

 2.43 1012 m  0.00243nm
31
8
mc  9.1110 kg  3.00 10 m / s 
h
6.63 1034 J s
proton:

 1.32 1015 m  1.32 fm
27
8
mc 1.67 10 kg  3.00 10 m / s 
(b) E 
hc

(i) electron: E 
(ii) proton: E 
1240eV nm
 5.10 105 eV  0.510MeV
0.00243nm
1240eV nm
 9.39 108 eV  939MeV
6
1.32 10 nm
3-43. Photon energy E  hf  hc /  allows us to rewrite Equation 3-25 as
hc hc
h


(1  cos  )
E2 E1 mc
Rearranging the above,
hc
h
hc

(1  cos  ) 
E2 mc
E1
Dividing both sides of the equation by hc yields
1
1
1 ( E1 / mc 2 )(1  cos  )  1

(1  cos  )  
E2 mc 2
E1
E1
Or
E2 
E1
( E1 / mc )(1  cos  )  1
2
3-44. (a) eV0  hf    hc /   
e  0.52V    hc / 450nm   
(i)
e 1.90V    hc / 300nm   
(ii)
68
Chapter 3 – Quantization of Charge, Light, and Energy
(Problem 3-44 continued)
Multiplying (i) by 450nm / e and (ii) by 300nm / e , then subtracting (ii) from (i) and
rearranging gives:
 300nm 1.90V    450nm  0.52V 

 2.24eV
e
150nm

e  300 109 m   4.14V 
hc
(b)
 1.90  2.24  h 
 6.63 1034 J s
8
e  300nm 
 3.00 10 m / s 
3-45. Including Earth’s magnetic field in computing y2, first show that y2 is given by
y2 
e  B 2 x1 x2 1 BE x22 



m E
2 E 
where the second term in the brackets comes from Fy  euBE  ma y and y 
1 2
a yt .
2
e  B 2 x1 x2 1 BE x22 

Thus, 1  
 The first term inside the brackets is the reciprocal of
m  Ey2
2 Ey2 
0.7  1011 C, Thomson’s value for e/m. Using Thomson’s data (B = 5.5 104 T ,
E  1.5  104V / m,
x1  5cm,
y2 / x2  8 / 110 ) and the modern value for e/m =
1.76  1011 C / kg and solving for BE.
1 BE Bx22
 8.20  1012. The minus sign means that B and BE are in opposite directions,
2 Ey2
which is why Thomson’s value underestimated the actual value.
BE 

 

m
 8.20  1012  2  1.5  104V / m 8 / 110 
5.5 10 T 8 10
4
2
69
2
 3.1  105 T  31T
Chapter 3 – Quantization of Charge, Light, and Energy
3-46. (a) Q  Ne and cM T  N
mu 2
where N = number of electrons, c = specific heat of
2
the cup, M = mass of the cup, and u = electron’s speed.
Q 2cM T

e
mu 2
N
(b)  
eB
mu

 u
e
Qu 2

m 2cM T
eB
m
Substituting u into the results of (a),
e Q  eB / m 

m
2cM T
2
Solving for e/m,
e 2cM  2 T

m
QB 2 2
3-47. Calculate 1/λ to be used in the graph.
1/λ (106/m)
5.0
3.3
2.5
2.0
1.7
V0 (V)
4.20
2.06
1.05
0.41
0.03
5
4
3
V0 (V)
2
1
0
-1
-2
1
2
3
4
5
1/λ (106/m)
(a) The intercept on the vertical axis is the work function  .   2.08eV .
70
Chapter 3 – Quantization of Charge, Light, and Energy
(Problem 3-47 continued)
(b) The intercept on the horizontal axis corresponds to the threshold frequency.
1
t
 1.65 106 / m
ft 
c
t
  3.00 108 m / s 1.65 106 / m   4.95 1014 Hz
(c) The slope of the graph is h/e. Using the vertical intercept and the largest experimental
point.
4.20V   2.08V 
h 1 V0


 4.19  1015 eV / Hz
8
6
e c  1 /  
3.00  10 m / s 5.0  10 / m  0



3-48. In the center of momentum reference frame, the photon and the electron have equal and
opposite momenta. p  E / c   pe .

The total energy is: E  Ee  E  pe2c 2  m2c 4

1/ 2

 E  E2  m2c 4

1/ 2
By conservation of momentum, the final state is an electron at rest, pe  0 . Conservation
of energy requires that the final state energy E  is
E   E  Ee


 mc  E  p c  mc
2


2 2
1/ 2
2
 mc 2  E   p 2 c 2  mc 2 



Squaring yields, mc 2

2
2

2 1/ 2


2
  E2  mc 2 



 2mc 2 E  E2  E2  mc 2
1/ 2

2
 mc 2 E  0. This can be true
only if E vanishes identically, i.e., if there is no photon at all.
3-49. Bragg condition:
m  2d sin .   (2)(0.28nm)(sin 20)  1.92  1010 m  0.192nm.
This is the minimum wavelength m that must be produced by the X ray tube.
m 
1.24  103
nm
V
or
V
1.24  103
 6.47  103V  6.47kV
0.192
71
Chapter 3 – Quantization of Charge, Light, and Energy




3-50. (a) E  100W  104 s  100 J / s  104 s  106 J
The momentum p absorbed is p 
(b)
 

E
106 J

 3.33  103 J s / m
8
c
3.00  10 m / s




p  m v f  vi  2  103 kg v f  0  3.3  103 J s / m
v 
3
3.33  10 J s / m
 1.67m / s
2  103 kg


2  103 kg 1.67m / s 
1 2
(c) E  mv f 
 2.78  103 J
2
2
2
The difference in energy has been (i) used to increase the object’s temperature and (ii)
radiated into space by the blackbody.
3-51. Conservation of energy: E1  mc2  E2  Ek  mc 2  Ek  E1  E2  hf1  hf 2
From Compton’s equation, we have: 2  1 
Thus,
h
1  cos  ,
mc
1 1
h
 
1  cos 
f 2 f1 mc 2
1
1
h
 
1  cos  
f2
f1 mc 2
f2 
f1mc 2
mc 2  hf1 1  cos 
Substituting this expression for f 2 into the expression for Ek (and dropping the subscript
on f1 ):
hfmc 2   hf  1  cos   hfmc 2
hfmc 2
Ek  hf  2


mc  hf 1  cos 
mc 2  hf 1  cos 
2
hf
mc 2
1
 hf 1  cos  
Ek has its maximum value when the photon energy change is maximum, i.e., when   
so cos  1. Then Ek 
hf
mc 2
1
2hf
72
Chapter 3 – Quantization of Charge, Light, and Energy
3-52. (a) mT  2.898  103 m K
(b) Equation 3-18:
 T
2.898  103 m K
 3.50  104 K
9
82.8  10 m
 70nm  /  ehc /70nmkT  1

u 82.8nm   82.8nm 5 /  ehc / 82.8nmkT  1
5
u  70nm 



6.63  1034 J s 3.00  108 m / s
hc

 5.88 and
where
 70nm  kT 70  109 m 1.38  1023 J / K 3.5  104 K



 70nm  /  e5.88  1

 0.929
u 82.8nm  82.8nm 5 /  e4.97  1
5
u  70nm 
hc
 4.97
82.8nm  kT
Similarly,

100nm  /  e4.12  1

 0.924
u 82.8nm  82.8nm 5 /  e4.97  1
5
u 100nm 
3-53. Fraction of radiated solar energy in the visible region of the spectrum is the area under
the Planck curve (Figure 3-6) between 350nm and 700nm divided by the total area. The
latter is 6.42  107W / m2 (see solution to Problem 3-25). Evaluating u     with
  525nm (midpoint of visible) and   700nm  350nm  350nm.
u     



8 6.63  1034 J s 3.00  108 m / s  525nm 



5

350nm 
 6.63  10 J s 3.00  10 m / s 
 1
exp 
23
 1.38  10 J / k  5800 K  525nm  

R   350  700  
34

8

 0.389 J / m3

c
u  3.00  108 m / s 0.389 J / m3 / 4  2.92  107W / m2
4



Fraction in visible = R  350  700 / R  2.92  107 W / m2 / 6.42  107 W / m2  0.455
3-54. (a) Make a table of f  c /  vs. V0 .

f 1014 Hz
V0 (V)

11.83
9.6
8.22
7.41
6.91
2.57
1.67
1.09
0.73
0.55
73
Chapter 3 – Quantization of Charge, Light, and Energy
(Problem 3-54 continued)
3
Li
2
Pb
1
0
2
−1
4
6
8
10
12
−2
−3
−4
The work function for Li (intercept on the vertical axis) is   2.40eV .
(b) The slope of the graph is h/e. Using the largest V0 and the intercept on the vertical


 4.97V  1.60  1019 C
h 2.57V   2.40V 
axis, 
or, h 
 6.89  1034 J s
14
14
11.53  10 Hz
e 11.53  10 Hz  0
(c) The slope is the same for all metals. Draw a line parallel to the Li graph with the
work function (vertical intercept) of Pb,   4.14eV . Reading from the graph, the
threshold frequency for Pb is 9.8  1014 Hz; therefore, no photon wavelengths larger



than   c / ft  3.00  108 m / s 9.8  1014 Hz  306nm will cause emission of
photoelectrons from Pb.
3-55. (a) Equation 3-18: u    
gives u    
(b)
8 hc 5
Letting C  8 hc and a  hc / kT
ehc /  kT  1
C 5
ea /   1


  5  1 ea /  a 2

du
d  C  5 
5 6 


 a/

 C
2
a/
d  d   ea /   1
e  1
e

1


6
6 a / 
C
 a a/
 C e
a


e  5 ea /   1  
 5 1  ea /    0
2 
2 


ea /   1  
ea /   1  





74





Chapter 3 – Quantization of Charge, Light, and Energy
(Problem 3-55 continued)
The maximum corresponds to the vanishing of the quantity in brackets.
Thus, 5 1  e a /   a


(c) This equation is most efficiently solved by trial and error; i.e., guess at a value for


a /  in the expression 5 1  e a /   a , solve for a better value of a /  ; substitute
the new value to get an even better value, and so on. Repeat the process until the
calculated value no longer changes. One succession of values is 5, 4.966310,
4.965156, 4.965116, 4.965114, 4.965114. Further iterations repeat the same value (to
seven digits), so we have
a
m
 4.965114 

hc
m kT

6.63  1034 J s 3.00  108 m / s
hc

(d) mT 
 4.965114 k  4.965114 1.38  1023 J / K

Therefore, mT  2.898  103 m K
3-56. (a) I 


Equation 3-5
P
1W 
1

2 
2
19
4 R
4 1m   1.602  10 J / eV

17
2
  4.97  10 eV / m s

(b) Let the atom occupy an area of  0.1nm  .
2



dW
2
 IA  4.97  1017 eV / m2 s  0.1nm  109 m / nm
dt
(c) t 

dW / dt


2
 4.97  103 eV / s
2eV
 403s  6.71 min
4.97  103 eV / s
3-57. (a) The nonrelativistic expression for the kinetic energy pf the recoiling nucleus is
p 2 15MeV / c  
1u

Ek 

 1.10  104 eV

2 
2m
2  12u  931.5MeV / c 
2
Internal energy U  15MeV  0.0101MeV  14.9899MeV
75
Chapter 3 – Quantization of Charge, Light, and Energy
(Problem 3-57 continued)
(b) The nucleus must recoil with momentum equal to that of the emitted photon, about
14.98 MeV/c.
p 2 14.98MeV / c  
1u

Ek 

 1.00  102 eV

2 
2m
2  12u
 931.5MeV / c 
2
E  U  Ek  14.9899MeV  0.0100MeV  14.9799MeV
3-58. Derived in Problem 3-47, the electron’s kinetic energy at the Compton edge is
Ek 
hf
1  mc 2 / 2hf
hf
E  520keV 
1   511keV  / 2hf
 520keV 
2  hf 
2
2hf  511keV
Thus,  hf   520  hf    520  511 / 2  0
2
2
2
520   520    2  520  511

  708keV
Solving with the quadratic formula: hf 
2
(only the + sign is physically meaningful). Energy of the incident gamma ray
hf  708keV .
hc

 708keV

6.63  10

34

J s 3.00  108 m / s
 708keV  1.60  10
16
J / keV

  1.76  10
3-59. (a) Ek  50keV and 2  1  0.095nm
hc
1

hc
2
 5.0  104 eV
1

1



12   0.095nm 
2hc
5  104 eV

1
5.0  104 eV

1  0.095
hc
21  0.095 5.0  104 eV

12  0.0951
hc
 0.095nm  hc

 1  5  104 eV  0

 12  0.045411  2.36  103  0
76
12
m  1.76 pm
Chapter 3 – Quantization of Charge, Light, and Energy
(Problem 3-59 continued)
Applying the quadratic formula,


2
0.04541   0.04541  4 2.36  103 


1 
2
1/ 2
1  0.03092nm and 2  0.1259nm
hc
(b) E1 
1

3-60. Let x 
kT



1240eV nm
 40.1 keV
0.03092nm

Eelectron  9.90keV
hf
in Equation 3-15:
kT

   e 
f n  A e nx  A e0  e x  e x


n 0
n 0
2
3
x


  A 1  y  y 2  y3 

 1
Where y  e x . This sum is the series expansion of
1  y 
1
, i.e., 1  y   1  y  y 2  y3 
1
. Then
f
 A 1  y   1 gives A  1  y.
1
n
Writing Equation 3-16 in terms of x and y.



n 0
n 0
n 0
E   En Ae En / kT  A nhfe nhf / kT  Ahf  ne nx
Note that
 ne
 nx
 ne

 nx
   d / dx   e nx . But  e nx  1  y  , so we have
1
d
d
dy 
1
2 
2
e nx   1  y   1  y      y 1  y 

dx
dx
 dx 
 
x
dy d e
Since

 e x   y.
dx
dx
Multiplying this sum by hf and by A  1  y  , the average energy is

E  hfA ne nx  hf 1  y  y 1  y  
2
n 0
hfy
hfe x

1  y 1  e x
Multiplying the numerator and the denominator by e x and substituting for x, we obtain
E
hf
e
hf / kT
1
, which is Equation 3-17.
77
Chapter 4 – The Nuclear Atom
4-1.
1
mn
1 
 1
 R  2  2  where R  1.097  107 m1 (Equation 4-2)
n 
m
The Lyman series ends on m =1, the Balmer series on m =2, and the Paschen series on
1
m =3. The series limits all have n  , so  0
n
1
1
 R  2   1.097  107 m1
L
1 
L  limit   1.097 107 m1  91.16 109 m  91.16nm
 1
 R  2   1.097  107 m1 / 4
B
2 
B  limit   4 / 1.097 107 m1  3.646 107 m  364.6nm
1
1
 R  2   1.097  107 m1 / 9
P
3 
1
P  limit   9 / 1.097  107 m1  8.204  107 m  820.4nm
4-2.
1
mn
1 
 1
 R  2  2  where m  2 for Balmer series (Equation 4-2)
n 
m
1
1.097  107 m1  1
1 

 2

9
2
379.1nm
10 nm / m  2
n 
1 1
109 nm / m
 2 
 0.2405
4 n
379.1nm 1.097  107 m1


1
 0.2500  0.2405  0.0095
n2
n2 
1
0.0095

n  1 / 0.0095
1/ 2
 10.3 
n  10  n  2
79
n  10
Chapter 4 – The Nuclear Atom
4-3.
1
mn
1 
 1
 R  2  2  where m  1 for Lyman series (Equation 4-2)
n 
m
1
1.097  107 m1 
1 

1 2 

9
164.1nm
10 nm / m  n 
1
109 nm / m

1

 1  0.5555  0.4445
n2
164.1nm 1.097  107 m1

n  1 / 0.4445
1/ 2

 1.5
No, this is not a hydrogen Lyman series transition because n is not an integer.
4-4.
1
mn
1 
 1
 R 2  2 
n 
m
(Equation 4-2)
For the Brackett series m = 4 and the first four (i.e., longest wavelength lines have n = 5,
6, 7, and 8.
 1 1
 1.097  107 m1  2  2   2.468  105 m1
45
4 5 
1
45 
1
 4.052  106 m  4052nm. Similarly,
5
1
2.68  10 m
46 
1
 2.625  106 m  2625nm
3.809  105 m1
47 
1
 2.166  106 m  2166nm
5
1
4.617  10 m
48 
1
 1.945  106 m  1945nm
5 1
5.142  10 m
These lines are all in the infrared.
4-5.
None of these lines are in the Paschen series, whose limit is 820.4 nm (see Problem 4-1)
and whose first line is given byL
1 1 
 R  2  2   34  1875nm. Also, none are in
34
3 4 
1
the Brackett series, whose longest wavelength line is 4052 nm (see Problem 4-4). The
Pfund series has m = 5. Its first three (i.e., longest wavelength) lines have n = 6, 7, and 8.
80
Chapter 4 – The Nuclear Atom
(Problem 4-5 continued)
1 1
 1.097  107 m1  2  2   1.341  105 m1
56
5 5 
1
56 
1
 7.458  106 m  7458nm. Similarly,
5
1
1.341  10 m
57 
1
 4.653  106 m  4653nm
5
1
2.155  10 m
58 
1
 3.740  106 m  3740nm
2.674  105 m1
Thus, the line at 4103 nm is not a hydrogen spectral line.
4-6.
(a) f   b2 nt (Equation 4-5)
For Au, n  5.90  1028 atoms / m3 (see Example 4-2) and for this
foil t  2.0 m  2.0  106 m.
2
kq Q
  2  79  ke
90  2  79 1.44eV nm 
b
cot

cot

m v 2
2
2 K
2
2 7.0  106 eV

5
 1.63 10 nm  1.63 10

f   1.63  1014 m
(b)
For   45,
14
m
 5.90 10
2

28


/ m3 2.0  106 m  9.8  105
b  45   b  90  cot 45 / 2  /  cot 90 / 2 
 b 90  tan 90 / 2  /  tan 45 / 2 
 3.92 105 nm  3.92  1014 m
and f  45   5.7  10 4
For   75,
b  75   b  90  tan 90 / 2  /  tan 75 / 2 
 2.12 105 nm  2.12  1014 m
and f  75   1.66  104
Therefore, f  45  75  5.7  104  1.66  104  4.05  104
81
Chapter 4 – The Nuclear Atom
(Problem 4-6 continued)
(c) Assuming the Au atom to be a sphere of radius r,
4 3
M
197 g / mole
r 

23
3
NA
6.02  10 atoms / mole 19.3g / cm3


 3
197 g / mole
r
23
 4 6.02  10 atoms / mole 19.3g / cm3







1/ 3
r  1.62 103 cm  1.62 1010 m  16.2nm
4-7.
N 
1
sin  / 2 
4

A
(From Equation 4-6), where A is the product of the two
sin  / 2 
4
quantities in parentheses in Equation 4-6.
N 10 
(a)
N 1 
N  30 
(b)
4-8.
N 1 
sin4  0.5 
sin4 15 

sin4  0.5 
sin4  5 
 1.01  104
 1.29  106
(Equation 4-3)
k 2e Ze
 1.44MeV fm  Z

cot

cot
2
m v
2
Ek
2

4-9.

A / sin4 1 / 2 
kq Q

cot
2
m v
2
b
=

A / sin4 10 / 2 
rd 
1.44MeV
fm  79 
7.7 MeV
cot
kq Q
ke2 2 79

Ek
1 / 2 m v 2
2
 8.5  1013 m
2
(Equation 4-11)
1.44MeV
For Ek  5.0MeV :
rd 
For Ek  7.7MeV :
rd  29.5 fm
For Ek  12MeV :
fm  2  79 
5.0MeV
rd  19.0 fm
82
 45.5 fm
Chapter 4 – The Nuclear Atom
4-10.
rd 
kq Q
ke2 2 79

Ek
1 / 2 m v 2
Ek 
4-11.
1.44MeV
fm  2 13
4 fm
xrms  N  
n
(Equation 4-11)
 9.4MeV
10  N  0.01   N  10 / 0.01   106 collisions
2
t
106 m
 10  104 layers
t 10 m
104 atomic layers is not enough to produce a deflection of 10 , assuming 1
collision/layer.
4-12. (a) f   b2 nt
(Equation 4-5)
For   25 (refer to Problem 4-6).
2  79  ke2

25  2  79 1.44eV
b
cot

2 K

2 7.0  10 eV
2
6
nm 

 25 
cot 

 2 
 7.33 105 nm  7.33 1014 m

f   7.33  1014 m
 5.90 10
2
Because N  f  N  1000
For   45, b 

28

N  1000 / 1.992  103  5.02  105

 2  79 1.44eV
f   3.92  1014 m

nm 
2 7.0  10 eV
6
 5.90 10
2
28

/ m3 2.0  106 m  1.992  103

 45 
cot 
 3.92  1014 m

 2 


/ m3 2.0  106 m  5.70  104


Because N   45  f  N  5.70  104 5.02  105  286
(b)
N  25  45  1000  286  714
(c)
For   75 , b  b   25 tan 25 / 2 /  tan 75 / 2  2.12  1014 m

f  1.992  103 2.12  1014 m
 / 7.33  10
2
 1.992  103  2.12 / 7.33  1.67  104
2
83
14
m

2
Chapter 4 – The Nuclear Atom
(Problem 4-12 continued)
For   90, b  b   25 tan 25 / 2 /  tan 90 / 2  1.63  1014 m

f  1.992  103 1.63  1014 m
 / 7.33  10
2
14
m

2
 1.992  103 1.63 / 7.33  9.85 105
2


N  f  N  9.85  105 5.02  105  49
N   75  90   84  49  35
n2 a0
4-13. (a) rn 
Z
r6 
(Equation 4-18)
62  0.053nm 
1


(b) r6 He 
4-14.
a0 
mke2
E1 
2
(Equation 4-19)
mk 2 e4
2 2
(from Equation 4-20)
 
mc 2 ke2
2  c
2
2
2
mc 2  ke2 
1 2 2


  mc 
2
2
 c
c
0.00243nm

 0.053nm
2 2 1 / 137 
 1
1 
 Z 2R  2  2 
 n f ni 



1
 0.95nm

c
c
1
1
h
1

 2


 2
 c
2
2
mcke
mc ke / c 2 mc ke / c 2
a0 
4-15.
62  0.053nm 
2


 1.91nm
E1 
(Equation 4-22)
84
1 2 2 5.11  105 eV
mc  
 13.6eV
2
2
2 137 
Chapter 4 – The Nuclear Atom
(Problem 4-15 continued)
1 1 
 n2  1 
 R  2  2  =R  i 2 
ni
 1 ni 
 ni 
1
ni 

ni2
 n i2 

91
.
17
nm


 2

1.0968  107 m n i2  1
 ni  1 
=
n i2
 
R ni2  1


4
9
 91.17nm   121.57nm 3   91.17nm   102.57nm
3
8
16
4   91.17nm   97.25nm
  91.17 nm
15
None of these are in the visible; all are in the ultraviolet.
2 
∞
4
3
2
λ, nm
80
4-16.
90
L  mvr  n
100
110
120
(Equation 4-17)
vE  2 r / 1y  2 r / 3.16 107 s
mE  5.98  1024 kg



n  m 2 r / 3.16  107 s r /  2 mr 2 / 3.16  107 s



2 5.98  1024 kg 1.50  1011 m

3.16  10 s 1.055  10
7
mv  n / r
130

34
J s


2
 2.54  1074
E   mv  / 2m   n / r  / 2m
2


2
(from Equation 4-17)



2
1.055  1034 J s 2.54  1074 1
  1
E   
 0.210  1040 J
 2nn  
2
11
24
r
2
m
 
1.50  10 m 5.98  10 kg
2


This would not be detectable.
n 
E 

  2.54  10   r   5.34  10  r 
 5.98  10 kg 
34
 2r  1.055  10 J s


2m  r 3 
1.50  1011 m
2
40
or  r  0.210  10

2
74
2
33
2
24
J / 5.34  1033 J / m  3.93  1075 m
The orbit radius r would still be 1.50  1011 m.
85
Chapter 4 – The Nuclear Atom
4-17.
f rev 

mk 2 Z 2e4
2 3n3
(Equation 4-29)
 
mc 2 Z 2 ke2
2 n3  c 
2
2
2
 ke2 
cZ 2
cZ 2 2


 h / mc  n3  c  c n3
3.00 10 m / s  1  1   8.22  10 Hz

 0.00243  10 m   2  137 
N  f t  8.22  10 Hz 10 s   8.22  10 revolutions
2
8
2
14
9
3
8
14
6
rev
4-18. The number of revolutions N in 10-8 s is:
N  108 s /  time/revolution   108 s /  circumference of orbit/speed 
N  108 s /  C / v   108 s /  2 r / v 
The radius of the orbit is given by:
r
2
n2 a0 4  0.0529nm 

Z
3
so the circumference of the orbit C  2 r is
C  2 42  0.0529nm  / 3  1.77nm  1.77  109 m
The electron’s speed in the orbit is given by

8.99  10 N m
/ mr  
9.11 10
9
v  kZe
2
2
2
 

kg 1.77  10 m 
/ C 2  3 1.60  1019 C
31
2
9
v  6.54  105 m / s
Therefore, N  108 s /  C / v   3.70  106 revolutions
In the planetary analogy of Earth moving around the sun, this corresponds to 3.7 million
“years”.
4-19. (a) au 
(b) E 
2
 ke2

 k 2e4
2
2
2
e
e
9.11  1031 kg

a

 0.0529nm   2.56  104 nm
 e ke2  0 1.69  1028 kg

  e k 2 e 4  
1.69  1028 kg

E

13.6eV   2520eV
e 2 2
e 0 9.11  1031 kg
86
Chapter 4 – The Nuclear Atom
(Problem 4-19 continued)
(c) The shortest wavelength in the Lyman series is the series limit ( ni  , n f  1). The
photon energy is equal in magnitude to the ground state energy  E .
 
hc 1240eV nm

 0.492nm
E
2520eV
(The reduced masses have been used in this solution.)
1/ 2
4-20.
E   Z  E0 / n
2
 n 2 E 
Z= 

 E0 
2
 22  5.39eV  


13.6eV


1/ 2
 1.26
4-21.
Energy (eV)
n=∞
n=4
n=3
0
-2
(d)
n=2
-4
(b)
(c)
-6
-8
-10
-12
n=1
(a)
-14
(a) Lyman limit, (b) H  line, (c) H line, (d) longest wavelength line of Paschen series
4-22. (a)
 1
1 
 R 2  2 
 n f ni 



1
For Lyman α:
EL 
hc
L

1 1
 1.097373  107 m1  2  2 

1 2 
1

L  121.5023nm
E
1240eV nm
 10.2056eV and pL  L  10.2056eV / c
121.5023nm
c
87
Chapter 4 – The Nuclear Atom
(Problem 4-22 continued)
Conservation of momentum requires that the recoil momentum of the H atom
pH  pL and the recoil energy EH is:
EH   pH  / 2mH   pH c  / 2mH c 
2
2
2

10.2056eV / c 

2 1.007825uc 2 931.50  106 eV / uc 2
 5.55  108 eV
EH
5.5  108 eV
(b)

 5  109
 EL  EH  10.21eV
4-23. (a) For C5+ (Z = 6) En  13.6
0
Z2
489.6
 2
2
n
n
n=∞
n=5
n=4
n=3
E∞ = 0 eV
E5 = -19.6 eV
E4 = -30.6 eV
E3 = -54.4 eV
E2 = -122.4 eV
E1 = -489.6 eV
-100
n=2
En (eV)
-200
-300
-400
n=1
-500
(b)  
hc
hc
1240eV nm


 18.2nm
E E3  E2  54.4   122.4  eV
(c) 18.2nm lies in the UV (ultraviolet) part of the EM spectrum.
88
2

Chapter 4 – The Nuclear Atom
4-24. (a) The reduced mass correction to the Rydberg constant is important in this case.
1

R  R 
1 m / M
En  hcR / n2

1
6
1
  R  2   5.4869  10 m

 
(from Equation 4-26)
(from Equations 4-23 and 4-24)



E1   1240eV nm  5.4869  106 m1 109 m / nm / 1  6.804eV
2
Similarly, E2  1.701eV and E3  0.756eV
(b) Lyman α is the n  2  n  1 transition.
hc

 E2  E1

 
hc
1240eV nm

 243nm
E2  E1 1.701eV   6.804eV 
Lyman β is the n  3  n  1 transition.
 
hc
1240eV nm

 205nm
E3  E1 0.756eV   6.804eV 
4-25. (a) The radii of the Bohr orbits are given by (see Equation 4-18)
r  n2 a0 / Z where a0  0.0529nm and Z  1 for hydrogen.
For n  600, r   600   0.0529nm   1.90  104 nm  19.0 m
2
This is about the size of a tiny grain of sand.
(b) The electron‟s speed in a Bohr orbit is given by
v2  ke2 / mr with Z  1
Substituting r for the n = 600 orbit from (a), then taking the square root,


 
2

v 2  8.99  109 N m2 1.609  1019 C / 9.11 1031 kg 19.0  106 m
v2  1.33  107 m2 / s 2
 v  3.65  103 m / s
For comparison, in the n = 1 orbit, v is about 2  106 m / s
89

Chapter 4 – The Nuclear Atom
4-26. (a)
1 
2 1
 R  Z  1  2  2 

 n1 n2 
1
1

1 
2 1
3   1.097  107 m1  42  1  2  2    6.10  1011 m  0.0610nm
 1 3 



1

1 
2 1
4   1.097  107 m1  42  1  2  2    5.78  1011 m  0.0578nm
 1 4 


(b) lim it
4-27.

1

2 1

  1.097  107 m1  42  1  2  0    5.42  1011 m  0.0542nm
1




1 
1 
2 1
2 1
 R  Z  1  2  2   R  Z  1  2  2  for K

1 2 
 n1 n2 
1
1/ 2


1/ 2




1
1
 

Z 1  
  R 1  1  
  0.0794nm  1.097  102 / nm  3 / 4  
  4  


Z  1  39.1  40 Zirconium


4-28. (a) Z  43; f 1 / 2  21 108 Hz1 / 2

f  4.4  1018 Hz
Z  61; f 1 / 2  30  108 Hz1 / 2

f  9.0  1018 Hz
Z  75; f 1 / 2  37 108 Hz1 / 2

f  1.4 1019 Hz
Note: f 1 / 2 for Z  61 and 75 are off the graph 4-19; however, the graph is linear and
extrapolation is easy.
(b) For Z = 43  
1
(Equation 4-37)
1 

R  Z  1 1  2 
 n 
2
where R  1.097  107 m1 and n  2

1
1.097  10 m 
7
1
1
 43  1 1  
 4
2
90
 6.89  1011 m  0.0689nm
Chapter 4 – The Nuclear Atom
(Problem 4-28 continued)
Similarly,
For Z = 61, λ = 0.0327nm
For Z = 75, λ = 0.0216nm
4-29.
n2 a0
(Equation 4-18)
Z
The n =1 electrons “see” a nuclear charge of approximately Z  1 , or 78 for Au.
rn 



r1  0.0529nm / 78  6.8  104 nm 109 m / nm 1015 fm / m  680 fm , or about 100 times
the radius of the Au nucleus.
4-30.
En  13.6
Z2
eV
n2
(Equation 4-20)
For Fe (Z = 26) E1
 26 
 13.6
2
12
 9.194keV
The fact that E1 computed this way (i.e., by Bohr theory) is approximate, is not a serious
problem, since the Kα x-ray energy computed from Figure 4-19 provides the correct
spacing between the levels.
The energy of the Fe Kα x-ray is:
E  Fe K   hf where f 1 / 2  12.2  108 Hz1 / 2


E  Fe K   6.626  1034 J s 12.2  108 Hz1 / 2

2
 9.862  1016 J  6.156keV
Therefore, E2  E1  E  K    9.194  6.156   3.038keV
The Auger electron energy E  K   E2  6.156  3.038  3.118keV
4-31.
E   me c 2 
511keV

1  2.25  10 / 3.00  10 m / s
8
8

2
 772.6keV
After emitting a 32.5 keV photon, the total energy is:
91
Chapter 4 – The Nuclear Atom
(Problem 4-31 continued)
E  740.1keV 
511keV
1 
2
v  1   511 / 740  


1/ 2
4-32. (a)  E1  E0 Z 2 / n2
 2  v 2 / c 2  1   511 / 740 

2
2
c  2.17  108 m / s
(Equation 4-20)
 13.6eV  74  1 / 1  7.25  104 eV  72.5keV
2
(b)
2
 E1  E0  Z    / n2  69.5  103 eV  13.6eV  74    / 1
2
 74   
2
2
2
 69.5  103 eV / 13.6eV
  74   69.5  103 eV / 13.6eV 
1/ 2
 2.5
4-33.
Element
Al
Ar
Sc
Fe
Ge
Kr
Zr
Ba
Z
13
18
21
26
32
36
40
56
E (keV)
1.56
3.19
4.46
7.06
10.98
14.10
17.66
36.35
6.14
8.77
10.37
13.05
16.28
18.45
20.64
29.62

f 1 / 2 108 Hz1 / 2

60
Z
40
20
0
0
5
10
15
20
25

30
f 1 / 2 108 Hz1 / 2
92

Chapter 4 – The Nuclear Atom
(Problem 4-33 continued)
slope =
58  10
 1.90  108 Hz 1 / 2
8
 30  4.8  10
slope (Figure 4-19) =
30  13
 2.13  108 Hz 1 / 2
8
5

7

10
 
The two values are in good agreement.
4-34. (a) The available energy is not sufficient to raise ground state electrons to the n =5 level
which requires 13.6 − 0.54 = 13.1eV. The shortest wavelength (i.e., highest energy)
spectral line that will be emitted is the 3rd line of the Lyman series, the
n = 4 → n = 1 transition. (See Figure 4-16.)
(b) The emitted lines will be for those transitions that begin on the n = 4, n = 3, or n = 2
levels. These are the first three lines of the Lyman series, the first two lines of the
Balmer series, and the first line of the Paschen series.
60
4-35.
60
15.7eV
50
E
(eV)
44.3
40
14.7eV
Average transition energy = 15.7 eV
30
29.6
16.6eV
20
13.0
10
93
Chapter 4 – The Nuclear Atom
4-36.
E 
hc


1240eV nm
 1.610eV . The first decrease in current will occur when the
790nm
voltage reaches 1.61V.
4-37. Using the results from Problem 4-24, the energy of the positronium Lyman α line is
E  E2  E1  1.701eV   6.804eV   5.10eV . The first Franck-Hertz current
decrease would occur at 5.10V, the second at 10.2V.
4-38. In an elastic collision, both momentum and kinetic energy are conserved. Introductory
physics texts derive the following expression when the second object (the Hg atom here)
 m  m2 
is initially at rest: v1 f   1
 v1i . The fraction of the initial kinetic energy lost by
 m1  m2 
the incident electron in a head-on collision is:
2
f 
KEei  KEef
KEei

v12i  v12f
v12i
 m  m2  2
v  1
 v
m1  m2  1i


v12i
2
1i


2
 0.511MeV  200uc 2 931.5MeV / uc 2
 m1  m2 
= 1 
  1 
 0.511MeV  200uc 2 931.5MeV / uc 2
 m1  m2 

 
 
2
= 1.10  105
If the collision is not head-on, the fractional loss will be less.
4-39. (a) Equation 4-24: En   E0 / n2  13.6 / n2 eV
 1
1
1 
 1
En1  En  13.6 
 2  eV  13.6  2  2  eV=2.89 104 eV
2
 46 45 
 (n  1) n 
(b) Ionization energy  En  13.6 / n2  13.6 / 452  6.72 103 eV
(c) E  hf  hc /   f  E / h  (2.89 104 eV)(1.60 1019 J/eV)/6.63 1034 J  s
f  6.97 1010 Hz
  c / f  (3.00 108 m/s)/(6.96 1010 Hz)  4.30 103 m=4.30mm
94
Chapter 4 – The Nuclear Atom
(Problem 4-39 continued)
(d) Equation 4-18: rn  n2 a0 / Z
r45  452 a0 /1  107 nm  1.07 104 mm , or 2025× the radius of the
For hydrogen:
hydrogen atom ground state.
1


7
1
4-40. (a) Equation 4-26: R  R 
 where R  1.0973732 10 m
 1 m / M 


1
Rd  R 

31
27
 1  9.1094 10 kg / 3.3436 10 kg 
Rd  1.0970743 107 m 1


1
Rt  R 

31
27
 1  9.1094 10 kg / 5.0074 10 kg 
Rt  1.0971736 107 m 1
(b) Equation 4-22:
 1 1 
 R  2  2  with Z  1 .
n


 f ni 
1
 1 1   1 1 5
The Balmer α transition is n  3  n  2 .  2  2    2  2  
n

 f ni   2 3  36
d  t 
36  1
1
   5.3978 1011 m  5.3978 102 nm

5  Rd Rt 
(c) Computed as in (b) above with RH  1.096762 107 m1 ,
H  t 
36  1
1
   2.4627 1010 m  2.4627 101 nm

5  RH Rt 
95
Chapter 4 – The Nuclear Atom
4-41.
N  I 0  2 b  db where b 
and db 
kq Q

cot
2
m v
2
(Equation 4-3)
kq Q 

 csc  d
2 
2m v 
2
2
 kq Q   1
 

N  I 0 2   2   cot  csc 2  d
2 
2
 m v   2
Using the trigonometric identities:
csc 2 
1
sin  / 2
2
and cot

2

sin
sin
sin


2
2
1  cos 1  cos  / 2   sin  / 2  2 sin2  / 2 


 kq Q   1  
sin
1
N  I 0 2   2    
d

2
2
 2 sin  / 2   sin  / 2  
m
v
2








2
and inserting 2e  q and Ze  Q,
2
 kZe2  sin d
N  I 0 2 
2 
4
 m v  sin  / 2 
4-42. Those scattered at   180 obeyed the Rutherford formula. This is a head-on collision
where the α comes instantaneously to rest before reversing direction. At that point its
kinetic energy has been converted entirely to electrostatic potential energy, so
k  2e  79e 
1
where r = upper limit of the nuclear radius.
m v 2  7.7 MeV 
2
r
r
4-43. (a)
k  2  79  e2
7.7 MeV
i  qf rev
=e

2  79 1.440MeV fm 
7.7 MeV
Z 2 mk 2 e4
e
2 3n3
  1
mc 2 ke 2
2
2
 c  1
2
1.602  10
=
19
2
3
 29.5 fm
(from Equation 4-28)
2
ec  ke 2 
ec 2


c
 h / mc   c 


C 3.00  1017 nm / s  1 2
 1.054  103 A


0.00243nm
 137 
96
Chapter 4 – The Nuclear Atom
(Problem 4-43 continued)
(b)
 emk 2 e4   2  e

=
3  
2 
 2
  mke  2m
  iA  i a02  
1.602  10 C 1.055  10
2  9.11  10 kg 
19
=
34
J s
31
  9.28  10
24
A m2
or

 
= 1.054  103 A  0.529  1010 m

2
 9.27  1024 A m2
4-44. Using the Rydberg-Ritz equation (Equation 4-2), set-up the columns of the spreadsheet to
carry out the computation of λ as in this example (not all lines are included here).
4-45.
C=m2 D=n2
1/λ
λ (nm)
0.96
10534572
94.92
16
0.9375
10287844
97.20
1
9
0.888889
9754400
102.52
2
1
4
0.75
8230275
121.50
2
6
4
36
0.222222
2438600
410.07
2
5
4
25
0.21
2304477
433.94
2
4
4
16
0.1875
2057569
486.01
2
3
4
9
0.138889
1524125
656.11
3
7
9
49
0.090703
995346.9
1004.67
3
6
9
36
0.083333
914475
1093.52
3
5
9
25
0.071111
780352
1281.47
3
4
9
16
0.048611
533443.8
1874.61
m
n
1
5
1
25
1
4
1
1
3
1
  1
1 
   R  2  2 
  n f ni  
Because R  ,
1
1/C−1/D
d
 
   R 2
d


dR / d   R / .    R
97
2

1
 1
1  dR

 2  2 
 n f ni  d 

 1
1 
 2  2 
 n f ni 
1
 R /        /  
Chapter 4 – The Nuclear Atom
(Problem 4-45 continued)
H 



me m p
me  m p
D 
me md
me  md
me  md  m p 
m m /  me  md 
m /  me  md 
D  H D

1  e d
1  d
1 
H
H
m p  me  md 
me m p /  me  m p 
m p /  me  m p 
If we approximate md  2mp and me
     /      656.3nm 
md , then



me
and
2m p
0.511MeV
 0.179nm
2  938.28MeV 
4-46. For maximum recoil energy for the Hg atoms, the collision is „head-on‟.
(a)
Before collision
Ek 
kinetic energy
1 2
mvei
2
pei  mvei
momentum
After collision
Ek 
1 2
mvef
2
EHg 
1
2
MvHg
2
pef  mvef
pHg  M vHg
Conservation of momentum requires:
mvei  mvef  MvHg  vHg 

m
vei  vef
M

Therefore, the maximum Hg recoil kinetic energy is given by:
2
2
1
1 m
m2 2
2
MvHg
 M    vei  vef  
vei  2vei vef  vef2
2
2 M 
2M
2
m

4vei2 since m M , vei  vef
2M
4m  1 2  4m

mvei  
Ek
M  2
 M

 
98

Chapter 4 – The Nuclear Atom
(Problem 4-46 continued)
(b) Since the collision is elastic, kinetic energy is conserved, so the maximum kinetic
energy gained by the Hg atom equals the maximum kinetic energy lost by the
electron. If Ek = 2.5eV, then the maximum lost is equal to:
9.11  1031 kg  2.5eV 
m
4  2.5eV   4
 2.7  105 eV
27
M
201
u
1
.
66

10
kg
/
u



4-47. (a) En   E0 Z 2 / n2

(Equation 4-20)
For Li++, Z = 3 and En  13.6eV  9  / n2  122.4 / n2eV
The first three Li++ levels that have the same (nearly) energy as H are:
n  3, E3  13.6eV
n  6, E6  3.4eV
n  9, E9  1.51eV
Lyman α corresponds to the n = 6 → n = 3 Li
To the n = 9 → n = 3 Li
++
++
transitions. Lyman β corresponds
transition.
(b) R  H   R 1 / 1  0.511MeV / 938.8MeV    1.096776  107 m1
R  Li   R 1 / 1  0.511MeV / 6535MeV    1.097287  107 m1
For Lyman α:
1

 R  H  1  2   1.096776  107 m1 109 m / nm  3 / 4   121.568nm

 2 

1

For Li++ equivalent:
1 1
1 1  2
 R  Li   2  2  Z 2  1.097287  107 m1 109 m / nm     3

3 6 
 9 36 

1
  121.512nm
  0.056nm
2
1
 I A nt   kZe2 
(Equation 4-6)
4-48. N   0 SC



2
 r
  2 EK  sin4 
2
3
where ASC  0.50cm
r  10cm t  106 m
99

Chapter 4 – The Nuclear Atom
(Problem 4-48 continued)
10.5g / cm  6.02 10
n  Ag  
3
23
atoms / mol

107.5 g / mol
 5.88  10 atoms / cm3  5.88 1028 atoms / m3
22
EK  6.0MeV

1
I 0  1.0nA  109 C / s 
 2 1.60  1019 C






  3.18  109 alphas / s


(a) At θ = 60°




 3.13  109  / s 0.50cm 2 5.88  1028 / m3 106
N  

102 cm 2

 



 9  109 N m 2 / C 2 1.60  1019 C 2  47   

1

=
 468 / s

13
 2  6.0 MeV  1.60  10 J / MeV
  4 60 

  sin 2 





60   4 120 

(b) At θ = 120°: N  N  60   sin4
/ sin
 52 / s
2  
2 

4-49.
En   E0 Z 2 / n2
(Equation 4-20)
For Ca, Z = 20 and E1  13.6eV  20  / 1  5.440keV
2
2
The fact that E1 computed this way is only approximate is not a serious problem because
the measured x-ray energies provide us the correct spacings between the levels.
E2  E1  3.69keV  5.440  3.69  1.750keV
E3  E2  0.341keV  1.750  0.341  1.409keV
E4  E3  0.024keV  1.409  0.024  1.385keV
These are the ionization energies for the levels. Auger electron energies E   En
where E  3.69keV .
Auger L electron: 3.69keV  1.750keV  1.94keV
Auger M electron: 3.69keV  1.409keV  2.28keV
Auger N electron: 3.69keV  1.385keV  2.31keV
100
Chapter 4 – The Nuclear Atom
4-50. (a) E  hc /   1240eV nm / 0.071nm  17.465keV
E  hc /   1240eV nm / 0.063nm  19.683keV
(b) Select Nb (Z = 41)
The Kβ Mo x-rays have enough energy to eject photoelectrons, producing 0.693 keV
electrons. The Kα Mo x-rays could not produce photoelectrons in Nb.
 180  
4-51. (a) b  R sin   R sin 
2



  R cos 2

(b) Scattering through an angle larger than θ corresponds to an impact parameter smaller
than b. Thus, the shot must hit within a circle of radius b and area πb2. The rate at
which this occurs is I 0 b 2  I 0 R 2 cos2

2


(c)    b    R cos    R 2
2

2
2
0
(d) An α particle with an arbitrarily large impact parameter still feels a force and is
scattered.
4-52. For He: En  13.6eV Z 2 / n2  54.4eV / n2
(Equation 4-20)
(a)
∞
0
5
4
3
-10
2
Energy (eV)
-20
-30
E1 = -54.4 eV
E2 = -13.6 eV
E3 = -6.04 eV
E4 = -3.04 eV
E5 = -2.18 eV
E∞ = 0 eV
-40
-50
1
101
Chapter 4 – The Nuclear Atom
(Problem 4-52 continued)
(b) Ionization energy is 54.5eV.
(c) H Lyman α:   hc / E  1240eV nm / 13.6eV  3.4eV   121.6nm
H Lyman β:   hc / E  1240eV nm / 13.6eV  1.41eV   102.6nm
He+ Balmer α:   hc / E  1240eV nm / 13.6eV  6.04eV   164.0nm
He+ Balmer β:   hc / E  1240eV nm / 13.6eV  3.40eV   121.6nm
  42.4nm   19.0nm
(The reduced mass correction factor does not change the energies calculated above
to three significant figures.)
(d) En  13.6eV Z 2 / n2 because for He+, Z = 2, then Z 2 = 22. Every time n is an even
number a 22 can be factored out of n 2 and cancelled with the Z 2 = 22 in the
numerator; e.g., for He+,
E2  13.6eV 22 / 22  13.6eV
(H ground state)
E4  13.6eV 2 / 4  13.6eV / 2
(H  1st excited state)
E6  13.6eV 22 / 62  13.6eV / 32
(H  2nd excited state)
2
2
2
etc.
Thus, all of the H energy level values are to be found within the He+ energy levels, so
He+ will have within its spectrum lines that match (nearly) a line in the H spectrum.
4-53.
Element
P
Ca
Co
Kr
Mo
I
Z
15
20
27
36
42
53
Lα λ(nm)
10.41
4.05
1.79
0.73
0.51
0.33
1.70
2.72
4.09
6.41
7.67
9.53

f 1 / 2 108 Hz




where f 1 / 2   3.00  108 m / s 109 nm / m /  
Slope =
1/ 2
50  15
 4.62  108 Hz 1
8
9
.
15

1
.
58

10
Hz


Slope (Figure 4-19) =
74  46
 4.67  108 Hz 1
8
14  8  10 Hz
102
Chapter 4 – The Nuclear Atom
(Problem 4-53 continued)
The agreement is very good.
50
40
Z
30
20
10

f 1 / 2 108 Hz1 / 2
1
2
3
4
5
6
7
8
9

10
The f 1 / 2  0 intercept on the Z axis is the minimum Z for which an Lα X-ray could
be emitted. It is about Z = 8.
4-54. (a) En  
ke2
ke2
 2
2rn
2n ro
hf  En  En 1  
En 1  
ke2
2  n  1 ro
2

ke2 
ke2




2n2 ro  2  n  12 ro 




2
2
ke2  1
1  ke2 n  n  2n  1

f 
 
2
2hro   n  12 n 2  2hro
n 2  n  1



(b) f rev 
ke2 2n  1
ke2

for n
2hro n2  n  12 ro hn3
v
2 r

2
f rev

1
v2
1 mv 2
1 ke2
ke2



4 2 r 2 4 2 mr r
4 2 mr r 2
4 2 mro3n6
103
Chapter 4 – The Nuclear Atom
(Problem 4-54 continued)
(c) The correspondence principle implies that the frequencies of radiation and revolution
are equal.
2
2
 ke2 
ke2
2
f 

 f rev
3 
2
3 6
4 mro n
 ro hn 
2
ke2  hn3 
h2
ro 




4 2 mn6  ke2 
4 2 mke2 mke2
2
which is the same as ao in Equation 4-19.
4-55.
kZe2 mv 2

r
r

kZe2  mv 

r2
mr
1/ 2
 kZe2 
v  

 mr 

2
(from Equation 4-12)
v
1  2
 2  kZe2    kZe2 
c 2  2  kZe2 
2

Therefore,

c  


  

1   2  mr 
 mr    mr 

1  kZe2 
  2

c  mao 
2

  0.0075Z 1 / 2

v  0.0075cZ 1 / 2  2.25  106 m / s  Z 1 / 2
 1
 kZe2
kZe2
2
E  KE  kZe / r  mc    1 
 mc 
 1 
2
r
r

 1  
2
2
And substituting   0.0075 and r  ao


1

E  511  10 eV
 1  28.8Z eV
2



 1   0.0075
3
 14.4eV  28.8Z eV  14.4Z eV
104
Chapter 4 – The Nuclear Atom
4-56. (The solution to this problem depends on the kind of calculator or computer you use and
the program you write.)
4-57.
Energy (eV)
23
22
Levels constructed from Figure 4-26.
21
20
0
4-58. Centripetal acceleration would be provided by the gravitational force:
Mm mv 2
FG  G 2 
r
r
L  mvr  n
rn 

n
m  GM / rn 
1/ 2
1/ 2
 GM 
M  proton mass and m  electron mass, so v  

 r 
r  n / mv or

The total energy is: E 
rn2 
n2 2 rn
n2 2
and,
r

n
m2GM
GMm2
1 2  GMm  1  GM
mv   
 m
2
r  2  r


 GMm  GMm
GMm
En  

2rn
2n 2 2
2
  G M
2
2
2n 2
11
N m2 / kg 2
GMm2
m3
G 2 M 2 m3  1 1 

2 2  22 32 
2
27
34
105
2
GMm

   2r

 1.67 10 kg  9.11 10
2 1.055  10 
2
ao 
2
The gravitational Hα line is: E  E2  E3 
 6.67  10
E 

2
31
kg
  0.1389
3
Chapter 4 – The Nuclear Atom
(Problem 4-58 continued)
 5.85 1098 J  3.66 1079 eV
E
5.85  1098 J
f 

 8.28  1065 Hz
34
h
6.63  10 J s
For the Balmer limit in each case,
E  3.66  1079 eV  0.250 / 0.1389   6.58  1079 eV
f  6.58  1079 eV / h  1.59 1064 Hz
These values are immeasurably small. They do not compare with the actual H values.
4-59. Refer to Figure 4-16. All possible transitions starting at n = 5 occur.
n = 5 to n = 4, 3, 2, 1
n = 4 to n = 3, 2, 1
n = 3 to n = 2, 1
n = 2 to n = 1
Thus, there are 10 different photon energies emitted.
ni
nf
fraction
no. of photons
5
4
1/4
125
5
3
1/4
125
5
2
1/4
125
5
1
1/4
125
4
3
1/ 4 1 / 3
42
4
2
1/ 4 1 / 3
42
4
1
1/ 4 1 / 3
42
3
2
1 / 2 1 / 4  1 / 4 1 / 3
83
3
1
1 / 2 1 / 4  1 / 4 1 / 3
83
2
1
1 / 2 1 / 4  1 / 4 1 / 3   1 / 4 1 / 3  1 / 4


250
Total = 1,042
Note that the number of electrons arriving at the n = 1 level (125+42+83+250) is 500, as it
should be.
106
Chapter 4 – The Nuclear Atom
4-60.
m k 2e4
Z2
En   Eo 2 where Eo 
n
2 2
m  1.88  1028 kg

Eo  4.50  1016 J  2.81 103 eV


Z2
Thus, for muonic hydrogen-like atom: En   2.81  10 eV 2
n
3
(a) muonic hydrogen
0
n=∞
n=5
n=4
n=3
-500
n=2
En (eV)
E∞ = 0 eV
E5 = -112 eV
E4 = -176 eV
E3 = -312 eV
E2 = -703 eV
E1 = - 2.81 103 eV
-1000
-1500
-2000
-2500
n=1
-3000
2
2
n2 2
n2
4 n
(b) rn 

 2.56  10
nm
mkZe2 mke2 Z
Z
For H, Z = 1: r1  2.56  104 nm
For He1+ , Z = 2: r1  1.28  104 nm
For Al12+ , Z = 13: r1  1.97  105 nm
For Au 78+ , Z = 79: r1  3.2  106 nm
107
(Equation 4-18)
Chapter 4 – The Nuclear Atom
(Problem 4-60 continued)
(c) Nuclear radii are between about 1 and 8  105 m , or 1  8  106 nm . (See Chapter
11.) The muon n = 1 orbits in H, He1+, and Al12+ are about roughly 10nm outside the
nucleus. That for Au78+ is very near the nucleus’ surface.
(d) hf  hc /   E2  E1
For H:  
1 
2

  hc /  E2  E1 
hc
 5.89  1010 m  0.589nm
3
2.81  10 1  1 / 4 

Similarly,
For He1+:   0.147nm
For Al12+:   0.00349nm
For Au78+:   9.44  105 nm
108

where En   2.81  103 eV
 Zn
2
2
Chapter 5 – The Wavelike Properties of Particles
5-1.



(b) v 

h
6.63  1034 J s

 6.6  1029 m / s  2.1  1021 m / y
3
2
m
10 kg 10 m



h
h
hc 1240MeV fm



 12.4 fm
p E/c E
100MeV
5-2.

5-3.
 hc 
p2
Ek  eVo 

2m 2mc 2 2
5-4.

1240eV nm 
1
Vo 
 940V
e 2 5.11  105 eV  0.04nm 2
2
h

p
h
2mEk
2

hc

(b) For a proton:  

(from Equation 5-2)
2mc 2 Ek
(a) For an electron:  
1240eV nm

 2  0.511  106 eV



4.5  103 eV 
1/ 2
1240eV nm

 2  983.3  106 eV

(c) For an alpha particle:  
5-5.

6.63  1034 J s 3.16  107 s / y
h
h
(a)   

 2.1  1023 m
3
p mv
10 kg 1m / y 
 4.5 10 eV 
3
1/ 2
 0.0183nm
 4.27  104 nm
1240eV nm

 2  3.728  109 eV

  h / p  h / 2mEk  hc / 2mc 2 1.5kT 
1/ 2
 4.5 10 eV 
3
1/ 2
 2.14  104 nm
(from Equation 5-2)
Mass of N2 molecule =
2  14.0031u 931.5MeV / uc 2  2.609  104 MeV / c 2  2.609  1010 eV / c 2


109
Chapter 5 – The Wavelike Properties of Particles
(Problem 5-5 continued)

5-6.
5-7.

1240eV nm




 2  2.609  1010 eV 1.5 8.617  105 eV / K  300 K  


h

p
h
hc

2mEk

2
2mc Ek
1/ 2
 0.0276nm
1240eV nm

 2 939.57  106 eV


 0.02eV 
1/ 2
 0.202nm
(a) If there is a node at each wall, then n   / 2  L where n  1, 2, 3,... or   2 L / n .
E  p 2 / 2m   hn / 2L  / 2m  h 2n 2 / 8mL2
(b) p  h /   hn / 2L
En
 hc 

2
2
n2
8mc 2 L2
For n = 1: E1 

1240eV
nm  1
8 938  106 eV
2
2
  0.01nm 
 2.05eV
2
For n = 2: E2  2.05eV  2   8.20eV
2
5-8.
(a)  / c  102 is a nonrelativistic situation, so

 / c   hc
Ek 
2mc 2 Ek
mc 2
2   c 
2


 hc mc    mc
2
0.511  106 eV
 
2 102
2
2
2Ek

1/ 2
 25.6eV
(b)  / c  0.2 is a relativistic for an electron, so   h  mu   u  h  m .
uc
1  u / c 
u c 
2
1  u / c 
2
2


h
 c
mc 
 
 c 

2

u c
c / 
1/ 2
1    /  2 
c


110
Chapter 5 – The Wavelike Properties of Particles
(Problem 5-8 continued)
u c
1 0.2   0.981
1  1 0.2 2 


   5.10
Ek  mc2   1  0.511MeV   1  2.10MeV
(c)  c  103
1 10 
u c
1  1 10  


3
3
2
1/ 2
 0.9999

  1000
Ek  mc2   1  0.511MeV  999   510MeV
5-9.
Ek  mc 2   1
p   mu
(a) Ek  2GeV
mc 2  0.938GeV
  1  Ek mc2  2GeV 0.938GeV  2.132 Thus,   3.132
Because,   1


1  u / c 
2
where u / c  0.948
h
h
hc


2
p  mc  u / c   mc  u / c 
1240eV nm
 4.45  107 nm  0.445 fm
6
 3.132 938  10 eV  0.948


(b) Ek  200GeV
  1  Ek mc2  200GeV 0.938GeV  213 Thus,   214 and u / c  0.9999

5-10.
1240eV nm
 6.18  103 fm
 214 938MeV  0.9999 
n  D sin
(Equation 5-5)
n n
hc
sin 

(see Problem 5-6)
D D 2mc 2 Ek
111
Chapter 5 – The Wavelike Properties of Particles
(Problem 5-10 continued)
 5.705eV 
1
1240eV nm


1/ 2
0.215nm  2 5.11  105 eV 
Ek
Ek


1/ 2


(a)
 5.705eV 
sin 
(b)
 5.705eV 
sin 
1/ 2
75eV
 0.659
  sin1  0.659   41.2
 0.570
  sin1  0.570   34.8
1/ 2
5-11.

100eV
h

p
h
2m p Ek
 0.25nm
Squaring and rearranging,
 hc  
1240eV nm 
h2
Ek 

 0.013eV
2
2
2m p 
2 m p c 2  2 2 938  106 eV  0.25nm 
2

2



n  D sin
 sin  n / D  1 0.25nm  /  0.304nm 
sin  0.822
   55
5-12. (a) n  D sin  D 

(b) sin 
n
nhc

sin sin 2mc 2 Ek
11240eV

nm 

 sin 55.6 2 5.11 105 eV  50eV 
1/ 2
11240eV nm 
n

 0.584
D  0.210nm   2 5.11  105 eV 100eV 1 / 2




  sin1  0.584   35.7
112
 0.210nm
Chapter 5 – The Wavelike Properties of Particles
5-13.
d  t cos 42
n  t  d  t 1  cos 42  0.30nm 1  cos 42
42°
For the first maximum n = 1, so   0.523nm
t
h
 
p
h
2mEk
 hc 
h2
Ek 

2
2m
2mc 2 2
2

d
Ek 
5-14.

1240eV
2 939  106 eV
nm 

2
 0.523nm 
 3.0  103 eV
D sin
(Equation 5-6)
n
For 54eV electrons λ =0.165nm and sin   0.165nm  n / 0.215nm  0.767n

For n = 2 and larger sin  1 , so no values of n larger than one are possible.
5-15.
sin  n / D
(Equation 5-6)
  h / p  h / 2mEk  hc / 2mc 2 Ek 
1240eV nm


 2 0.511  106 eV  350eV 


1/ 2
 0.0656nm
sin  n  0.0656nm  /  0.315nm   0.208n
For n = 1,  = 12°. For n = 2,  = 24.6°. For n = 3,  = 38.6°. For n = 4,  = 56.4°.
This is the largest possible  . All larger n values have sin  1.
1
1

 105 s  10 s
1
f 100, 000s
1
1
1
 f 

 1.59  104 Hz
(b) f t 
2
2t 2  105 s
5-16. (a) t 
113
Chapter 5 – The Wavelike Properties of Particles
5-17. (a)
y  y1  y2
 0.002m cos 8.0 x / m  400t / s   0.002m cos  7.6 x / m  380t / s 
1
1

 2  0.002m  cos  8.0 x / m  7.6 x / m    400t / s  380t / s 
2
2

1
1

 cos  8.0 x / m  7.6 x / m    400t / s  380t / s 
2
2

 0.004m cos  0.2 x / m  10t / s   cos  7.8x / m  390t / s 
(b) v 

(c) vs 
k

390 / s
 50m / s
7.8 / m

20 / s

 50m / s
k 0.4 / m
(d) Successive zeros of the envelope requires that 0.2x / m   , thus

2
x 
 5 m with k  k1  k2  0.4m1 and x 
 5 m.
0.2
k
5-18. (a) v  f  Thus,

dv
df
dv
df
 2 d
 f 
, multiplying by , 
  f  2
v
d
d
d
d
2 d 
 2 d
dv
 v
Because k  2 / , dk    2 /  2  d  and
2 d 
d
d
dv
 vs  v  
dk
d
(b) v decreases as λ decreases, dv/dλ is positive.
5-19. (a) c  f    / T
 T   / c  2  102 m / 3  108 m / s  6.7  1011 s / wave


m  3.73  10   74.6m
The number of waves = 0.25 s / 6.7  1011 s / wave  3.73  103
Length of the packet =    # of waves   2  102
(b)

3

f  c /   3  108 m / s / 2  102 m  1.50  1010 Hz
(c)  t  1 
  1 / t  1 / 0.25  106 s  4.0  106 rad / s  637kHz
114
Chapter 5 – The Wavelike Properties of Particles
5-20.
 t  1 
  1 / t  1 / 0.25s  4.0rad / s or f  0.6Hz
5-21.
 t  1 
 2f  t  1
5-22. (a)  
h

p
h
2mEk
Thus, t  1 /  2  5000Hz   3.2  105 s
hc

2mc 2 Ek

1240eV nm


 2 0.511  106 eV  5eV 


1/ 2
 0.549nm
d sin   / 2 For first minimum (see Figure 5-17).
d

2 sin

0.549nm
 3.15nm slit separation
2 sin 5
(b) sin 5  0.5cm / L where L = distance to detector plane L 
0.5cm
 5.74cm
2 sin 5
5-23. (a) The particle is found with equal probability in any interval in a force-free region.
Therefore, the probability of finding the particle in any interval ∆x is proportional to
∆x. Thus, the probability of finding the sphere exactly in the middle, i.e., with
∆x = 0 is zero.
(b) The probability of finding the sphere somewhere within 24.9cm to 25.1cm is
proportional to ∆x =0.2cm. Because there is a force free length L = 48cm available
to the sphere and the probability of finding it somewhere in L is unity, then the
probability that it will be found in ∆x = 0.2cm between 24.9cm and 25.1cm (or any
interval of equal size) is: Px  1 / 48 0.2cm   0.00417cm.
5-24. Because the particle must be in the box
Let u   x / L;
 A  L /   sin
2
0
L
0
0
2
2
 * dx  1   A sin  x / L  dx  1
x  L  u   and dx   L /   du , so we have
x  0  u  0;

L

2
udu  A  L /    sin2 udu  1
2
0


u sin 2u 
  L /   A2  / 2   LA2 / 2  1
 L /   A2  sin2 udu   L /  A2  

4 0
2
0
 A2  2 / L  A   2 / L 
1/ 2
115


Chapter 5 – The Wavelike Properties of Particles
5-25. (a) At x  0 :
(b) At x   :
(c) At x  2 :
Pdx    0, 0  dx  Ae0 dx  A2 dx
2
2
Pdx  Ae
2
/ 4 2
Pdx  Ae4
2
2
/ 4 2
2
dx  Ae1 / 4 dx  0.61A2 dx
2
2
dx  Ae1 dx  0.14 A2 dx
(d) The electron will most likely be found at x = 0, where Pdx is largest.
5-26. (a) One does not know at which oscillation of small amplitude to start or stop counting.
N
t
f 
(b)  
N
1

t
t
x
2 2 N
2 n 2
and k 

, so k 

N

x
x
x
5-27. E t 
5-28. xp 
f 
 E  / t 

1.055  1034 J s
 6.6  109 eV
7
19
10 s 1.609  10 J / eV


p  mv 
2x
1.055  1034 J s
 v 

 5.3  1027 m / s
3
2
3
2mx
2  10 kg 10  10 m
2

5-29.
E t 


E  / t 

6.58  1016 eV s
 1.99  1021 eV
4
3.823d 8.64  10 s / d


The energy uncertainty of the excited state is ∆E, so the α energy can be no sharper than
∆E.
5-30.
xp 
 p  h
 p  h / . Because   h / p, p  h / ; thus, p  p.
5-31. For the cheetah p  mv  30kg  40m / s   1200kg m / s. Because p  p
(see Problem 5-30), x  / p  50 J s / 1200kg m / s  4.2  102 m  4.2cm
116
Chapter 5 – The Wavelike Properties of Particles
5-32. Because c  f  for photon,   c / f  hc / hf  hc / E, so
hc
E


and p 
1240eV nm
 2.48  105 eV
5.0  103 nm
E 2.48  105 eV

 8.3  107 eV s / m
c
3  108 m / s
For electron:
p 
h
4.14  1015 eV s

 8.3  104 eV s / m
x
5.0  1012 m
Notice that ∆p for the electron is 1000 times larger than p for the photon.
5-33. (a) For
48
Ti:
E  upper state  
E  lower state  
t
t

1.055  1034 J s
 4.71  1010 MeV
14
13
1.4  10 s 1.60  10 J / MeV

1.055  1034 J s
 2.20  1010 MeV
12
13
3.0  10 s 1.60  10 J / MeV




E  total   EU  EL  6.91 1010 MeV
ET 6.91  1010 MeV

 5.3  1010
E
1.312MeV
(b) For Hα: EU 
1.055  1034 J s
 6.59  108 eV
8
19
10 s 1.60  10 J / eV


and EL  6.59  108 eV also.
ET  1.32  107 eV is the uncertainty in the Hα transition energy of 1.9eV.
5-34.
 t  1 
2f t  1
For the visible spectrum the range of frequencies is f   7.5  4.0   1014  3.5  1014 Hz
The time duration of a pulse with a frequency uncertainty of f is then:
t 
1
1

 4.5  1016 s  0.45 fs
14
2f 2  3.5  10 Hz
117
Chapter 5 – The Wavelike Properties of Particles
5-35. The size of the object needs to be of the order of the wavelength of the 10MeV neutron.
  h / p  h /  mu.  and u are found from:
Ek  mnc2   1 or   1  10MeV / 939MeV
  1  10 / 939  1.0106  1 / 1  u 2 / c 2 
1/ 2
or u  0.14c
h
hc
1240eV nm


 9.33 fm
2
 mu  mc  u / c  1.0106  939  106 eV  0.14 


Nuclei are of this order of size and could be used to show the wave character of 10MeV
neutrons.
Then,  


5-36. (a) E  135MeV , the rest energy of the pion.
(b) E t 
2
6.58  1016 eV s
t 

 2.44  1024 s
6
2E 2  135  10 eV
5-37.
ps
L  rps
s
φ
r

s
r
s
r
L   rps  s / r   ps s 
L  r p
 
In the Bohr model, L  n and may be known to within L  0.1 .
Then   /  0.1
  10rad.
This exceeds one revolution, so that  is completely
unknown.
5-38.
E  hf
Et  h

E  hf E

f t  1 where t  0.85ns
118
Chapter 5 – The Wavelike Properties of Particles
(Problem 5-38 continued)
f  1 / 0.85ns  1.18  109 Hz
For   0.01nm
f  c/ 
3.00  108 m / s  109 nm / m
 3.00  1019 Hz
0.01nm
f
1.18  109 Hz

 3.9  1011
19
f
3.00  10 Hz
5-39.
Et 
 t 
2
2E
16
6.58  10 eV s
t 
 1.32  1024 s
6
2  250  10 eV
5-40. In order for diffraction to be observed, the aperture diameter must be of the same order of
magnitude as the wavelength of the particle. In this case the latter is

h
6.63 1034 J  s

 1.66 1033 m
3
p (4 10 kg)(100 m/s)
The diameter of the aperture would need to be of the order of 1033 m. This is many,
many orders of magnitude smaller than even the diameter of a proton or neutron. No
such apertures are available.
5-41. The kinetic energy of the electron needed must be no larger than 0.1 nm. The minimum
kinetic energy of the electrons needed is then given by:
h2
2me  2

h
h

p
2me E
E
(6.6334 J  s) 2
 2.411017 J  151eV
31
9
2
2(9.1110 kg)(0.110 nm)
 E
119
Chapter 5 – The Wavelike Properties of Particles
5-42. (a) For a proton or neutron:
xp 
v 
and p  mv assuming the particle speed to be non-relativistic.
2

2mx
1.055  1034 J s
 3.16  107 m / s  0.1c (non-relativistic)
27
15
2 1.67  10 kg 10 m





1.67  1027 kg 3.16  107 m / s
1 2
(b) Ek  mv 
2
2

2
 8.34  1013 J  5.21MeV
(c) Given the proton or neutron velocity in (a), we expect the electron to be relativistic,
in which case, Ek  mc 2   1 and
p 
2x
  mv
 v 
2mx
For the relativistic electron we assume v  c
 
2mcx

1.055  1034 J s
 193
2 9.11  1031 kg 3.00  108 m / s 1015 m





Ek  mc 2   1  9.11 1031 kg 3.00  108 m / s
E  hf  
5-43. (a) E 2  p2c2  m2c4
v

k
(b) vs 


k
2

d d

dk dk
 192  1.58 10
2
p  h/  /k
k 2 c 2  m2 c 4
 c 1  m2c 2 /
k
k 2c 2  m2c 4

k

2
k2  c
c2k
k 2c 2  m2c 4
2

5-44.
c2k


c2 k


c2 p
u
E
2 y 1 2 y

(Equation 5-11)
x 2 v 2 t 2
(by Equation 2-41)
y3  C1 y1  C2 y2
 1  2 y1 
 1  2 y2 
 2 y3
 2 y1
 2 y2

C

C

C

C


1
2
1
2 2
2
2
2 
x 2
x 2
x 2
 v t 
 v t 
120
2 
2
11
J  98MeV
2
k 2c 2  m2c 4
Chapter 5 – The Wavelike Properties of Particles
(Problem 5-44 continued)

1 2
1  2 y3
C
y

C
y



1 1
2 2
v 2 t 2
v 2 t 2
5-45. The classical uncertainty relations are
 t  2f t  1 (Equation 5-18)
and x 
(a) f 
2
2
(Equation 5-20)
1
1

 0.0541Hz
2t 2  3.0s 
(b) Length of the wave traing L  vt , where v = speed of sound in air = 330m/s.
L   330m / s  3.0s   990m
(c)  
2
where x = length of the wave train  990m from (b)
2x
 0.13m   2.72  106  2.72 m
 
2  990m 
2
and   0.13m from (d) .
(d) v  f 

5-46. (a) n   / 2   L

v  330m / s 

 0.13m  13cm
f  2500 Hz 
   2L / n. Because   h / p  h / 2mE , then
h2
h2
h2 n2
E


2m 2 2m  L / n 2 8mL2
h2 n2
If E1  h / 8mL , then En 
 n 2 E1
2
8mL
2
2
 hc  
1240eV nm 
h2
(b) For L  0.1nm, E1 

2
2 2
8mL 8mc L 8 0.511  106 eV  0.1nm 2
2
2

E1  37.6eV and En  37.6n 2eV
121

Chapter 5 – The Wavelike Properties of Particles
(Problem 5-46 continued)
n
E
(eV)
1000
5
800
600
4
400
3
200
0
(c)
f  E / h
2
1
0

L
c /   E / h


x
hc
E
For n  2  n  1 transition, E  112.8eV and  
1240eV nm
 11.0nm
112.8eV
(d) For n  3  n  2 transition, E  188eV and  
1240eV nm
 6.6nm
188eV
(e) For n  5  n  1 transition, E  903eV and  
1240eV nm
 1.4nm
903eV
5-47. (a) For proton: E1 
 hc 
2
8m p c 2 L2
1240MeV fm 
E1 
2
8  938MeV 1 fm 
from Problem 5-46.
2
 205MeV and En  205n2 MeV
 E2  820MeV and E3  1840MeV
(b) For n  2  n  1 transition,  =
hc 1240MeV fm

 2.02 fm
E
615MeV
(c) For n  3  n  2 transition,  =
hc 1240MeV fm

 1.22 fm
E
1020MeV
(d) For n  3  n  1 transition,  =
hc 1240MeV fm

 0.76 fm
E
1635MeV
122
Chapter 5 – The Wavelike Properties of Particles
5-48. (a) E 
2
/ 2mL2 (Equation 5-28) and E 
2
/ 2mA2
(b) For electron with A  1010 m :
E
 c
2
2mc 2 A2


197.3eV
nm 

2
1
2 0.511  10 eV 10 nm
6

2
 3.81eV
For electron with A  1cm or A  102 m :

 / 10 nm  3.8110 eV
1.055 10 J s 

2 100  10 g  10 kg / g  2  10 
E  3.81eV 101
(c) E 
2
2
7
16
34
2
2
2mL
3
2
3
2
2
 1.39  1061 J  8.7  1043 eV
5-49. p  mv  m  0.0001 500m / s   0.05m
For proton: xp 



x  / p  6.58  1016 eV s /  0.05m / s  938  106 eV

 1.40  1023 m  1.40 108 fm




For bullet: x  1.055  1034 J s /  0.05m / s  10  103 kg  2.1 1031 m
5-50.
2 y 1 2 y

(Equation 5-11) where y  f   and   x  vt.
x 2 v t 2
y f 
y 2 f  2   2 f 


and





x  x
x 2  x 2 x  2 x
y f 
y 2 f  2   2 f 


and





t  t
t 2  t 2 t  2 t
 2
Noting that 2  0,
x

 1,
x
 2

 0, and
 v, we then have:
2
t
t

f
2 f
1  f
2 f
 0  1  2  1  2   0   v   2    v  


v  


2 f 2 f

 2  2
123
Chapter 5 – The Wavelike Properties of Particles
5-51. (a)   h / p The electrons are not moving at relativistic speeds, so
  h / mv  6.63  1034 J s /  9.111031 kg 3 106 m / s   2.43 1019 m  0.243nm
(b) The energy, momentum, and wavelength of the two photons are equal.
1 2
1
1

mv  mc 2  mc 2 v 2 / c 2  mc 2  mc 2  v 2 / c 2  1
2
2
2


E

1
 0.511  106 eV  3  106 / 3  108
2

(c)





 1  0.511MeV

2
p  E / c  0.511MeV / c
(d)   hc / E  1240eV nm / 0.511106 eV  2.43 103 nm
5-52. (a) Q  mp c 2  mnc 2  m c 2
 1.007825uc2  1.008665uc2  139.6MeV
 938.8MeV  939.6MeV  139.6MeV  140.4MeV
E  140.4MeV
(b) Et 

t  / E  6.58  1016 eV s / 140.4  106 eV  4.7  1024 s


(c) d  ct  3  108 m / s 4.7  1024 s  1.4  1015 m  1.4 fm
5-53. hf   mc 2
  
hf
1

2
2
mc
1 v
c2
1/ 2
2
2
  mc 2 2 


mc
1 v 2  
 v  1  
 
c 
c  hf 
hf  



2
Expanding the right side, assuming mc
hf ,
2
2
4
v
1  mc 2  1  mc 2 
 1 
  
 
c
2  hf  8  hf 
and neglecting all but the first two terms,
2
v
1  mc 2 
 1 
 Solving this for m and inserting deBroglie’s assumptions that
c
2  hf 
v
 0.99 and   30m, m is then:
c


1  0.99  2 6.63  1034 J s
m
 1.04  1044 kg
8
3.00  10 m / s  30m 
1/ 2


124
Chapter 5 – The Wavelike Properties of Particles
5-54. (a)
m
xp 

mxvx 
1/ 2
 2y 
1
y0  gt 2  t   0 
2
 g 

vx  / mx
1/ 2
 2y 
1
X  vx t  vx  0 
2
 g 
1/ 2
 2y 
2  0
g 
 
mx
y0
1/ 2
 2y 
X  2vx  0 
 g 
∆x
(b) If also yp y 
 v y  / my and
t  v y / g  / mg y so, X 
5-55.
1
X  x  t  t  where vy  g t or
2
2 
1/ 2
2 y0 / g   / mg y 


mx 
1
3
m v 2  kT
2
2
vrms


23
3kT  3 1.381  10 J / K  300 K  



m
 56u 1.66  1027 kg / u

f   fo 1  v / c 


1/ 2

hf   hf o 1  v / c 
E  hf   hf o  hf o v / c 
1eV  366m / s   1.2  106 eV
3.0  108 m / s
This is about 12 times the natural line width.
10 eV  366m / s   1.2eV
E  hf v / c 
6
o
 366m / s
3.0  108 m / s
This is over 107 times the natural line width.
125
Chapter 5 – The Wavelike Properties of Particles
5-56. recoil    E / c
(a) Erecoil 
1eV 
Erecoil
2


2 56uc 2
 
 recoil
2
2m

E2
2mc 2
uc 2
 9.6  1012 eV
6
931.5  10 eV
This is about 10-4 times the natural line width estimated at 10-7eV.
(b) Erecoil
1MeV 


2
2 56uc 2

uc 2
 9.6eV
931.5  106 eV
This is about 108 times the natural line width.
126
Chapter 6 – The Schrödinger Equation
6-1.
d
d 2
 kAekx t  k  and
 k 2
2
dx
dx
Also,
d
  . The Schrödinger equation is then, with these substitutions,
dt
 2 k 2  / 2m  V   i  . Because the left side is real and the right side is a pure
Imaginary number, the proposed  does not satisfy Schrödinger’s equation.
6-2.
For the Schrödinger equation:

2 

 ik  and
 k 2 . Also,
 i .
2
x
t
x
Substituting these into the Schrödinger equation yields:
2
k 2  / 2m  V    , which is true, provided  
2
k 2 / 2m  V , i.e., if E  Ek  V .
For the classical wave equation: (from Equation 6-1)
2 
2 
2


k

and
also
  2 . Substituting into Equation 6-1
2
2
x
t
(with  replacing E and v replacing c) k 2   1/ v2  2  , which is true for
From above:



v   / k.
6-3.
(a)
d
d 2
  x / L2  and
dx
dx 2


 x  x
  2   2
 L  L
x2
1
 1



  2

2 
4
L
L
 L 
Substituting into the time-independent Schrödinger equation,
2 2
2
2


x

  V  x   E 


4
2 
2mL2
 2mL 2mL 
Solving for V(x), V  x  
2 2
2
2 2


x
x
1




 kx 2


2
4
2
4
2mL  2mL 2mL  2mL
2
2
127
Chapter 6 – The Schrödinger Equation
(Problem 6-3 continued)
where k 
2
/ mL4 . This is the equation of a parabola centered at x = 0.
V(x)
0
x
(b) The classical system with this dependence is the harmonic oscillator.
6-4.
(a) Ek  x   E  V  x  
2
/ 2mL2 
x / 2mL4 
2 2

2

/ 2mL2 1  x 2 / L2

(b) The classical turning points are the points where E  V  x  or Ek  x   0. That occurs
when x2 / L2  1, or when x   L.
(c) For a harmonic oscillator V  x   m 2 x 2 / 2, so
2
x2
  2 x2 / 2   2 
2mL4
Thus, E 
6-5.
2
/ m2 L4    / mL2
1


 2 

2mL  mL  2 2
2
2
(a)   x, t   A sin  kx  t 

  A cos  kx  t 
t
i

 i  A cos  kx  t 
t
2
 k 2 A sin  kx  t 
x 2
 2 2  k 2 A


sin  kx  t   i
2
2m x
2m
t
128
Chapter 6 – The Schrödinger Equation
(Problem 6-5 continued)
(b)   x, t   A cos  kx  t   iA sin  kx  t 
i

 i  A sin  kx  t   i 2  A cos  kx  t 
t
  A cos  kx  t   i  A sin  kx  t 

2 2
2
k A

cos  kx  t  
2
2m x
2m
2
k2
 A cos  kx  t   iA sin  kx  t  
2m 

t
i
(a)
ik 2 A
sin  kx  t 
2m
2

6-6.
2
2
if
k2
  it does. (Equation 6-5 with V = 0)
2m
For a free electron V(x) = 0, so 
2
d 2
d 2

E


  2.5  1010 
2
2
2m dx
dx

2

Substituting into the Schrödinger equation gives: 2.5  1010

2

2

and, since E  Ek  p 2 / 2m for a free particle, p 2  2m 2.5  1010

p  2.5  1010
(b)


/ 2m   E

2
2
 2.64  1024 kg m / s




E  p 2 / 2m  2.64  1024 kg m / s /  2  9.11 1031 kg  3.82  1018 J

2


 3.82  1018 J 1 / 1.60  1019 J / eV  23.9eV
(c)
6-7.
  h / p  6.63 1034 J s 2.64 1024 kg m / s  2.511010 m  0.251nm
  x   Ce x
(a)
2
/ L2
and E  0
d 2
 V  x   0
2m dx 2
d
2x
d 2  4 x 2 2 
  2   x  and

 2 
dx
L
dx 2  L4
L 
2
2
 4x2 2 
 2 x2




V
x


0
so
V
x

 1
And 




 4
2 
2 
2
2m  L
L 
mL  L


2
129

/ 2m and
Chapter 6 – The Schrödinger Equation
(Problem 6-7 continued)
V(x)
(b)
x
6-8.
a
a
a
a
 i  kx t 
i kx t 
2
e 
dx  1
   dx  A  e
a
A
2
 dx  A x
2
a
a
 A2  2a   1
a
 A
1
 2a 
1/ 2
Normalization between −∞ and +∞ is not possible because the value of the integral is
infinite.
6-9.
(a) The ground state of an infinite well is E1  h2 / 8mL2   hc  / 8mc 2 L2
2
For m  m p , L  0.1nm : E1 
(b) For m  m p , L  1 fm : E1 


1240MeV
8 938.3  106 eV
1240MeV
2
  0.1nm
fm 

fm 
2
 0.021eV
2
8 938.3  106 eV 1 fm 
2
 205MeV
6-10. The ground state wave function is (n = 1)  1  x   2 / L sin  x / L  (Equation 6-32)
The probability of finding the particle in ∆x is approximately:
P  x  x 
2
2x 2   x 
x 
sin2 
x 
sin 


L
L
 L 
 L 
130
Chapter 6 – The Schrödinger Equation
(Problem 6-10 continued)
2  0.002 L  2   L 
L

(a) For x  and x  0.002 L, P  x  x 
sin 
 0.004 sin2  0.004

2
L
2
 2L 
2  0.002 L  2  2 L 
2L
2
(b) For x 
and P  x  x 
sin 
 0.004 sin2
 0.0030

3
L
3
 3L 
(c) For x  L and P  x  x  0.004 sin2   0
6-11. The second excited state wave function is (n = 3)  3  x   2 / L sin  3 x / L 
(Equation 6-32). The probability of finding the particle in ∆x is approximately:
P  x  x 
2
 3 x 
sin2 
 x
L
 L 
(a) For
2  0.002 L  2  3 L 
L
3
x  and x  0.002 L, P  x  x 
sin 
 0.004 sin2
 0.004

2
L
2
 2L 
2L
 6 L 
2
(b) For x 
and P  x  x  0.004 sin2 
  0.004 sin 2  0
3
3
L


 3 L 
(c) For x  L and P  x  x  0.004 sin2 
 0.004 sin2 3  0

 L 
6-12.
2
1
n2 2 2
 1 2   2mL   mvL 
2
E  mv 2 
(Equation
6-24)
n

mv
2
  2 2     
2
2mL2



 
n
mvL

10

9


kg 103 m / s 102 m
 1.055  10
34
J s

  3  10
19


p  0.0001 p   0.0001 10 kg 10
6-13. (a) x  0.0001L   0.0001 102 m  106 m
9
(b)
xp
10 m10

6
16
1.055  10
kg m / s
34
J s
3

m / s  1016 kg m / s
  9  10
11
131
2
Chapter 6 – The Schrödinger Equation
6-14
(a) This is an infinite square well with width L. V(x) = 0, and E  Ek  p 2 / 2m. From
uncertainty principle: Ekmin  pmin  p  / x  / L and
2
Emin  pmin
/ 2m 
2
/ 2mL2  h2 / 8 2 mL2
(b) The solution to the Schrödinger equation for the ground state is:
1/ 2
 1  x    2 / L  sin  x / L 
d 2 1
   2 
and
    
2
dx
 L L
1/ 2
2
 
sin  x / L       1
L
2
h2
 


E

or
E

1
1
1
2m  L 
8mL2
The result in (a) is about 1/10 of the computed value, but has the correct
2
2
So,
dependencies on h, m, and L.
6-15. (a) For the ground state, L   / 2, so   2L.
(b) Recall that state n has n half-wavelengths between x = 0 and x = L, so for n = 3,
L  3 / 2, or   2L / 3.
6-16.
(c)
p  h /   h / 2L in the ground state.
(d)
p 2 / 2m  h2 / 4L2 / 2m  h2 / 8mL2 , which is the ground state energy.
En 
h2 n2
h2
and

E

E

E

n 2  2n  1
n
n 1
n
2
2
8mL
8mL



or, En   2n  1
 3 h 
so, L  

 8mc 
1/ 2

h2
hc

2
8mL

 3 hc 

2 
 8mc 
1/ 2
1/ 2
 3  694.3nm 1240eV nm  


6


8
0
.
511

10
eV



6-17. The uncertainty principle requires that E 

 0.795nm
2
2mL2
for any particle in any one-dimensional
box of width L (Equation 5-28). For a particle in an infinite one-dimensional square well:
En 
n2 h2
8mL2
132
Chapter 6 – The Schrödinger Equation
(Problem 6-17 continued)
h2
 0. This violates Equation 5-28 and, hence, the
8mL2
For n = 0, then E0 must be 0 since
exclusion principle.
6-18. (a) Using Equation 6-24 with L = 0.05 nm and n = 92, the energy of the 92nd electron in
the model atom is E92 given by:
En  n 2
2
2
 E92  (92)2
2mL2
(6.63 1034 J  s)2
8(9.111031 kg)(0.05 109 m)2
1eV
 1.28 106 eV  1.28 MeV
19
1.60 10 J
(b) The rest energy of the electron is 0.511 MeV. E92  2.5  the electron’s rest energy.
E92  2.04 1013 J 
6-19. This is an infinite square well with L = 10cm.


2.0  103 kg  20nm / y 
h2 n2 1 2
En 
 mv 
2
8mL2 2
2 3.16  107 s / y
n2 



  20 10 m   0.1m 
2  3.16  10 s   6.63  10 J s 
2
8 2.0  103 kg
7
n

9
2
2
34


3.16 10 s  6.63 10 J s 
2 2.0  103 kg 20  109 m  0.1m 
34
7
6-20. (a)  5  x    2 / L 
1/ 2
2
2
2
 3.8  1014
sin  5 x / L  dx
0.4 L
P
  2 / L  sin 5 x / L  dx
2
0.2 L
Letting 5 x / L  u, then 5 dx / L  du and x  0.2L  u  
and x  0.4L  u  2 , so
133
Chapter 6 – The Schrödinger Equation
(Problem 6-20 continued)
2
x

 sin 2 x 

2
L
1
 2  L 



2
P   
sin2 udu   

 



4
5
 L  5  
 L  5  



2
(b) P   2 / L  sin2
5  L / 2 
L
 0.01L   0.02 where 0.01L  x
1240MeV fm 
E1 
2
8  0.511MeV 10 fm 
2
6-21. (a) For an electron:
1240MeV fm 
E1 
2
8  938.3Mev 10 fm 
 3.76  103 MeV
2
(b) For a proton:
 2.05MeV
(c) E21  3E1 (See Problem 6-16)
For the electron: E21  3E1  1.13  104 MeV
For the proton: E21  3E1  6.15MeV
6-22.
F  dE / dL comes from the impulse-momentum theorem F t  2mv where t  L / v.
So, F
mv2 / L
E / L. Because E1  h2 / 8mL2 , dE / dL  h2 / 4mL3 where the minus
sign means “on the wall”. So F  h / 4mL 
2
3
The weight of an electron is mg  9.11 1031
 6.63 10
34
J s

2

 m 
kg  9.8m / s   8.9  10 N
31
4 9.11  10 kg 10
2
10
30
3
 1.21  107 N
which is
minuscule by comparison.
2
n x
sin
L
L
L
 n x   m x 
To show that  sin 
 sin  L  dx  0
L

 

0
6-23.  n  x  
Using the identity 2sin A sin B  cos  A  B   cos  A  B , the integrand becomes


1
cos  n  m   x / L   cos  n  m   x / L 
2
134
Chapter 6 – The Schrödinger Equation
(Problem 6-23 continued)
L sin  n  m   x / L
and similarly for the second

 n  m
Term with (n + m) replacing (n – m). Since n and m are integers and n ≠ m, the sines both
The integral part of the first term is
vanish at the limits x = 0 and x = L.
 n x   m x 
  sin 
 sin  L  dx  0 for n  m.
L

 

0
L
6-24. (a)
4
x
L
(b)
P
x
L
6-25. Refer to MORE section “Graphical Solution of the Finite Square Well”. If there are only
two allowed energies within the well, the highest energy E2  V0 , the depth of the well.
From Figure 6-14, ka   / 2 , i.e., ka 
2mE2
a  2
where a  1 / 2 1.0 fm   0.5 fm and m  939.6MeV / c 2 for the neutron.
Substituting above, squaring, and re-arranging, we have:
 
E2  V0   
2
 2  2 939.6MeV / c 2  0.5 fm 
2
2


   c 
V0 
2
8  939.6MeV   0.5 fm  106 nm / fm 
2
2
V0  2.04  108 eV  204MeV
135
  197.3eV
2


8 939.6  106 eV
nm 
2
 0.5 10
6
nm

2
Chapter 6 – The Schrödinger Equation
0.5eV and for a finite well also En
6-26. Because E1
n2 E1 , then n = 4 is at about 8eV, i.e.,
near the top of the well. Referring to Figure 6-14, ka
2mE
ka
L
L
2
2 / 7.24 109
6-27. For V2
E
7.24 109 m
1
L
2 .
2
8.7 10 10 m 0.87nm
V1 : x1 is where V
0
:
V1 and x2 and x2 is where V1
V2
From
to 0 and x2 to
is exponential
0 to x1:
is oscillatory; Ek is large so p is large and λ is small; amplitude is small because
Ek is large, hence v is large.
x1 to x2:
is oscillatory; Ek is small so p is small and λ is large; amplitude is large
because Ek is small, hence v is small.
6-28. (a)
6
5
4
3
2
1
x
-10
+10
0
136
Chapter 6 – The Schrödinger Equation
(Problem 6-28 continued)
6-29.
*
3
px
L
x
s dx
2
3 x
sin
L
L i x
0
L
2
Li
Let
3
i x
3 x
L
sin
0
3 x
L
y Then x
cos
0
(Equation 6-48)
2
3 x
sin
dx
L
L
3 x
L
y
0,
3
dx
L
x
L
y
3 , and
3
dx
L
dy
dx
L
dy
3
Substituting above gives:
2 L
Li 3
px
2
Li
3
sin y cos ydy
0
3
L
3
sin y cos ydy
0
2
sin2 y
Li
2
3
2
0 0
Li
0
0
Reconiliation: px is a vector pointing half the time in the +x direction, half in the –x
direction. Ek is a scalar proportional to v 2, hence always positive.
6-30. For n = 3,
3
2/ L
1/ 2
sin 3 x / L
L
(a)
x 2 / L sin2 3 x / L dx
x
0
137
Chapter 6 – The Schrödinger Equation
(Problem 6-30 continued)
3 x / L, then x
Substituting u
x
0
u
0 and x
L
u
Lu / 3 and dx
L/3
du. The limits become:
3
3
2/ L L/3
x
u sin2 udu
1/ 3
0
2
2/ L L/3
2
2 / L 1/ 3
u2
4
u sin 2u
4
2
3
/4
cos 2u
8
3
0
L/2
L
x2
(b)
x 2 2 / L sin2 3 x / L dx
0
Changing the variable exactly as in (a) and noting that:
3
2
u3
6
2
u sin udu
0
u2
4
1
1
3 18
We obtain x 2
L2
2
3
u cos 2u
4
1
sin 2u
8
0
0.328L2
6-31. (a) Classically, the particle is equally likely to be found anywhere in the box, so
L
P(x) = constant. In addition,
P x dx 1 so P x
1/ L.
0
L
(b)
L
x / L dx
x
L / 2 and x
2
x 2 / L dx
0
0
2
6-32.
d2
2m dx 2
L2 / 3
1
d
2m i dx
V x
d
i dx
1
pop pop
2m
Multiplying by
x
E
x
x
(Equation 6-18)
E V x
x
E V x
and integrating over the range of x,
138
Chapter 6 – The Schrödinger Equation
(Problem 6-32 continued)
2
pop
dx
2m
E V x
p2
2m
E V x
dx
p2
or
2m E V x
x does not vanish and vice versa.
For the infinite square well V(x) = 0 wherever
0 and p 2
Thus, V x
6-33.
x
L2
3
2
x
x
L2
2 2
2
2mE
2m
n2 2 2
2mL2
L2
3
x
L2
2 2
L2
4
1/ 2
for n 1
L2
L
2
(See Problem 6-30.) And x
2
2 2
1
L
12
1/ 2
1
0.181L
2
2
2 2
p
2
x
p
P
6-34.
0
x
x
2
p
A0e
A02 xe
x
A02
x
0.)
2
(See Problem 6-32)
1/ 2
2
0
L2
m x2 /
L
. And
dx Letting u 2
ue
m x2 /
/m
/m
u2
du
x
0.181L
p
/L
m x2 /
1
udu
1/ 2
/m
and x
u
xdx; limits are unchanged.
0 (Note that the symmetry of V(x) would also tell us that
dx
3/ 2
3/ 2
u 2e
0.568
1/ 4
m /
where A0
2 xdx . And thus, m /
A02 x 2 e
2 A02
0
2
/m
A02
p
m x2 / 2
m /
2udu
x2
and
L2
u2
du
/4
2 A02
m /
/m
1/ 2
139
3/ 2
/m
u 2e
3/ 2
u2
du
/2
/ 2m
Chapter 6 – The Schrödinger Equation
6-35.
p2
2m
1
m
2
n 1/ 2
x
2
m 2
x2
m
2 2
p2
m
1
6-36. (a)
/2
p 2 / 2m
or
2m
x,t
0
. For the ground state (n = 0),
m /
1/ 4
x2
and
m x2 / 2
e
m
p2
m
1
e
p2
m2 2
1
2
/ 2m
1
m
2
p2
i t/2
2
(b) px op
p
i x
0
A0 m x /
x
2
x, t
0
1/ 4
e
m x2 / 2
e
0
i x
i t/2
2
0
x
2
m x/
A0
p2
2
A02 m /
2
A02 m /
m x/
Letting u
p2
2
m x/
m /
m x2 /
1 e
m x2 /
1/ 2
e
m
2
1/ 2
2
m /
m /
m x2 /
m x2 /
1/ 2
u 2e
2
u 2e
u2
1/ 2
1/ 2
m /
m
u2
du
0
2
m x2 / 2
e
e
i t/2
dx
dx
e
du
e
m x2 / 2
x , then
A02 m /
A02 m /
x, t dx
e
u2
du
0
1/ 2
2
/2
140
4
2
u2
du
dx
(See Problem 6-34)
Chapter 6 – The Schrödinger Equation
6-37.
0
x
m x2 / 2
C0e
(a)
0
2
x
(Equation 6-58)
2
dx 1
2
C0
(b) x 2
0
m
(c)
6-38.
1
x
C1 xe
(a)
1
3
1
1
4
with
x
2
x2
dx
3
m /
with
1
m
2
1
2m
m x2 /
2
2
1
4
1
2 m3
2
3
3
dx
C1
with
2
2I 2
m /
3
3
1/ 4
3
dx
m /
1
2m
2
4m3
C1
x
m x2 /
2
C1 x 2e
C1
x
1
2
2
(Equation 6-58)
C1
(b)
e
1
m
2
x2
dx 1
2
=
3
m
m x2 / 2
2
x
m
3
2
m
x2
2I2
1
m
2
V x
2
1/ 4
dx
1 m
2
C0
m
m
2
x2
2I0
dx
2
C0
C0
m x2 /
C0 e
x
3
4m3
3
3
1/ 2
e
m x2 /
141
dx
0
1
4
Chapter 6 – The Schrödinger Equation
(Problem 6-38 continued)
x
(c)
2
x
2
2
1
4m3
dx
x
6-39. (a)
3
1
m
2
x p
m
2
1
m
2
(1) E
5
/ x
2
0
2
En
2m
n 1/ 2
1
En
limn
En
En
2
2
2
2
1
m
2
x2
2
3
8
5
where
m /
2
3
2m
3
4
/ 8mA2
E02 2mA2 4
2
4
m
2
En
E02
4 Ek
4 Ek
1
/4
2 Ek
n 3/ 2
En n 3 / 2 n 1
n 1/ 2
En En
x 2 dx
/ x2 is computed in Problem 6-36(b). Using that quantity,
Ek
En
2
/ 2 also E0
2
6-40.
1/ 2
3
3
2
A2
m x2 /
/ 2A
Because E0
(2)
4m3
h / 2 A / 2m
2
e
3
2m
5
1
m
2
x2
p
p 2 / 2m
(b) Ek
5
3
1/ 2
3
3
2I4
3
(d) V x
4m3
1/ 2
3
3 m3
2
2
limn
1 n 1/ 2
1
n 1/ 2
0. In agreement with the correspondence principle.
142
Chapter 6 – The Schrödinger Equation
6-41. (a) Harmonic oscillator ground state is n = 0.
x
0
A0e
m x2 / 2
(Equation 6-58)
A02
Therefore, for x: x
Let m x 2 /
y2
m x2 /
xe
x
dx
m y
dx
m dy
A02
Substituting above yields: x
m
ye
y2
dy
0
by inspection of Figure 6-18, integral tables, or integration by parts.
For x 2 :
x2
A02
m x2 /
x 2e
dx
Substituting as above yields: x 2
A02
The value of the integral from tables is
x2
Therefore,
A02
2
m
3/ 2
m
y 2e
y2
dy
2.
3/ 2
(b) For the 1st excited state, n = 1,
A1 m / xe
x
1
A12
x
x3 m
e
m x2 / 2
m x2 /
(Equation 6-58)
dx
Changing the variable as in (a),
x
A12 m
A12
3/ 2
m
m
y 3e
y2
dy
y 3e
y2
m
1/ 2
dy
0
by inspection of Figure 6-18, integral tables, or integration by parts.
x2
A12
x4 m
e
m x2 /
dx
changing variables as above yields:
143
Chapter 6 – The Schrödinger Equation
(Problem 6-41 continued)
x2
A12 m
A12
2
m
3/ 2
m
y 4e
y 4e
y2
y2
m
1/ 2
dy
The value of the integral from tables is 3
x2
Therefore,
6-42. (a)
A12 3
2 /T
2 f
1
E0
2
4
2 / 1.42s
34
1.055 10
dy
4.
3/ 2
m
4.42rad / s
J s 4.42rad / s / 2
2.33 10
34
J
499.9 mm
500.0 mm
(b) A
500.0
2
499.9
2
10mm
0.1mm
E
n 1/ 2
1 / 2m
2
A
2
A
n 1 / 2 1 / 2 0.010kg 4.42rad / s 10 2 m
2.1 1028 or n
(c)
6-43.
0
f
x
1.055 10
34
J s
2.1 1028
0.70 Hz
2
x2 / 2
A0e
2
1
x
A1
m
xe
m x2 / 2
From Equation 6-58.
Note that
0
is an even function of x and
It follows that
0
1
dx
0
144
1
is an odd function of x.
Chapter 6 – The Schrödinger Equation
2
6-44. (a) For x > 0,
So, k2
(b) R
k22 / 2m V0
1/ 2
2mV0
2
k1 k2
2
E
k12 / 2m
. Because k1
k1
k2
2
2V0
1/ 2
4mV0
, then k2
k1 / 2
(Equation 6-68)
2
2
1 1/ 2
1 1/ 2
0.0294, or 2.94% of the incident particles are
reflected.
(c) T
1 R 1 0.0294 0.971
(d) 97.1% of the particles, or 0.971 106
9.71 105 , continue past the step in the +x
direction. Classically, 100% would continue on.
6-45
(a) Equation 6-76:
T
16
E
E
1
e
V0
V0
2 a
where
2 2m p (V0
E) /
and a barrier width .
2 a
2(938 MeV/c 2 )(50 44)MeV / 6.58 10
2
(b) decay rate
N
44 MeV
44 MeV
1
e
50 MeV
50 MeV
T
16
T
0.577
22
MeV s
10
15
1.075
N T where
2 44 MeV 1.60 10 13 J/MeV
1.67 10 27 kg
v proton
2R
0.577 4.59 1022 s
decay rate
(c) In the expression for T, e
1.075
e
2.150
1/2
1
1
2 10 15 m
2.65 1022 s
, and so T
0.577
4.59 1022 s
So, k2
2
k22 / 2m V0
6mV0
1/ 2
E
2
k12 / 2m
. Because k1
T
0.197 . The decay
2V0
4mV0
145
1/ 2
, then k2
1
1
rate then becomes 9.05 10 21 s 1 , a factor of 0.34× the original value.
6-46. (a) For x > 0,
1.075
3 / 2k1
Chapter 6 – The Schrödinger Equation
(Problem 6-46 continued)
(b) R
k1
R
k2
k1
2
k2
k1
2
k2
k1
2
2
2
k2
2
3/ 2
1
1
3/ 2
0.0102
Or 1.02% are reflected at x = 0.
(c) T
1 R 1 0.0102
0.99
(d) 99% of the particles, or 0.99 106
9.9 105, continue in the +x direction.
Classically, 100% would continue on.
6-47. (a)
E
4eV
2m V0
9eV =V0
E /
2 0.511 106 eV / c2 5eV /
0.6 nm = a
5.11 106 eV
2260eV
197.3eV nm
and
Since
a
0.6nm 11.46nm 1
a is not
eV
/
c
11.46nm
1
6.87
1, use Equation 6-75:
The transmitted fraction
T
Recall that sinh x
T
1
sinh2 a
1
4 E V0 1 E V0
ex
e
x
1
81
sinh2 6.87
80
1
2,
81 e6.87 e
1
80
2
6.87
2
1
4.3 10
(b) Noting that the size of T is controlled by
6
is the transmitted fraction.
a through the sinh2 a and increasing T
implies increasing E. Trying a few values, selecting E = 4.5eV yields T
or approximately twice the value in part (a).
146
8.7 10
6
Chapter 6 – The Schrödinger Equation
6-48.
B
E1 / 2
E
E V0
1/ 2
E V0
1/ 2
1/ 2
A
For E
V0 , E V0
denominator are complex conjugates. Thus, B
hence T
6-49.
A B
1 R
1/ 2
2
is imaginary and the numerator and
2
A and therefore R
B
2
A
2
0.
C and k1 A k1B
k 2C
Substituting for C, k1 A k1 B
B
k1
k1
A
k1 k2
A C
k1 k2
(Equations 6-65a & b)
k2 A k2 B and solving for B,
k2 A B
k2
A, which is Equation 6-66. Substituting this value of B into Equation 6-65(a),
k2
A
k1
k2 k1 k2
k1 k2
or C
2k1
, which is Equation 6-67.
k1 k2
6-50. Using Equation 6-76,
6-51.
T
16
T
16
R
1,
E
E
1
e
V0
V0
2.0
6.5
k1
k2
k1
k2
1
2 a
2.0
e
6.5
where E
2 10.87 0.5
2.0eV , V0
6.5 10
5
6.5eV , and a
0.5nm.
(Equation 6-75 yields T
6.6 10 5 .)
2
and T
2
1 R
(Equations 6-68 and 6-70)
(a) For protons:
k1
2mc 2 E / c
k2
2mc 2 E V0 / c
R
1.388 0.694
1.388 0.694
40MeV / 197.3MeV fm 1.388
2 938MeV
2
2 938MeV 10MeV / 197.3MeV fm
0.694
2.082
2
0.111 And T
147
1 R
0.889
0.694
Chapter 6 – The Schrödinger Equation
(Problem 6-51 continued)
(b) For electrons:
1/ 2
0.511
1.388
938
k1
0.0324
0.0324 0.0162
0.0324 0.0162
R
k2
0.511
0.694
938
1/ 2
0.0162
2
0.111 And T
1 R
0.889
No, the mass of the particle is not a factor. (We might have noticed that
m could
be canceled from each term.
6-52. (a) En
E1
(b)
n2 h2
8mL2
hc
The ground state is n = 1, so
2
1240MeV fm
8 mc 2 L2
2000
2
8 938.3MeV 1 fm
2
204.8MeV
(c)
1844 MeV
E21
21
n =3
21
1500
En
(MeV)
(d)
1000
hc /
32
1240MeV fm
819 205 MeV
1240MeV fm
1844 819 MeV
2.02 fm
1.21 fm
819 MeV
(e)
500
205 MeV
0
148
31
1240 MeV fm
1844 205 MeV
0.73 fm
Chapter 6 – The Schrödinger Equation
2
x
2 / L sin2 x / L.
6-53. (a) The probability density for the ground state is P x
The probability of finding the particle in the range 0 < x < L/2 is:
L/2
P
2L
L
P x dx
0
L/3
(b) P
2L
L
P x dx
0
/2
sin2 udu
2
sin2 udu
2
4
0
/3
sin 2 / 3
4
6
0
1
where u
2
0
1
3
x/L
3
4
0.195
(Note: 1/3 is the classical result.)
3L / 4
(c) P
2L
L
P x dx
0
3 /4
2 3
8
sin2 udu
0
sin 3 / 2
4
3
4
1
2
0.909
(Note: 3/4 is the classical result.)
6-54.
(a) En
So,
2
n2 h2
8mL2
En
1
and En
En
n2
En
2
n
En
En
1
En
n 1 h2
1
8mL2
2n 1 n 2
n2
2n 1
n2
2 1/ n
For large n,1 / n
n
(b) For n = 1000 the fractional energy difference is
2
1000
0.002
2 and
0.2%
(c) It means that the energy difference between adjacent levels per unit energy for large
n is getting smaller, as the correspondence principle requires.
6-55. The n = 2 wave function is
2
Ek
op
2
2m x 2
Therefore,
2
Ek
2
x
2
2 x
sin
and the kinetic energy operator
L
L
x
2
2
2
2 x
sin
L
L
x dx
2
2m x 2
2
2
2m x
2
2
2 x
sin
dx
L
L
149
Chapter 6 – The Schrödinger Equation
(Problem 6-55 continued)
L
2
2
L
2m
0
2
2
L 2m
Let
2 x
L
2
L
y, then x
2 L
2 x
dx
L
y
0 and x
0
0
sin ydy
0
6-56. (a) The requirement is that
x
x
or
2
2
0
2
2
L 2m
Therefore, Ek
2
L
2
2 x
dx
L
L
2
2
L 2m
2
sin 2 y
4
y
2
2
sin
sin2
Substituting above gives: Ek
2
2
2 x 2
sin
L
L
2
2
2
2
L
0
dy
dx
L
dy
2
2
L
2
sin2 ydy
0
0 0
x
h2
2mL2
x
x . This can only be true if:
x .
x
(b) Writing the Schrödinger equation in the form
of this 2nd order differential equation are:
where k
2 dx
L
2 and
4 2 2
2mL2
L
2
x
y
d2
dx 2
2mE
2
, the general solutions
A sin kx and
x
x
2mE . Because the boundaries of the box are at x
A cos kx
L / 2, both
solutions are allowed (unlike the treatment in the text where one boundary was at
L / 2 provided that an integral number
x = 0). Still, the solutions are all zero at x
L / 2 and x
of half wavelengths fit between x
n
x
2/ L
n
x
2/ L
1/ 2
1/ 2
L / 2. This will occur for:
cos n x / L when n 1, 3, 5,
. And for
sin n x / L when n
.
2, 4, 6,
The solutions are alternately even and odd.
(c) The allowed energies are: E
2
k 2 / 2m
150
2
n /L
2
/ 2m
n 2 h 2 / 8mL2 .
Chapter 6 – The Schrödinger Equation
6-57.
0
(a)
x2 / 2 L2
Ae
d 0
dx
d 1
dx
So,
x 2 / 2 L2
x / L2 Ae
And
1/ L
d2 1
dx 2
0
1/ L d
2 x / L3
and
x/L d
0
L
1
/ dx
0
1/ L d
0
2
/ 2m 3m / L3
x3 / L5
0
2
will make
1
1
0
E
0
0
x 2 / 2mL4 , the Schrödinger equation
2 3
x / 2mL5
0
x/L
0
E
0
L
0
x
2
2 2 2n x
x sin
dx
L
L
2 L
L n
2 L
L n
2
3
x/L
0
or,
. Thus, choosing E appropriately
3
2
/ 2mL2 , or three times the ground state energy.
plotted looks as below. The single node indicates that
x2
0
/ dx 2
1
is the first excited state.
(The energy value in [b] would also tell us that.)
6-58.
x/L
a solution.
(b) We see from (a) that E
(c)
x3 / L5
3 2 x / 2mL3
simplifying:
x/ L d2
/ dx
Recalling from Problem 6-3 that V x
becomes
x 2 / 2 L2
x / L2 Ae
L
/ dx
x / L3
0
d 0
dx
L
n
u3
6
Letting u
n x / L, du
n
u 2 sin2 udu
0
u2
4
n
1
sin 2u
8
u cos 2u
4
151
0
n / L dx
Chapter 6 – The Schrödinger Equation
(Problem 6-58 continued)
3
2 L
L n
6-59.
T
16
2m V0
n
4
0
6
E
E
1
e
V0
V0
(a)
3
n
2 a
2 m0c 2 V0
E
T
16
10
25
16
10
25
x, t
6-60. (a) For
d2
dx 2
1
k2
1
1
29.68
10
e
25
2.968
A sin kx
and
k2
A sin kx
2m
4.95 10
1
1nm.
c
197.3eV nm 19.84nm
19.84nm
a
2
25eV , and
E
19.84;
1nm
10
e
25
0.1nm :
(b) For a
T
19.84nm
a
L2
2n 2
where E 10eV , V0
2 0.511 106 eV 15eV
And
L2
3
0
2 a
1
29.68
13
0.1nm
1.984
0.197
t
A cos kx
t
t so the Schrödinger equation becomes:
2
V x A sin kx
t
t
i
Because the sin and cos are not proportional, this
for
x, t
A cos kx
(b) For
x, t
A cos kx
d2
dx 2
k2
2m
k2
and
t
cos kx
cannot be a solution. Similarly,
t , there are no solutions.
i kx
t
t
i sin kx
i
. And the Schrödinger equation becomes:
t
Ae
2
V x
t
2
for
152
k 2 / 2m V .
, we have that
Chapter 6 – The Schrödinger Equation
6-61.
V
mgz
E
Classically allowed:
0 < z < z0
Ek = E - mgz
0
z0
z
The wave function will be concaved toward the z axis in the classically allowed region
and away from the z axis elsewhere. Each wave function vanishes at z = 0 and as z → ∞.
The smaller amplitudes are in the regions where the kinetic energy is larger.
6-62. Writing the Schrödinger equation as: Ek
have: Ek
x
E V x
x
x
2
V x
x
E
x from which we
/ 2m d 2 / dx 2 . The expectation value of
153
Chapter 6 – The Schrödinger Equation
(Problem 6-62 continued)
Ek is Ek
Ek
x dx. Substituting Ek
x
x from above and reordering
2
multiplied quantities gives: Ek
6-63. (a)
p x
x dx .
m v x
/m a
v
d2
2m dx 2
x
34
1.055 10
v 1.6 108 m / s
9.11 10
J s
31
10
12
m
0.39c
(b) The width of the well L is still an integer number of half wavelengths, L
and deBroglie’s relation still gives: L
nh / 2 p . However, p is not given by:
E2
2mEk , but by the relativistic expression: p
p
nhc
Substituting this yields: L
2 E2
En
nhc
2L
hc
4 L2
(c) E1
/2 ,
n
2
mc 2
E2
1/ 2
mc 2
mc 2
2
2
1/ 2
c.
nhc / 2 L
2
1/ 2
2
mc 2
2
1/ 2
2
mc 2
2
1/ 2
1240eV nm
3
4 10 nm
2
2
0.511 106 eV
2
8.03 105 eV
(d) Nonrelativistic:
E1
h2
8mL2
hc
2
2
1240eV nm
2
8 mc L
6
2
3
8 0.511 10 eV 10 nm
2
3.76 105 eV
E1 computed in (c) is 2.14 times the nonrelativistic value.
6-64. (a) Applying the boundary conditions of continuity to
and d / dx at x = 0 and x = a,
where the various wave functions are given by Equation 6-74, results in the two pairs
of equation below:
154
Chapter 6 – The Schrödinger Equation
(Problem 6-64 continued)
At x
0:
At x
a : Feika
A B C D and ikA ikB
a
Ce
De
a
C
and ikFeika
D
Ce
a
De
a
Eliminating the coefficients C and D from these four equations, a straightforward
2
but lengthy task, yields: 4ik A
2
a
ik e
ik e
a
Feika
The transmission coefficient T is then:
2
F
T
A
2
4ik
2
2
eika
1
e
2
Recalling that sinh
2
a
ik e
ik e
a
and noting that
e
complex conjugates, substituting k
2mE
ik and
and
2m V0
ik are
E
, T then
1
can be written as T
sinh2 a
1
4
a
(b) If
E
E
1
V0
V0
1, then the first term in the bracket on the right side of the * equation in part
(a) is much smaller than the second and we can write:
F
A
ik
Or T
6-65.
2
II
ik a
4ik e
16
2
C e
E
V0
2 a
F
A
and T
2
1
E
e
V0
2
6
2
k 2e
2
2 a
k2
2
2 a
(Equation 6-72)
2
Where C
2 E1 / 2
2
2m V0
E1 / 2
E
E V0
1/ 2
A
2 0.5V0
2
0.5V0
2 mp c 2 20MeV
155
1/ 2
2
1/ 2
0.5V0
1/ 2
2.000
Chapter 6 – The Schrödinger Equation
(Problem 6-65 continued)
2 938.3MeV
x ( fm)
20MeV
e
197.3MeV fm
0.982 fm
2
2 x
II
2
C e
1
0.1403
0.5612
2
0.0197
0.0788
3
2.76×10-3
1.10×10-2
4
3.87×10-4
1.55×10-3
5
5.4×10-5
2.2×10-4
156
1
2 x
Chapter 7 – Atomic Physics
7-1.
2
n
2mL
2
En1 n2 n3 
E311 
2
1
2
2
3
2mL
 n22  n32

(Equation 7-4)

2
 12  12  11E0 where E0 
2
2

2
2
2mL2



E222  E0 22  22  22  12E0 and E321  E0 32  22  12  14E0
The 1st, 2nd, 3rd, and 5th excited states are degenerate.
E
(×E0)
14
321
12
222
311
10
221
8
6
211
4
111
2
0
 2  n12
2 2
n22 n32 
  2 n22 n32 

 =
 
 n1   
2m  L12 L22 L23  2mL12 
4
9 
2
7-2.
En1 n2 n3
n1  n2  n3  1 is the lowest energy level.
E111  E0 1  1 / 4  1 / 9   1.361E0 where E0 
The next nine levels are, increasing order,
157
2
2
2mL12
(Equation 7-5)
Chapter 7 – Atomic Physics
(Problem 7-2 continued)
7-3.
n1
n2
n3
E  E0 
1
1
2
1.694
1
2
1
2.111
1
1
3
2.250
1
2
2
2.444
1
2
3
3.000
1
1
4
3.028
1
3
1
3.360
1
3
2
3.472
1
2
4
3.778
(a)  n1 n2 n3  x, y, z   A cos
nz
n1 x
ny
sin 2 sin 3
L
L
L
(b) They are identical. The location of the coordinate origin does not affect the energy
level structure.
7-4.
7-5.
 111  x, y, z   A sin
x
 121  x, y, z   A sin
x
 113  x, y, z   A sin
x
En1 n2 n3 
L1
L1
L1
sin
sin
sin
y
2 L1
y
L1
y
2 L1
z
sin
sin
3L1
z
3L1
sin
 112  x, y, z   A sin
x
 122  x, y, z   A sin
x
L1
L1
sin
sin
y
2 L1
y
L1
sin
2 z
3L1
sin
2 z
3L1
z
L1
 2  n12
2 2
n32 
n22
  2 n22 n32 
 2


=
 n1   
2m  L1  2 L1 2  4 L1 2  2mL12 
4 16 


2
2 2

n2 n2 

En1 n2 n3   n12  2  3  where E0 
4 16 
2mL12

158
(from Equation 7-5)
Chapter 7 – Atomic Physics
(Problem 7-5 continued)
(a)
n1
n2
n3
E  E0 
1
1
1
1.313
1
1
2
1.500
1
1
3
1.813
1
2
1
2.063
1
1
4
2.250
1
2
2
2.250
1
2
3
2.563
1
1
5
2.813
1
2
4
3.000
1
3
1
3.313
(b) 1,1,4 and 1,2,2
7-6.
7-7.
 111  x, y, z   A sin
x
 113  x, y, z   A sin
x
 114  x, y, z   A sin
x
 123  x, y, z   A sin
x
 124  x, y, z   A sin
x
E0 
2
2
2mL2
L1
L1
L1
L1
L1
sin
sin
sin
sin
sin
y
2 L1
y
2 L1
y
2 L1
y
L1
y
L1
z
 112  x, y, z   A sin
x
3 z
4 L1
 121  x, y, z   A sin
x
z
 122  x, y, z   A sin
x
3 z
4 L1
 115  x, y, z   A sin
x
z
 116  x, y, z   A sin
x
sin
4 L1
sin
sin
sin
sin
L1
L1
1.055  10 J s  
kg  0.10  10 m  1.609  10
2
34


2 9.11  1031
9
E311  E111  E  11E0  3E0  8E0  301eV
159
2
L1
L1
L1
L1
L1
sin
sin
sin
sin
sin
y
2 L1
y
L1
y
L1
y
2 L1
y
2 L1
sin
sin
J / eV

2 L1
z
4 L1
sin
z
2 L1
sin
5 z
4 L1
sin
3 z
2 L1
2
19
z
 37.68eV
Chapter 7 – Atomic Physics
(Problem 7-7 continued)
E222  E111  E  12E0  3E0  9E0  339eV
E321  E111  E  14E0  3E0  11E0  415eV
7-8.
(a) Adapting Equation 7-3 to two dimensions (i.e., setting k3 = 0), we have
 n n  A sin
1 2
n1 x
ny
sin 2
L
L
(b) From Equation 7-5, En1 n2 
2
2
2
2mL
n
2
1
 n22

(c) The lowest energy degenerate states have quantum numbers n1 = 1, n2 = 2, and n1 = 2,
n2 = 1.
7-9.
 0, 1, 2
(a) For n = 3,
(b) For
 0, m  0 . For
 1, m  1, 0,  1 . For
 2, m  2,  1, 0,  1,  2 .
(c) There are nine different m-states, each with two spin states, for a total of 18 states for
n = 3.
7-10. (a) For
4

L
 1  4  5  20
m 4
min  cos1
(b) For
4
20
 min  26.6
2
L 6
m 2
min  cos1

2
6
 min  35.3



7-11. (a) L  I  105 kg m2  2  735min1 1min / 60s   7.7  104 kg m2 / s
(b) L 

 1  7.7  104 kg m2 / s
160
Chapter 7 – Atomic Physics
(Problem 7-11 continued)
 7.7  10 kg m / s 
  1 
1.055  10 J s 
4
2
34
2
2
7.7  104 kg m2 / s

 7.3  1030
34
1.055  10 J s
7-12. (a)
+1
1
L  2
0
−1
(b)
+2
+1
2
L  6
0
−1
−2
161
Chapter 7 – Atomic Physics
(Problem 7-12 continued)
(c)
+4
+3
+2
4
+1
L  20
0
−1
−2
−3
−4
(d) L 
7-13.

 1
(See diagrams above.)
L2  L2x  L2y  L2z  L2x  L2y  L2  L2z 
(a)
L
2
x
 L2y

(b)
L
 L2y

2
x
min
max




 6  22
 6  02
(c) L2x  L2y   6  1
2
5
2
2
2
2
6

 1
2
 m

L
(b) For  1,
m  1, 0,  1

 6  m2

2
2
Lx and Ly cannot be determined separately.
 1  2  1.49  1034 J s
(a) For  1,
2
2
(d) n = 3
7-14

162
Chapter 7 – Atomic Physics
(Problem 7-14 continued)
(c)
Z
+1ћ
L  2
0
−1ћ
(d) For  3,
L

 1  12  3.65  1034 J s and m  3,  2,  1, 0,1, 2, 3
Z
3ћ
2ћ
1ћ
0
−1ћ
−2ћ
−3ћ
7-15.
L= r× p
dL dr
dp

 p+ r
dt dt
dt
dr
dp
 p  v  mv = mv  v  0 and r 
 r  F . Since for V =V(r), i.e., central forces,
dt
dt
F is parallel to r, then r  F = 0 and
7-16. (a) For
(b) For
dL
0
dt
 3, n = 4, 5, 6, … and m = −3, −2, −1, 0, 1, 2, 3
 4, n = 5, 6, 7, … and m = −4, −3, −2, −1, 0, 1, 2, ,3 ,4
163
Chapter 7 – Atomic Physics
(Problem 7-16 continued)
(c) For
 0, n = 1 and m = 0
(d) The energy depends only on n. The minimum in each case is:
E4  13.6eV / n2  13.6eV / 42  0.85eV
E5  13.6eV / 52  0.54eV
E1  13.6eV
7-17. (a) 6 f state: n  6,  3
(b) E6  13.6eV / n2  13.6eV / 62  0.38eV
(c) L 
(d) Lz  m

 1  3  3  1  12  3.65  1034 J s
Lz  3 ,  2 ,  1 , 0, 1 , 2 , 3
7-18. Referring to Table 7-2, R30 = 0 when

2r
2r 2 

0
1 
2 
 3a0 27a0 
Letting r a0  x, this condition becomes x2  9 x  13.5  0
Solving for x (quadratic formula or completing the square), x = 1.90, 7.10.
Therefore, r a0  1.90, 7.10 . Compare with Figure 7-10(a).
2
 kZe2  
7-19. Equation 7-25: En   

2

 2n
Using SI units and noting that both Z and n are unitless, we have:
2
 (N  m2 / C2 )  C2 
J
  kg
J

s


2
 N  m2 
Cancelling the C and substituting J  N  m on the right yields J  
  kg .
N

m

s


2
164
Chapter 7 – Atomic Physics
(Problem 7-19 continued)
2
m
Cancelling the N and m gives J     kg  J , since kg  m2 / s2 are the units of kinetic
s
energy.
7-20. (a) For the ground state n = 1,
 100  R10Y00 
(b)  2 
2
a03
e r / a0
 0, and m = 0.
1
4

2
4 a02
e r / a0 
4 a03
at r  a0
1 2 r / a0
1
e
 3 e2 at r  a0
3
 a0
 a0
(c) P  r    2  4 r 2 
4 2
e at r  a0
a0


7-21. (a) For the ground state, P  r  r   2 4 r 2 r 
4r 2 2 r / a0
e
r
a03
For r  0.03a0 , at r  a0 we have P  r  r 
(b) For r  0.03a0 , at r  2a0 we have P  r  r 
7-22.
2e1
4a02 2
e  0.03a0   0.0162
a03
4  2a0 
3
0
a
2
e 4  0.03a0   0.0088
P  r   Cr 2e2 Zr / a0 For P(r) to be a maximum,
  2Z  2 Zr / a0

dP
2Zr  a0

 C r 2  
 2re2 Zr / a0   0  C 
 r  e2 Zr / a0  0
e

dt
a0  Z

  a0 

This condition is satisfied with r = 0 or r = a0 /Z. For r = 0, P(r) = 0 so the maximum
P(r) occurs for r = a0 /Z.
165
Chapter 7 – Atomic Physics
  2
7-23.
2
2 2
 d      r sin drd d  1
0 0 0

 4  r dr  4 C
2 2
2
210
0
2

 Zr  2  Zr / a
0  a0  r e 0 dr  1

 Z 2 r 4   Zr / a0
2
 4 C210
0  a02  e dr  1
Letting x  Zr / a0 , we have that r  a0 x / Z and dr  a0dx / Z and substituting
these above,
2 
4 a03C210
4 x
 d  Z 3 0 x e dx
2
Integrating on the right side

x e
4 x
dx  6
0
1/ 2
Solving for C
7-24.  200
Z

 
32  a0 
1
P  r  r   200
2

3/ 2
2
210
yields:
2
210
C

r   r / 2 a0
1   e
 a0 
 Z3 
Z3


C


210
3 
24 a03
 24 a0 
(Z = 1 for hyrdogen)
2
1 1
r 
4 r r 
1   e r / a0 4 r 2 r
3 
32 a0  a0 
2



(a) For r  0.02a0 , at r  a0 we have
P  r  r 
4 1
1
2
1  1 e1a02  0.02a0    0  e1  0.02   0
3 
32 a0
8
(b) For r  0.02a0 , at r  2a0 we have
P  r  r 
4 1
1
2
1 e2 a02  0.02a0   1 e2  0.02   3.4  104
3 
32 a0
8
166
Chapter 7 – Atomic Physics
7-25.  210  C210
Zr  Zr / 2 a0
e
cos
a0
P  r    210
2
 4 r   4 r
 4 C210
2
2
Z
2
(Equation 7-34)
2
C210
2
Z 2 r 2  r / a0
e
cos2 
a02

/ a02 r 4e r / a0 cos2 
 Ar 4e r / a0 cos2 
where A  4 C210
7-26.  200
1

 
32  a0 
1
2
Z
3/ 2
(a) At r  a0 ,  200
2

/ a02 , a constant.

r   r / 2 a0
2 
e
2a0 

(Z = 1 for hyrdogen)
 1
0.606  1 
1 / 2


 3   2  1 e
 
32  a0 
32  a0 
1
 1  1 0.368 1
 3 e 
32 a03
32  a0 
1
(b) At r  a0 ,
 200 
(c) At r  a0 ,
P  r    200
2
3/ 2
2
 4 r 
2
4 0.368a02 0.368


32
a03
8a0
7-27. For the most likely value of r, P(r) is a maximum, which requires that (see Problem 7-25)
  Z

dP
 A cos2   r 4    e Zr / a0  4r 3e Zr / a0   0
dr
  a0 



For hydrogen Z = 1 and A cos2  r 3 / a0  4a0  r  e r / a0  0 . This is satisfied for r = 0
and r = 4a0. For r = 0, P(r) = 0 so the maximum P(r) occurs for r = 4a0.
7-28. From Table 7-1, Y21  ,  
From Table 7-2, R32  r  
15
sin cos ei
8
r 2  r 3a0
e
2
81 30a02 a0
4
167
Chapter 7 – Atomic Physics
(Problem 7-28 continued)
 321  r, ,  
4
1
2
81 30a03 a0
15 2  r 3a0
r e
sin cos ei
8
To be sure that  is normalized, we do the usual normalization as follows, where C is the
normalization constant to be compared with the coefficient above.
  2
  2
0 0 0
0 0 0
2

2
4 2 r 3a0
C32
sin2  cos2  r 2 sin dr d d  1
1       dr  C32 1    r e
2
 d  2 , we have
Noting that
0
2 C
2
32 1

r e
6 2 r 3 a0
0

dr  sin3  cos2  d  1
0
Evaluating the integrals (with the aid of a table of integrals) yields:
4
2
4

2 C32
a07     1
1  1.23  10
15 


C32 1  0.00697
a07
This value agrees with the coefficient of  32 1 above.
7-29.  100 
2
e r / a0
4 a
3
0
Because  100 is only a function of r, the angle derivatives in Equation 7-9 are all zero.
d

dr
r2
d

dr
2
e r / a0
4 a
3
0
 1  2  r / a0
 r e
4 a03  a0 
2
d  2 d 
r

dr  dr 
1 d  2 d
r
r 2 dr  dr

 1  2  1 
   r     2r  e r / a0


4 a03  a0    a0 



2
 1   2 1   r / a0
Substituting into Equation 7-9,
   e
4 a03  a0   r a0 
2
168
Chapter 7 – Atomic Physics
(Problem 7-29 continued)

 1
2 
 2
 100  V 100  E 100
2  a0 a0 r 
2
For the 100 state r  a0 and 2 a0    2 / k or a0  1 / k, so
 1
2   1
2 
1
2
 2
   2  2    2  k
a0
 a0 a0 r   a0 a0 
Thus, 
2
k2
2
2 2
 1
2 
k
and we have that
 2

2  a0 a0 r  2
2
 V  E, satisfying the Schrödinger equation.
7-30. (a) Every increment of charge follows a circular path of radius R and encloses an area
 R 2 , so the magnetic moment is the total current times this area. The entire charge
Q rotates with frequency f   / 2 , so the current is
i  Qf  q / 2
  iA   Q / 2   R2   Q R2 / 2
L  I 
g
1
MR 2
2
2M  2MQ R 2 / 2

2
QL
QMR 2 / 2
(b) The entire charge is on the equatorial ring, which rotates with frequency f   / 2 .
i  Qf  Q / 2
  iA   Q / 2   R2   Q R2 / 2
g
2M  2MQ R 2 / 2

 5 / 2  2.5
QL
QMR 2 / 5
169
Chapter 7 – Atomic Physics
7-31. Angular momentum S  I   2 / 5 mr 2  v / r  or
v   5 / 2  S 1 / mr   5S / 2mr  5  3 / 4 
1/ 2
1.055  10
2  9.11  10 kg 10
5 3 / 4
1/ 2

34
31
J s
15
m

/ 2mr
  2.51  10
11
m / s  837c
 0 , so two lines due to spin of the single s electron would
7-32. (a) The K ground state is
be seen.
 0 with two s electrons whose spins are opposite resulting
(b) The Ca ground state is
in S = 0, so there will be one line.
(c) The electron spins in the O ground state are coupled to zero, the orbital angular
momentum is 2, so five lines would be expected.
(d) The total angular momentum of the Sn ground state is j = 0, so there will be one line.
7-33.
Fz  ms g L B  dB / dz   mAg az
(From Equation 7-51)
and az  mS g L B  dB / dz  / mAg
Each atom passes through the magnet’s 1m length in t = (1/250)s and cover the additional
1m to the collector in the same time. Within the magnet they deflect in the z direction an
amount z1 given by: z1  1 / 2  az t 2  1 / 2  ms g L B  dB / dz  / mAg  1 / 250  and leave
2
the magnet with a z-component of velocity given by vz  az t . The additional z deflection
in the field-free region is z2  vz t  az t 2 .
The total deflection is then z1  z2  0.5mm  5.0  104 m.
5.0  104 m  z1  z2   3 / 2  az t 2   3 / 2  ms g L B  dB / dz  / mAg  1 / 250 or,
2



5.0  104 m  250  mAg
dB

dz
ms g L  B
2
  2
5.0 10 m  250s  1.79 10 kg   2  0.805T / m

3 1 / 2 1  9.27  10 J / T 
4
1
2
25
24
170
Chapter 7 – Atomic Physics
7-34. (a) There should be four lines corresponding to the four mJ values −3/2, −1/2, +1/2, +3/2.
(b) There should be three lines corresponding to the three m values −1, 0, +1.
7-35. (a) For the hydrogen atom the n = 4 levels in order of increasing energy are:
4 2S1 / 2 , 4 2 P1 / 2 , 4 2 P3 / 2 , 4 2 D3 / 2 , 4 2 D5 / 2 , 4 2 F5 / 2 , 4 2 F7 / 2
(b) 2j + 1
7-36. For
 2,

L
 1  6  2.45 ,
j   1 / 2  3 / 2, 5 / 2 and J 
For j  3 / 2,
J
3 / 23 / 2  1
 15 / 4  1.94
For j  5 / 2,
J
5 / 25 / 2  1
 35 / 4  2.96
j  j  1
7-37. (a) j   1 / 2  2  1 / 2  5 / 2 or 3 / 2
(b) J 
5
 5 / 2  1  2.96
2
j  j  1 
or
3
3 / 2  1  1.94
2
(c) J  L  S J Z  LZ  SZ  m  ms  m j where m j   j,  j  1,..., j  1, j. For
j = 5/2 the z-components are 5 / 2,  3 / 2,  1/ 2,  1/ 2,  3 / 2,  5 / 2. For j = 3/2
the z-components are 3 / 2,  1 / 2,  1 / 2,  3 / 2.
7-38. (a) d5 / 2
L
  cos1
s  3/ 4
ms =1/2
θ
(b) d3 / 2
L
θ
  cos1
ms
=−1/2
s  3/ 4
171
1/ 2
3/ 4
 54.7
 1 / 2   125.3
3/ 4
Chapter 7 – Atomic Physics
7-39.
m
n
−3, −2, −1, 0, 1, 2, 3
−2, −1, 0, 1, 2
−1, 0, 1
0
−2, −1, 0, 1, 2
−1, 0, 1
0
−1, 0, 1
0
3
2
1
0
2
1
0
1
0
4
3
2
ms
1 / 2 for each m state
1 / 2 for each m state
1 / 2 for each m state
7-40. (a) L  L1  L2

1

2
, 
1

 1,...,
2
1

2
 1  1, 1  1  1, 1  1  2, 1, 0
(b) S  S1  S2
s   s1  s2  ,  s1  s2  1,..., s1  s2  1 / 2  1 / 2 , 1 / 2  1 / 2   1, 0
(c) J  L  S
j    s ,
For

 s  1 ,...,  s
 2 and s  1, j  3, 2, 1
 2 and s  0, j  2
For
 1 and s  1, j  2, 1, 0
 1 and s  0, j  1
For
 0 and s  1, j  1
 0 and s  0, j  0
(d) J1  L1  S1
J 2  L2  S2
(e) J  J1  J 2
j1 
j2 
1
 1 / 2  3 / 2, 1 / 2
2
 1 / 2  3 / 2, 1 / 2
j   j1  j2  ,
 j1  j2  1,...,
For j1  3 / 2 and j2  3 / 2, j  3, 2, 1, 0
j1  3 / 2 and j2  1 / 2, j  2, 1
For j1  1 / 2 and j2  3 / 2, j  2, 1
j1  1 / 2 and j2  1 / 2, j  1, 0
These are the same values as found in (c).
172
j1  j2
Chapter 7 – Atomic Physics
7-41. (a)
E  hf  f  E / h  (4.372 106 eV)(1.602 1019 J/eV) / 6.63 1034 J  s=1.056 109 Hz
(b) c  f     c / f  (3.00 108 m/s) / (1.056 109 Hz)  0.284m  28.4cm
(c) short wave radio region of the EM spectrum
7-42. (a) E3 / 2 
hc

E3 / 2 
Using values from Figure 7-22,
1239.852eV nm
 2.10505eV
588.99nm
E1 / 2 
1239.852eV nm
 2.10291eV
589.59nm
(b) E  E3 / 2  E1 / 2  2.10505eV  2.10291eV  2.14  103 eV
(c) E  2 B B  B 
7-43.  12    x1, x2   C sin

E
2.14  103 eV

 18.5T
2 B 2 5.79  104 eV / T

 x1
L
sin
2 x2
L
Substituting into Equation 7-57 with V = 0,
  2 12  2 12   2 
2 


1

4



 

 2  12  E 12
2m  x12
x22   2m 
L 
2
Obviously,  12 is a solution if E 
7-44.

En 
n 2 2 2
2mL2
5 2 2
2mL2
Neutrons have antisymmetric wave functions, but if spin is ignored then
one is in the state n = 1 state, but the second is in the n = 2 state, so the minimum energy


is: E  E1  E2  12  22 E1  5E1 where
E1
 hc 

2
2
2mc 2 L2
197.3  2

2
2  939.6  2.0 
2
 51.1MeV
173
E  5E1  255MeV
Chapter 7 – Atomic Physics
7-45. (a) For electrons: Including spin, two are in the n = 1 state, two are in the n = 2 state, and
one is in the n = 3 state. The total energy is then:
E  2 E1  2 E2  E3
where E1
 hc 

2
2
2me c 2 L2
where En 


n2 2 2
2mL2
197.3
2

2

2 0.511  106 1.0 
 

E  2 E1  2 22 E1  32 E1  19 E1
2
 0.376eV
E  19 E1  7.14eV
(b) Pions are bosons and all five can be in the n = 1 state, so the total energy is:
E  5E1 where E1 
0.376eV
 0.00142eV
264
E  5E1  0.00712eV
7-46. (a) Carbon: Z  6; 1s 2 2s 2 2 p 2
(b) Oxygen: Z  8; 1s 2 2s 2 2 p 4
(c) Argon: Z  18; 1s 2 2s 2 2 p6 3s 2 3 p6
7-47. Using Figure 7-34:
Sn (Z = 50)
1s 2 2s 2 2 p6 3s 2 3 p6 3d 10 4s 2 4 p6 4d 10 5s 2 5 p 2
Nd (Z = 60)
1s 2 2s 2 2 p6 3s 2 3 p6 3d 10 4s 2 4 p6 4d 10 5s 2 5 p6 4 f 4 6s 2
Yb (Z = 70)
1s 2 2s 2 2 p6 3s 2 3 p6 3d 10 4s 2 4 p6 4d 10 4 f 14 5s 2 5 p6 6s 2
Comparison with Appendix C.
Sn: agrees
Nd: 5 p6 and 4 f 4 are in reverse order
Yb: agrees
7-48. Both Ga and In have electron configurations  ns   np  outside of closed shells
2
 n  1, s   n  1, p   n  1, d 
2
6
10
. The last p electron is loosely bound and is more easily
removed than one of the s electrons of the immediately preceding elements Zn and Cd.
174
Chapter 7 – Atomic Physics
7-49. The outermost electron outside of the closed shell in Li, Na, K, Ag, and Cu has
 0. The
ground state of these atoms is therefore not split. In B, Al, and Ga the only electron not in
a closed shell or subshell has
 1, so the ground state of these atoms will be split by the
spin-orbit interaction.
7-50.
En  
Z eff2 E1
Z eff  n
n2
(Equation 7-25)
 En
5.14eV
3
 1.84
E1
13.6eV
7-51. (a) Fourteen electrons, so Z = 14. Element is silicon.
(b) Twenty electrons, so Z = 20. Element is calcium.
7-52. (a) For a d electron,
(b) For an f electron,
 2, so Lz  2 ,  1 , 0, 1 , 2
 3, so Lz  3 ,  2 ,  1 , 0, 1 , 2 , 3
7-53. Like Na, the following atoms have a single s electron as the outermost shell and their
energy level diagrams will be similar to sodium’s: Li, Rb, Ag, Cs, Fr.
The following have two s electrons as the outermost shell and will have energy level
diagrams similar to mercury: He, Ca, Ti, Cd, Mg, Ba, Ra.
7-54. Group with 2 outer shell electrons: beryllium, magnesium, calcium, nickel, and barium.
Group with 1 outer shell electron: lithium, sodium, potassium, chromium, and cesium.
7-55. Similar to H: Li, Rb, Ag, and Fr. Similar to He: Ca, Ti, Cd, Ba, Hg, and Ra.
175
Chapter 7 – Atomic Physics
7-56.
n
j
4
0
1/2
4
1
1/2
4
1
3/2
5
0
1/2
3
2
3/2
3
2
5/2
5
1
1/2
5
1
3/2
4
2
3/2
4
2
5/2
6
0
1/2
4
3
5/2
4
3
7/2
Energy is increasing downward in the table.
7-57. Selection rules:   1 j  1, 0
Transition
∆ℓ
∆j
Comment
4S1/2 → 3S1/2
0
0
ℓ - forbidden
4S1/2 → 3P3/2
+1
+1
allowed
4P3/2 → 3S1/2
−1
−1
allowed
4D5/2 → 3P1/2
−1
−2
j – forbidden
4D3/2 → 3P1/2
−1
−1
allowed
4D3/2 → 3S1/2
−2
−1
ℓ - forbidden
5D3/2 → 4S1/2
−2
−1
ℓ - forbidden
5P1/2 → 3S1/2
−1
0
allowed
7-58. (a) E1  13.6eV  Z  1  13.6eV  74  1  7.25  104 eV  72.5keV
2
2
(b) E1  exp   69.5keV  13.6eV  Z     13.6eV  74  1
2

74    69.5  103 eV / 13.6eV

1/ 2
 71.49
  74  71.49  2.51
176
2
Chapter 7 – Atomic Physics
7-59.
j  1, 0
 no j  0  j  0
(Equation 7-66)
The four states are 2 P3 / 2 , 2 P1 / 2 , 2 D5 / 2 , 2 D3 / 2
Transition
∆ℓ
∆j
Comment
D5/2 → P3/2
−1
−1
allowed
D5/2 → P1/2
−1
−2
j - forbidden
D3/2 → P3/2
−1
0
allowed
D3/2 → P1/2
−1
−1
allowed
7-60. (a) E  hc / 
E  3P1 / 2   E  3S1 / 2  
1240eV nm
 2.10eV
589.59nm
E  3P1 / 2   E  3S1 / 2   2.10eV  5.14eV  2.10eV  3.04eV
E  3D   E  3P1 / 2  
1240eV nm
 1.52eV
818.33nm
E  3D   E  3P1 / 2   1.52eV  3.04eV  1.52eV  1.52eV
(b) For 3P :
Z eff  3
3.04eV
 1.42
13.6eV
For 3D :
Z eff  3
1.52eV
 1.003
13.6eV
(c) The Bohr formula gives the energy of the 3D level quite well, but not the 3P level.
7-61. (a) E  gm j B B
(Equation 7-73) where s  1 / 2,  0 gives j  1 / 2 and
(from Equation 7-73) g  2.

m j  1 / 2.

E   2  1 / 2  5.79  105 eV / T  0.55T   3.18  105 eV
The total splitting between the m j  1 / 2 states is 6.37  105 eV .
(b) The m j  1 / 2 (spin up) state has the higher energy.
(c) E  hf  f  E / h  6.37  105 eV / 4.14  1015 eV s  1.54  1010 Hz
This is in the microwave region of the spectrum.
177
Chapter 7 – Atomic Physics
7-62.
E
hc

 E 
7-63. (a) E 
dE
hc
2
   2     E
d

hc


e
B  5.79  105 eV / T  0.05T   2.90  106 eV
2m
(b)  
2
hc
 579.07nm   2.90  106 eV 
2
E 
1240eV nm
 7.83  104 nm
(c) The smallest measurable wavelength change is larger than this by the ratio
0.01nm 7.83 104 nm  12.8. The magnetic field would need to be increased by this
same factor because B  E  . The necessary field would be 0.638T.
7-64.

 13.6eV  Z

2   5.39eV
En  13.6eV Zeff2 n2
E2
2
eff
2
Zeff  2  5.39 13.6 
1/ 2
7-65.  100
1 Z

 
  a0 
 1.26
3/ 2
e Zr / a0
*
P  r   4 r 2 100
 100
(Equations 7-30 and 7-31)
(Equation 7-32)
Z 3  Zr / a0 4Z 3 2 2 Zr / a0
 4 r
e
 3 r e
 a03
a0
2


4Z 3 3 2 Zr / a0
r   rP  r  dr   3 r e
dr
a0
0
0
3
a0   2Zr  2 Zr / a0
a
3a

d  2Zr / a0   0  3!  0

e

4Z 0  a0 
4Z
2Z
178
Chapter 7 – Atomic Physics
7-66. (a) E 

 1
2
: 0
 1: 1
E : 0
2I
30E1
E5 = 30E1
20E1
E4 = 20E1
1
2
2
3
1E1 6E1
3
4
4
5
12E1 20E1
5
…
6
…
30E1 …
E3 = 12E1
10E1
E2 = 6E1
2
E1 =
0
/I
E0 = 0
(b) E 1  E 
2
  1  2  
2I 


2
  1  2 
2I 
The values of
 1
 
2
I

 1    1 E1
 0, 1, 2, … yield all the positive integer multiples of E1.
2
2 c
1
2 2
(c) I  m p r 2  E1 


2
2
I
mp r
mpc2r 2
2

(d)  
2 197.3eV nm 
2
938.28  10 eV   0.074nm 
6
2
 1.52  102 eV
hc 1.24  106 eV nm

 8.18  105 m  81.8 m
E1
1.52  102 eV
7-67. (a) Fz  ms g L B  dB / dz  (From Equation 7-51)
From Newton’s 2nd law, Fz  mH az  ms g L B  dB / dz 



az  ms g L  dB / dz  / mH  1 / 2 1 9.27  1024 J / T  600T / m  / 1.67  1027 kg
 1.67  106 m / s 2
179

Chapter 7 – Atomic Physics
(Problem 7-67 continued)

(b) At 14.5km / s  v  1.45  104 m / s, the atom takes t1  0.75m / 1.45 104 m / s

 5.2 105 s to traverse the magnet. In that time, its z deflection will be:


z1  1 / 2  az  t12  1 / 2  1.67  106 m / s 2 5.2  105 s

2
 2.26  103 m  2.26mm
Its vz velocity component as it leaves the magnet is vz  az t1 and its additional z
deflection before reaching the detector 1.25m away will be:

z2  vz t2   az t1  1.25m 1.45  104 m / s 





 1.67  106 m / s 2 5.2  105 s 1.25 1.45 104 m / s

 7.49  103 m  7.49mm
Each line will be deflected z1  z2  9.75mm from the central position and, thus,
separated by a total of 19.5mm = 1.95cm.

7-68. min  cos1 m

cosmin 

or, sin min  1 
2
 1  with m  .

 1 . Thus, cos2 min 
2

 1


 1 

2
2

 1

2

 1  1  sin2 min
  2
  1
1/ 2
 1 
And, sinmin  

 1
For large , min is small.
1/ 2
 1 
Then sinmin  min  

 1

1
 
1/ 2
7-69. (a) E1  hf  hc / 1  1240eV nm / 766.41nm  1.6179eV
E2  hf  hc / 2  1240eV nm / 769.90nm  1.6106eV
(b) E  E1  E2  1.6179eV  1.6106eV  0.0073eV
(c) E / 2  gm j B B  B 
E
0.0073eV

 63T
2 gm j B 2  2 1 / 2  5.79  105 eV / T

180

Chapter 7 – Atomic Physics
7-70.
P r 
4Z 3 2 2 Zr / a0
re
a03
(See Problem 7-65)
For hydrogen, Z = 1 and at the edge of the proton r  R0  1015 m. At that point, the
exponential factor in P(r) has decreased to:
e2 R0 / a0  e

2 1015
 0.52910
10
m
  e3.7810   1  3.78  105  1
5
Thus, the probability of the electron in the hydrogen ground state being inside the nucleus,
to better than four figures, is:
r0
4r 2
P r   3
a0
P   P  r dr 
0
R0
4r 2
4
 3
3
a0
a0

0
R0
4 r3
r
dr

0
a03 3


j  j  1  s  s  1 
7-71. (a) g  1 
2 j  j  1
For 2 P1 / 2 :
g  1
j  1 / 2,

 1

g  1
 9.0  1015
 1, and s  1 / 2
2  1 / 2 1 / 2  1
j  1 / 2,
3
(Equation 7-73)
1 / 2 1 / 2  1  1 / 2 1 / 2  1  11  1
For 2S1 / 2 :
0
3
4 1015 m
4  R03 
 3 
a0  3  3 0.529  1010 m

R0
2
 1
3/ 4  3/ 4  2
 2/3
3/ 2
 0, and s  1 / 2
1 / 2 1 / 2  1  1 / 2 1 / 2  1  0
2  1 / 2 1 / 2  1
 1
3/ 4  3/ 4
2
3/ 2
The 2 P1/ 2 levels shift by:
E  gm j B B 
2 1
1
  B B   B B

3 2
3
The 2 S1/ 2 levels shift by:
 1
E  gm j B B  2     B B   B B
 2
181
(Equation 7-72)
Chapter 7 – Atomic Physics
(Problem 7-71 continued)
To find the transition energies, tabulate the several possible transitions and the
corresponding energy values (let Ep and Es be the B = 0 unsplit energies of the two
states.):
Transition
Energy
P1 / 2,1 / 2  S1 / 2,1 / 2
1
2


 E p  3 B B    Es  B B   E p  Es  3 B B


P1 / 2, 1 / 2  S1 / 2,1 / 2
1
4


 E p  3 B B    Es  B B   E p  Es  3 B B


P1 / 2,1 / 2  S1 / 2, 1 / 2
1
4


 E p  3 B B    Es  B B   E p  Es  3 B B


P1 / 2, 1 / 2  S1 / 2, 1 / 2
1
2


 E p  3 B B    Es  B B   E p  Es  3 B B










Thus, there are four different photon energies emitted. The energy or frequency
spectrum would appear as below (normal Zeeman spectrum shown for comparison).
anomolous
normal
(b) For 2 P3 / 2 :
g  1
j  3 / 2,
 1, and s  1 / 2
3 / 2  3 / 2  1  1 / 2 1 / 2  1  11  1
2  3 / 2  3 / 2  1
 1
15 / 4  3 / 4  2
 4/3
30 / 4
These levels shift by:
E  gm j B B 
4 1
2
  B B   B B

3 2
3
182
E 
4 3
 B B  2 B B
3  2 
Chapter 7 – Atomic Physics
(Problem 7-71 continued)
Tabulating the transitions as before:
Transition
Energy
P3 / 2, 3 / 2  S1 / 2,1 / 2
E
P3 / 2, 3 / 2  S1 / 2, 1 / 2
forbidden, m j  2
P3 / 2,1 / 2  S1 / 2,1 / 2
2
1


 E p  3 B B    Es  B B   E p  Es  3 B B


P3 / 2,1 / 2  S1 / 2, 1 / 2
2
5


 E p  3 B B    Es  B B   E p  Es  3 B B


P3 / 2, 1 / 2  S1 / 2,1 / 2
2
5


 E p  3 B B    Es  B B   E p  Es  3 B B


P3 / 2, 1 / 2  S1 / 2, 1 / 2
2
1


 E p  3 B B    Es  B B   E p  Es  3 B B


P3 / 2, 3 / 2  S1 / 2,1 / 2
forbidden, m j  2
P3 / 2, 3 / 2  S1 / 2, 1 / 2
E
p



 2B B   Es  B B   E p  Es  B B







p




 2B B   Es  B B   E p  Es  B B
There are six different photon energies emitted (two transitions are forbidden); their
spectrum looks as below:
anomolous
normal
7-72. (a) Substituting   r,  into Equation 7-9 and carrying out the indicated operations
yields (eventually)

2
2
  r,   2 / r 2  1 / 4a02  
2
2
  r,   2 / r 2   V  r,   E  r, 
183
Chapter 7 – Atomic Physics
(Problem 7-72 continued)
Canceling   r,  and recalling that r 2  4a02 (because  given is for n = 2) we
have 
 1 / 4a   v  E
2
2
2
0
The circumference of the n = 2 orbit is: C  2  4a0   2  a0   / 4  1 / 2k.
Thus, 
(b) or
2 2
1 
k



V

E

V  E

2 
2   4 / 4k 
2
2
p2
 v  E and Equation 7-9 is satisfied.
2m
2

 r  r / a
2
2
0  dx   A  a0  e 0 cos  r sin drd d  1
2
2
2

2
 r 
A    e r / a0 r 2 dr  cos2  sin d  d  1
a
0 0 
0
0

2
Integrating (see Problem 7-23),
 
A2 6a03  2 / 3 2   1
A2  1 / 8a03  A  1 / 8a03
7-73.
   g L B L
(Equation 7-43)
(a) The 1s state has
The 2p state has
 0, so it is unaffected by the external B.
 1, so it is split into three levels by the external B.
(b) The 2 p  1s spectral line will be split into three lines by the external B.
(c) In Equation 7-43 we replace B with k  e / 2mk , so
kz   11 e / 2mk   B  me / mk 
(From Equation 7-45)
Then E  B  me / mk  B




 5.79  105 eV / T  0.511 106 MeV / c 2 / 497.7MeV / c2  1.0T 
 5.94  108 eV
184
Chapter 7 – Atomic Physics
(Problem 7-73 continued)




hc
E (From Problem 7-62) where λ for the (unsplit) 2p → 1s transition
is given by
  hc Ek and Ek  E2  E1  13.6eV  mk / me 1  1 / 4  9.93  103 eV
and   1240eV nm 9.93 103 eV  0.125nm

and
7-74.

E    B 


0.125nm 5.94  108 eV
1240eV nm
12
ke2
S L where, for n  3, r  a0 n 2  9a0
r 3m mc 2

For 3P states S L 
E 
  5.98  10


2
  6.58 10
9  0.053nm   0.511  10 eV 
2
1.440eV nm 3.00  108 m / s  109 nm / m
3
For 3D states S L 
2
6
16
eV s

2
 1.60  104 eV
2
/3
E  1.60 104 eV / 3  0.53 104 eV
 = B  L  2 S 
7-75. (a) J = L + S
J 
(Equation 7-71)
  B  L  2 S  /    L + S 


  B  L L  2 S S  3S L 
J
J
J
J

B
J
L
2
 2S 2  3S L

(b) J 2  J J   L + S   L + S   L L + S S + 2 S L  S L 
(c)  J  
B 
(d) Z   J

1 2
J  L2  S 2
2

3

L2  2S 2  J 2  L2  S 2    B 3J 2  S 2  L2

J
2
2 J


JZ

J

  B 3J 2  S 2  L2 Z   B 3J 2  S 2  L2
J
2 J
J
2 J





185



Chapter 7 – Atomic Physics
(Problem 7-75 continued)

J 2  S 2  L2  J Z
   B 1 

2J 2


(e) E  Z B
(Equation 7-69)

j  j  1  s  s  1 
   B B 1 
2 j  j  1

 gm j B B

 1 
 mj

(Equation 7-72)

j  j  1  s  s  1 
where g  1 
2 j  j  1


 1 


(Equation 7-73)
7-76. The number of steps of size unity between two integers (or half-integers) a and b is b – a.
Including both values of a and b, the number of distinct values in this sequence is
b – a + 1. For F = I + J, the largest value of f is I + J = b. If I < J, the smallest value of
f is J – I = a. The number of different values of f is therefore (I + J) – (J – I) + 1 = 2I + 1.
For I > J, the smallest value of f is I – J = a. In that case, the number of different values
of f is (I + J) – (I – J) + 1 = 2J + 1. The two expressions are equal if I = J.
7-77. (a)  N 
B

e
 5.05  1027 J / T
2m p
2km  2km  2.8 N  2km  2.8 N 


r3
r3
a03



2 107 H / m  2.8  5.05  1027 J / T
 0.529  10

10
m

3
  0.0191T

(b) E  2B B  2 5.79  104 eV / T  0.0191T   2.21 106 eV
(c)  
hc 1.24  106 eV m

 0.561m  56.1cm
E
2.21  106 eV
186
Chapter 8 – Statistical Physics
1/ 2
8-1.
(a) vrms
3RT  3 8.31J / mole K  300 K  



M
 2 .0079  103 kg / mole 

 1930m / s



2
2 1.0079  103 kg / mole 11.2  103 m / s
Mvrms
(b) T 

3R
3 8.31J / mole K 
8-2.
8-3.
2
 1.01  104 K
2 13.6eV 
2 Ek

 1.05  105 K
5
3k
3 8.617  10 eV / K
(a) Ek 
3
kT
2
(b) Ek 
3
3
kT  8.67  105 eV / K 107 K  1.29keV
2
2
T 






3RT
M
vrms 
(a) For O2: vrms 
(b) For H2: vrms 
1/ 2
3 8.31J / K mol  273K 
32  103 kg / mol
3 8.31J / K mol  273K 
2  103 kg / mol
  J / mole K  K  


kg / mole


1/ 2
 461m / s
 1840m / s
1/ 2
 kg m2 / s 2 


kg


8-4.
 3RT 
 M 


8-5.
3
3
(a) EK  n  RT  1 mole  8.31J / mole K  273  3400 J
2
2
 m/ s
(b) One mole of any gas has the same translational energy at the same temperature.
187
Chapter 8 – Statistical Physics
8-6.
v2
v
2
1 2
v n v dv
N0
m
2 kT
4
I4
v
2
vrms
8-7.
3
8
1/ 2
m
2 kT
4
v2
m
2 kT
4
3/ 2
v2
v 4e
dv where
m / 2kT
0
3/ 2
I 4 where I 4 is given in Table B1-1.
3
8
5/ 2
3/ 2
3
8
1/ 2
1/ 2
m / 2kT
5/ 2
2kT
m
5/ 2
3kT
m
3RT
mN A
3RT
M
3RT
M
v
8kT
m
8 1.381 10
vm
2kT
m
2 1.381 10
n v
4 N m / 2 kT
23
27
1.009u 1.66 10
At the maximum:
23
dn
dv
v 2e
27
mv 2 / kT
1/ 2
2220m / s
kg / u
(Equation 8-28)
0
4 N m / 2 kT
0
ve
mv 2 / 2 kT
2510m / s
kg / u
J / K 300 K
1.009u 1.66 10
3/ 2
1/ 2
J / K 300 K
3/ 2
2v v 2
mv / kT e
mv 2 / 2 kT
2 mv 2 / kT
The maximum corresponds to the vanishing of the last factor. (The other two factors give
minima at v = 0 and v = ∞.) So 2 mv 2 / kT
188
0 and vm
2kT / m
1/ 2
.
Chapter 8 – Statistical Physics
8-8.
8-9.
n v dv
4 N
dn
dv
A v2
A
2mv3
2kT
m
2 kT
2vm
2kT
2v e
3/ 2
v 2e
2v e
mv2 / 2 kT
mv 2 / 2 kT
mv2 / 2 kT
dv
(Equation 8-8)
The v for which dn / dv
0 is vm .
0
Because A = constant and the exponential term is only zero for v → ∞, only the quantity
in [] can be zero, so
or v 2
2kT
m
vm
2mv3
2kT
2kT
m
2v
0
(Equation 8-9)
8-10. The number of molecules N in 1 liter at 1 atm, 20°C is:
N
1 1g mol / 22.4
N A molecules / g mol
Each molecule has, on the average, 3kT/2 kinetic energy, so the total translational kinetic
189
Chapter 8 – Statistical Physics
(Problem 8-10 continued)
23
6.02 1023 3 1.381 10 J / K 293K
22.4
2
energy in one liter is: KE
8-11.
n2
n1
g2e
g1e
e
T
E2 / kT
g2
e
g1
E1 / kT
g2
g1
E2 E1 / kT
E2
163J
E2 E1 / kT
n1
n2
E1 / kT
E2
ln
E1
g2
g1
n1
n2
10.2eV
8.617 10 5 eV / K ln 4 106
k ln g 2 / g1 n1 / n2
7790 K
4 10 3 eV
8-12.
n2
n1
g2
e
g1
E2 E1 / kT
3
e
1
8.617 10 5 eV / K 300 K
2.57
8-13. There are two degrees of freedom, therefore,
8-14.
Cv
2 R/2
cv
3R / M
(a) Al: cv
(b) Cu: cv
(c) Pb: cv
R, C p
R
R
3 1.99cal / mole K
27.0 g / mole
3 1.99cal / mole K
62.5 g / mole
3 1.99cal / mole K
207 g / mole
2 R, and
2R / R
0.221cal / g K
2.
0.215cal / g K
0.0955cal / g K
0.0920cal / g K
0.0288cal / g K
0.0305cal / g K
The values for each element shown in brackets are taken from the Handbook
of Chemistry and Physics and apply at 25° C.
190
Chapter 8 – Statistical Physics
8-15.
2 N
n( E )
kT
3/ 2
E1 / 2e
E / kT
dn
dE
0
At the maximum:
(Equation 8-13)
2 N
kT
1/ 2
E
1
2
E / kT
e
1
E
2
3/ 2
1/ 2
E1 / 2
1
kT
e
E / kT
E / kT
The maximum corresponds to the vanishing of the last factor. (The vanishing of the other
two factors corresponds to minima at E = 0 and E = ∞.)
1/ 2 E / kT
0
E 1/ 2kT.
3/ 2
m
4 N
2 kT
8-16. (a) n v
4 N
3/ 2
v 2e
v 2e
v 2 / vm2
n v
v
v 2 m / 2 kT
2kT / m
where vm
2
4N 1 v
vm vm
N
(Equation 8-8)
3/ 2
m
2kT
4 N v2
e
vm3
mv 2 / 2 kT
v / vm
e
2
4N A 1 v
vm vm
1.36 1022 v vm
2
(b)
N
1.36 1022 0 e
(c)
N
1.36 1022 1 e
(d)
N
1.36 1022 2 e
(e)
N
1.36 1022 8 e
2
0
2
e
2
v / vm
e
v / vm
2
0.01vm
2
0
1
5.00 1021
2
2
2
8
2
9.96 1020
2
1.369 10
191
4
(or no molecules most of the time)
Chapter 8 – Statistical Physics
mk 2e4 1
2 2 n2
8-17. For hydrogen: En
13.605687
eV using values of the constants accurate
n2
to six decimal places.
E1
13.605687eV
E2
3.401422eV
E2
E1
10.204265eV
E3
1.511743eV
E3
E1
12.093944eV
(a)
(b)
n2
n1
g2
e
g1
n3
n1
g3
e
g1
n2
n1
0.01 4e
E2 E1 / kT
E3 E1 / kT
(c)
n3
n1
10.20427 / 0.02586
18
e
2
4e
12.09394 / 0.02586
10.20427 / kT
10.20427 / kT
T
8
e
2
e
ln 0.0025
12.09394 / 8.61734 10
5
19,760
9e
4 10
468
10.20427 / kT
172
9 10
203
0
0
0.0025
5.99146
10.20427eV
5.99146 8.61734 10 5 eV K
9e
395
19, 760 K
0.00742
0.7%
B field
8-18.
m
1
no field
E
m
E
m
hf
0
1
E
ground state
Neglecting the spin, the 3p state is doubly degenerate:
levels equally populated.
E
hf
hc /
1.8509eV
670.79nm
192
0,1 hence, there are two m = 0
Chapter 8 – Statistical Physics
(Problem 8-18 continued)
e B
2me
E
2.315 10 4 eV
(a) The fraction of atoms in each m-state relative to the ground state is: (Example 8-2)
n1
n
e
1.8511 / 0.02586
n0
n
2 e
n0
n1
e
71.58
e
1.8509 / 0.02586
1.8507 / 0.02586
10
71.57
2e
71.56
e
31.09
8.18 10
2 10
10
31.08
31.08
32
1.64 10
8.30 10
31
32
(b) The brightest line with the B-field “on” will be the transition from the m = 0 level, the
center line of the Zeeman spectrum.
With that as the “standard”, the relative
intensities will be: 8.30 / 16.4 / 8.18
N
h3
V 2 2 me kT
8-19. (a) e
N
V
e
2 2 me kT
(Equation 8-44)
3/ 2
3/ 2
e
h3
2 2 me c 2 kT
hc
5.11 105 eV
2
1240eV nm
8-20. (a) e
O2
3/ 2
3/ 2
3
2.585 10 2 eV
1 2
N
h3
V 2 MkT
0.51/ 1.00 / 0.50
3
1/ 2
107 nm
1cm
3
2.51 1019 / cm3
(Equation 8-44)
3
hc
NA
VM 2 Mc 2 kT
3/ 2
6.022 1023 / mole
22.4 103 cm3 / mole
1.24 10 4 eV cm
2
3
32uc 2 931.5 106 eV / u 8.617 10 5 eV / K 273K
1.75 10 7
193
Chapter 8 – Statistical Physics
(Problem 8-20 continued)
(b) At temperature T, e
7
1.75 10
1
273K
1.75 10
T
7
1.75 10
O2
2/3
3/ 2
/ T 3/ 2
7
273
T 3/ 2
3/ 2
/ T 3/ 2
1.75 10
7
273K
3/ 2
8.5mK
273K
8-21. Assuming the gasses are ideal gases, the pressure is given by: P
2N E
for classical,
3 V
FD, and BE particles. PFD will be highest due to the exclusion principle, which, in effect,
limits the volume available to each particle so that each strikes the walls more frequently
than the classical particles. On the other hand, PBE will be lowest, because the particles
tend to be in the same state, which in effect, is like classical particles with a mutual
attraction, so they strike the walls less frequently.
1
8-22. (a) f BE
e e
E / kT
1
e
E / 5800 k
1
1
1
For
e E / 5800 k
0.35V ,
1
0.35 / kT
1
0.5
0.5
= 0 and f BE
e0.35 / kT
3
0.35eV 8.62 10 5 eV / K T
T
8-23.
h
p
ln 2
0.347eV
(b) For E
e
5800 K
2
E 5800 K 8.62 10 5 eV / K
E
1, at T
= 0 and f BE
0.35eV
ln 3 8.62 10 5 eV / K
h
h
2m E
2m 3kT / 2
ln 3
3700 K
h
3mkT
1/ 2
The distance between molecules in an ideal gas V / N
194
1/ 3
is found from
Chapter 8 – Statistical Physics
(Problem 8-23 continued)
PV
nRT
nRT N A N A
and equating this to
V /N
NkT
above, kT / P
1/ 3
kT / P
h
1/ 3
3mkT
h3
kT
P
3mkT
3/ 2
2/5
Ph
T
3/ 2
k 3mk
8-24.
N0
N
1
(a) For T
(b) For T
(c) For T
(d) For T
T
TC
N0
N
TC / 2
N0
N
TC / 4
N0
N
TC / 8
N0
N
which for small
TC
J s
kg 1.38 10
23
2/5
3
4.4 K
5/ 2
J/K
(Equation 8-52)
3TC / 4
h2
2mk 2
27
34
3/ 2
3/ 2
8-25. For small values of
8-26.
101kPa 6.63 10
3 2 1.67 10
1/ 2
P
h3
k 3mk
and solving for T , yields: T 5 / 2
3
1/ 3
,e
1
3T
4TC
1
T
2TC
1
T
4TC
1
T
8TC
1
3/ 2
0.351
3/ 2
0.646
3/ 2
0.875
3/ 2
0.956
2
/ 2!
values becomes: N0 1
2/3
N
2.315 V
(Equation 8-48)
The density of liquid Ne is 1.207 g/cm3, so
195
and N0
1
N0
1
e
1
N0 e
1 or N0
1
1
1
Chapter 8 – Statistical Physics
(Problem 8-26 continued)
1.207 g / cm3 6.022 1023 molecules / mol 106 cm2 / m3
N
V
T
3.601 1028 / m3
20.18 g / mol
34
6.626 10
27
2 20u 1.66 10
20
Thus, TC at which
2
J s
23
kg / u .381 10
2/3
3.601 1028 m3
2 2.315
J /K
0.895K
Ne would become a superfluid is much lower than its freezing
temperature of 24.5K.
8-27. Power per unit area R arriving at Earth is given by the Stefan-Boltzmann law: R
is Stefan’s constant. For a 5% decrease in the Sun’s temperature,
where
T4
R 0.95T
R T
0.95T
R T
8-28.
E
T
hf
e
hf / kT
(a) For T
10hf / k;
(b) For T
hf / k;
(c) For T
0.1hf / k;
hf
hf
kT / 10
kT
hf
E
10kT
According to equipartition E
8-29.
CV
e hf / kT
CV
0.95
1
4
4
0.186 , or a decrease of 18.6%.
(Equation 8-60)
1
hf
3N A k
kT
4
2
1
1 / 10
e
hf
1
e
1
kT
1.718
1
hf
E
10
e
1
kT / 10
0.1051
0.951kT
0.582kT
10kT
2.20 104
As T
2
, hf / kT gets small and
1 hf / kT
hf
3N A k
kT
2
1 hf / kT
hf / kT
4.54 10 4 kT
kT in each case.
ehf / kT
ehf / kT
hf
E
2
3N A k
The rule of Dulong and Petit.
196
3N A R / N A
3R
T4
Chapter 8 – Statistical Physics
8-30.
CV
3R
2
hf
kT
ehf / kT
ehf / kT
Writing hf / kT
CV
3R Af
1
(Equation 8-62)
2
h / kT
Af where A
e Af
2
2
e Af
CV
3R / e Af
e Af
8-31.
CV
3R
3 8.31
ln
f
hf
kT
ehf / kT
e
hf / kT
1
3R 1
2
e1
e1 1
2
22.95K / K mol
, so
8.97 1011 Hz
13
2.46 1012 Hz
(Equation 8-62)
2
At the Einstein temperature TE
CV
2
(From Figure 8-13)
1 / 2.40 10
13.8
2
13
13.8 J / K mol
For Si, CV 200 K
1 / e Af
(From Figure 8-13)
1 / 2.40 10
20.1
2
ln 3R / CV 1 / A
f
20.1J / K mol
3 8.31
ln
f
Af
0 and e Af dominates Af
3R / CV
For Al, CV 200 K
e
1
Because Af is “large”, 1 / e Af
200 K ,
when T
1
2
eR Af
2e Af
13
2.40 10
hf / k,
3R 0.9207
3 8.31J / K mol 0.9207
5.48cal / K mol
3/ 2
8-32. Rewriting Equation 8-69 as
V
E1 / 2
8mc 2
n E
2
hc
2
e
E EF / kT
1
Set up the equation on a spreadsheet whose initial and final columns are E(eV) and
n(E)/V (eV•nm3)-1, respectively.
197
Chapter 8 – Statistical Physics
(Problem 8-32 continued)
E eV
n E /V
eV nm3
4.5
14.4
4.6
14.6
4.7
14.5
4.8 (=EF)
7.46
4.9
0.306
5
0.0065
5.1
0.00014
1
The graph of these values is below.
From the graph, about 0.37 electrons/nm3 or 3.7 1026 electrons / m3 within 0.1eV below
EF have been excited to levels above EF.
198
Chapter 8 – Statistical Physics
8-33. The photon gas has the most states available, since any number of photons may be in the
ground state. In contrast, at T = 1K the electron gas’s available states are limited to those
within about 2kT
2 8.62 10 5 eV K 1K
1.72 10 4 eV of the Fermi level. All
other states are either filled, hence unavailable, or higher than kT above the Fermi level,
hence not accessible.
8-34. From the graph.
TE Au
TE Be
136 K
575K
TE Al
243K
TE Diamond
off the graph (well over 1000K)
8-35. Approximating the nuclear potential with an infinite square well and ignoring the
Coulomb repulsion of the protons, the energy levels for both protons and neutrons are
given by En
n2h2
8mL2 and six levels will be occupied in
protons and six levels with 12 neutrons.
EF protons
5
2
1240MeV fm
2
8 1.0078u 931.5MeV / u 3.15 fm
199
2
516MeV
22
Ne , five levels with 10
Chapter 8 – Statistical Physics
(Problem 8-35 continued)
6
EF neutrons
2
1240 MeV fm
2
8 1.0087u 931.5MeV / u 3.15 fm
3 / 5 EF
E protons
742 MeV
310MeV
3 / 5 EF
E neutrons
2
445MeV
As we will discover in Chapter 11, these estimates are nearly an order of magnitude too
large. The number of particles is not a large sample.
8-36.
E1
h 2 / 8mL2 . All 10 bosons can be in this level, so E1 total
8-37. (a) f FD E
1
e
E EF / kT
e
E EF / 0.1EF
e
E EF / 0.5 EF
(Equation 8-68)
1
1
(b) f FD E
1
1
10 E EF / EF
e
1
1
1
1
e
2 E EF / EF
200
1
10h 2 / 8mL2 .
Chapter 8 – Statistical Physics
8-38.
NO
N
3/ 2
T
Tc
1
N
(a) O
N
N
(b) O
N
(Equation 8-52)
3/ 2
1
Tc / 2
Tc
1
Tc / 4
Tc
3/ 2
1
1
2
1
1
4
3/ 2
0.646
3/ 2
0.875
n2 h2
8-39. For a one-dimensional well approximation, En
8mL2 . At the Fermi level EF,
n=N/2, where N = number of electrons.
2
N / 2 h2
EF
2
h2 N
32m L
8mL2
where N / L = number of electrons/unit length,
i.e., the density of electrons. Assuming 1 free electron/Au atom,
N
L
EF
6.02 1023 electrons / mol 19.32 g / cm3 102 cm / m
1/ 3
NA
M
3
197 g / mol
2
6.63 10
34
J s
3.81 109 m
32 9.11 10
31
kg 1.602 10
19
1
1/ 3
3.81 109 m
1
2
1.37eV
J / eV
This is the energy of an electron in the Fermi level above the bottom of the well. Adding
the work function to such an electron just removes it from the metal, so the well is
1.37eV
4.8eV
8-40. (a) At T
6.2eV deep.
850 K
v
vrms
vm
8kT
m
3kT
m
1/ 2
1/ 2
2kT
m
4
3
2
/2
2 1.3807 10
23
J / K 850 K
6.94 10
1/ 2
vm
207.5m / s
1/ 2
vm
225.2m / s
201
25
kg
1/ 2
183.9m / s
Chapter 8 – Statistical Physics
(Problem 8-40 continued)
The times for molecules with each of these speeds to travel across the 10cm diameter
of the rotating drum is:
0.10m
183.9m / s
t vm
5.44 10 4 s
t v
0.10m
207.5m / s
4.82 10 4 s
t vrms
0.10m
225.2m / s
4.44 10 4 s
The drum is rotating at 6250rev/min = 104.2rev/s or 9.600 10 3 s / rev. The fraction
of a revolution made by the drum while molecules with each of these three speeds are
crossing the diameter is:
5.44 10 4 s
9.600 10 3 s / rev
for vm :
0.05667rev
for v :
4.82 10 4 s
9.600 10 3 s / rev
0.05021rev
for vrms :
4.44 10 4 s
9.600 10 3 s / rev
0.04625rev
Assuming that point A is directly opposite the slit s2 when the first (and fastest)
molecules enter the drum, molecules with each of the three speeds will strike the plate
at the following distances from A: (The circumference of the drum C
vm :
0.05667 rev 0.314159m / rev
0.01780m 1.780cm
v :
0.05021rev 0.314159m / rev
0.01577m 1.577cm
vrms :
0.04625rev 0.314159m / rev
0.01453m 1.453cm
0.10 m .)
(b) Correction is necessary because faster molecules in the oven will approach the oven’s
exit slit more often than slower molecules, so the speed distribution in the exit beam
is slightly skewed toward higher speeds.
(c) No. The mean speed of N2 molecules at 850K is 710.5m/s, since they have a smaller
mass than Bi2 molecules.
Repeating for them the calculations in part (a),
N2
molecules moving at vm would strike the plate only 0.4cm from A. Molecules moving
at v and vrms would be even closer to A.
202
Chapter 8 – Statistical Physics
8-41.
1
m vescape
2
EK ( escape)
0
1 2 3
mv v e
2
0
1 2
mv F v dv
2
mv 2 / 2 kT
v 3e
mv 2 / 2 kT
3
m
dv
1 I5
m
2 I3
dv
0
1
m
2
8-42. (a) f u du
1
(b)
8-43.
Ce
2
/2
E / kT
du
f u du
Ce
Ce
2C
1/ 2
C
kT / A
C
Au 2
vx
2kT
du
du
2C e
/ 2 where
du
A / kT
Au 2 A / kT e
A A / kT 2
1
A A / kT kT / A
2
3/ 2
e
Au 2 / kT
A / kT
Au 2 f u du
1/ 2
m
2kT
where
(from Equation 8-5)
A A / kT 2I 2
m / 2 kT
f vx
2kT
m
Au 2 / kT
Au 2 / kT
2CI 0
E
m
mvx2 / 2 kT
/4
Au 2 / kT
du
3/ 2
where
A / kT
1
kT
2
(from Equation 8-6)
vx f vx dvx
0
vx m / 2 kT
1/ 2
e
mvx2 / 2 kT
vx m / 2 kT
dvx
0
2 m / 2 kT
1/ 2
vx e
mvx2 / 2 kT
dvx
0
2 m / 2 kT
1/ 2
I1
with
m / 2kT
203
1/ 2
e
mvx2 / 2 kT
dvx
Chapter 8 – Statistical Physics
(Problem 8-43 continued)
1/ 2
2 m / 2 kT
8-44.
1
f FD
e
E EF / kT
E EF / kT
1
f FD
N
1
EF , e
For E
8-45.
1
E / kT
e e
e
e
2kT
m
1/ 2
1
1 and
4
EF / kT
e
3/ 2
2me
h
V
e
=
E / kT
E1 / 2 e
3
EF
kT
where
1
E EF / kT
2kT
m
E / kT
1
e e E / kT
dE
fB
(Equation 8-43)
0
Considering the integral, we change the variable: E / kT
E
kTu 2 , E1 / 2
E1 / 2 e
E / kT
kT
dE
1/ 2
2 kT
0
u, and dE
3/ 2
u 2e
u2
kT 2u du. So,
du
0
/ 4, so
The value of the integral (from tables) is
N
u 2 , then
e
4
3/ 2
2me
V 2 kT
h3
8-46. (a) N
ni
3/ 2
or e
4
f 0 E0
f1 E1
2 2me kT
3/ 2
Nh3
(with g0
g1
1)
i
Ce0
So, C
(b)
E
Ce
N
1 e
0 n0
/ kT
/ kT
/ kT
n1
N
C 1 e
Ce
N
/ kT
N e
1 e
/ kT
/ kT
204
N
e
1 e
/ kT
/ kT
V
, which is Equation 8-44.
Chapter 8 – Statistical Physics
(Problem 8-46 continued)
/ kT
As T
0,
e
As T
,
e
(c) CV
1/ e
/ kT
1/ e
d N E
dE
dT
1 e
2
Nk
kT
/ kT
2
/ kT
e
0, so E
e
/ kT
2
e
/ kT
1 e
/ kT
0
0, so E
/2
/ kT
d N e
dT 1 e
dT
N 2
kT 2
/ kT
/ kT
/ kT
1 e
/ kT
2
(d)
T
/k
CV
Nk
8-47. (a) p
k
L
2
x
n12
k
2
y
n22
1
k
2
z
0.1
0.25
0.5
1.0
2.0
3.0
0.005
0.28
0.42
0.20
0.06
0.03
1/ 2
n32
1/ 2
n1
L
2
n2
L
N
L
205
2
n3
L
2
1/ 2
Chapter 8 – Statistical Physics
(Problem 8-47 continued)
E
pc
c N
L
(b) Considering the space whose axes are n1, n2 , and n3 . The points in space correspond
to all possible integer values of n1, n2 , and n3 , all of which are located in the all
positive octant. Each state has unit volume associated with it. Those states between
N and N + dN lie in a spherical shell of the octant whose radius is N and whose
thickness is dN. Its volume is 1 / 8 4 N 2 dN . Because photons can have two
polarizations (spin directions), the number of possible state is
2
1 / 8 4 N 2 dN
N 2 dN .
(c) This number of photon states has energy between E and E+dE, where N
EL / hc.
The density of states g(E) is thus:
g E dE
number of photon states at E
dE
number of photon states at N
dN
N 2dN
EL
c
8 L3
2
c
2
L
c dE
8 L3
2
3
E dE
hc
3
E 2 dE
The probability that a photon exists in a state is given by:
f BE E
1
e e
E / kT
1
1
e
E / kT
1
(Equation 8-24)
The number of photons with energy between E and E+dE is then:
n E dE
f BE E g E dE
3
L / hc E 2 dE
8
e E / kT
1
(d) The number of photons per unit volume within this energy range is n E dE / L3.
Because each photon has energy E, the energy density for photons is:
u E dE
8 E 3dE
3
E n E dE / L
hc
3
e E / kT
1
which is also the density of photons with wavelength between λ and λ+dλ, where
206
Chapter 8 – Statistical Physics
(Problem 8-47 continued)
hc / E
d
u
E
d
dE
dE
d
hc / . So,
2
hc
dE
E2
u E dE
hc
8
dE
3
hc /
hc
dE
3
hc /
ehc /
kT
207
hc
2
2
1
d
d
8 hc 3d
ehc / kT 1
Chapter 9 – Molecular Structure and Spectra
9-1.
(a) 1
19
23
eV
eV

  1.609  10 J  6.022  10 molecules 
 1



molecule  molecule  
eV
mole


J  1cal

  96472
mole 

 4.184 J
cal
kcal

  23057 mole  23.06 mole

eV

 23.06kcal / mole 
(b) Ed   4.27
  98.5kcal / mole
molecule 

 1eV / molecule 
eV

 1eV / molecule 
(c) Ed  106
  1.08eV / molecule
molecule 

 96.47kJ / mole 
9-2.
Dissociation energy of NaCl is 4.27eV, which is the energy released when the NaCl
molecule is formed from neutral Na and Cl atoms. Because this is more than enough
energy to dissociate a Cl2 molecule, the reaction is exothermic. The net energy release is
4.27eV – 2.48eV = 1.79eV.
9-3.
From Cs to F: 3.89eV – 3.40eV = 0.49eV
From Li to I: 5.39eV – 3.06eV = 2.33eV
From Rb to Br: 4.18eV – 3.36eV = 0.82eV
9-4.
Ed  U C  
CsI :
NaF :
LiI :

ke2
 Eion
r0
ke2
1.440eV nm
 Eion  
  3.89eV  3.06eV   Ed  3.44eV
r0
0.337nm

ke2
1.440eV nm
 Eion  
  5.14eV  3.40eV   Ed  5.72eV
r0
0.193nm
ke2
1.440eV nm

 Eion  
  5.39eV  3.06eV   Ed  3.72eV
r0
0.238nm
While Ed for CsI is very close to the experimental value, the other two are both high.
Exclusion principle repulsion was ignored.
209
Chapter 9 – Molecular Structure and Spectra
9-5.
(a) Total potential energy: U  r   
ke2
 Eex  Eion
r
(Equation 9-1)
ke2
1.44eV nm
attractive part of U  r0   

 5.39eV
r0
0.267nm
(b) The net ionization energy is:
Eion   ionization energy of Rb    electron affinity of Cl 
 4.18eV  3.62eV  0.56eV
Neglecting the exclusion principle repulsion energy Eex ,
dissociation energy  U  r0   5.39eV  0.56eV  4.83eV
(c) Including exclusion principle repulsion,
dissociation energy  4.37eV  U  r0   5.39eV  0.56eV  Eex
Eex  5.39eV  4.37eV  0.56eV  0.46eV
9-6.
Uc  
ke2
1.440eV nm
 Eion  
  4.34eV  3.36eV   4.13eV
r0
0.282nm
The dissociation energy is 3.94eV.
Ed  U c  Eex  3.94eV  4.13eV  Eex
Eex  0.19eV at r0  0.282nm
9-7.
Eex 
A
rn
(Equation 9-2)
0.19eV 
A
 0.282nm 
n
At r0 the net force on each ion is zero, so we have (from Example 9-2)
U c  r0 
r0
n

ke2
nA n A n
 18.11eV / nm  n1   n   0.19eV 
2
r0
r0
r0 r0
r0
18.11eV / nm  0.282nm   26.9  27
0.19eV
A  Eex r0n   0.19eV  0.282nm   2.73  1016 eV nm27
27
210
Chapter 9 – Molecular Structure and Spectra
9-8.
Ed  3.81eV per molecule of NaBr (from Table 9-2)


1eV / molecule  1eV / molecule  1.609  1019 J / eV 
 6.02 10
23

molecules / mol / 1cal / 4.186 J   23.0kcal / mol
Ed  NaBr    3.81eV / molecule  23.0kcal / mol  / 1eV / molecule   87.6kcal / mol
9-9.
For KBr : U C 
1.440eV nm
  4.34eV  3.36eV   4.13eV
0.282nm
Ed  3.94eV  UC  Eex  4.13eV  Eex
Eex  0.19eV
For RbCl : U C 
1.440eV nm
  4.18eV  3.62eV   4.60eV
0.279nm
Ed  4.37eV  UC  Eex  4.60eV  Eex
Eex  0.23eV
9-10.
H 2 S , H 2Te, H3 P, H3Sb
9-11. (a) KCl should exhibit ionic bonding.
(b) O2 should exhibit covalent bonding.
(c) CH4 should exhibit covalent bonding.
9-12. Dipole moment pionic  er0
(Equation 9-3)


 1.609  1019 C  0.0917nm 
 1.47 1020 C nm 109 m / nm
 1.47  1029 C m
if the HF molecule were a pure ionic bond. The measured value is 6.64 1029 C m , so

the HF bond is 6.40  1030 C m
 1.47 10
211
29

C m  0.44 or 44% ionic.
Chapter 9 – Molecular Structure and Spectra
9-13.
pionic  er0  1.609 1019 C  0.2345 109 m 
(Equation 9-3)
 3.757 1029 C m, if purely ionic.
The measured value should be:


pionic  measured   0.70 pionic  0.70 3.757  1029 C m  2.630  1029 C m
9-14.
pionic  er0  1.609 1019 C  0.193 109 m 
(Equation 9-3)
 3.09  1029 C m
The measured values is 2.67  1029 C m, so the BaO bond is
 2.67 10
29
Cm
 3.09 10
29

C m  0.86 or 86% ionic.
9-15. Silicon, germanium, tin, and lead have the same outer shell configuration as carbon.
Silicon and germanium have the same hybrid bonding as carbon (their crystal structure is
diamond, like carbon); however, tin and lead are metallic bonded. (See Chapter 10.)
9-16.
p = p1  p2 and p  6.46  1030 C m and p  p1 cos 52.25  p2 cos 52.25
If bonding were ionic, pionic  er0  1.609 1019 C  0.0956 109 m   1.532 1029 C m
p1  actual   p / 2  cos 52.25  6.46  1030 C m / 2 cos 52.25   5.276 1030 C m
Ionic fraction = fraction of charge transferred =
5.276  1030 C m
 0.34 or 34%
1.532  1029 C m
9-17. U   k 2 p12 / r 2 (Equation 9-10)
(a) Kinetic energy of N2  0.026eV , so when U  0.026eV the bond will be broken.
1.110
0.026eV 
r
6
1.110

37
37

m C 2 / N 9  109 N m2 / C 2
  6.46 10
2
30
Cm

2
r6

  6.46  10
J / eV 
m C 2 / N 9  109 N m2 / C 2

0.026eV 1.60  10
r  6.7  1010 m  0.67nm
212
19
2
30
Cm

2
 8.94  1056 m6
Chapter 9 – Molecular Structure and Spectra
(Problem 9-17 continued)
(b) U 
ke2
1.440eV nm
 U  0.026eV 
 r  55nm
r
r
(c) H2O-Ne bonds in the atmosphere would be very unlikely. The individual molecules
will, on average, be about 4nm apart, but if a H2O-Ne molecule should form, its
U  0.003eV at r  0.95nm , a typical (large) separation. Thus, a N2 molecule with
the average kinetic energy could easily dissociate the H2O-Ne bond.
9-18. (a) E  0.3eV  hc /   1240eV nm /     1240eV nm / 0.3eV  4.13  103 nm
(b) Infrared
(c) The infrared is absorbed causing increased molecular vibrations (heat) long before it
gets to the DNA.
9-19. (a) NaCl is polar. The Na+ ion is the positive charge center, the Cl− ion is the negative
charge center.
(b) O2 is nonpolar. The covalent bond involves no separation of charges, hence no
polarization of the molecule.
9-20. For N 2
E0 r  2.48  104 eV 
2.48  104 eV  2 I  
r02 
2
/ 2I where I 
1 2
mr0 and m  14.0067u
2
2
2
 2.48 10
4

eV 14.0067u 


2

1.055  1034 J s
r0  
 2.48  104 eV 1.60  1019 J / eV 14.0067u  1.66  1027 kg / u




 1.611010 m  0.161nm
213






1/ 2
Chapter 9 – Molecular Structure and Spectra
9-21.
E0 r 
E0 r 
2
(Equation 9-14) where I 
2I
2
2
0
mr

 c
197.3eV
2
2 2
0
mc r

16uc   931.5  10
2
9-22. For Co: f  6.42 1013 Hz
EV   v  1 / 2  hf
1 2
mr0 for a symmetric molecule.
2
6
nm 
eV / uc
2
2
  0.121nm 
2
 1.78  104 eV
(See Example 9-6)
(Equation 9-20)
(a) E1  E0  3hf / 2  hf / 2  hf


 4.14  1015 eV s 6.42  1013 Hz

 0.27eV
(b)
n1
 E  E / kT
 e  1 0
n0
0.01  e
(from Equation 8-2)


 0.27  / 8.62105 T


ln  0.01    0.27eV  8.62  105 eV / K T
T
  0.27eV 

ln  0.01 8.62  105 eV / K

T  680K
9-23. For LiH: f  4.22  1013 Hz
(from Table 9-7)



(a) EV   v  1 / 2  hf  E0  hf / 2  4.14  1015 eV s 4.22  1013 Hz / 2
E0  0.087eV
(b)  

(c) f 
m1m2
m1  m2
(Equation 9-17)
 7.0160u 1.0078u   0.8812u
 7.0160u   1.0078u 
1
2
K

(Equation 9-21)
214
Chapter 9 – Molecular Structure and Spectra
(Problem 9-23 continued)






K   2 f     2  4.22  1013 Hz  0.8812  1.66  1027 kg / u
2
2
K  117 N / m
(d) En  n2 h2 8mr02  r02  n2 h2 8mEn
r0  h 8mE0 
1/ 2
r0 
6.63  1034 J s


8  0.8812u  1.66  1027 kg / u  0.087eV  1.60  1019 J / eV 


1/ 2
r0  5.19  1011 m  0.052nm
1.0078u   0.504u
mm
9-24. (a) For H2:   1 2 
m1  m2 2 1.0078u 
2
14.0067u   7.0034u

2 14.0067u 
2
(b) For N2:
(c) For CO:  
12.0111u 15.9994u   6.8607u
(d) For HCl:  
9-25. (a)  
12.0111u  15.9994u
1.0078u 35.453u   0.980u
1.0078u  35.453u
 39.1u  35.45u   18.6u
m1m2

m1  m2
39.1u  35.45u
(b) E0 r 
2
2I
E0 r 
 r0 
(Equation 9-14) I   r02
2
2 r02

 c
2
2 c 2 r02
c
 2 c E 
1/ 2
2
V

r 
2
0
 c
2
2 c 2 EV
197.3eV nm




1/ 2
 2 10.6uc 2 931.5  106 eV / uc 2 1.43  105 eV 


r0  0.280nm
215
Chapter 9 – Molecular Structure and Spectra
9-26.
1
2
f 
K
(Equation 9-21)

(a) For H35Cl: μ = 0.980u and f  8.97  1013 Hz .

K   2 f     2  8.97  1013 Hz
2
2
(b) For K79Br:  
2
1
2
K

2
27

kg / u  517 N / m .
 39.102u  78.918u   26.147u and f  6.93 1012 Hz
m1m2

m1  m2
39.102u  78.918u

K   2 f     2  6.93  1012 Hz
9-27. 1. f 
  0.980u  1.66 10
2
  26.147u  1.66 10
2
27

kg / u  82.3N / m
(Equation 9-21) Solving for the force constant, K  (2 f )2 
2. The reduced mass  of the NO molecule is

mN mO
(14.01u)(16.00 u)

 7.47 u
mN  mO 14.01u  16.00 u
3. K  (2  5.63 1013 Hz)2  7.47 u 1.66 1027 kg/u  1.55 103 N/m
( Note: This is equivalent to about 8.8 lbs/ft, the force constant of a moderately strong
spring.)
9-28.
E0r 
2
2I Treating the Br atom as fixed,


I  mH r02  1.0078u  1.66  1027 kg / u  0.141nm 
E0 r 

1.055  10
34

J s

2
2

2 1.0078u  1.66  1027 kg / u  0.141nm  109 m / nm
2
 1.67  1022 J  1.04  103 eV
E 

 1 E0r for  0, 1, 2,
(Equation 9-13)
216

2
Chapter 9 – Molecular Structure and Spectra
(Problem 9-28 continued)
14
The four lowest states have energies:
E0  0
E
3
12
10
3
eV

10
3
E1  2E0r  2.08  10 eV
8
3
E2  6E0r  6.27  10 eV
2
6
E3  12E0r  12.5  103 eV
4
9-29.
2
1
0
0
E  hf where f  1.05  1013 Hz for Li. Approximating the potential (near the bottom)
2  2
with a square well, E  2  1  22  1   2  hf
 2  mr0

For Li2: r02 

3 2
1
3

2 2 f 
4 f
 3 
1.055  1034 J s
r   
 4  1.05  1013 Hz  6.939u  1.66  1027 kg / u



 4.53  1011 m  0.045nm
9-30.
E0 r 
2
2I
where I   r02
(Equation 9-14)
For K35Cl:  
 39.102u  34.969u   18.46u
For K37Cl:  
 39.102u  34.966u   19.00u
39.102u  34.969u
39.102u  34.966u
r0  0.267nm for KCl.
217




1/ 2
Chapter 9 – Molecular Structure and Spectra
(Problem 9-30 continued)


E0 r K Cl 
35

1.055  10
34
J s


2
2 18.46u  1.66  1027 kg / u 0.267  109 m

2
 2.55 1024 J  1.59 105 eV


E0 r K 37Cl 

1.055  10
34
J s


2
2 19.00u  1.66  1027 kg / u 0.267  109 m

2
 2.48 1024 J  1.55 105 eV
E0r  0.04  105 eV
9-31. (a) NaF – ionic
(b) KBr – ionic
(c) N2 – covalent
9-32.
E0,1


 1
I
(d) Ne – dipole-dipole
2

2
 r02
(a) For NaCl: r0  0.251nm (from Table 9-7)

m  Na  m


 22.9898 34.9689  13.8707u
m  Na   m  35Cl   22.9898   34.9689 
E0,1 
35
Cl

1.055 10
13.8707u  1.66 10
27

kg / u  0.251  10 m 
34
J s
2
9
2
E0,1  7.67  1024 J  4.80  105 eV
(b) E0,1  hf  f  E0,1 / h 
7.67  1024 J
6.63  1034 J s
f  1.16  1010 Hz
  c / f   3.00  108 m / s 1.16  1010 Hz   0.0259nm  2.59cm
218
Chapter 9 – Molecular Structure and Spectra
9-33. (a)   2400nm  E  hc / 2400nm 
E2  E1  3.80eV
1240eV nm
 0.517eV
2400nm
E3  E2  0.500eV
E4  E3  2.9eV
E5  E4  0.30eV
The E3  E2 and E5  E4 transitions can occur.
(b) None of these can occur, as a minimum of 3.80eV is needed to excite higher states.
(c)   250nm  E  1240eV nm 250nm  4.96eV . All transitions noted in (a) can
occur. If the temperature is low so only E1 is occupied, states up to E3 can be
reached, so the E2  E1 and E3  E2 transitions will occur, as well as E3  E1 .
(d) E4  E3  2.9eV  hc  or   1240eV nm 2.9eV  428nm
E4  E2  3.4eV  hc  or   1240eV nm 3.4eV  365nm
E4  E1  7.2eV  hc  or   1240eV nm 7.2eV  172nm
9-34.
A21
 ehf / kT  1
B21u  f 
(Equation 9-42)
For the Hα line λ=656.1nm
At T = 300K,
hf
hc
1240eV nm


 73.1
kT  kT  656.1nm  8.62  105 eV / K  300 K 


ehf / kT  1  e73.1  1  5.5  1031
Spontaneous emission is more probable by a very large factor!
9-35.
n  E1 
n  E0 
e E1 / kT
  E0 / kT i.e., the ratio of the Boltzmann factors.
e
For O2: f  4.74  1013 Hz and



E0  hf 2  4.14  1015 eV s 4.74  1013 Hz 2  0.0981eV
E1  3hf 2  0.294eV


At 273K, kT  8.62  105 eV / K  273K   0.0235eV
219
Chapter 9 – Molecular Structure and Spectra
(Problem 9-35 continued)
n  E1 
n  E0 

e0.0294 / 0.0235 e12.5
 4.17  2.4  104
0.0981 / 0.0235
e
e
Thus, about 2 of every 10,000 molecules are in the E1 state.
Similarly, at 77K,
9-36.

E
n  E0 
 1.4  1013
 1 E0r for  0, 1, 2,
Where E0 r 
E0 r 
n  E1 
2
2I
(Equation 9-13)
and I   r02 with   m / 2

1.055 10
2 18.99u  1.66  10
27

kg / u  0.14  10 m 
34
J s
2
9
2
 1.80  1023 J  1.12  104 eV
(a) E0  0
E1  2E0r  2.24  104 eV
E1  E0  2.24  104 eV
E2  6E0r  6.72  104 eV
E2  E1  4.48  104 eV
E3  12E0r  13.4  104 eV
E3  E2  6.72  104 eV
14
3
12
E
10
4
eV

10
8
2
6
4
1
0
2
0
220
Chapter 9 – Molecular Structure and Spectra
(Problem 9-36 continued)
E  hc     hc E
(b)   1
For E1  E0 :  
1240eV nm
 5.54  106 nm  5.54nm
4
2.24  10 eV
For E2  E1 :  
1240eV nm
 2.77  106 nm  2.77nm
4.48  104 eV
For E3  E2 :  
1240eV nm
 1.85  106 nm  1.85nm
4
6.72  10 eV



9-37. (a) 10MW  107 J / s  E  107 J / s 1.5  109 s  1.5  102 J
(b) For ruby laser:   694.3nm, so the energy/photon is:
E  hc   1240eV nm 694.3nm  1.786eV
1.5  10 J 
Number of photons =
1.786eV  1.60  10
2
19
9-38.
J / eV

 5.23  106
4mW  4  103 J / s
E  hc  
1240eV nm
 1.960eV per photon
632.8nm
4  103 J / s
 1.28  1016 / s
Number of photons =
19
1.960eV  1.60  10 J / eV



9-39. (a) sin  1.22 / D  1.22 600  109 m
 10 10 m  7.32 10
2
6
   7.32 106 radians
  S / R where S  diameter of the beam on the moon and R  distance to moon.



S  R  3.84  108 m 7.32  106 radians  2.81 103 m  2.81km

(b) sin  1.22 600  109 m

 1m  7.32 10

7

radians
S  R  3.84  108 m 7.32  107 radians  281m
221
Chapter 9 – Molecular Structure and Spectra
9-40. (a)
n  E2 
n  E1 

e E2 / kT
E  E / kT
 e 2 1 
 E1 / kT
e
At T  297 K,
E2  E1  hc   1240eV nm 420nm  2.95eV


kT  8.61 105 eV / K  297 K   0.0256eV
n  E2   n  E1  e2.95 / 0.0256  2.5  1021 e115  2  1029  0


(b) Energy emitted = 1.8  1021  2.95eV / photon   5.31  1021 eV  850 J
9-41. (a) Total potential energy: U  r   
ke2
 Eex  Eion
r
the electrostatic part of U  r  at r0 is 
ke2
1.44eV nm

 6.00eV
r0
0.24nm
(b) The net ionization energy is:
Eion   ionization energy of Na    electron affinity of Cl 
 5.14eV  3.62eV  1.52eV
dissociation energy of NaCl =4.27eV (from Table 9-2)
4.27eV  U  r0   6.00eV  1.52eV  4.67eV  Eex
Eex  6.00eV  4.27eV  1.52eV  0.21eV
(c) Eex 
A
rn
(Equation 9-2)
At r0  0.24nm, Eex  0.21eV
At r0  0.14nm, U  r   0 and Eex 
At r0 : Eex  0.21eV 
A
 0.24nm 
At r  0.14nm : Eex  8.77eV 
n
ke2
 Eion  8.77eV
r
 A   0.21eV  0.24nm 
A
 0.14nm 
n
 A  8.77eV  0.14nm 
Setting the two equations for A equal to each other:
 0.24nm 
n
 0.14nm 
2
2
 0.24   8.77eV

 
 0.14   0.21eV
n

  1.71  41.76

222
n
n
Chapter 9 – Molecular Structure and Spectra
(Problem 9-41 continued)
n log1.71  log 41.76
n  log 41.76  log1.71  6.96
A  0.21eV  0.24nm   0.21eV  0.24nm 
n
9-42. (a)

sin  1.22 / D  1.22 694.3  109 m
6.96
 1.02  105 eV nm6.96
 0.01m  8.47 10
5
   8.47  105 radians
(b) E photon  hc /   1240eV nm / 694.3nm  1.786eV / photon
For 1018 photons / s :


Etotal  1.786eV / photon  1.602  1019 J / eV 1018 photons / s

 0.286 J / s  0.286W
Area of spot A is: A   d 2 / 4   8.47cm  / 4
2
2
and E  Etotal / A  0.286W /  8.47cm  / 4  5.08  103W / cm2


9-43. (a) U att  
ke2 1.440eV nm

 5.39eV
r
0.267nm
(b) To form K  and Cl  requires Eion  4.34eV  3.61eV  0.73eV
 ke2

Ed  U C    
 Eion   5.39eV  0.73eV  4.66eV
 r

(c) Eex  4.66eV  4.43eV  0.23eV at r0
9-44.
E0 r 

2
2I
where I   r02 with r0  0.267nm and
 39.102u  35.453u   18.594u
m1m2

m1  m2
39.102u  35.453u
E0 r 

1.055 10
2 18.594u  1.66  10
27

kg / u  0.267  10 m 
34
J s
2
9
223
2
 2.53  1024 J  1.58  105 eV
Chapter 9 – Molecular Structure and Spectra
9-45. (a) Ed 
kp1
where p1  qa, being the separation of the charges  q and  q of the dipole
x3
(b) U   p E and p  E  p =  E
So the individual dipole moment of a nonpolar molecule in the field produced by p1 is
p2   Ed   kp1 x3 and U   p2 Ed    kp1  x6
2
Fx  

dU
d
   k 2 p12
dx
dx

x6   6 k 2 p12 x7
9-46. 1. The energies E of the vibrational levels are given by Equation 9-20:
1
E  (  )hf for   0, 1, 2, 3,
2
The frequencies are found from Equation 9-21 and requires first that the reduced mass
for each of the molecules be found using Equation 9-17.
mH mH
(1.01u)(1.01u)

 0.51u
2
mH  mH 1.01u  1.01u
HD  0.67 u
H 
Similarly,
D  1.01u
2
2. The vibrational frequencies for the molecules are then:
f H2 
Similarly,
1
2
K
H

2
1
2
580 N/m
 1.32 1014 Hz
27
(0.51u)(1.66 10 kg/u)
f HD  1.15 1014 Hz
f D2  9.36 1013 Hz
3. The energies of the four lowest vibrational levels are then:
For H 2 :
1
1
hf H2  (6.63 1034 J  s)(1.32 1014 Hz)  4.38 1020 J
2
2
4.38 1020 J
E0 
 0.27 eV
1.60 1019 J/eV
E0 
224
Chapter 9 – Molecular Structure and Spectra
(Problem 9-46 continued)
E1  0.82 eV
Similarly,
E2  1.37 eV
E3  1.91eV
For HD:
1
1
hf HD  (6.63 1034 J  s)(1.15 1014 Hz)  3.811020 J
2
2
3.811020 J
E0 
 0.24 eV
1.60 1019 J/eV
And again similarly,
E1  0.72 eV
E0 
E2  1.19 eV
E3  1.67 eV
For D 2 :
1
1
E0  hf D2  (6.63 1034 J  s)(9.36 1013 Hz)  3.10 1020 J
2
2
3.10 1020 J
E0 
 0.19 eV
1.60 1019 J/eV
And once again similarly,
E1  0.58eV
E2  0.97 eV
E3  1.36 eV
4. There are three transitions for each molecule:
  3   2;   2   1;   1   0
For H 2 : E  hf  hc /     hc / E  1240eV  nm / E eV
E  E 3  E 2  (1.91  1.37)eV  0.54eV
32  1240eV  nm / 0.54eV  2.30 103 nm
Similarly,
21  2.25 103 nm
10  2.25 103 nm
For HD:
E  E 3  E 2  (1.67  1.19)eV  0.48eV
32  1240eV  nm / 0.48eV  2.58 103 nm
225
Chapter 9 – Molecular Structure and Spectra
(Problem 9-46 continued)
21  2.64 103 nm
And again similarly,
10  2.58 103 nm
For D 2 :
E  E 3  E 2  (1.36  0.97)eV  0.39eV
32  1240eV  nm / 0.39eV  3.18 103 nm
And once again similarly,
21  3.18 103 nm
10  3.18 103 nm


9-47. (a) E3  hc   1240eV nm  0.86mm  106 nm / mm  1.44  103 eV


 1240eV nm  2.59mm  10 nm / mm   4.79  10
E2  1240eV nm 1.29mm  106 nm / mm  9.61 104 eV
E1
6
4
eV
These are vibrational states, because they are equally spaced. Note the v  0 state
at the ½ level spacing.
18
v3
16
14
v2
12

Ev 104 eV

10
v 1
8
6
4
v0
2
0
226
Chapter 9 – Molecular Structure and Spectra
(Problem 9-47 continued)
(b) Approximating the potential with a square well (at the bottom),
E1  4.79  104 eV  n 2
2
2
2 mr02

 



2

22  12  2 1.055  1034 J s
r0  
 2  28.01u  1.66  1027 kg / u 4.79  104 eV 1.60  1019 J / eV







1/ 2
 2.15  1010 m  0.215nm
9-48. Using the NaCl potential energy versus separation graph in Figure 9-23(b) as an example
(or you can plot one using Equation 9-1):
The vibrational frequency for NaCl is 1.14  1013 Hz (from Table 9-7) and two vibrational
levels, for example v = 0 and v = 10 yield (from Equation 9-20)
E0  1 / 2hf  0.0236eV
E10  11 / 2hf  0.496eV
above the bottom of the well. Clearly, the average separation for v10 > v0.
2
9-49. (a) E0 r 
2 r02

E0 r 
E 
where r0  0.128nm for HCl and
1.0079u  35.453u   0.980u
m1m2

m1  m2
1.0079u  35.453u

1.055 10
2  0.980u  1.66  10

E0  0
27
34
J s


2
9
kg / u 0.128  10 m

2
 2.089  1022 J  1.303  103 eV
 1 E0 r
E1  2E0r  2.606  103 eV
E01  E1  E0  2.606  103 eV
f 01   E01 h 
E2  6E0r  7.82  103 eV
E12  E2  E1  5.214  103 eV
2.606  103 eV
 0.630  1012 Hz
4.136  1015 eV s
227
Chapter 9 – Molecular Structure and Spectra
(Problem 9-49 continued)
f12   E12 h 
5.214  103 eV
 1.26  1012 Hz
4.136  1015 eV s
f01  f  f01  6.884  1014 Hz  0.63  1012  6.890  1014 Hz; 6.878  1014 Hz
01  c f01  435.5nm; 436.2nm
f02  f  f02  6.884  1014 Hz  1.26  1012  6.897  1014 Hz; 6.871 1014 Hz
02  c f02  435.0nm; 436.6nm
(b) From Figure 9-29: f01  0.6  1012 Hz and f12  1.2  1012 Hz
The agreement is very good!
Ev   v  1 / 2  hf
9-50. (a) Li2 :



E1   3 / 2  4.14  1015 eV s 1.05  1013 Hz  0.0652eV  6.52  102 eV
E 

 1 E0 r


E1  2 8.39  105 eV  1.68  104 eV
(b) K 79 Br :
Ev   v  1 / 2  hf



E1   3 / 2  4.14  1015 eV s 6.93  1012 Hz  4.30 102 eV
E 

 1 E0 r


E1  2 9.1 106 eV  1.8  105 eV
9-51.
  HCl   0.980u
(See solution to Problem 9-49)
From Figure 9-29, the center of the gap is the characteristic oscillation frequency f :
f  8.65  1013 Hz  E  0.36eV

K   2  8.65  1013 Hz
2
Thus, f 
  0.980u  1.66 10
2
228
27
1
2

K

or K   2 f  
kg / u  480 N / m
2
Chapter 9 – Molecular Structure and Spectra
9-52.
n En
g En e
En / kT
694.3nm
21
E2
E2
E1
hc
1.7860eV
E1
1240eV nm
694.3nm
21
0.0036eV
1.7860eV
1.7896eV
Where E2 is the lower energy level of the doublet and E2 is the upper.
Let T = 300K, so kT = 0.0259eV.
(a)
n E2
g E2
n E1
g E1
n E2
1
e
2
n E1
1.7896 / 0.0259
(b) If only E2
2
e
4
E2 E1 / kT
e
1.7896 / 0.0259
5.64 10
1
e
2
69
4.91 10
31
31
E1 transitions produce lasing, but E2 and E2 are essentially equally
populated, in order for population inversion between levels E2 and E1 , at least 2/3
rather than 1/2) of the atoms must be pumped. The required power density (see
Example 9-8) is:
p
9-53. (a) Ev
so I
(c) I
h/ 4
r 2 , where
mH mCl
mH mCl
(Equation 9-20)
E
2
J s 4.32 1014 Hz
3 3 10 s
hf / 2
1,
34
3
v 1 / 2 hf
For v = 0, E0
(b) For
2 2 1019 cm3 6.63 10
2 N hf
3 ts
f
6.63 10
2
/I
34
J s 8.66 1013 Hz / 2
h f
6.63 10 34 J s
4 2 6 1011 Hz
2.8 10
is given by Equation 9-17.
0.973u
r
0.132nm
229
47
kg m 2
0.179eV
1273W / cm3
Chapter 9 – Molecular Structure and Spectra
9-54. (a)
dU
dr
U0
12a12 r
For Umin,
dU / dr
(b) For U
U min ,
r
13
6a 6 r
2
0, so
12a12 r
a then U min
(c) From Figure 9-8(b): r0
7
6
U0
12a6
a
a
0
12
0.074nm ( a )
2
U0
r
a
a
6
a
6
1 2 U0
32.8eV
6
r / r0
r0 / r
0.85
7.03
−5.30
+56.7
0.90
3.5
−3.8
−9.8
0.95
1.85
−2.72
−28.5
1.00
1
−2.0
−32.8
1.05
0.56
−1.5
−30.8
1.10
0.32
−1.12
−26.2
1.15
0.19
−0.86
−22.0
1.20
0.11
−0.66
−18.0
2 r0 / r
230
a
6
(d)
12
r
U
U0
Chapter 9 – Molecular Structure and Spectra
ke2
r
9-55. (a) U r
Eex
Eion
(Equation 9-1)
4.27eV and r0
For NaCl, Ed
Eion Na
Eaff Cl
Eex
4.27
ke2
1.52
0.236
Ar
n
0.31eV
0.31eV
9-56. For H
4.27eV
0.31eV
ke2
r02
25.85eV / nm
n A
r0 r0n
25.85eV / nm 0.236nm / 0.31eV
Solving for n: n
Ed
(Equation 9-2)
Following Example 9-2,
A
(Table 9-1).
5.14 3.62 1.52eV and U r0
Eion
(b) Eex
0.236nm
0.236nm
H system, U r
20
8.9 10
ke2
r
14
n
0.31eV
r0
19.7
20
eV nm20
Eion
There is no Eex term, the two electrons of H are in the n = 1 shell with opposite spins.
Eion
ionization energy for H
electron affinity for H = 13.6eV
1.440eV nm
12.85eV
r
U r
dU r
dr
For U r to have a minimum and the ionic H
0.75eV
12.85eV .
1.440
r2
H molecule to be bound, dU / dr
As we see from the derivative, there is no non-zero or finite value of r for which this
occurs.
9-57. (a)
u 35
u 37
35
1.007825u 34.968851u
0.979593u
1.007825u 34.968851u
1.007825u 36.965898u
0.981077u
1.007825u 36.965898u
1.52 10
3
(b) The energy of a transition from one rotational state to another is:
231
0.
Chapter 9 – Molecular Structure and Spectra
(Problem 9-57 continued)
E,
2
1
f
hf
(Equation 9-15)
1 h2
hI
f
4
2
4
2
1 h
2
0
h r
2
2
r
1 h
r02
f
f
4
1.53 10
r02
1
2
0
1 h
4
2
4
1 h
df
d
f
f
(c)
/I
1
1
3
2
1
r02
from part (b). In Figure 9-29 the
f between the 35Cl
lines (the taller ones) and the 37Cl lines is of the order of
0.01 1013 Hz, so f / f
9-58. (a) For CO :
I
r02
m1m2
m1 m2
0.113
6.861u 1.66 10
2
E0 r
r0
0.0012, about 20% smaller than
2I
(b) E
1.055 10
34
2 1.454 10
46
27
J s
0
E1
2E0r
4.78 10 4 eV
E2
6E0r
1.43 10 3 eV
E3
12E0r
2.87 10 3 eV
E4
20E0r
4.78 10 3 eV
E5
30E0r
7.17 10 3 eV
12.0112u 15.9994u
12.0112u 15.9994u
kg / u 0.113 10
9
2
1.454 10
46
2
kg m
2
3.827 10
1 E0 r
E0
/ .
232
23
J
2.39 10 4 eV
6.861u
kg m2
Chapter 9 – Molecular Structure and Spectra
(Problem 9-58 continued)
(c) (See diagram)
E54
7.17 4.78
10 3 eV
2.39 10 3 eV
E43
4.78 2.87
10 3 eV
1.91 10 3 eV
E32
2.87 1.43
10 3 eV
1.44 10 3 eV
E21
1.43 0.48
10 3 eV
0.95 10 3 eV
E10
4.78 10 4 eV
8
7
5
6
Ev
10 3 eV
5
4
4
3
3
2
2
1
1
0
0
hc / E
(d)
54
43
32
21
10
1240eV nm
2.39 10 3 eV
5.19 105 nm
0.519mm
1240eV nm
1.91 10 3 eV
6.49 105 nm
0.649mm
1240eV nm
1.44 10 3 eV
8.61 105 nm
0.861mm
1240eV nm
0.95 10 3 eV
13.05 105 nm 1.31mm
1240eV nm
4.78 10 4 eV
25.9 105 nm
2.59mm
All of these are in the microwave region of the electromagnetic spectrum.
233
Chapter 9 – Molecular Structure and Spectra
n 1, v 1,
#2
9-59.
1
# 3 n 1, v
#1 n 1, v
0,
0
(a) E 1
1
hf H
2 2
1 E0 r
E 2
3
hf H
2 2
2 E0 r since
1
E 3
1
hf H
2 2
6 E0 r since
2
1
hf H since
2 2
A
E 2
E 1
3
hf H 2
2
2 E0 r
B
E 2
E 3
3
hf H
2 2
2 E0 r
Re-writing [A] and [B] with E0r
A1
hf H 2
B1
hf H 2
2
2
0
1
hf H 2
2
h 1.356 1014 Hz
1
hf H
2 2
2
6 E0 r
h 1.246 1014 Hz
/ 2I :
h 1.356 1014 Hz
I
2
0,
I
h 1.246 1014 Hz
Subtracting [B1] from [A1] and cancelling an h from each term gives:
3h 4
I
2
2
4
3h
0.110 1014 Hz
2
0
(b) I
r
r0
I/
r0
0.110 1014 Hz
I
4.58 10
4.58 10
7.40 10 11 m
kg m2
1.007825
For H 2 :
1/ 2
48
2
2 1.007825
48
kg m2
0.503912u
0.5039u 1.66 10
27
kg / u
1/ 2
0.0740nm in agreement with Table 9-7.
Canceling an h from [B1] and substituting the value of I from (a) gives:
2
I 1.246 1014 Hz
f H2
2h 4
f H2
1.32 1014 Hz also in agreement with Table 9-7.
234
2
Chapter 10 – Solid State Physics
10-1. U  r0   
ke2  1 
1
r0  n 
(Equation 10-6)
ke2  1 
E  U  r0   
1
r0  n 
1
1 Ed r0  741kJ / mol  0.257nm  1eV / ion pair



 0.7844
n  ke2
1.7476 1.44eV nm 
96.47kJ / mol
n
1
 4.64
1  0.7844
10-2. The molar volume is
M

 2 N A r03
 M  
74.55 g / mole
r0  

23
3
 2 N A    2 6.022  10 / mole 1.984 g / cm


10-3. The molar volume is

M

1/ 3
 3.15  108 cm  0.315nm
 2 N A r03
M
42.4 g / mole

3
23
2 N A r0 2 6.022  10 / mole 0.257  107 cm
10-4. (a) U att  






ke2
r0
(Equation 10-1)
ke2  1 
1
(b) Ed  U  r0   
r0  n 

3
 2.07 g / cm3
(Equation 10-6)
 1
 8.01eV  1    7.12eV / ion pair
 9
 96.47kJ / mol   1cal 
  7.12eV / ion pair  

  164kcal / mole
 1eV / ion pair   4 / 186 J 
235
Chapter 10 – Solid State Physics
(Problem 10-4 continued)
(c) 1 
1 Ed r0 165.5kcal / mol  0.314nm  4.186 J  1eV / ion pair 



 0.8960
n  ke2
1.7476 1.44eV nm 
1cal  96.47kJ / mol 
Therefore, n 
1
 9.62
1  0.8960
10-5. Cohesive energy (LiBr)


1mol
1eV

= 788  103 J / mol 
 8.182ev / ion pair

23
19 
 6.02  10 ion pairs   1.60  10 
 4.09eV / atom
This is about 32% larger than the value in Table 10-1.
10-6. Molecular weight Na = 22.990
Molecular weight Cl = 35.453
 the NaCl molecule is by weight 0.3934 Na and 0.6066 Cl.
Since the density of NaCl = 2.16 g/cm3, then

mol of Na / cm3   0.3934 2.16 g / cm3
  2.990g / mol   0.03696 mol / cm
3
mol of Cl / cm3  0.03696 mol / cm3 , also
since there is one ion of each element per molecule.
Number of Na ions / cm3  0.03696 N A
Number of Cl ions / cm3  0.03696 N A


Total number of ions / cm3   0.07392  6.022  1023  4.45  1022
Nearest neighbor distance = equilibrium separation r0 .

1
r0  
 4.45  1022 ions / cm3 102 cm / m





3


1/ 3

236
 2.88  1010 m  0.288nm
Chapter 10 – Solid State Physics
10-7.
r0  KCl   0.315nm  3.15  1010 m


N ions / m3  1 / r03  3.20  1028 / m3  3.20  1022 / cm3
Half of the ions in 1cm3 are K and half are Cl , so there are 1.60 1022 / cm3 of each
element.
This number of ions equals:
1.60  1022 ions
1.60  1022 ions

 0.02657 mol
NA
6.022  1023 / mol
This is the moles of each ion in 1 cm3.
Molecular weight of K = 39.102 g/mol.
Molecular weight of Cl = 35.453 g/mol.


Weight of Cl/cm3 =  35.453g / mol   0.02657mol / cm   0.942 g / cm
Weight of K/cm3 =  39.102 g / mol  0.02657mol / cm3  1.039 g / cm3
3
3
 density of KCl  1.039  0.942  g / cm3  1.98 g / cm3
10-8. (a)
8.0
7.0
Cohesive
Energy
(eV)
6.0
5.0
4.0
3.0
2.0
1.0
500
1000
1500
2000
2500
Melting point, K
(b) Noting that the melting points are in kelvins on the graph,
Co melting point = 1768 K, cohesive energy = 5.15 eV
Ag melting point = 1235 K, cohesive energy = 3.65 eV
Na melting point = 371 K, cohesive energy = 1.25 eV
237
Chapter 10 – Solid State Physics
2
2
2
2
2 2
10-9. U att  ke2  





 a 2a 3a 4a 5a 6a
2 1 2 1

U att  ke2  2  1     
3 3 5 3







The quantity in parentheses is the Madelung constant α. The 35th term of the series (2/35)
is approximately 1% of the total, where α = 4.18.
10-10. (a)  

(b)
me v
ne2

(Equation 10-13)
9.1110
8.47  10
28

electrons / m 1.60  10
1/ 2
 100 K 


 300 K 
100  300
19
2
9
7
m
(from Equation 10-9)
1/ 2
100

 1.23  10
C   0.4  10 m 
kg 1.17  105 m / s
3
v   kT / me 
v
31

1
3
3  7.00  108  m


4 103 A
I
I
10-11. (a) j  

A  d 2 / 4  1.63  103 m


2
 479 A / m3
(from Equation 10-10)
I
d
479 A / m2


 3.53  108 m / s
(b) vd 
28
3
19
Ane ne
8.47  10 / m 1.602  10 C



 3.53  106 cm / s
10-12. (a) There are na conduction electrons per unit volume, each occupying a sphere of


volume 4 rs3 3 : na  4 rs3 / 3  1
rs3 
3
4 na

rs   3 / 4 na 
1/ 3
238
Chapter 10 – Solid State Physics
(Problem 10-12 continued)

3
(b) rs  
 4 8.47  1028 / m3





1/ 3
 1.41  1010 m  0.141nm
10-13. (a) n   N A / M for 1 electron/atom
10.5g / cm  6.022 10
n
3
23
/ mole
107.9 g / mole
19.3g / cm  6.022 10
n
3
(b)
23
/ mole
196.97 g / mole
  5.86 10
  5.90 10
22
22
/ cm3
/ cm3
Both agree with the values given in Table 10-3.
10-14. (a) n  2 N A / M for two free electrons/atom


2 1.74 g / cm3 6.022  1023 / mole
n
24.31g / mole


2 7.1g / cm3 6.022  1023 / mole
(b) n 
65.37 g / mole
  8.62  10
  13.110
22
22
/ cm3  8.62  1028 / m3
/ cm3  13.1  1028 / m3
Both are in good agreement with the values in Table 10-3, 8.61 1028 / m3 for Mg and
13.2 1028 / m3 for Zn.
10-15. (a)  
me v
ne2

(Equation 10-13)
1


ne2
me v
(Equation 10-13)
9.1110 kg 1.08 10 m / s 

 1.22  10
8
.
47

10
m
1
.
602

10
C
0
.
37

10
m




31
28

1


3
5
19
2
1
1
 8.17  106   m 
7
1.22  10  m
239
9
7
m
Chapter 10 – Solid State Physics
(Problem 10-15 continued)
(b)
v 
8kT
 me
(Equation 10-9)
  200K     300K  v  200K 
v  300K 
   300K  200K / 300K 
1/ 2


 1.22  107  m  200 K / 300K 
  200 K  
1/ 2
1
1
1

 1.00  107   m 
8
  200 K  9.96  10  m
(c)  100K     300K  v 100K 

v  300K 

 1.22  107  m 100K / 300K 
 100 K  
10-16. E 
3
EF
5
 9.96  108  m
1/ 2
 7.04  108  m
1
1
1

 1.42  107   m 
8
 100 K  7.04  10  m
(Equation 10-22)
(a) for Cu:
E 
3
 7.06eV   4.24eV
5
(b) for Li:
E 
3
 4.77eV   2.86eV
5
EF
 hc 

2
N
 3
 
2 
2mc  8 V V 
1240eV
2/3


3
2
28
3
nm   3 5.90  10 m  109 m  



 
8
2 5.11  105 eV 
 1nm  


240
1/ 3
 5.53eV
Chapter 10 – Solid State Physics
10-17. A long, thin wire can be considered one-dimensional.
 hc   N 
h2  N 
EF 

32m  L 
32mc 2  L 
2
2

2
For Mg: N / L  8.61 1028 / m2

1/ 3
1240eV nm  10 m / nm  8.61 10

32  0.511  10 eV 
2
9
EF
(Equation 10-15)
28
/ m3

2/3
 1.87eV
6
10-18. (a) For Ag:
h 2  3N 
EF 
2m  8 V 
2/3
1240eV nm 10 m / nm 

2  0.511  10 eV 
9
6
2
 3  5.86  1028 m3 


8


2/3
 5.50eV
For Fe: Similarly, EF  11.2eV
(b) For Ag: EF  kTF
TF 
(Equation 10-23)
EF
5.50eV

 6.38  104 K
k
8.617  105 eV / K
For Fe: Similarly, TF  13.0  104 K
Both results are in close agreement with the values given in Table 10-3.
10-19. Note from Figure 10-11 that most of the excited electrons are within about 2kT above the
Fermi energy EF , i.e., E  2kT . Note, too, that kT
EF , so the number N of excited
electrons is: N  N  EF  n  EF  E  N  EF 1 / 2  2kT   N  EF  kT and
8 V  2m 
N
3  h2 
3/ 2
EF3 / 2
(from Equation 10-20)
Differentiating Equation 10-19 gives: N  EF  
 V  8m 
N
2 

Then,
N
8 V
3
3/ 2
1/ 2
 EF kT 3
h 
 kT EF
3/ 2
2
 2m 
3/ 2
 2  EF
h 
2
241
 V  8m 
2  h2 
3/ 2
EF1/ 2
Chapter 10 – Solid State Physics
(Problem 10-19 continued)
EF for Ag  5.35eV , so
1/ 2
 2E 
10-20. uF   F 
 me 
N 3
 8.617  105 eV / K  300 K  5.53eV  0.0070  0.1%
2
N


1/ 2
 2E 
 c  F2 
 me c 
(Equation 10-24)
 2  3.26eV  
(a) for Na: uF  3.00  10 m / s 

5
 5.11  10 eV 


8
1/ 2
 1.07  106 m / s
 2  5.55eV  
(b) for Au: uF  3.00  10 m / s 

5
 5.11  10 eV 


8
 2 10.3eV  
(c) for Sn: uF  3.00  10 m / s 

5
 5.11  10 eV 

10-21.  
meuF
ne2

8

(Equation 10-25)
1/ 2
 1.40  106 m / s
1/ 2
 1.90  106 m / s
meuF
ne2 
9.11 10 kg 1.07 10 m / s 
(a) for Na:  
 2.65 10 m 1.609 10 C   4.2 10
31
28
3
6
19
2
8
m

 3.42 108 m  34.2nm
9.11 10 kg 1.40  10 m / s 
(b) for Au:  
5.90  10 m 1.609  10 C   2.04  10
31
28
3
6
19
2
8
m

 4.14  108 m  41.4nm
9.1110 kg 1.90 10 m / s 

14.8 10 m 1.609 10 C  10.6 10
31
(c) for Sn:
28
3
6
19
 4.31 108 m  43.1nm
242
2
8
m

Chapter 10 – Solid State Physics
10-22. Cv  electrons  
2
Cv  electrons  
2
assuming T
T
2
2
R
T
TF
R
kT
because EF  kTF .
EF
(Equation 10-30)
TE (rule of Dulong and Petit):
0.10  3 2  EF
 k
2

 0.60  7.06eV 
 8.617  10 eV / K 
5
2
Cv due to the lattice vibrations is 3R,
2
2
R
kT
 0.10  3R 
EF
 4.98  103 K
This temperature is much higher than the Einstein temperature for a metal such as copper.
10-23. U 
 kT 
3
NEF   N 
 kT
5
 EF 
(Equation 10-29)
 kT
3
Average energy/electron = U N  EF   
5
 EF

3
 2  kT 
kT

E


F
5
4 EF

2
For copper: E  7.06eV , so
At T = 0K: U N 
3
 7.06eV   4.236eV
5

5
3
 2 8.61  10 eV / K
At T = 300K: U N   7.06eV  
5
4
7.06
 300K 
2
2
 4.236eV
The average energy/electron at 300K is only 0.0002eV larger than at 0K, a consequence of
the fact that 300K is very small compared to the TF for Cu (81,600K). The classical value
of U N   3 / 2 kT  0.039eV , is far too small.
10-24. Cv  electrons  
2
2
R
T
TF
(Equation 10-30)
Melting temperature of Fe = 1811K (from Table 10-1)
TF for Fe  13  104 K (from Table 10-3)
The maximum Cv for the Fe electrons, which is just before Fe melts, is:
243
Chapter 10 – Solid State Physics
(Problem 10-24 continued)
Cv  electrons  
2
 1811K 
R
 0.0219 R
2  13  104 K 
The heat capacity of solids, including Fe, is 3R (rule of Dulong and Petit, see Section 8-1).
Cv  electrons 

Cv
10-25. P 
   M  B



 kT
9.285  10
P
1.38  10
24
23
10-26.  
0.0291R
 0.0073
3R
0 M

B
J /T

(Equation 10-35)
  2.0T   6.7  10
J / K  200 K 
3
0  2
kT
2
 N  1  J   1 
 units   2  3    
 A  m  T   J 


NJ 2
NJ
NJ
NJA2 m2



A2 m3T 2 J A2 m3 Wb / m 2 A2 m3  N / Am 2 A2 m3 N 2
J
Nm

 1 dimensionless
Nm Nm
10-27. E  hc 
(a) For Si:   hc E  1240eV nm 1.14eV  1.088  103 nm  1.09  106 m  1.09  103 nm
(b) For Ge: Similarly,   1.722  103 nm  1.72  106 m  1.72  103 nm
(c) For diamond: Similarly,   1.77  102 nm  1.72 107 m
10-28. (a) For Ge: All visible light frequencies (wavelengths) can excite electrons across the
0.72eV band gap, being absorbed in the process. No visible light will be transmitted
through the crystal.
244
Chapter 10 – Solid State Physics
(Problem 10-28 continued)
(b) For insulator: No visible light will be absorbed by the crystal since no visible
frequencies can excite electrons across the 3.6eV band gap. The crystal will be
transparent to visible light.
(c) Visible light wavelengths are 380-720nm corresponding to photon energies
E  hf  hc /   1240eV nm / 
  380nm  E  3.3eV
  720nm  E  1.7eV
Visible light photons will be absorbed, exciting electrons for band gaps  3.3eV .


10-29. (a) E  hc   1240eV nm 3.35 m  103 nm /  m  0.37eV
(b) E  kT  0.37eV T  0.37eV / k  0.37eV / 8.617  105 eV / K  4300K


2.33g / cm3 100nm  107 cm / nm
mN A VN A
10-30. (a) N 


M
M
28 g / mol
  6.02  10
3
23
/ mol
 5.01107 Si atoms


(b) E  13eV 4  5.01 107  6.5  108 eV
2
1  ke2  m 1
10-31. (a) E1   
 2
2
2
  1

(Equation 10-43)


2
 9  109 N m2 / C 2 1.60  1019 C 2 
31
1
   0.2  9.11  10 kg
E1   
2
2
2
11.8
1.055  1034 J s


 3.12  1021 J  0.0195eV
Ionization energy = 0.0195eV
(b)


r1  a0 1 me m 
2
(Equation 10-44)
 0.0529nm 1 / 0.2 11.8  3.12nm
(c) Eg ( Si )  1.11eV at 293K
245



Chapter 10 – Solid State Physics
(Problem 10-31 continued)
E1 / Eg  0.0195 / 1.11  0.0176 or about 2%.
2
1  ke2  m 1
10-32. (a) E1   
 2
2
2
  1
(Equation 10-43)



2
 9  109 N m2 / C 2 1.60  1019 C 2 
 0.34  9.11 1031 kg
1 

E1  

2
2
2
15.9 
1.055  1034 J s




 2.92  1021 J  0.0182eV
(b)


r1  a0 1 me m 
2
(Equation 10-44)
 0.0529nm 1 / 0.34 15.9   2.48nm
10-33. Electron configuration of Si: 1s 2 2s 2 2 p6 3s 2 3 p 2
(a) Al has a 3s 2 3 p configuration outside the closed n = 2 shell (3 electrons), so a p-type
semiconductor will result.
(b) P has a 3s 2 3 p3 configuration outside the closed n = 2 shell (5 electrons), so an n-type
semiconductor results.
 T  0.01eV / 8.617  105 eV / K  116K
10-34. E  kT  0.01eV
10-35. (a) VH  vd Bw 
dBw iB

nq
qnt
(Equation 10-45 and Example 10-9)
The density of charge carriers n is:
n
(b) N 
 20 A 0.25T 
iB

 7.10  1028 carriers/m3
19
3
6
qtVH
1.60  10 C 0.2  10 m 2.2  10 V

 NA
M


5.75 10 kg / m  6.02 10

3
3
118.7kg / mol
26
/ mol

  2.92 10
28
Each Sn atom contributes n / N  7.10  1028 / 2.92  1028  2.4 charge carriers
246
Chapter 10 – Solid State Physics


10-36. I net  I 0 eeVb / kT  1
(Equation 10-49)
(a) eeVb / kT  10, so eVb / kT  ln 10 . Therefore,
1.60 10 C V 1.38110 J / K  300K   ln 10
V  1.38  10 J / K   300K  ln 10  1.60  10 C   0.0596V  59.6mV
19
23
b
23
19
b
(b) eeVb / kT  0.1
Vb  0.0596V ln 0.1 / ln10   0.0596  59.6mV


10-37. I net  I 0 eeVb / kT  1
(Equation 10-49)
(a) eeVb / kT  5, so eVb / kT  ln  5. Therefore,
Vb 
kT ln  5
1.38  10 J / K   200K  ln 5  0.0278V  27.8mV

1.60  10 C 
23
19
e
(b) eeVb / kT  0.5
eVb / kT  ln  0.5
Vb 
kT ln  0.5

1.38 10 J / K   200K  ln  0.5  0.0120V  12.0mV

1.60  10 C 
23
19
e

10-38. I net  I 0 eeVb / kT  1
(Equation 10-49)
Assuming T = 300K,
I  0.2V   I  0.1V 
I  0.1V 


I0 e 
e 0.2V  / kT

 
 1  I0 e 
I0 e 
e 0.1V  / kT
e 0.1V  / kT

1
e
1
e 0.2V  / kT

e
e 
e 0.1V  / kT
e 0.1V  / kT

1
 47.7
10-39. (a) From Equation 10-49, exp  eVb / kT   10
Taking ln of both sides and solving for Vb,
1.38 10 J / K   77 K  ln 10  0.0153 volts  15.3mV
  kT / e  ln 10  
1.60  10 C 
23
Vb
19
247
Chapter 10 – Solid State Physics
(Problem 10-39 continued)
(b) Similarly, for exp  eVb / kT   1; Vb  0


 I net =1mA 10  1  9mA
(c) For (a): I net  I 0 eeVb / kT  1
For (b): I net  0
10-40. E  hc / 
For   484nm
E  Egap
Egap  hc / 484nm  1240eV nm / 484nm  2.56eV
10-41. M  Tc  constant
(Equation 10-55)
First, we find the constant for Pb using the mass of natural Pb from Appendix A,
Tc for Pb from Table 10-6, and α for Pb from Table 10-7.
constant   207.19u 
0.49
 7.196K   98.20
For
206
Pb : Tc  constant / M   98.20  205.974u 
For
207
Pb : Tc  constant / M   98.20  206.976u 
For
208
Pb : Tc  constant / M   98.20  207.977u 
10-42. (a) Eg  3.5kTc
0.49
 7.217 K
0.49
 7.200K
0.49
 7.183K
(Equation 10-56)


Tc for I is 3.408K, so, Eg  3.5 8.617  105 eV / K  3.408K   1.028  103 eV
(b) Eg  hc / 
  hc / Eg  1240eV nm / 1.028  103 eV  1.206  106 nm
 1.206  103 m  1.206mm
10-43. (a) Eg  3.5kTc

For Sn : Tc  3.722K

Eg  3.5 8.617  105 eV / K  3.722 K   0.0011eV
248
Chapter 10 – Solid State Physics
(Problem 10-43 continued)
(b) Eg  hc / 
  hc / Eg  1240eV nm / 6  104 eV  2.07  106 nm  2.07  103 m
Eg  T  / Eg  0   0.95 where Eg  0   3.5kTc (Equation 10-56)
10-44. At T / Tc  0.5
So, Eg T   0.95  3.5 kTc  3.325kTc
(a) For Sn :
(b) For Nb :
(c) For Al :
(d) For Zn :


E T   3.325 8.617  10 eV / K   9.25K   2.65  10
E T   3.325 8.617  10 eV / K  1.175K   3.37  10
E T   3.325 8.617  10 eV / K   0.85K   2.44  10
Eg T   3.325 8.617  105 eV / K  3.722K   1.07  103 eV
5
3
g
5
4
g
5
g
10-45. BC T  BC  0   1  T / TC 
2
4
eV
eV
2
(a) BC T  BC  0   0.1  1  T / TC 
T / TC 
eV
2
 1  0.1  0.9
T / TC  0.95
(b) Similarly, for BC T  BC  0   0.5
T / Tc  0.71
(c) Similarly, for BC T  BC  0   0.9
T / Tc  0.32
10-46. 1. Referring to Figure 10-56, notice that the length of the diagonal of a face of the cube is
r; therefore, a 2  a 2  (4r )2  a  8 r
2. Volume of the unit cube Vcube  a3  83/2 r 3
249
Chapter 10 – Solid State Physics
(Problem 10-46 continued)
3. The cube has 8 corners, each occupied by ⅛ of an atom’s volume. The cube has 6
faces, each occupied by ½ of an atom’s volume. The total number of atoms in the unit
cube is then 8×⅛ + 6×½ = 4. Each atom has volume 4 r 3 / 3 . The total volume
occupied by atoms is then Vatoms  4  (4 r 3 / 3) .
4. The fraction of the cube’s volume occupied by atoms is then:
Vatoms 4  (4 r 3 )

 0.74
Vcube
83/2 r 3
10-47. TF for Cu is 81,700K, so only those electrons within EF  kT of the Fermi energy could
be in states above the Fermi level. The fraction f excited above EF is approximately:
f  kT / EF  T / TF
(a) f  300K / 81, 700K  3.7 103
(b) f  1000K / 81, 700K  12.2 103
10-48.
−e
●
−6
−e
●
−4
+e
●
−5
−e
●
−2
+e
●
−3
−e
●
0
+e
●
−1
+e
●
1
−e
●
2
+e
●
3
−e
●
4
+e
●
5
−e
●
6
r0
(a) For the negative ion at the origin (position 0) the attractive potential energy is:
2ke2  1 1 1 1 1
V 
1     
r0  2 3 4 5 6
(b) V  
ke2
, so the Madelung constant is:
r0


1
1
1
1
1
  2 1      
2 3 4 5 6



250



Chapter 10 – Solid State Physics
(Problem 10-48 continued)
Noting that ln 1  x   x 
x 2 x3 x 4
  
2
3
4
ln  2   1 
,
1 1 1
  
2 3 4
and   2 ln  2   1.386
10-49. Cv 
TF 
2
R
2
2
R
2
T
TF

(Equation 10-30)
T
3.74  10 J / mol K T
and EF  kTF 


4
2
2
R

kT
3.74  104 J / mol K T

 2 8.314 J / mol K  1.38  1023 J / K 

2 3.74  104 J / mol K



 1.51 1018 J 1 / 1.60  1019 J / eV  9.45eV
10-50. (a) N 
EF
EF
 g  E  dE   AE
0
(b) N  
1/ 2
dE  A  2 / 3 E
EF
0
EF

EF  kT
  2 A / 3 EF3 / 2
3/ 2
0
AE1 / 2 dE   2 A / 3  EF3 / 2   EF  kT 

3/ 2
  2 A / 3  EF3 / 2  EF3 / 2 1  kT / EF 

3/ 2
Because kT




EF for most metals,
1  kT / EF 
3/ 2
 1   3 / 2  kTEF1 / 2
N    2 A / 3  EF3 / 2  EF3 / 2   3 / 2  kTEF1 / 2   AkTEF1 / 2
AkTEF1 / 2
N
3kT


The fraction within kT of EF is then f 
3/ 2
N  2 A / 3  EF
2 EF
(c) For Cu EF  7.04eV ; at 300K ,
f 
3  0.02585eV 
2  7.04eV 
251
 0.0055
Chapter 10 – Solid State Physics

  0.72eV / pair   1.63 10
 1.33  10 eV / photon   0.72eV / pair   1.85  10
10-51. (a) N1  1.17  106 eV / photon
N2
6
6
pairs / photon
6
pairs / photon
(b) N1  N1  1.23  103  N1 / N1  7.8  104
N2  N2  1.36  103  N2 / N2  7.4  104
(c) Energy resolution
E N

E
N
 E1 / E1  0.078%
E2 / E2  0.074%
10-52.  
meff v
(Equation 10-13)
n e2

v  3kT / meff

1/ 2


 3 1.38  1023 J / K  300 K  


0.2 9.11  1031 kg



1/ 2

 2.61  105 m / s
Substituting into λ:



0.2 9.11  1031 kg 2.61  105 m / s
10
22
m
3
5  10
3

 m 1.60  10
19

C

2
 3.7  108 m  37nm
 2  7.06eV  


31
 9.11  10 kg 
1/ 2
For Cu: uF   2 EF / me 
1/ 2
 1.57  106 m / s
n  8.47  1028 m3 and   1.7  108  m (Example 10-5)
Substituting as above,   3.9 108 m  39nm
The mean free paths are approximately equal.
252
Chapter 10 – Solid State Physics
10-53. Compute the density of Cu (1) as if it were an fcc crystal and (2) as if it were a bcc crystal.
The result closest to the actual measured density is the likely crystalline form.
1. The unit cube of an fcc crystal of length a on each side composed of hard, spherical
atoms each of radius r contains 4 atoms. The side and radius are related by a  8 r
and the volume of the cube is Vcube  a3  83/2 r 3 . (See problem 10-46 and
Appendix B3.) The density of fcc Cu would be:
mCu 4( M / N A )

Vcube
83/2 r 3
where M = molar mass (atomic weight) and N A = Avagadro’s number.
d
d
4(63.55g/mol)
 8.90g/cm3
8
3
23
8 (1.28 10 cm) (6.022 10 / mol)
3/2
2. For a bcc crystal (refer to Appendix B3), let the length of a side = a, the diagonal of a
face = f, and the diagonal through the body = b. Then from geometry we have
a 2  f 2  b2  (4r )2 and a 2  a 2  f 2 . Combing these yields:
4r
3
The unit cube has 8 corners with ⅛ of a Cu atom at each corner plus 1 Cu atom at the
a 2  f 2  3a 2  (4r )2  a 
center, or 8×⅛ + 1 = 2 atoms. The density of bcc Cu is then:
d
2( M / N A )
m
2(63.55g/mol)  33/2


 8.17 g/cm3
3
8
3
23
3
Vcube (4r / 3)
4 (1.28 10 cm) (6.022 10 /mol)
Based on the above density calculations, metallic Cu is most likely a face-centered
cubic crystal.
253
Chapter 10 – Solid State Physics
10-54. (a) For small Vb (from Equation 10-49)
eeVb / kT  1  eVb / kT , so I  I 0eVb / kT  Vb / R
R  Vb
 I0eVb / kT   kT


eI 0  0.025eV e  109 A  25.0M 
(b) For Vb  0.5V ;
R  Vb I  0.5 109 A  500M 
(c) For Vb  0.5V ;
I  109 A e0.5 / 0.025  1  0.485 A


Thus, R  Vb I  0.5 / 0.485  1.03
(d)
10-55. a0 
eI 0 eVb / kT
dI

e
dVb kT 
dVb kT  eVb / kT

e
 25M e20  0.0515
dI
eI 0
Rac 
 0 h2
 o h2
 2


 me e2   me eff e2  me eff ke2
silicon: a0 

kg  9  10 N m

/ C 1.602  10
12 1.055  1034 J s

0.2 9.11  1031
9
2
2
2
19
C

2
 3.17 109 m  3.17nm
This is about 14 times the lattice spacing in silicon (0.235nm)
germanium: : a0 

kg  9  10 N m
16 1.055  1034 J s

0.10 9.11  1031
9
2

2

/ C 2 1.602  1019 C

 8.46  109 m  8.46nm
This is nearly 35 times the lattice spacing in germanium (0.243nm)
2
1  ke2  m 1
10-56. (a) En   
 2 2
2
  n
(Equation 10-44)

6
2
2
1  1.440eV nm   0.015 0.511  10 eV / c

 
2  6.582  1016 eV s  
182

or
254
  1
 n
2
2
Chapter 10 – Solid State Physics
(Problem 10-56 continued)


9
2
2
19

1 2 9  10 N m / C 1.60  10 C
 
2
6.63  1034 J s



2




2

 0.015 9.11  1031 kg

182

  1
 n
2
1.01  1022
J  6.28  104 eV / n2
n2
 donor ionization energy = 6.28  104 eV
(b) r1  a0
me

m
(Equation 10-45)
 0.0529nm 1 / 0.01518
r1  63.5nm
(c) Donor atom ground states will begin to overlap when atoms are 2r1  127nm apart.
3
 109 nm / m 
20
3
donor atom concentration  
  4.88  10 m
 127nm / atom 
10-57. (a)  
meuF
ne2
(Equation 10-25)
So the equation in the problem can be written as   m  i . Because the impurity
increases m by 1.1 108  m, i  1.1 108  m and
i 

meuF
8
ne 1.18  10  m
2
and uF   2EF / me 
1/ 2

where n  8.47  1028 electrons / m3 (from Table 10-3)
 1.57  106 m / s. Therefore,
9.11 10 kg 1.57  10 m / s 

8.47  10 / m 1.60  10 C  1.1 10
31
i
28
6
19
3
(b)   1 / na r 2 and d  2r
2
8
m

 6.00  108 m  60.0nm
(Equation 10-12)
So we have d 2  4 / nii where ni  1% of n  8.47 1026 / m3

 

d 2  4 / 8.47  1028 / m3  6.60  108 m  2.28  1022 m2  d  0.0151nm
255
Chapter 10 – Solid State Physics
10-58. (a) The modified Schrödinger equation is:

d  2 dR  r    ke2

r
  
2mr 2 dr 
dr   r
2

 1 
 R  r   ER  r 
2m r 
2
 2
The solution of this equation, as indicated following Equation 7-24, leads to solutions
of the form: Rn  r   a0 e r / a0 n r 1Ln  r / a0  , where a0 

 2 / ke2m 
2
(b) By substitution into Equation 7-25, the allowed energies are:
2
E1
1  ke2  m
1  ke2  
En   


where
E



 m
1
2    n2
n2
2  
(c) For As electrons in Si: m  0.2me (see Problem 10-31) and   Si   11.8,



2
 9  109 N m2 / C 2 1.60  1019 C 2 
31
1 
 0.2 9.11  10 kg
E1  

2
2
2
11.8
1.055  1034 J s


 3.12  1021 J  0.0195eV


Energy  102 eV
n
0
5
4
1.0
3
2
2.0
3.0
4.0
10-59. U  
F 
ke2
r0
 r 1  r n 
 0  0 
 r n  r  
(Equation 10-5)
 n  1 ke
dU
  Kr yields K  
dr
r03
2
(a) For NaCl:   1.7476, n  9.35, and r0  0.282nm and
256
1
Chapter 10 – Solid State Physics
(Problem 10-59 continued)

m  Na  m  Cl 
 22.99u  35.45u   13.95u
m  Na   m  Cl   22.99u    35.45u 

2
K
1    n  1 ke 



M 2  13.95u  r03 
1/ 2
1
f 
2


9
2
2
19

1  1.7476  8.35 9  10 N m / C 1.60  10 C

3
27
9
2 
13
.
95
u

1
.
66

10
kg
/
u
0
.
282

10
m






2
1/ 2




 1.28  1013 Hz

(b)   c / f  .00  108 m / s / 1.28  1013 Hz  23.4 m
This is of the same order of magnitude as the wavelength of the infrared absorption
bands in NaCl.
10-60. (a) Electron drift speed is reached for:
dv
 0  vd  eE / m
dt
(b) Writing Ohm’s law as j   E and j  vd ne (from Equation 10-11)
j  eE ne / m  E ne2 / m, which satisfies Ohm’s law because j  E. Thus,
   ne2 / m and   1 /   m /  ne2.
10-61. (a) For r, s, and t all even  1

r  s t
 1 and the ion’s charge at that location is:

1 1.60  1019 C  1.60  1019 C.
Similarly, for any permutation of
r, s even; t odd:  1
r  s t
 1, ion charge = 1.60  1019 C.
r even; s, t odd:  1
r  s t
 1, ion charge = 1.60  1019 C.
r, s, and t all odd:  1
r  s t
 1, ion charge = 1.60  1019 C.
257
Chapter 10 – Solid State Physics
(Problem 10-61 continued)
(b) U  
ke2
r
If the interatomic distance r = a, then a cube 2a on each side
4
4
2
4
4
4
4 
U  ke2  
 




2a a
2a
2a
3a
3a 
a
U 
ke2
 2.1335 where   2.1335.
a
Similarly, for larger cubes (using spreadsheet). The value of α is approaching 1.7476
slowly.
10-62. (a) M        
M


M


    

e B / kT  e  B / kT
  tanh   B / kT 
e B / kT  e  B / kT
and M   tanh   B / kT 
(b) For  B

0 M
B
kT , T

0 and tanh   B / kT    B / kT
0  B
BkT

0  2
kT
258
Chapter 11 – Nuclear Physics
11-1.
Isotope
Protons
Neutrons
F
9
9
Na
11
14
V
23
28
Kr
36
48
120
Te
52
68
148
Dy
66
82
175
W
74
101
222
Rn
86
136
18
25
51
84
11-2. The momentum of an electron confined within the nucleus is:
p  / x  1.055  1034 J s / 1014 m

 1.055  1020 J s / m  1 / 1.602  1013 J / MeV

 6.59  108 MeV s / m
The momentum must be at least as large as p, so pmin  6.59  108 MeV s / m and the
electron’s kinetic energy is:



Emin  pminc  6.59  108 MeV s / m 3.00  108 m / s  19.8MeV .
This is twenty times the observed maximum beta decay energy, precluding the existence
of electrons in the nucleus.
11-3. A proton-electron model of 6 Li would consist of 6 protons and 3 electrons. Protons and
electrons are spin −1/2 (Fermi-Dirac) particles. The minimum spin for these particles in
the lowest available energy states is 1 / 2 , so 6 Li  S  0  cannot have such a structure.
259
Chapter 11 – Nuclear Physics
14
11-4. A proton-electron model of
N would have 14 protons and 7 electrons. All are Fermi-
Dirac spin-1/2 particles. In the ground state the proton magnetic moments would add to a
small fraction of the proton magnetic moment of 2.8 N , but the unpaired electron would
give the system a magnetic moment of the order of that of an electron, about 1  B .
Because  B is approximately 2000 times large than  N , the
14
N magnetic moment would
be about 1000 times the observed value, arguing against the existence of electrons in the
nucleus.
11-5. The two proton spins would be antiparallel in the ground state with S  1/ 2  1/ 2  0. So
the deuteron spin would be due to the electron and equal to 1 / 2 . Similarly, the proton
magnetic moments would add to zero and the deuteron’s magnetic moment would be 1  B .
From Table 11-1, the observed spin is 1
(rather than 1 / 2 found above) and the
magnetic moment is 0.857  N , about 2000 times smaller than the value predicted by the
proton-electron model.
11-6.
Isotopes
(a)
18
(b)
208
(c)
120
17
F
Pb
206
Sn
119
F
Isotones
19
Pb
210
Sn
118
F
16
Pb
207
Sn
121
N
17
O
Tl
209
Sb
122
Bi
Te
11-7.
Nuclide
(a)
14
8
O6
(b)
63
28
(c)
236
93
Isobars
14
6
C8
Ni35
63
29
Np143
236
92
Cu34
U144
260
Isotopes
14
7
63
30
236
94
N7
Zn33
Pu142
16
8
O8
60
28
235
93
Ni32
Np142
Chapter 11 – Nuclear Physics
11-8.

mass  Au  A 1.66  1027 kg / u


volume   4 / 3  R 3   4 / 3  R0 A1 / 3
where R0  1.2 fm  1.2  1015 m


A 1.66  1027 kg / u
mass
density =

 2.29  1017 kg / m3
3

15
volume  4 / 3  1.2  10 m A
11-9.


B  ZM H c2  NmN c2  M Ac2
(a)
9
4
Be5
(Equation 11-11)

 

  0.062443uc  931.5MeV / uc 
B  4 1.007825uc 2  5 1.008665uc 2  9.012182uc 2
 0.062443uc 2
2
2
 58.2MeV
B / A  58.2MeV / 9 nucleons  6.46MeV / nucleon
(b)
13
6
C7

 

  0.104250uc  931.5MeV / uc 
B  6 1.007825uc 2  7 1.008665uc 2  13.003355uc 2
 0.104250uc 2
2
2
 91.1MeV
B / A  91.1MeV / 13 nucleons  7.47 MeV / nucleon
57
26
(c)
Fe31




B  26 1.007825uc 2  31 1.008665uc 2  56.935396uc 2


 0.536669uc 2  0.536669uc 2 931.5MeV / uc 2
 499.9MeV
B / A  499.9MeV / 57 nucleons  8.77 MeV / nucleon
11-10. R  R0 A1 / 3 where R0  1.2 fm
O  R  1.2 fm 16 
1/ 3
(a)
16
(b)
56
(c)
197
(d)
238
(Equation 11-3)
 3.02 fm
Fe  R  1.2 fm  56 
1/ 3
 4.58 fm
Au  R  1.2 fm 197 
 6.97 fm
U  R  1.2 fm  238
 7.42 fm
1/ 3
1/ 3
261


3
Chapter 11 – Nuclear Physics

11-11. (a) B  M
3

He c 2  mn c 2  M

4

He c 2
 3.016029uc2  1.008665uc2  4.002602uc2


 0.022092uc 2 931.5MeV / uc 2  20.6MeV
(b) B  M
 Li  c
6
2
 Li  c
 mn c 2  M
7
2
 6.015121uc2  1.008665uc2  7.016003uc2


 0.007783uc 2 931.5MeV / uc 2  7.25MeV
(c) B  M
 N c
13
2
 mn c 2  M
 N c
14
2
 13.005738uc2  1.008665uc2  14.003074uc2


 0.011329uc 2 931.5MeV / uc 2  10.6MeV
11-12. B  a1 A  a2 A2 / 3  a3 Z 2 A1 / 3  a4  A  2Z  A1  a5 A1 / 2  c 2


2
(This is Equation 11-13
on the Web page www.whfreeman.com/tiplermodernphysics6e.) The values of the ai in
MeV/c2 are given in Table 11-3 (also on the Web page).
For 23Na:
B  15.67  23  17.23  23

 184.9MeV
M

2/3
 0.75 11  23
2

Na c 2  11m p c 2  12mnc 2  B
23


1 / 3
 93.2  23  2  11  23  0  23
2
1
(Equation 11-14 on the Web page)


 11 1.007825uc 2  12 1.008665uc 2   184.9 MeV
M



Na  23.190055u  184.9MeV / c 2 1 / 931.5MeV / c 2 u
23

 23.190055u  0.198499u  22.991156u
This result differs from the measured value of 22.989767u by only 0.008%.
11-13. R  1.07  0.02  A1 / 3 fm (Equation 11-5)
(a)
16
(b)
63
O:
Cu :
R  1.4 A1 / 3 fm (Equation 11-7)
R  1.07 A1 / 3  2.70 fm and R  1.4 A1 / 3  3.53 fm
R  1.07 A1 / 3  4.26 fm and R  1.4 A1 / 3  5.57 fm
262
1 / 2
 c2

Chapter 11 – Nuclear Physics
(Problem 11-13 continued)
(c)
208
11-14. U 
R  1.07 A1 / 3  6.34 fm and R  1.4 A1 / 3  8.30 fm
Pb :

3 1 e2 2
2
Z   Z  1
5 4 0 R

(Equation 11-2)
where Z = 20 for Ca and U  5.49MeV from a table of isotopes (e.g., Table of Isotopes
8th ed., Firestone, et al, Wiley 1998).

3 1
e2
2
R
Z 2   Z  1
5 4 0 RU





 0.6 8.99  109 N m2 / C 2 1.60  1019 C 2.02  192 / 5.49  106 eV
 6.13  1015 m  6.13 fm
11-15. (a) R  R0et  R0e
at t  0 :
ln 2t / t1 / 2
(Equation 11-19)
R  R0  4000 counts / s
at t  10s :
R  R0e
ln 210 s  / t1 / 2
1000  4000e
1/ 4  e
 ln 2 10 s  / t1 / 2
 ln 2 10 s  / t1 / 2
ln 1 / 4    ln 2 10s  / t1/ 2  t1/ 2  5.0s
(b) at t  20s :
11-16. R  R0e
ln 2t / 2 min
R   4000counts / s  e
at t  0 :
ln 2  20 s  / 5 s
 200counts / s
R  R0  2000counts / s
(a) at t  4 min :
R   2000counts / s  e
(b) at t  6 min :
R   2000counts / s  e
(c) at t  8 min :
R   2000counts / s  e
 ln 2  4 min  / 2 min
 500counts / s
ln 2  6 min  / 2 min
 250counts / s
ln 2  8 min  / 2 min
 125counts / s
263

Chapter 11 – Nuclear Physics
11-17. R  R0et  R0eln 2t / t1/ 2
(a) at t  0 :
(Equation 11-19)
R  R0  115.0 decays / s
at t  2.25h :
R  85.2 decays / s
85.2 decays / s  115.0 decays / s  e
   2.25 h 
85.2 / 115.0   e  2.25h
ln  85.2 / 115.0     2.25h 
   ln 85.2 / 115.0  / 2.25h  0.133h1
t1 / 2  ln 2 /   ln 2 / 0.133h1  5.21h
(b)
dN
 N
dt

dN0
 R0   N0
dt
(from Equation 11-17)


N0  R0 /   15.0 atoms / s  / 0.133h 1 1h / 3600s 
 3.11  106 atoms
11-18. (a)
226
t1 / 2  1620 y
Ra
R
dN
ln 2
ln 2 N A m
 N 
N
dt
t1 / 2
t1 / 2 M



ln 2 6.022  1023 / mole 1g 
1620 y   3.16  10
7

s / y  26.025 g / mole 
1Ci  3.7  1010 s 1, or nearly the same.
(b) Q  M

226

Ra c 2   M

222

Rn c 2  M

4

He c 2 
 226.025402uc 2   222.017571uc 2  4.002602uc 2 


 0.005229uc 2  0.005229uc 2 931.5MeV / uc 2
 4.87MeV
264

 3.61  1010 s 1
Chapter 11 – Nuclear Physics
11-19. (a) R  
dN
 ln 2 t / t
 R0e   1 / 2
dt
when t  0,
R  R0  8000 counts / s
when t  10 min,
e
(from Equation 11-19)
R  1000 counts / s  8000 counts / s  e
10 ln 2  / t1 / 2
10 ln 2  / t1 / 2
 1000 / 8000  1 / 8
10 ln( 2) / t1 / 2  ln 1 / 8 
t1 / 2 
10 ln  2 
ln 1 / 8
 3.33 min
Notice that this time interval equals three half-lives.
(b)   ln  2  / t1 / 2  ln  2  / 3.33 min = 0.208 min-1
(c) R  R0e
 t  ln 2  / t1 / 2
 R0et Thus, R  8000 counts / s  e
0.2081
 6500 counts / s
11-20. (a) and (b)
(c) Estimating from the graph, the next count (at 8 min) will be approximately 220
counts.
265
Chapter 11 – Nuclear Physics
11-21.
62
Cu is produced at a constant rate R0 , so the number of 62Cu atoms present is:


N  R0 /  1  e t (from Equation 11-26). Assuming there were no 62Cu atoms initially
present. The maximum value N can have is R0 /   N 0 ,

N  N 0 1  e  t

0.90 N 0  N 0 1  e
e
 t  ln 2  / t1 / 2

 t  ln 2  / t1 / 2

 1  0.90  0.10
t ln  2  / t1 / 2  ln  0.10 
t  10 ln  0.10  / ln  2   33.2 min
11-22. (a) t1 / 2  ln  2  /   ln  2  / 9.8  1010 y 1  7.07  108 y (Equation 11-22)
(b) Number of
235
U atoms present is:



106 g 6.02  1023 atoms / mol
1.0 gN A
N

 2.56  1015 atoms
M
235 g / mol



dN
  N  9.8  1010 y 1 1 / 3.16  107 s / y 2.56  1015 atoms
dt
 0.079 decays / s
(c) N  N0et
(Equation 11-18)


N  2.56  1015 e


 9.81010 y 1 106 y
  2.558 1015
11-23. (a) t1 / 2  ln  2  /   ln  2  / 0.266 y 1  2.61y (Equation 11-22)
(b) Number of N atoms in 1 g is:


23
1.0 gN A 1g  6.02  10 atoms / mol
N

 2.74  1022 atoms
M
22 g / mol




dN
  N  0.266 y 1 1 / 3.16  107 s / y 2.74  1022 atoms
dt
 2.3  1014 decays / s  2.3  1014 Bq
266


(Equation 11-17)
Chapter 11 – Nuclear Physics
(Problem 11-23 continued)
(c) 
dN
  N 0 e  t
dt
(Equation 11-19)




 0.266 y 1 1 / 3.16  107 s / y 2.74  1022 e
 0.266 y  3.5 y 
 9.1  1013 decays / s  9.1  1013 Bq
(d) N  N0et
(Equation 11-18)


N  2.74  1022 e
11-24. (a)
22


 0.266 y 1  3.5 y 
 1.08  1022
Na has an excess of protons compared with 23 Na and would be expected to decay
by β+ emission and/or electron capture. (It does both.)
(b)
24
Na has an excess of neutrons compared with 23 Na and would be expected to decay
by β− emission. (It does.)
11-25. logt1 / 2  AE1 / 2  B
for t1 / 2  1010 s,
for t1 / 2  1s,


(Equation 11-30)
E  5.4MeV
E  7.0MeV
log 1010  A  5.4 
1 / 2
from Figure 11-16
B
(i) 10  0.4303A  B
log 1  A  7.0 
1 / 2
B
(ii) 0  0.3780 A  B  B  0.3780 A
Substituting (ii) into (i),
10  0.4303A  0.3780 A  0.0523 A,
267
A  191, B  0.3780 A  72.2
Chapter 11 – Nuclear Physics
11-26.
237
Np
144
233
P
142
a
229
140
A
c
229
T
h
138
225
R
a
225
136
Ac
221
F
134
r
N
217
A
132
c
213
217
i
n
B
130
209
Tl
R
213
P
128
o
209
P
126
b
80
82
84
86
88
92
Z
11-27.
232
90
Th 
QM

228
88
Ra  

Th c 2  M
232

228

Ra c 2  M

4

He c 2
 232.038051uc2  228.031064uc2  4.002602uc2


 0.004385uc2 931.50MeV / uc 2  4.085MeV
The decay is a 2-particle decay so the Ra nucleus and the α have equal and opposite
momenta.
  2m E  Ra  2mRa ERa where E  ERa  4.085MeV
2m E  2M Ra ERa  2M Ra  4.085  E 
E 
228.031064  4.085MeV 
M Ra
  0.983 4.085MeV   4.01MeV
 4.085MeV  
M Ra  m
228.031064  4.002602
268
Chapter 11 – Nuclear Physics
11-28.
7
4
Be3  73 Li4  ve
(a) Yes, the decay would be altered. Under very high pressure the electrons are
“squeezed” closer to the nucleus. The probability density of the electrons,
particularly the K electrons, is increased near the nucleus making electron capture
more likely, thus decreasing the half-life.
(b) Yes, the decay would be altered. Stripping all four electrons from the atom renders
electron capture impossible, lengthening the half-life to infinity.
11-29.
67
E .C .
Ga 
QM

67
Zn  ve

Ga c 2  M
67

67

Zn c 2
 66.9282uc2  66.972129uc2


 0.001075uc2 931.50MeV / uc 2  1.00MeV
11-30.
72
Zn  72 Ga     ve
QM

72

Zn c 2  M

72

Ga c 2
 71.926858uc2  71.926367uc2


 0.000491uc2 931.50MeV / uc 2  0.457MeV  457keV
11-31.
233
Np 
232
Np  n and
For n emission: Q  M

233
233
Np 
232
Up

Np c 2  M

232

Np c 2  mnc 2
 233.040805uc2  232.040022uc2  1.008665uc2
 0.007882uc2
Q  0 means M  products   M 
For p emission: Q  M

233

Np c 2  M

233


U c2  mnc 2
232
 233.040805uc2  232.037131uc2  1.008665uc2
 0.004991uc2
269

Np ; prohibited by conservation of energy.
Chapter 11 – Nuclear Physics
(Problem 11-31 continued)
Q  0 means M  products   M 
233


Np ; prohibited by conservation of energy.
11-32.
286
247
280
235
174
124
80
61
30
286
−
280
6
−
247
39
33
−
235
50
45
12
−
174
112
106
73
61
−
124
162
156
123
111
50
−
80
206
200
167
155
94
44
−
61
225
219
186
174
113
63
19
−
30
256
250
217
205
144
91
50
31
−
0
286
280
247
235
174
124
80
61
30
0
−
Tabulated γ energies are in keV. Higher energy α levels in Figure 11-19 would add
additional columns of γ rays.
11-33. 8 Be  2
QM
 Be c
8
2
M

4

He c 2
 8.005304uc2  2  4.002602  uc 2


 0.000100uc2 931.50MeV / uc2  0.093MeV  93keV
11-34.
80
Br  80 Kr     ve and
For   decay: Q  M

80
80
Br 

Br c 2  M
80

Se     ve and
80
80
E.C .
Br 

Kr c 2
 79.918528uc2  79.916377uc2


 0.002151uc2 931.50MeV / uc 2  2.00MeV
270
80
Se  ve
Chapter 11 – Nuclear Physics
(Problem 11-34 continued)
For   decay: Q  M

80

Br c 2  M

80

Se c2  2mec 2
 79.918528uc2  79.916519uc2  2  0.511MeV 


 0.002009uc2 931.50MeV / uc 2  1.022MeV  0.85MeV
For E.C.: Q  M

80

Br c 2  M

80

Se c 2
 79.918528uc2  79.916519uc2


 0.002009uc2 931.50MeV / uc 2  1.87MeV
11-35. R  R0 A1 / 3 where R0  1.2 fm
For 12C : R  1.2 12 
1/ 3
(Equation 11-3)
 2.745 fm  2.745 1015 m and the diameter = 5.490 10 15 m


9.00  109 1.6  1019 C
ke2
Coulomb force: FC  2 
2
r
5.490  1015 m
Gravitational force: FG  G

m2p
r
2



2

 7.65 N
6.67  1011 1.67  1027 kg
5.490 10
15
m


2
 6.18  1036 N
2

The corresponding Coulomb potential is: UC  FC  r  7.65N 5.490  1015 m

 4.20  1014 J 1.60 1013 J / MeV
 0.26MeV
The corresponding gravitational potential is:


UG  FG  r  6.18  1036 N 5.490  1015 m



 3.39  1050 J 1 / 1.60 1013 J / MeV  2.12 1037 MeV
The nuclear attractive potential exceeds the Coulomb repulsive potential by a large margin
(50MeV to 0.26MeV) at this separation. The gravitational potential is not a factor in
nuclear structure.
271
Chapter 11 – Nuclear Physics
11-36. The range R of a force mediated by an exchange particle of mass m is:
R  / mc
(Equation 11-50)
mc2  c / R  197.3MeV fm / 5 fm  39.5MeV
m  39.5MeV / c2
11-37. The range R of a force mediated by an exchange particle of mass m is:
R  / mc
(Equation 11-50)
mc2  c / R  197.3MeV fm / 0.25 fm  789MeV
m  789MeV / c 2
11-38.
Nuclide
Last proton(s)
Last neutron(s)
 2s1 / 2 
0
1/2
1d3 / 2 
2
3/2
1
3/2
3
7/2
Si15
1d5 / 2 
Cl20
1d3 / 2 
71
31
Ga40
 2 p3 / 2 
3
 2 p1 / 2 
59
27
Co32
1 f7 / 2 
7
 2 p3 / 2 
73
32
Ge41
 2 p3 / 2 
29
14
37
17
6
j
4
2
4
4
1g9 / 2 
4
9/2
2
3/2
4
9/2
33
16 17
S
 2s1 / 2 
2
1d3 / 2 
81
38
Sr49
 2 p3 / 2 
1g9 / 2 
6
9
The nucleon configurations are taken directly from Figure 11-35, and the
are those of the unpaired nucleon.
272
and j values
Chapter 11 – Nuclear Physics
11-39.
Isotope
Odd nucleon
Predicted μ μ N
29
14
Si
neutron
−1.91
27
17
Cl
proton
+2.29
71
31
Ga
proton
+2.29
59
27
Co
proton
+2.29
73
32
Ge
neutron
−1.91
S
neutron
−1.91
Sr
neutron
−1.91
33
16
87
38
11-40. Nuclear spin of 14 N must be 1= because there are 3 mI states, +1, 0, and 1.
11-41.
36
S,
53
Mn,
82
Ge,
88
Sr,
94
Ru,
131
In,
145
Eu
11-42.
3
H
3
14
He
273
N
14
C
Chapter 11 – Nuclear Physics
(Problem 11-42 continued)
n
11-43.
3
2
He,
15
40
20
Ca,
p
n
15
N
60
28
Ni,
124
50
204
82
Sn,
p
n
O
16
p
O
Pb
11-44. (a)
1p1/2
1p3/2
1s1/2
n
p
13
N
(b) j = ½ due to the single unpaired proton.
(c) The first excited state will likely be the jump of a neutron to the empty neutron level,
because it is slightly lower than the corresponding proton level. The j = 1/2 or 3/2,
depending on the relative orientations of the unpaired nucleon spins.
274
Chapter 11 – Nuclear Physics
(Problem 11-44 continued)
(d)
1p1/2
1p3/2
1s1/2
n
p
First excited state. There are several diagrams possible.
11-45.
30
14
Si
j0
37
17
Cl
j  3/ 2
55
27
Co
j  7/2
90
40
Zr
j0
107
49
In
j  9/ 2
 H c  M  H c  M  H c
 2  2.014102uc   3.016049uc  1.007825uc
 0.004330uc  931.5MeV / uc   4.03MeV
Q  M  He  c  M  H  c  M  He  c  M  H  c
11-46 (a) Q  M
 H c
2
2
M
2
2
3
2
3
1
2
2
2
(b)
2
2
2
2
2
2
4
2
1
2
 3.016029uc2  2.014012uc2  4.002602uc2  1.007825uc2


 0.019704uc2 931.5MeV / uc 2  18.35MeV
275
Chapter 11 – Nuclear Physics
(Problem 11-46 continued)
 Li  c
(c) Q  M
6
2
 mnc 2  M
 H c
3
2
M

4

He c 2
 6.01512uc2  1.008665uc2  3.016049uc2  4.002602uc2


 0.005135uc 2 931.5MeV / uc 2  4.78MeV
 H c
11-47. (a) Q  M
3
2
 
 M 1H c 2  M
 He c
3
 mN c 2
2
 3.016049uc2  1.007825uc2  3.016029uc2  1.0078665uc2


 0.000820uc2 931.5MeV / uc 2  0.764MeV
(b) The threshold for this endothermic reaction is:
Eth 

(c) Eth 
11-48.
14
mM
Q
M
(Equation 11-61)
3.016049uc 2  1.007825uc 2
0.764MeV  3.05MeV
1.007825uc 2
1.007825uc 2  3.016049uc 2
3.016049uc 2
0.764MeV  1.02MeV
N  2 H  16O
O  16O  
O  14 N  2 H
16
O  15 N  p
16
16
Possible products:
O  12C  
16
11-49. (a)
O  15O  n
16
C  , p  15 N
12
QM
 Cc
12
2
M

4

He c 2  M
 N c
15
2
 mpc2
 12.000000uc2  4.002602uc2  15.000108uc2  1.007825uc2


 0.005331uc2 931.5MeV / uc2  4.97 MeV
(b)
O  d, p  17O
16
QM
 O c
16
2
M
 H c
2
2
M
 O c
17
2
 mpc2
 15.994915uc2  2.014102uc2  16.999132uc2  1.007825uc2


 0.002060uc2 931.5MeV / uc 2  1.92MeV
276
Chapter 11 – Nuclear Physics
75
11-50. The number of
N
As atoms in sample N is:



1cm  2cm  30 m  104 cm /  m 5.73g / cm3 6.02  1023 atoms / mol
V  NA

M
74.9216 g / mol
 2.76  1020
75
As atoms
75
The reaction rate R per second per
R I
As atoms is:
(Equation 11-62

 4.5  1024 cm2
75

As 0.95  1013 neutrons / cm2 s

 4.28  1011 s 1
Reaction rate = NR



 .76  1020 atoms 4.28  1011 / s atom  1.18  1010 / s
11-51. (a)
23
Ne  p, n  23 Na
22
(b)
11
B  , p  14C
(c)
29
Si  , d  31P
11-52. (a)
14
(e)
14
11-53.
C
N
(b) n
160
(f)
Ne  d, n  23 Na
20
14
N  n, p  14C
13
32
P  p, d  31P
32
Er
C  d , p  14C
Ni
(d) α
(g) 3 H
(i) p
(c)
58
Q
 m p  mn  md
c2
 1.007276u  1.008665u  2.013553u
 0.002388u
(See Table 11-1.)


Q   0.002388u  931.5MeV / uc 2 c 2  224MeV
277
F  , n  23 Na
Si  n, d  31P

Chapter 11 – Nuclear Physics
11-54. P 
dW
dN
E
dt
dt
dN P
500  106 J / s
1eV

 

6
dt
E 200  10 eV / fission  1.60  1019 J

19
  1.56  10 fissions / s

11-55. The fission reaction rate is:
R  N   R  0 k N
(see Example 11-22 in More section)
k N  R  N  / R  0
N log k  log  R  N  / R  0 
N
log  R  N  / R  0  
log k
(a) For the reaction rate to double R  N   2 R  0  :
(b) For R  N   10 R  0  :
(c) For R  N   100 R  0  :
N
N
log10
 24.2
log1.1
N
log100
 48.3
log1.1
(d) Total time t  N 1ms   N ms : (a) 7.27ms
(b) 24.2ms
(e) Total time t  N 100ms   100 N ms : (a) 0.727ms
11-56.
235
92
U143
log 2
 7.27
log1.1
134
 101
40 Zr61  52 Te82  n


 101 Nb  133 Sb  2n
 41 60 51 82
n
 101
132
 43 Tc58  49 In83  3n

 102
130
 45 Rh57  47 Ag83  4n
278
(c) 48.3ms
(b) 2.42ms
(c) 4.83ms
Chapter 11 – Nuclear Physics
J  1MeV

 1 fusion 
11-57. 500MW   500 
 1.78  1014 fusions / s


13
s  1.60  10 J  17.6MeV 

Each fusion requires one 2 H atom (and one 3 H atom; see Equation 11-67) so 2 H must be
provided at the rate of 1.78  1014 atoms / s.
238
11-58. The reactions per
R I
U atom is:
(Equation 11-62)
 1m2 
 0.02  1024 cm2 atom 5.0  1011 n / m2  4 2   1.00  1018 / atom
 10 cm 


238
The number N of
N

U atoms is:
 5.0 g   6.02  1023 atoms / mol 
Total
 1.26  1022
238.051g / mol
238
U atoms
239
U atoms produced = RN



 1.00  1018 / atom 1.26  1022 atoms  1.26  104
 
 
11-59. Q1  M 1H c 2  M 1H c 2  M
 H c
2
239
U atoms
2
 1.007825uc2  1.007825uc2  2.014102uc2


 0.001548uc2 931.50MeV / uc 2  1.4420MeV
Q2  M

2

 
He c 2  M 1H c 2  M
 He  c
3
2
 2.014102uc2  1.007825uc2  3.016029uc2


 0.005898uc2 931.50MeV / uc 2  5.4940MeV

 
 2  3.016029uc   4.002602uc  2 1.007825uc 
 0.013806uc  931.50MeV / uc   12.8603MeV
Q3  M
 He c
3
2
M
2
2
 He c
3
2
M

4
He c 2  2m 1H c 2
2
2
2
Q  Q1  Q2  Q3  1.4420MeV  5.4940MeV  12.8603MeV  19.80MeV
279
Chapter 11 – Nuclear Physics
11-60. Total power = 1000MWe / 0.30  3333MW


 3.33  109 J / s 1MeV / 1.60  1013 J  2.08  1022 MeV / s
(a) 1 day  8.64  104 s


Energy/day  2.08  1022 MeV / s 8.64  104 s / day  1.80  1027 MeV / day
The fission of 1kg of
235
1kg
(b) 1kg
235
U provides 4.95 1026 MeV (from Example 11-19)


U / day  1.80  1027 MeV 4.95  1026 MeV / kg  3.64kg / day
U / year  3.64 kg
235
U / day  365days / year   1.33  103 kg / year
235
(c) Burning coal produces


3.15  107 J / kg 1MeV / 1.60  1013 J  1.97  1020 MeV / kg coal
Ratio of kg coal needed per kg of
For 1 day: 3.64kg
235

235
U is:
4.95  1026 MeV / kg 235U
 2.51  106
20
1.97  10 MeV / kg coal

U 2.51  106  9.1  106 kg This is about 10,000 tons/day,
the approximate capacity of 100 railroad coal hopper cars.
For 1 year: 9.12  106 kg / day  365days / year   3.33 109 kg / year
11-61.   H 2O   1000kg / m3, so
(a) 1000kg :



106 g 6.02  1023 molecules H 2O / mol  2H / molecule  0.00015 2 H
18.02 g / mol
 1.00  1025 2 H atoms
Each fusion releases 5.49MeV.


 5.49  10 MeV 1.60  10
Energy release = 1.00  1025  5.49MeV   5.49  1025 MeV
25
13

J / MeV  8.78  1012 J
(b) Energy used/person (in 1999) = 3.58  1020 J 5.9 109 people
 6.07  1010 J / person y
280

Chapter 11 – Nuclear Physics
(Problem 11-61 continued)
Energy used per person per hour = 6.07  1010 J / person y 
1y
8760h
 6.93  106 J / person h
At that rate the deuterium fusion in 1m3 of water would last the “typical” person
8.80  1012 J
 1.27  106 h  145 y
6
6.93  10 J / person h
11-62. (a) Q  M

235

U c 2  mnc 2  M


Cd c 2  M
120

110

Ru c 2  5mnc 2
 235.043924uc2  1.008665uc2  119.909851uc2  109.913859uc2

5 1.008665uc 2



 1.186uc2 931.5MeV / uc 2  1.10  103 MeV
(b) This reaction is not likely to occur. Both product nuclei are neutron-rich and highly
unstable.
11-63. The original number N0 of 14 C nuclei in the sample is:


N0  15g  6.78  1010 nuclei / g  1.017  1012 where the number of
of C was computed in Example 11-27. The number N of
N  N 0 e  t  N 0 e

ln 2 t / t1 / 2 

 1.017  1012 e

C nuclei per gram
C present after 10,000y is:
(Equation 11-18)
ln 210000 / 5730 
R   N  ln 2 t1 / 2  N
14
14
 3.034  1011
(Equation 11-19)

 ln 2 5730 y  1y 3.16  107 s 3.034  1011

 1.16 decays / s
11-64. If from a living organism, the decay rate would be:
15.6decays / g min 175g   20.230decays / min
(from Example 11-27)
The actual decay rate is: 8.1decays / s  60s / min   486decays / min
281
Chapter 11 – Nuclear Physics
(Problem 11-64 continued)
2
486decays / min
1
 2   20, 230decays / min (from Example 11-27)
 
2n  20, 230 / 486
n ln  2   ln  20, 230 / 486 
n  ln  20, 230 / 486  ln(2)  5.379 half lives
Age of bone =  5.379 half lives  5730 y / half life   30,800 y
t
11-65.
t1 / 2

87
t1 / 2
ln 1  N D / N P 
ln  2 
(Equation 11-92)

Rb  4.88  1010 y and N P / N D  36.5
t
4.88  1010 y
ln 1  1 / 36.5  1.90  109 y
ln  2 
11-66. The number of X rays counted during the experiment equals the number of atoms of
interest in the same, times the cross section for activation  x , times the particle beam
intensity, where
I = proton intensity = 250nA   eC / proton   1.56  1012 protons / s
1
 x  650b  650  1024 cm2
m  mass  0.35mg / cm2  0.00001  3.5  106 mg / cm2
n  number of atoms of interest  mN A / A
t  exposure time
detector efficiency  0.0035
  overall efficiency  0.60  detector efficiency
N  I x
mN A
t
A


250  109 C / s
24
2
N 
 650  10 cm
19
 1.60  10 C / proton 

282

Chapter 11 – Nuclear Physics
(Problem 11-66 continued)



 0.35  103 g / cm2  0.00001 6.02  1023 mol 1


80 g / mol

 


 15 min  60s/min  0.60  0.0035
 5.06 104 counts in 15 minutes
t
11-67.
t1 / 2

232
NP

232
ND

208
t1 / 2
ln 1  N D / N P 
ln  2 
(Equation 11-92)

Th  1.40  1010 y

  1.066 10

4.11g 6.02  1023 atoms / mol

0.88 g 6.02  1023 atoms / mol
Th 
Pb 
232.04 g / mol

208 g / mol
22
  2.547  10
atoms
21
atoms
N D N P  2.547  1021 1.066  1022  0.2389
t
11-68. f 
1.40  1010 y
ln 1  0.2389  4.33  109 y
ln  2 
E 2   z  p B

h
h
(a) For Earth’s field: f 



2  2.79 N   3.15  108 eV / T 1 N  0.5  104 T

4.14  1015 eV s
 2.13  103 Hz  2.12 kHz


(b) For B = 0.25T: f  2.12  103 Hz 0.25T 0.5  104 T  1.06  107 Hz  10.6 MHz


(c) For B= 0.5T: f  2.12  103 Hz 0.5T 0.5  104 T  2.12  107 Hz  21.2 MHz
283
Chapter 11 – Nuclear Physics
11-69. (a) N

14
12
C
3
12
C

12 10

6

3 1.60  10
(b) mass
19
C

 1.50  1016
C ratio  1500 1.50  1016  1013
1.50 10
C
15
12

C / s 10 min  60s / min 


atoms / min  75 min 12  1.66  1027 kg

0.015
 1.49  107 kg  1.49  104 g  0.149mg
(c) The
14
C
12
sample
living
C ratio in living C is 1.35  1012.
14
C
14
C
C
1013
0.10  1 



12
12
C
1.35  10
1.35  2 
12
n
where n = # of half-lives elapsed. Rewriting as (see Example 11-28)
2n 
1.35
 13.5
0.10
n ln  2  ln 13.5
 n  ln 13.5 / ln  2   3.75
age of sample = 3.75t1 / 2  3.75  5730 y   2.15 104 y
11-70. If from live wood, the decay rate would be 15.6 disintegrations/g•min. The actual rate is
2.05 disintegrations/g•min.
2
2.05decays / g min
1
 2   15.6decays / g min (from Example 12-13)
 
2n  15.6 / 2.05
n ln  2   ln 15.6 / 2.05
n  ln 15.6 / 2.05 ln  2   2.928 half lives of 14 C
Age of spear thrower = nt1 / 2   2.928 5730 y   16,800 y
284
Chapter 11 – Nuclear Physics
11-71. Writing Equation 11-14 as:
M Z, A c2
Zm2p
a1 A a2 A2 / 3 a3 A
A Z mnc 2
1/ 3 2
z
a4 A 2Z
2
A
1
a5 A
1/ 2
c2
and differentiating,
M
Z
mp
mn
0
mp
mn
2a3 A
0
mp
mn
4a4
mp
Z
mn
2a3 A
1/ 3
2a3 A
4a4
8a4 A
1/ 3
1/ 3
Z
Z
2a4 A 2Z
1/ 3
where a3
1
1
8a4 A 1Z
4a4
2a3 A
2 A
8a4 A
1
Z
0.75 and a4
93.2
1.008665 1.007825 931.5MeV / uc 2
(a) For A = 27: Z
2 0.75 27
1/ 3
8 93.2 27
4 93.2
1
13.2
Minimum Z = 13
(b) For A = 65: Computing as in (a) with A = 65 yields Z = 31.5. Minimum Z = 29.
(c) For A = 139: Computing as in (a) with A = 139 yields Z = 66. Minimum Z = 57.
11-72. (a) R
0.31 E 3 / 2
(b) R g / cm2
(c) R cm
0.31 5MeV
R cm
3/ 2
3.47cm
3.47cm 1.29 10 3 g / cm3
R g / cm2
4.47 10 3 g / cm2
4.47 10 3 g / cm2
2.70 g / cm3
1.66 10 3 cm
11-73. For one proton, consider the nucleus as a sphere of charge e and charge density
c
3e 4 R3. The work done in assembling the sphere, i.e., bringing charged shell dq
up to r, is: dU c
Uc
k
2
c
16 2 R5
15
k
c
4 r3
3
c
4 r 2 dr
1
and integrating from 0 to R yields:
r
3 ke2
5 R
For two protons, the coulomb repulsive energy is twice Uc, or 6ke 2 5 R.
285
Chapter 11 – Nuclear Physics
11-74. The number N of
dN
dt
144
Nd atoms is: N
N
11-75. R
ln 2
R0e
144 g / mol
2.25 1023 atoms
dN dt N
2.36s
t1 / 2
53.94 g 6.02 1023 atoms / mol
1
ln 2 1.05 10
t ln 2 / t1 / 2
2.25 1023
23
s
1
1.05 10
6.61 1023 s
23
s
1
2.09 1015 y
(from Equation 11-19)
(a) At t = 0: R = R0 = 115.0 decays/min
At t = 4d 5h = 4.21d: R = 73.5 decays/min
ln 2 4.21d / t1 / 2
73.5decays / min = 115.0decays / min e
73.5 / 115.0
e
ln 73.5 / 115.0
ln 2 4.21d / t1 / 2
ln 2 4.21d / t1 / 2
ln 2 2.41d / ln 73.5 / 115.0
t1 / 2
(b) R 10decays / min = 115.0decays / min e
t
ln 10 / 115.0 6.52d ln 2
t
ln 2 t / 6.52 d
23.0d
2.5decays / min = 115.0decays / min e
(c) R
6.52d
ln 2.5 / 115.0 6.52d ln 2
ln 2 t / 6.52 d
36.0d
(because t = 0)
This time is 13 days (= 2t1/2) after the time in (b).
11-76. For
227
For
223
Th : t1 / 2
Ra : t1 / 2
18.72d
(nucleus A)
11.43d
(nucleus B)
At t = 0 there are 106 Th atoms and 0 Ra atoms
(a) N A
NB
N 0 Ae
N0 A
B
At
A
(Equation 11-18)
e
At
e
Bt
N0Be
Bt
A
286
(Equation 11-26 on the Web page)
Chapter 11 – Nuclear Physics
(Problem 11-76 continued)
ln 2 15 d / 18.72 d
106 e
NA
5.74 105
106 ln 2 18.72d
NB
ln 2 1 / 11.43d 1 / 18.72d
(b) N A
N0 A
At
N B means N 0 Ae
B
ln 2 15 / 18.72
e
A
e
At
e
e
ln 2 15 / 11.43
0
2.68 105
Bt
A
Cancelling N0A and rearranging,
B
A
1 e
A
t
B
A
ln 2
A
0.0370d
18.82d
B
ln
A
1
ln 2
1
B
B
A
11.43d
0.0606d
1
t
A
B
ln
t
A
1
A
B
0.0606 0.0370
1
0.0370
ln
A
0.0370 0.0606
43.0d
/
11-77. (a)
6.582 10
hf
(b) Er
2
16
eV s 0.13 10 9 s
0.12939MeV
2Mc 2
2M
4.71 10 8 MeV
191
5.06 10 6 eV
2
(Equation 11-47)
I c2
4.71 10 2 eV
(c) (See Section 1-5)
The relativistic Doppler shift Δf for either receding or approaching is:
f
v
c
f0
E
v
c
v
c
E
h f
hf 0
E
5.06 10 6 eV
3.00 108 m / s
0.12939MeV 106 eV / MeV
287
0.0117m / s
1.17cm / s
Chapter 11 – Nuclear Physics
11-78. B
ZM 1H c 2
For 3 He :
B
Nmnc 2
M Ac 2
2 1.007825uc 2
1.008665uc 2
3.016029uc 2
0.008826uc 2 931.5MeV / uc 2
For 3 H :
B 1.007825uc 2
7.72MeV
2 1.008665uc 2
3.016029uc 2
0.009106uc 2 931.5MeV / uc 2
R
Uc
R0 A1 / 3
1.2 fm31 / 3
ke2 / R
8.48MeV
15
1.730 fm 1.730 10
8.32 105 eV
ke / R eV
m
0.832MeV
or about 1/10 of the binding
energy.
11-79. For 47Ca :
B
M
46
Ca c 2
mn c 2
M
47
Ca c 2
45.953687uc2 1.008665uc2 46.954541uc2
0.007811uc 2 931.5MeV / uc 2
7.28MeV
For 48Ca :
B
M
47
Ca c 2
mn c 2
M
48
Ca c 2
45.954541uc2 1.008665uc2
47.952534uc2
0.010672uc 2 931.5MeV / uc 2
9.94MeV
Assuming the even-odd
Ca to be the “no correction” nuclide, the average magnitude of
47
the correction needed to go to either of the even-even nuclides 46Ca or
48
Ca is
approximately B – average binding energy of the odd neutron,
9.94MeV
7.28MeV / 2
MeV and for
8.61MeV . So the correction for 46Ca is 8.16 – 7.28 = 0.88
Ca is 9.94 – 8.16 = 1.78 MeV, an “average” of about 1.33 MeV. The
48
estimate for a5 is then: a5 A 1 / 2
1.33MeV
a5
30% below the accepted empirical value of a5 = 12.
288
1.33 / 48 1 / 2
9.2 . This value is about
Chapter 11 – Nuclear Physics
11-80. For a nucleus with I > 0 the α feels a centripetal force
mv2 / r
Fc
dV / dr where r
distance of the
from the nuclear center. The
ln r and becomes larger (i.e., more negative) as r
corresponding potential energy V
increases. This lowers the total energy of the α near the nuclear boundary and results in a
wider barrier, hence lower decay probability.
11-81. (a) R
R0 A1 / 3 where R0
R
141
R
92
(b) V
1.2 f
Ba
1.2 fm 10
Kr
1.2 fm 10
15
15
(Equation 11-3)
m / fm 141
m / fm 92
1/ 3
1/ 3
6.24 10
5.42 10
15
8.998 109 N m2 / C 2 56 1.60 10
kq1q2 / r
6.24 10
2.49 108 eV
15
m 5.42 10
15
15
m
m
19
C 36 1.60 10
m 1.60 10
19
19
C
J / eV
249MeV
This value is about 40% larger than the measured value.
11-82. (a) In the lab, the nucleus (at rest) is at x = 0 and the neutron moving at vL is at x.
xCM
V
M 0
M
mx
m
mx
M m
vL
x
and V
dt
dxm
dt
m x / dt
M
m
mvL
M m
(b) The nucleus at rest in the lab frame moves at speed V in the CM frame before the
collision.
In an elastic collision in the CM system, the particles reverse their
velocities, so the speed of the nucleus is still V, but in the opposite direction.
(c) In the CM frame in the nucleus velocity changes by 2V. This is also the change in the
lab system where the nucleus was initially at rest. It moves with speed 2V in the lab
system after the collision.
(d)
1
M 2V
2
1/ 2
2mvL
1
M
2
M m
1 2
mvL
2
289
4mM
M
m
2
Chapter 11 – Nuclear Physics
(Problem 11-82 continued)
Before collision: Ei
1 2
mvL
2
1 2
mvL
2
After collision: E
1 2
mvL
2
4mM
M
m
11-83. At the end of the two hour irradiation the number of
R0
N
For
32
t
1 e
M
P and
56
m
2
Mn atoms are given by
I (Equation 11-62)
P:
24
0.180 10
N0
R0t1 / 2
1 e
ln 2
1.29 10
56
32
from Equation 11-26 where R0
R0
For
4mM
Ei 1
2
cm2 1012 neutrons / cm2 s
1.80 10
ln 2 t / t1 / 2
13
1.80 10
13
32
P atoms / s per 31 P
/ s 3600s / h 342.2h
ln 2
9 32
1 e
ln 2 2 h / 342.2 h
P atoms / 31 P atom
Mn :
R0
N0
13.3 10
1.33 10
24
cm2 1012 neutrons / cm2 s
11
/ s 3600s / h 2.58h
1 e
ln 2
7.42 10
8 56
1.33 10
11
56
Mn atoms / s per 55 Mn
ln 2 2 h / 2.58 h
Mn atoms / 55 Mn atom
(a) Two hours after the irradiation stops, the activities are:
dN
dt
For
N 0e
32
56
N 0 ln 2
t1 / 2
e
ln 2 t / t1 / 2
P:
dN
dt
For
t
1.29 10
9
ln 2
4
14.26d 8.64 10 s / d
e
ln 2 2 h / 342.2 h
Mn :
290
7.23 10
16
decays / 31 P atom
Chapter 11 – Nuclear Physics
(Problem 11-83 continued)
8
7.42 10
dN
dt
ln 2
2.58h 3600s / h
e
ln 2 48 h / 2.58 h
1.39 10
17
decays / 56 Mn atom
The total activity is the sum of these, each multiplied by the number of parent atoms
initially present.
11-84. Q
E
N
200MeV / fission.
7.0 1019 J
NQ
N 200MeV / fission 1.60 10
7.0 1019 J
200MeV / fission 1.60 10
235
Number of moles of
13
2.19 1030 6.02 1023
Fissioned mass/y = 3.63 106 moles 235 g / mole
That is 3% of the mass of the
235
J / MeV
2.19 1030 fissions
J / MeV
U needed = N / N A
13
8.54 108 g
3.63 106 moles
8.54 105 kg
U atoms needed to produce the energy consumed.
Mass needed to produce 7.0 1019 J
8.54 105 kg / 0.03
2.85 107 kg.
Since the energy conversion system is 25% efficient:
Total mass of
11-85. The number of
87
235
U needed = 1.14 108 kg.
Sr atoms present at any time is equal to the number of
have decayed, because
87
Rb nuclei that
Sr is stable.
N Sr
N 0 Rb
N Rb
0.010
N 0 Rb N Rb
N Sr
N Sr
87
N Rb
N Rb
N Sr
N Rb
N 0 Rb N Rb
1N
1 1.010
and also
N Rb N0 Rb
ln 2 t
t1 / 2
t
e
ln 2 t / t1 / 2
1 / 1.010
ln 1 / 1.010
t1 / 2 ln 1 / 1.010 / ln( 2)
4.9 1010 y ln 1 / 1.010 / ln( 2)
291
7.03 108 y
Chapter 11 – Nuclear Physics
11-86. (a) Average energy released/reaction is: 3.27 MeV
P
E
t
4W
4J / s
N 3.65MeV 1.60 10
4J / s
3.65MeV / reaction 1.60 10
N
13
4.03MeV / 2
13
3.65MeV
J / MeV
J / MeV
6.85 1012 reactions / s
Half of the reactions produce neutrons, so 3.42 1012 neutrons / s will be released.
0.10 3.42 1012
(b) Neutron absorption rate
3.42 1011 neutrons / s
Energy absorption rate =
0.5MeV / neutron 3.42 1011 neutrons / s 1.60 10
13
J / MeV
2.74 10 2 J / s
Radiation dose rate =
2.74 10 2 J / s / 80kg
3.42 10 2 rad / s 4
3.42 10 3 rad / s
100rad / J / kg
0.137 rem / s
493 rem / h
(c) 500 rem, lethal to half of those exposed, would be received in:
500rem / 492rem / h
11-87. R t
N0 I 1 e
For Co: N 0
For Ti: N 0
t
1.02h
(Equation 11-86)
35Bq
19 10
24
cm
2
12
3.5 10 / s cm
2
1 e
1.319 10
115 Bq
0.15 10
11-88. The net reaction is: 5 2 H
24
cm
2
3
12
3.5 10 / s cm
He
4
He
1
H
2
1 e
0.120 2
2
2.00 1017 atoms
1.03 1015 atoms
n 25MeV
Energy release / 2 H
5MeV (assumes equal probabilities)
4 water
2 1.007825
4000 g
6
15.994915 g / mol
222.1 moles
4 of water thus contains 2(222.1) moles of hydrogen, of which 1.5 10 4 is 2 H , or
Number of 2 H atoms =
292
Chapter 11 – Nuclear Physics
(Problem 11-88 continued)
2 222.1 moles 6.02 1023 atoms / mol 1.5 10
Total energy release = 4.01 1022 5MeV
4
4.01 1022
2.01 1023 MeV
3.22 1010 J
Because the U.S. consumes about 1.0 1020 J / y, the complete fusion of the 2 H in 4
of water would supply the nation for about 1.01 10 2 s 10.1ms
2hc / Mc 2
11-89. (a)
hc
hc
E
2
2
2
hc
E 2 2hc
E
hc Mc 2
Ep
2E 2
Mc 2
E 2 Mc 2 E p / 2
E
Ef
E
Ei
M
4mM
Ei 1
m
Mc 2 E p / 2
E
M
m
2
1 m/M
2
5.7MeV 938.28MeV / 2
(b) E
1/ 2
Ei
2
4m / M
4mM
Ei
E2
hc
4mM
Ei
M
m
2
which is Equation 11-82 in More section.
1/ 2
51.7 MeV
(c)
O 14N (M)
x
vL
CM
neutron (m)
The neutron moves at vL in the lab, so the CM moves at v
vL mN / mN
M
toward
the right and the 14N velocity in the CM system is v to the left before collision and v to
the right after collision for an elastic collision. Thus, the energy of the nitrogen
nucleus in the lab after the collision is:
E
14
N
1
M 2v
2
2
2Mv
2
mvL
2M
m M
293
2
Chapter 11 – Nuclear Physics
(Problem 11-89 continued)
2Mm mvL2
m M
4Mm
2
1 2
mvL
2
2
m M
4 14.003074u 1.008665u
1.008665u 14.003074u
5.7 MeV
2
= 1.43 MeV
14.003074uc 2 931.5MeV / uc 2 1.43MeV / 2
(d) E
1/ 2
96.5MeV
11-90. In the lab frame:
photon
E
hv
p
hv / c
deuteron,
M at rest
pc
E /c
In CM frame:
For E
photon
deuteron, M
E
p
EK
1 / 2Mv 2
p
2MEK
pc
E/c
p 2 / 2M
pc in CM system means that a negligible amount of photon energy goes to
recoil energy of the deuteron, i.e.,
p2
2M
E
pc
pc
E
2Mc 2
or
pc
2Mc
2
pc
2
2 1875.6MeV
pc
2Mc 2
3751.2 MeV
(see Table 11-1)
In the lab, that incident photon energy must supply the binding energy B = 2.22 MeV plus
the recoil energy EK given by:
EK
p 2 / 2M
2.22 MeV
2
pc / 2Mc 2
2
B / 2Mc 2
2
2 1875.6 MeV
0.0013MeV
294
Chapter 11 – Nuclear Physics
(Problem 11-90 continued)
So the photon energy must be E
2.22MeV
0.001MeV
2.221MeV , which is much
less than 3751 MeV.
ZM 1H c 2
11-91. (a) B
For 7 Li :
Nmn c 2
M Ac 2
3 1.007825uc 2
B
(Equation 11-11)
4 1.008665uc 2
7.016003uc 2
0.042132uc 2 931.5MeV / uc 2
For 7 Be :
4 1.007825uc 2
B
39.25MeV
3 1.008665uc 2
0.040367uc 2 931.5MeV / uc 2
7.016928uc 2
37.60MeV
B 1.65MeV
For 11B :
B
5 1.007825uc 2
6 1.008665uc 2
11.009305uc 2
0.0081810uc 2 931.5MeV / uc 2
For 11C :
B
6 1.007825uc 2
76.21MeV
5 1.008665uc 2
11.011433uc 2
0.078842uc 2 931.5MeV / uc 2
B
73.44MeV
2.77MeV
For 15 N :
7 1.007825uc 2
B
8 1.008665uc 2
0.123987uc 2 931.5MeV / uc 2
For 15O :
B
8 1.007825uc 2
115.5MeV
7 1.008665uc 2
0.120190uc 2 931.5MeV / uc 2
B
(b)
B
a3
For A
15.000108uc 2
15.003065uc 2
112.0MeV
3.54MeV
Z2 A
7; Z
For A 11; Z
1/ 3
4:
6:
a3
a3
a3
B
Z2 A
1.65MeV 7 7
1/ 3
2.77 MeV 11 11
295
1/ 3
0.45MeV
1/ 3
0.56MeV
Chapter 11 – Nuclear Physics
(Problem 11-91 continued)
For A 15; Z
8:
1/ 3
3.54 MeV 15 15
a3
0.58MeV
0.53MeV
a3
0.75MeV .
These values differ significantly from the empirical value of a3
11-92. (a) Using M Z
Z
mn
mp
2a3 A
1
2A
0 from Problem 11-71.
4
4
1/ 3
8a4 A
mn
mp
1
(b) & (c) For A
a3 A
29 :
1
where a3
2a4
Z
93.2MeV / c 2
a4
A 1 mn m p 4a4
2 1 a3 A2 / 3 4a4
4a4
1/ 3
0.75MeV / c 2 ,
29 1
2
1.008665 1.007276 931.5 / 4 93.2
1 0.75 29
The only stable isotope with A = 29 is
29
14
2/3
14
/ 4 93.2
Si
For A =59: Computing as above with A = 59 yields Z = 29. The only stable
isotope with A = 59 is
59
27
Co
For A = 78: Computing as above with A = 78 yields Z = 38.
stable. Stable isotopes with A = 78 are
78
34
The only stable isotope with A = 119 is
119
50
Se and
78
38
78
38
Sr is not
Kr .
Sn .
For A = 140: Computing as above with A = 140 yields Z = 69.
stable. The only stable isotope with A = 140 is
140
69
Tm is not
140
58
Ce .
The method of finding the minimum Z for each A works well for A ≤ 60, but
deviates increasingly at higher A values.
296
Chapter 11 – Nuclear Physics
11-93. (a)
(b)
j = 3/2
n
1d5/2
1d5/2
1p1/2
1p1/2
1p3/2
1p3/2
1s1/2
1s1/2
n
p
11
Ground state B
(c)
j = 5/2
1d5/2
1p1/2
1p3/2
1s1/2
n
2nd excited state 11B
j = 1/2
p
297
1st excited state 11B
p
Chapter 11 – Nuclear Physics
(Problem 11-93 continued)
(d)
17
n
O
Ground state (j = 5/2)
1d5/2
1d5/2
1p1/2
1p1/2
1p3/2
1p3/2
1s1/2
1s1/2
p
n
(e)
2s1/2
1d5/2
1p1/2
1p3/2
1s1/2
n
17
p
O
nd
2 excited state (j = 1/2)
298
17
O
1st excited state (j = 1/2)
p
Chapter 11 – Nuclear Physics
11-94. (a) Data from Appendix A are plotted on the graph. For those isotopes not listed in
Appendix A, data for ones that have been discovered can be found in the reference
sources, e.g., Table of Isotopes, R.B. Firestone, Wiley – Interscience (1998). Masses
for those not yet discovered or not in Appendix A are computed from Equation 11-14
(on the Web). Values of M(Z, 151) computed from Equation 11-14 are listed below.
Because values found from Equation 11-14 tend to overestimate the mass in the
higher A regions, the calculated value was adjusted to the measured value for Z = 56,
the lowest Z known for A = 151 and the lower Z values were corrected by a
corresponding amount. The error introduced by this correction is not serious because
the side of the parabola is nearly a straight line in this region. On the high Z side of
the A = 151 parabola, all isotopes through Z = 70 have been discovered and are in the
reference cited.
Z
N
M Z , 151 [Equation 11-14]
50
101
152.352638
151.565515
51
100
152.234612
151.447490
52
99
152.122188
151.335066
53
98
152.015365
151.228243
54
97
151.914414
151.127292
55
96
151.818525
151.031403
56
95
151.728507
150.941385*
* This value has been measured.
299
M Z , 151
[adjusted]
Chapter 11 – Nuclear Physics
(Problem 11-94 continued)
(b) The drip lines occur for:
protons: M Z , 151
M Z 1, 150
neutrons: M Z , 151
M Z , 150
mp
mn
0
0
Write a calculator or computer program for each using Equation 11-14 (on Web page)
and solve for Z.
11-95. (a) M Z , A
Zmp
Nmn
a1 A a2 A2 / 3
a3 A
1/ 3
a4 A 2Z
2
A
1
a5 A
1/ 2
from Equation 11-14 on the Web page.
15.67 310
M 126, 310
126m p 184mn
17.23 310
0.75 126
2
310
313.969022u
(b) For β− decay: (126, 310) → (127, 310) + β− + ve
Computing M(127, 310) as in (a) yields 314.011614u.
300
1/ 3
93.2 310 2 126
12 310
M 126, 310
2/3
1
2
310
12 310
1/ 2
Chapter 11 – Nuclear Physics
(Problem 11-95 continued)
Q
M 126, 310 c 2
M 127, 310 c 2
313.969022uc2 314.011614uc2
0.042592u 931.5MeV / uc 2
39.7 MeV
Q < 0, so β− decay is forbidden by energy conservation.
For β+ decay: (126, 310) → (125, 310) + β+ + ve
Computing M(125, 310) as in (a) yields 313.923610u.
Q
M 126, 310 c 2
M 125, 310 c 2
2mec 2
313.969022uc2 313.923610uc2 1.022MeV
41.3MeV
β+ decay and electron capture are possible decay modes.
For α decay: (126, 310) → (124, 310) + α
Computing M(124, 310) as in (a) yields 309.913540u.
313.969022uc 2
Q
309.913540uc 2
4.002602uc 2
49.3MeV
α decay is also a possible decay mode.
11-96. (a) If the electron’s kinetic energy is 0.782MeV, then its total energy is:
E
E2
p
mec2
0.782MeV
pc
E2
2
2
me c 2
me c 2
0.511MeV
1.293MeV
(Equation 2-32)
1/ 2
2
c
2
1.293MeV
0.782MeV
0.511MeV
2
1/ 2
c
1.189MeV / c
(b) For the proton p 1.189MeV / c also, so
Ekin
p 2 / 2m
pc
1.189MeV
2
2
2mc 2
2 938.28MeV
301
7.53 10 4 MeV
0.753keV
Chapter 11 – Nuclear Physics
(Problem 11-96 continued)
(c)
11-97.
7.53 10 4 MeV
100
0.782MeV
dN
dt
Rp
(a)
N
N
Rp
(Equation 11-17)
N 0e
Rp /
N
0.0963%
t
1 e
Rp
t
Rp e
t
Rp 1 e
(from Equation 11-17)
R p / , its maximum value
At t = 0, N(0) = 0. For large t, N t
(b) For dN / dt
Rp
N
N
t
0
N
Rp /
Rp / ln 2 / t1 / 2
100s 1 / ln 2 / 10 min
8.66 104
11-98. (a) 4n + 3 decays chain
100s
1
60s / min / ln 2 / 10 min
62
Cu nuclei
235
92
U143
207
82
Pb125 There are 12 α decays in the chain. (See graph
below.)
302
Chapter 11 – Nuclear Physics
(b) There are 9 β− decays in the chain. (See graph below.)
(Problem 11-98 continued)
(c) Q
235
U c2
M
207
M
235.043924uc 2
Pb c 2
7M
4
206.975871uc 2
0.049839uc 2 931.50MeV / uc 2
He c 2
7 4.002602 uc 2
46.43MeV
(d) The number of decays in one year is:
dN
dt
N0e
t
ln 2 / t1 / 2
where
ln 2 / 7.04 108 y
1kg 1000 g / kg 6.02 1023 atoms / mol
N0
235 g / mol
dN
dt
9.85 10
10
y
1
2.56 1024 e
1y
9.85 10
10
y
1
2.56 1024 atoms
2.52 1015 decays / y
Each decay results in the eventual release of 46.43 MeV, so the energy release
per year Q is: Q
2.52 1015 decays / y 46.43MeV / decay
1.17 1017 MeV / y 1.60 10
13
J / MeV
1.87 104 J / y 1cal / 4.186 J
4.48 103 cal / y
The temperature change ΔT is given by:
Q
cm T or T
Q / cm where m 1kg 1000g
and the specific heat of U is c
T
4.48 103 cal / y
0.0276cal / g C .
0.0276cal / g C 1000 g
303
162 C
Chapter 11 – Nuclear Physics
(Problem 11-98 continued)
235
U
α
144
231
Th
β−
142
231
Pa
α
237
140
Ac
β−
α
223
138
α
β−
136
219
α
215
N
β
α
−
215
Po
β−
215
At
211
α
Pb
β
128
−
211
Bi
β−
α
126
Ra
Rn
α
130
223
219
Bi
132
211
Po
207
Tl
α
β
−
207
Pb
124
80
82
84
86
88
90
Z
11-99. The reactions are:
(1) 1H
1
(2) 1H
2
H
H
2
H
3
ve
He
followed by
(3) 3 He
3
Th
α
At
134
227
α
Fr
He
4
He
1
H
1
H or
304
92
Chapter 11 – Nuclear Physics
(Problem 11-99 continued)
(4) 1H
3
4
He
He
ve
(a) From 2(1) + 2(2) + (3):
61 H
2 2H
2 3He
2 2H
2 3He
4
He 2 1H
2
2ve
2
Cancelling 2 1H , 2 2 H , and 2 3 He on both sides of the sum,
41 H
4
He 2
2ve
2
From (1) + (2) + (4):
41 H
2
3
H
2
He
H
3
4
He
He 2
2ve
Cancelling 2 H , and 3He on both sides of the sum,
41 H
(b) Q
4
He 2
4m 1 H c 2
M
4 938.280MeV
2ve
4
He c 2
2mec 2
3727.409MeV
2 0.511MeV
24.7MeV
(c) Total energy release is 24.7 MeV plus the annihilation energy of the two β+:
energy release
24.7 MeV
2 me c 2
24.7 MeV
2 1.022MeV
26.7MeV
Each cycle uses 4 protons, thus produces
26.7MeV / 4 6.68MeV / proton.
Therefore, 1H (protons) are consumed at the rate:
dN
dt
4 1026 J / s
1eV
6
6.68 10 eV 1.60 10
P
E
19
J
3.75 1038 protons / s
The number N of 1H nuclei in the Sun is:
N
1 / 2 2 1030 kg
1.673 10 27 kg
M
M
1
H
5.98 1056 protons
which will last at the present consumption rate for
t
N
dN / dt
5.98 1056 protons
3.75 1038 protons / s
1.60 1018 s
1y
3.16 107 s
1.60 1018 s
5.05 1010 y
305
Chapter 11 – Nuclear Physics
11-100. At this energy, neither particle is relativistic, so
2
pHe
2mHe
EHe
2mHe EHe
mHe
EHe
En
pn2
2mn
En
pHe
2mn 17.7 MeV
2mn En
17.7 MeV mn
mn EHe
1.008665u 17.7 MeV
2.002602u 1.008665u
17.7 MeV
pn
EHe
EHe
Therefore, EHe
R 0
kN
(b) Because k
1.005 62.5 1
100
R N
1.005 1
1.005
0.00498
0.05
235
f
U
0.05 584b
a
R 0
100
neutron flux, the fractional change in flux necessary is equal to the
11-102. (a) For 5% enrichment:
f
R 0
62.5 generations
100 137%
fractional change in k:
k 1
k
17.7 MeV
k N (from Example 11-22 in More section)
100 where R N / R 0
1
mn
14.1MeV
Percentage increase in energy production =
1
mHe
5s
0.08s / gen
11-101. (a) The number N of generations is: N
R N
mn
3.56MeV
17.7 3.56 MeV
EHe
17.7MeV
En
0.05
235
a
0.05 97b
0.95
238
f
0.95 0
U
0.95
U
29.2b
238
a
0.95 2.75b
306
U
7.46b
Chapter 11 – Nuclear Physics
(Problem 11-102 continued)
2.4 29.2b
f
2.4
k
f
1.91 (Equation 11-68 in More section)
29.2b 2.46b
a
(b) For 95% enrichment:
0.95
f
235
f
0.95 584b
a
0.95
235
a
238
0.05
U
f
0.05 0
0.95 97b
554.8b
238
0.05
U
U
a
0.05 2.75b
U
92.3b
R 0 kN.
The reaction rate after N generations is R N
For the rate to double R N
N 5%
ln 2 / ln 1.91
N 95%
kN
2 R 0 and 2
N
ln 2 / ln k .
1.07 generations
ln 2 / ln 2.06
0.96 generations
Assuming an average time per generation of 0.01s
t 5%
1.07 10 2 s
0.96 10 2 s
t 95%
Number of generations/second = 1/seconds/generation
In 1s: N 5%
93.5 and N 95%
104
One second after the first fission:
R 5%
R 0 kN
1 1.91
93.5
1.9 1026
Energy rate = 1.9 1026 fissions / s 200MeV / fission
3.8 1028 MeV / s 1.60 10
6.1 1015 J / s
R 95%
R 0 kN
13
J / MeV
6.1 1015W
1 2.06
104
4.4 1032
Energy rate = 4.4 1032 fissions / s 200MeV / fission
8.8 1034 MeV / s 1.60 10
1.4 1022 J / s 1.4 1022W
307
13
J / MeV
Chapter 12 – Particle Physics
12-1. (a) Because the two pions are initially at rest, the net momentum of the system is zero,
both before and after annihilation. For the momentum of the system to be zero after
the interaction, the momentum of the two photons must be equal in magnitude and
opposite in direction, i.e., their momentum vectors must add to zero. Because the
photon energy is E  pc, their energies are also equal.
(b) The energy of each photon equals the rest energy of a   or a   .
E  m c 2  139.6MeV (from Table 12-3)
(c) E  hf  hc /  Thus,  
hc 1240MeV fm

 8.88 fm
E
139.6MeV
12-2. (a) E  m c2  m c 2  2285MeV  139.6MeV  2424.6MeV
(b) E  2mp c2  2  938.28MeV   1876.56MeV
(c) E  2m c 2  2 105.66MeV   211.32MeV
12-3. (a)
e
e
e
e
γ
Z
e
e
e
e
(b)
γ
e
γ
e
e
γ
e
γ
309
e+
e
Chapter 12 – Particle Physics
12-4. (a)
e
e
γ
e
e
(b) See solution to Problem 12-3(a).
(c)
e
e
e
12-5. (a)
32
P  32S  e assuming no neutrino
QM
 P c
32
2
M
 S c
32
(electron’s mass is included in that of
2
32
S)
 31.973908uc2  31.972071uc2



 0.001837uc2 931.5MeV / uc 2  1.711MeV
To a good approximation, the electron has all of the kinetic energy
Ek  Q  1.711MeV
(b) In the absence of a neutrino, the
32
S and the electron have equal and opposite
momenta. The momentum of the electron is given by:
 pc 
2

 E 2  mec 2

2
(Equation 2-32)

  m c 

  m c 
 Ek  me c 2
 Q  mec 2
2
2
2
e
2
2
e
2
 Q 2  2Qmec 2
310
Chapter 12 – Particle Physics
(Problem 12-5 continued)
The kinetic energy of the
32
S is then:
 pc   Q 2  2Qmec 2
p2
Ek 

2M 2Mc 2
2M 32 S c 2
2
 
1.711MeV 


2
 2 1.711MeV  0.511MeV 

2 31.972071uc 2 931.5MeV / uc 2

 7.85  105 MeV  78.5eV
(c) As noted above, the momenta of the electron and
32
S are equal in magnitude and
opposite direction.
 pc 
2
 Q2  2Qmec 2  1.711MeV   2 1.711MeV  0.511MeV 
2
2
p  1.711MeV   2 1.711MeV  0.511MeV 


1/ 2
c
 2.16MeV / c
12-6. (a) A single photon cannot conserver both energy and momentum.
(b) To conserver momentum each photon must have equal and opposite momenta so that
the total momentum is zero. Thus, they have equal energies, each equal to the rest
energy of a proton: E  mp c 2  938.28MeV
(c) E  hv  hc /    
hc 1240MeV fm

 1.32 fm
E
938.28MeV
3.00  108 m / s
(d) v  
 2.27  1023 Hz
15
 1.32  10 m
c
12-7. (a) Conservation of charge: +1 +1 → +1 −1 +1 −1 = 0. Conservation of charge is
violated, so the reaction is forbidden.
(b) Conservation of charge: +1 +1 → +1 −1 = 0. Conservation of charge is violated, so
the reaction is forbidden.
311
Chapter 12 – Particle Physics

12-8. (a)   2 / m 5    2 / m 5

a b
c
Since the units of  must be seconds s, we have
5
 N m2 
2
s  kg  
  e
2
C


1
 
5
 1 
 
 c
5
a b
c
having substituted the internal units of   ke2 / c. Re-writing the units, noting that
J  N m, gives: s 
1
C10
1
 5 10  10
kg N m
C
 J 5 s5 m5 


s5


a b
c
Noting that J  kg m2 / s 2 and cancelling yields,
s   kg 
a 1
 m 2 a  b  s  a b
Since a  1 must be 0, a  1, and since  a  b  1, b  2.
   2 / mc2 5
(b)  


2 1.055  1034 J s 137 
9.11 10
31

5
kg 3.00  108 m / s

2
 1.24  1010 s  0.124ns
12-9. (a) Weak interaction
(b) Electromagnetic interaction
(c) Strong interaction
(d) Weak interaction
12-10.  0     is caused by the electromagnetic interaction;       v is caused by the
weak interaction. The electromagnetic interaction is the faster and stronger, so the  0
will decay more quickly; the   will live longer.
12-11. (a) allowed; no conservation laws violated.
(b) allowed; no conservation laws violated.
(c) forbidden; doesn’t conserve baryon number.
(d) forbidden; doesn’t conserve muon lepton number.
312
Chapter 12 – Particle Physics
12-12. (a) Electromagnetic interaction
(b) Weak interaction
(c) Electromagnetic interaction
(d) Weak interaction
(e) Strong interaction
(f) Weak interaction
12-13. For neutrino mass m = 0, travel time to Earth is t = d/c, where d =170,000c•y. For
neutrinos with mass m  0, t   d / v  d /  c, where   v / c.
t  t   t 
 
2 
1
1  2
 d 1  
d1
  1  

c
 c  
(Equation 1-19)
1
1

2
1 
1   1   
1  
1
1
 2 since   1
 1    2
2
Substituting into t,
t 
d 1 


c  2 2 
d  mc 2 
t  

2c  E 

E   mc 2   2  E / mc 2
2
(Equation 2-10)
2
  t  2cE 2 
mc  


d


2

1/ 2
 2tE 2 


 d /c 

1/ 2

2

2 12.5s  10  106 eV

 170, 000c y / c  3.16  107 s / y


m  22eV / c2
313





1/ 2
 21.6eV
Chapter 12 – Particle Physics
12-14. The  and  are members of a isospin multiplet, two charge states of the  hadron.
Their mass difference is due to electromagnetic effects. The   and   are a particleantiparticle pair.
12-15. (a)
νμ
μ−
νe
W−
W+
μ+
π−
e+
μ−
ντ
W−
νμ
τ−
νμ
12-16. (See Table 12-8 in More Section.)
(a) 30 MeV
(b) 175 MeV
(c) 120 MeV
12-17. (a) mp c 2   mn  me  c 2 Conservation of energy and lepton number are violated.


(b) mn c 2  mp  m c 2 Conservation of energy is violated.
(c) Total momentum in the center of mass system is zero, so two photons (minimum) must
be emitted. Conservation of linear momentum is violated.
(d) No conservation laws are violated. This reaction, p p annihilation, occurs.
(e) Lepton number before interaction is +1; that after interaction is −1. Conservation of
lepton number is violated.
(f) Baryon number is +1 before the decay; after the decay the baryon number is zero.
Conservation of baryon number is violated.
314
Chapter 12 – Particle Physics
12-18. (a) The u and u annihilate via the EM interaction, creating photons.


u
u
u
(b) Two photons are necessary in order to conserve linear momentum.
(c)


W
d
u
12-19. (a) The strangeness of each of the particles is given in Table 12-6.
S  1 The reaction can occur via the weak interaction.
(b) S  2 This reaction is not allowed.
(c) S  1 The reaction can occur via the weak interaction.
12-20. (a) The strangeness of each of the particles is given in Table 12-6.
S  2 The reaction is not allowed.
(b) S  1 This reaction can occur via the weak interaction.
(c) S  0 The reaction can occur via either the strong, electromagnetic, or
weak interaction.
12-21. (a) n  n
(b) n  p
1 1
T3     1
2 2
1 1
T3     0
2 2
T 1
T  1 or 0
315
Chapter 12 – Particle Physics
(Problem 12-21 continued)
1 3

2 2
(c)    p
T3  1 
(d)    n
T3  1 
(e)    n
T3  1 
T
1
3

2
2
1 1

2 2
3
2
T
T
3
2
1
3
or
2
2
12-22. (a)    e   Electron lepton number changes from 0 to 1; violates conservation of
electron lepton number.
(b)  0  e  e  ve  ve Allowed by conservation laws, but decay into two photons via
electromagnetic interaction is more likely.
(c)    e  e     v Allowed by conservation laws but decay without the electrons
is more likely.
(d) 0       Baryon number changes from 1 to 0; violates conservation of baryon
number. Also violates conservation of angular momentum, which changes from 1/2 to
0.
(e) n  p  e  ve Allowed by conservation laws. This is the way the neutron decays.
12-23. (a)   0   
   0   
(b)   p   0
  n   
(c) 0  p   
0  n   0
(d)  0    
(e)       v
 0  e  e  e  e
      0
12-24.    p   0     
Because s have B = 0 and p has B = 1, conservation of B requires the  to have B = 1.
  0   
The   has B = 0, so conservation of B requires that the 0 have B = 1.
316
Chapter 12 – Particle Physics
12-25. (a) 0  0  0  0
S is conserved.
(b) 2  0  1
S is not conserved.
(c) 1  1  0
S is conserved.
(d) 0  0  0  1
S is not conserved.
(e) 3  2  0
S is not conserved.
12-26. Listed below are the baryon number, electric charge, strangeness, and hadron identity of
the various quark combinations from Table 12-8 and Figure 12-21.
Quark Structure
Baryon Number
Electric Charge (e)
Strangeness
Hadron
(a)
uud
+1
+1
0
p
(b)
udd
+1
0
0
n
(c)
uuu
+1
+2
0
Δ++
(d)
uss
+1
0
−2
0
(e)
dss
+1
−1
−2

(f)
suu
+1
+1
−1

(g)
sdd
+1
−1
−1

Note that 3-quark combinations are baryons.
12-27. Listed be below are the baryon number, electric charge, strangeness, and hadron identity
of the various quark combinations from Table 12-9 and Figure 12-21.
Quark Structure
Baryon Number
Electric Charge (e)
Strangeness
Hadron
(a)
ud
0
+1
0

(b)
ud
0
−1
0

(c)
us
0
+2
+1

(d)
ss
0
0
+1
0
(e)
ds
0
0
−1
 0
* forms  and   along with uu and dd
317
Chapter 12 – Particle Physics
12-28.
u
K0
d
d
c
g
u
b
g
K0
d
b

12-29. (a) T3 = 0 (from Figure 12-20a)
(b) T = 1 or 0 just as for ordinary spin.
(c) uds
B = 1/3 + 1/3 + 1/3 = 1
S = 0 + 0 + −1 = −1
C = 2/3 – 1/3 – 1/3 = 0
The T = 1 state is the Σ0.
The T = 0 state is the Λ0.
12-30. The +2 charge can result from either a uuu, ccc, or ttt quark configuration. Of these, only
the uuu structure also has zero strangeness, charm, topness, and bottomness. (From Table
12-5.)
12-31. The range R is R  c / mc2 (Equation 11-50). Substituting the mass of W + (from Table
12-4),
1.055  10 J s 3.00  10 m / s   2.44  10
R
81GeV / c 1.60  10 J / GeV 
34
2
8
18
10

12-32.
u
d

 0        v
W
 ds   du 
v
s
d
0
318
m  2.44  103 fm
Chapter 12 – Particle Physics
12-33.

p
u
0 
p
u
d
d


 uds   uud   ud 
Z0
Weak decay
u
12-34. n  p   
d
s
0
Q  mn c2  mp c 2  m c 2
  939.6  938.3  139.6 MeV
 138.3MeV
This decay does not conserve energy.

p
12-35.
u
0 
p

d
u
u
d

 uds   uud   ud 
Z0
Strong decay
u
d
s
12-36. (a) B = 1, S = −1, C = 0, B  0
(b) Quark content is: uds
12-37. (a) The   has charge +1, B = 0, and S = +1 from Table 12-6. It is a meson (quarkantiquark) structure. us produces the correct set of quantum numbers. (From Table
12-5.)
319
Chapter 12 – Particle Physics
(Problem 12-37 continued)
(b) The  0 has charge 0, B = 0, and S = +1 from Table 12-6. The quark-antiquark
structure tp produce these quantum numbers is d s . (From Table 12-5.)
12-38. (a) Being a meson, the D+ is constructed of a quark-antiquark pair. The only combination
with charge = +1, charm = +1 and strangeness = 0 is the cd . (See Table 12-5.)
(b) The D−, antiparticle of the D+, has the quark structure c d .
12-39. The  0 decays via the electromagnetic interaction whose characteristic time is
1020 s.
The  and  both decay via the weak interaction. The difference between these two
being due to their slightly different masses.
12-40. If the proton is unstable, it must decay to less massive particles, i.e., leptons. But leptons
have B = 0, so p  e  ve would have 1 = 0 + 0 = 0 and B is not conserved. The
lepton numbers would not be conserved either; a “leptoquark” number would be
conserved.


12-41. V  H 2O   0.75V  0.75 4 R 2 R , where R(Earth) = 6.37  106 m and
R  1km  103 m.

V  H 2O   0.75  4 6.37  106 m


 10   3.82 10
2
3
17

m3

M  H 2O   V  H 2O    3.82  1017 m3 1000kg / m3  3.82  1020 kg
Number of moles (H2O) = 3.82  1023 g / 18.02 g / mole  2.12  1022 moles
Number of H2O molecules = N A   # of moles 


 6.02  1023 molecules / mole 2.12  1022 moles

 1.28  1046 molecules H2O
Each molecule contains 10 protons (i.e., 2 in H atoms and 8 in the oxygen atom), so the
number of protons in the world’s oceans is N  1.28  1047.
320
Chapter 12 – Particle Physics
(Problem 12-41 continued)
The decay rate is
dN
  N where   1 /   1 / 1032 y  1032 y 1
dt


 1032 y 1 1.28  1047 protons

 1.28  1015 proton decays / y  4  107 decays / s
12-42. (a) p  e  0  ve


 
Q  mp c 2  M 0 c 2  mec 2 MeV
  938.3  1116  0.511 MeV  178MeV
Energy is not conserved.
(b) p     
Spin (angular momentum)
1
 0  1  1. Angular momentum is not conserved.
2
(c) p      0
Spin (angular momentum)
1
 0  0  0. Angular momentum is not conserved.
2
12-43. n, B = 1, Q = 0, spin = 1/2, S = 0
Quark strucure
u
d
d
B
1/3
+1/3
+1/3
=1
Q
2/3
−1/3
−1/3
=0
spin
1/2↑
1/2↑
1/2↓
= 1/2
0
+0
+0
=0
S
n , B = −1, Q = 0, spin = 1/2, S = 0
u
d
d
B
−1/3
−1/3
−1/3
= −1
Q
−2/3
+1/3
+1/3
=0
spin
1/2↑
1/2↑
1/2↓
= 1/2
0
+0
+0
=0
Quark strucure
S
321
Chapter 12 – Particle Physics
(Problem 12-43 continued)
(b)  0 , B = 1, Q = 0, spin = 1/2, S = −2
Quark strucure
u
s
s
B
1/3
+1/3
+1/3
=1
Q
2/3
−1/3
−1/3
=0
spin
1/2↑
1/2↓
1/2↑
= 1/2
0
−1
−1
= −2
S
(c)   , B = 1, Q = 1, spin = 1/2, S = −1
Quark strucure
u
u
s
B
1/3
+1/3
+1/3
=1
Q
2/3
2/3
−1/3
=0
spin
1/2↑
1/2↓
1/2↑
= 1/2
0
+0
−1
= −1
S
(d)  , B = 1, Q = −1, spin = 3/2, S = −3
Quark strucure
s
s
s
B
1/3
+1/3
+1/3
=1
Q
−1/3
−1/3
−1/3
= −1
spin
1/2↑
1/2↑
1/2↑
= 3/2
−1
−1
−1
= −3
S
(e)   , B = 1, Q = −1, spin = 1/2, S = −2
Quark strucure
u
d
d
B
1/3
+1/3
+1/3
=1
Q
−1/3
−1/3
−1/3
= −1
spin
1/2↑
1/2↓
1/2↑
= 1/2
0
−1
−1
= −2
S
322
Chapter 12 – Particle Physics
12-44. (a)
Quark strucure
d
d
d
B
1/3
+1/3
+1/3
=1
Q
−1/3
−1/3
−1/3
= −3/2
spin
1/2
1/2
1/2
= 3/2, 1/2
0
+0
+0
=0
u
c
B
1/3
−1/3
=0
Q
2/3
−2/3
=0
spin
1/2
1/2
= 1, 0
0
+0
=0
u
b
B
1/3
−1/3
=0
Q
2/3
+1/3
=1
spin
1/2
1/2
= 1, 0
0
+0
=0
s
s
s
B
−1/3
−1/3
−1/3
= −1
Q
1/3
+1/3
+1/3
=1
spin
1/2
1/2
1/2
= 3/2, 1/2
1
+1
+1
=3
S
(b)
Quark strucure
S
(c)
Quark strucure
S
(d)
Quark strucure
S
12-45. The Z 0 has spin 1. Two identical spin 0 particles cannot have total spin 1.
12-46. (a) The final products (p, γ, e−, neutrinos) are all stable.
(b) 0  p  e  ve  v  v
(c) Conservation of charge: 0  1  1  0  0  0  0
Conservation of baryon number: 1  1  0  0  0  0  1
323
Chapter 12 – Particle Physics
(Problem 12-46 continued)
Conservation of lepton number:
(i) for electrons: 0  0  1  1  0  0  0
(ii) for muons: 0  0  0  0  1  1  0
Conservation of strangeness: 2  0  0  0  0  0  0
Even though the chain has S  2, no individual reaction in the chain exceeds
S  1, so they can proceed via the weak interaction.
(d) No, because energy is not conserved.
12-47.  2, 1, 0, 1, 0   c u u
 0, 1,  2, 1, 0  c s s
 0, 0, 1, 0,  1  b s
 0,  1, 1, 0, 0  s d u
 0, 1,  1, 1, 0  c s d
 1, 1,  3, 0, 0  s s s
12-48. (a)
t1  x / u1 t2  x / u2  t  t2  t1 
x x

u2 u1
 u  u  x u
t  x  1 2   2
c
 u1u2 
(b)
E
mc 2
1  u 2 / c2
(Equation 2-10)
1/2
2
(mc 2 )2
(mc 2 )2
u   mc 2  
2
2
E 
 1 u / c 
  1  
 
1  u 2 / c2
E2
c   E  


Expanding the right side of the equation in powers of (mc 2 / E )2 and keeping only
2
the first term yields
324
Chapter 12 – Particle Physics
(Problem 12-48 continued)
u
1  mc 2 
 1 

c
2 E 
(c)
2
2
 1  mc 2 2
1  mc 2  
u1  u2  c 1  
 1 
 
2  E2  
 2  E1 

c(mc 2 )2  1
1  c(mc 2 ) 2  E12  E22 

 2  2
 2 2 
2
2
 E2 E1 
 E1 E2 
u  u1  u2 
c(2.2 eV / c 2  c 2 ) 2  (20 MeV) 2  (5 MeV) 2 
 (20 MeV) 2 (5 MeV) 2 
2



c(2.2 eV) 2  (106 MeV/eV) 2 
375

2
2
2
2
 (20) (5) (MeV) 
2
12
(2.2) (375) 10
c
 2.7 105 m/s
2
2
2(20) (5)

And therefore, t 
1.7 105 c  y  9.46 1015 m / c  y  2.7 105 m/s
 0.48s
(3.0 108 m/s)2
12-49. (a)    e  ve  v
Electron lepton number: 0  1  1  0  0
Muon lepton number: 1  0  0  1  1
Tau lepton number: 0  0  0  0  0
(b)       v  v
Electron lepton number: 0  0  0  0  0
Muon lepton number: 0  1  1  0  0
Tau lepton number: 1  0  0  1  1
(c)       v
Electron lepton number: 0  0  0  0
Muon lepton number: 0  1  1  0
Tau lepton number: 0  0  0  0
325
Chapter 12 – Particle Physics
12-50.  0    
hf

E 2   pc   m c 2
2

In lab:
2


0
hf
Conservation of momentum requires that each carry half of the initial momentum, hence

the total energy: 2  hf / c  cos  p
cos 


2
2
hf  E / 2   pc   m c 2 


1/ 2
p
p

1/ 2
2
2
2  hf / c 
2  pc   m c 2 
2c



pc


 pc 2  m c 2 



2
1/ 2


850MeV
 850MeV 2  135MeV 2 


1/ 2
 0.9876
  cos1  0.9876  9.02
12-51. (a) 0  p   
Energy: 1116MeV   938  140  MeV  38MeV conserved.
Electric charge: 0  1  1  0 conserved.
Baryon number: 1  1  0  1 conserved.
Lepton number: 0  0  0  0 conserved.
(b)   n  p 
Energy: 1197MeV   940  938 MeV  681MeV not conserved.
Electric charge: 1  0  1  1 conserved.
Baryon number: 1  1  1  0 not conserved.
Lepton number: 0  0  0  0 conserved.
This reaction is not allowed (energy and baryon conservation violated).
326
2
Chapter 12 – Particle Physics
(Problem 12-51 continued)
(c)    e  ve  v
Energy: 105.6MeV  0.511MeV  105.1MeV conserved.
Electric charge: 1  1  0  0  1 conserved.
Baryon number: 0  0  0  0  0 conserved.
Lepton number:
(i) electrons: 0  1  1  0  0 conserved.
(ii) muons: 1  0  0  1  1 conserved.
12-52. (a) The decay products in the chain are not all stable. For example, the neutron decays via
n  p  e  ve
Only the e+ and e− are stable.
(b) The net effect of the chain reaction is:   p  3e  e  3ve  2v  2v
(c) Charge: 1  1  3  1  1 conserved
Baryon number: 1  1  0  0  0  0  0  0  1 conserved.
Lepton number:
(i) electrons: 0  0  3  1  3  1  0  0  0 conserved
(ii) muons: 0  0  0  0  0  0  2  2  0 conserved
Strangeness: 3  0  0  0  0  0  0  0  0 not conserved
Overall reaction has S  3; however, none of the individual reactions exceeds
S  1, so they can proceed via the weak interaction.
12-53. The proton and electron are free particles. The quarks are confined, however, and cannot
be separated. The gluon clouds give the u and d effective masses of about 330 MeV/c2,
about 1/3 of the proton’s mass.
12-54. (a) 0  p   
 
Ekin   M 0  mp  m  c 2
 1116MeV / c 2  938.3MeV / c 2  139.6MeV / c 2  c 2
 38.1MeV
327
Chapter 12 – Particle Physics
(Problem 12-54 continued)
(b) Because the  0 decayed at rest, the p and   have momenta of equal magnitudes and
opposite direction.
mp v p  m v  mp / m  v / v p
1
m v2 m  m 2 m
938.3
p
p
2

 


 6.72

Ekin  p  1 m v 2 m p  m 
m 139.6
2 p p
Ekin  
(c) Ekin  Ekin  p   Ekin    Ekin  p   6.72Ekin  p   7.72Ekin  p   38.1MeV
Ekin  p   38.1MeV / 7.71MeV  4.94MeV
Ekin    6.72Ekin  p   33.2MeV
12-55. 0  0  
(a) ET for decay products is the rest energy of the  0 , 1193MeV.
(b) The rest energy of 0  116MeV , so E  1193MeV  1116MeV  77MeV
and p  E / c  77MeV / c .
(c) The  0 decays at rest, so the momentum of the  0 equals in magnitude that of the
photon.
 


2
Ekin 0  p2 / 2M      77MeV / c  / 2 1116MeV / c 2 
 2.66MeV
small compared to E
(d) A better estimate of E and p are then E  77MeV  2.66MeV  74.3MeV and
p  74.3MeV / c.
12-56. (a) t  t2  t1 
t 
x x x  u1  u2 
 
Note that u1u2  c 2
u2 u1
u1u2
x  u1  u2 
c
2

xu
c2
328
Chapter 12 – Particle Physics
(Problem 12-56 continued)
(b) E 
mc
2
1  u 2 / c2

1  m c 2
u
o
(Equation 2-10). Thus,  
c 
E2


2




1/ 2
1  m c2 
 1  o 
2 E 
2
2
 1  m c 2 2
1  mo c 2  
o
(c) u1  u2  c 1  
 1 
 
2 E  
 2  E1 

2
2

c mo c 2
c  mo c 2  c  mo c 2 
 
  
 
2  E2  2  E1 
2

 


2
 E12  E22 
 2 2 
 E1 E2 

2
2
2
6
6
c  20eV   20  10 eV  5  10 eV 

 20  106 eV 2 5  106 eV 2 
2



2
2
2
c  20eV  
20    5


  20 2  52 106 eV
2






  7.5  1012 c
2



12
xu 170, 000c y  7.5  10 C
T  2
 1.28  106 y  40.3s
c
c2
(d) If the neutrino rest energy is 40eV, then u  3.00 1011 c and t  161s. The
difference in arrival times can thus be used to set an upper limit on the neutrino’s mass.
12-57.    e  ve  v
      v  v
   d  u  v
The last decay is the most probable (three times as likely compared to each of the others)
due to the three possible quark colors.
329
Chapter 12 – Particle Physics
12-58. (i )
(ii )
dp
 ev  B
dt
dE
 0 which follows from the fact that the Lorentz force is  v; therefore, v  v =
dt
constant and thus   v   constant.
Equation (i) then becomes:
For circular orbits
dp
dv
 m
 ev  B
dt
dt
m v 2
 evB or, re-writing a bit,
R
m v  p  eBR and pc  ceBR 
1GeV
 0.35BR GeV
1.60  1019 J
and finally, p  0.35BR GeV / c
330
Chapter 13 – Astrophysics and Cosmology
vW
13-1.
vW  vE  4km / s. Assuming Sun’s rotation to be
uniform, so that vW  vE , then vW  vE  2km / s.
r
Because v  2 / T , vE  2 r / T or
ω
T
vE
13-2.



2 6.96  105 km
2 r

 2.19  106 s  25.3 days
vE
2km / s

11
1.99  1030
2GM 2 2 6.67  10
U 

R
6.96  108

2
J  7.59  1041 J
The Sun’s luminosity L  3.85  1026W
 tL 
U
L

7.59  1041 J
 1.97  1015 s  6.26  107 years
26
3.85  10 J / s
13-3. The fusion of 1H to 4 He proceeds via the proton-proton cycle. The binding energy of
4
He is so high that the binding energy of two 4 He nuclei excees that of 8 Be produced in
the fusion reaction: 4 He  4 He  8 Be   and the 8 Be nucleus fissions quickly to two
4
He nuclei via an electromagnetic decay. However, at high pressures and temperatures a
very small amount is always present, enough for the fusion reaction:
8
Be  4 He  12C   to proceed. This 3- 4 He fusion to 12C produces no net 8 Be and
bypasses both Li and B, so their concentration in the cosmos is low.
13-4. The Sun is 28, 000c y from Galactic center = radius of orbit

15
2 r 2 28, 000c y  9.45  10 m / c y
 time for 1 orbit 

v
2.5  105 m / s
 6.65  1015 s  2.11  108 yr
331

Chapter 13 – Astrophysics and Cosmology
1 H atom / m3
13-5. Observed mass (average)
missing mass
500 photons / cm 3
9 1.67 10
kg / m3
1.56 10
26
27
kg / m3
10% of total mass (a)
kg / m3
500 106 photons / m 3, so the mass of each photon would be
1.50 10 26 kg / m 3
500 106 photons / m 3
3.01 10
3.01 10 35 kg
1.60 10 19 J / eV
or mv
27
1.67 10
35
kg
m2
s2
c2
16.9eV / c 2
13-6.
5
4
T
3
4
10 K
2
1
0
0
1
2
3
4
M /M
13-7.
1c s
c 1s
1c min
1c h
1c day
3.00 108 m / s 1s
c 1min 60s / min
6
7
8
3/ 8
3.00 108 m 3.00 105 km
3.00 105 km 60s 1.80 107 km
c 1h 3600s / h 1.08 109 km
c 24h 3600s / h
13-8. (a) See Figure 13-16. 1AU
2.59 1010 km
1.496 1011 m. R 1pc when
332
1", so R
1AU
1"
Chapter 13 – Astrophysics and Cosmology
(Problem 13-8 continued)
or R
3600 "
1
1AU
1"
180
rad
3.086 1016 m
9.45 1015 m / c y
1 pc
3.26c y
0.01", R 100 pc and the volume of a sphere with that radius is
(b) When
4 3
R
3
V
3.086 1016 m 1 pc
4.19 106 pc3. If the density of stars is 0.08 / pc3, then the number
of stars in the sphere is equal to 0.08 / pc3 4.19 106 pc3
13-9.
L
4 r2 f
m1
m2
4 rp2 f p and LB
Thus, Lp
rp2 f p
rB2 f B
rB2
12 pc,
Because rp
2.5 log f1 / f 2
4 rB2 f B and Lp
0.30
rp f p / f B
rB
f p / fB
1/ 2
13-10. (a) M
0.3M
Te
3300K
(b) M
3.0M
Te
13, 500K
(c) R
M
R
R
M
Similarly, R3.0
3.0 R
tL
M
3
(d) tL 0.3
or tL 0.3
tL
0.3M
2.00
12 2
17.0 pc
5 10 2 L
L
L 102 L
1.93 1025W
3.85 1028W
R /M
M
0.3M
R0.3
LB
rp2 f p / f B
1.16 0.41
2.5
log f p / f B
3.4 105 stars.
3
2.09 108 m
0.3R
2.09 109 m
3
M
0.3M
tL / M
3
tL M 3
tL M 3 0.3M
37tL . Similarly, tL 3.0
333
3
0.3
0.04tL
3
tL
Chapter 13 – Astrophysics and Cosmology
S
R
13-11. Angular separation
100 106 km
100c y
distance between binaries
distance Earth
1011 m
100c y 3.15 107 s / y
1.057 10 7 rad
6.06 10 6 degrees 1.68 10 9 arcseconds
13-12. Equation 13-18:
56
26
13 24 He 4n. m56 Fe
Fe
Energy required: 13 m4 He
1u
4mn
931.49432 MeV / c 2
Equation 13-19:
4
2
He
Energy required: 2m1H
13-13.
55.939395u, m4 He
2n
2mn
m4 He
0.020277u
(a) r
1.5 c y
; assuming constant expansion rate,
2
m1H
Age of Shell
(b) Lstar
R
M
Te
M 1/ 2
R
mstar ,
M
Te star
or Rstar
Lstar
L
28.3MeV
1.5 c y / 2
2.4 104 m / s
Te
M 1/ 2
Te
L
star
2.95 1011 s
9400 y
1.4Te
M4
L
M4
L /M4
Te
1/ 2
M star
,
1/ 2
M
Using either the Te or L relations, Rstar
1.007825
12L
Te / M 1 / 2 ,
R /M ,
Rstar
120 MeV
2 1H
1.5 c y
M
1.008665u.
0.129104u.
m56 Fe
0.129104u
24 km / s
R
4.002603u, mn
Lstar
M star
R
M
1/ 2
1.86 R
334
L
4
M star
4
M
Te star
Te
2
R
2
1.4 R
1.96 R
Chapter 13 – Astrophysics and Cosmology
2GM / c2
13-14. RS
(Equation 13-24)
2 6.67 10
(a) Sun RS
(b) Jupiter mJ
318mE
1.99 1030 / c2
2.9 103 m 3km
2.8m
RS
8.86 10 3 m
(c) Earth RS
13-15. M
11
9mm !
2M
(a) (Equation 13-22) R 1.6 1014 M
(b) 0.5rev / s
1
I
2
2
2
MR 2
5
I
d
2
1 2
2M
2 5
1.01 104
d
1
/ day
108
where
2 8.0 1038 J
d
8
1.99 1030 1011
(b) Its radius would be RS
11
1.8 1042 kg
2GM / c 2 (Equation 13-24).
1.8 1042 / c2
2.6 1015 m 17, 000 AU
72, 000 km / s.
(a) v
Hr
r
v
H
72, 000km / s
21.2 km / s 106 c y
3.40 109 c y
(b) From Equation 13-29 the maximum age of the galaxy is:
1/ H
4.41 1017 s
L 1.85 1025W
10% of total
(a) Mass of a central black hole = 9 1.99 1041
13-17. v
8.0 1038 J
1011 stars of average mass M , therefore the visible mass =
1.99 10 41 kg
2 6.67 10
2
1.85 1025 J / s
5
10 d 8.64 10 s / d
13-16. Milky Way contains
RS
1.01 10 4 m
where for a sphere
I d
2
1/ 3
1.6 1014 2 M
rad / s
I
(c) d
1/ 3
1.4 1010 y
335
Chapter 13 – Astrophysics and Cosmology
(Problem 13-17 continued)
1/ H
r/v
1/ H
1/ H
r/v
1/ H
r
r
10%
so the maximum age will also be in error by 10%.
13-18. The process that generated the increase could propagate across the core at a maximum rate
of c, thus the core can be at most 1.5 y 3.15 107 s / y 3.0 108 m / s 1.42 1016 m
9.45 104 AU in diameter. The Milky Way diameter is
60, 000c y
3.8 109 AU .
13-19. Combining Hubble’s law (Equation 13-28) and the definition of the redshift (Equation 1327) yields
z
Hor
c
0
r 5 106 c y
(a)
Hor
1
c
0
21.7
km
5 106 c y
1 656.3nm
s 106 c y 3 108 m/s
656.5nm
(b) r
50 10 6 c y
Similarly,
(c) r
(d) r
658.7nm
500 106 c y
Similarly,
680.0nm
5 109 c y
Similarly,
0
893.7nm
336
Chapter 13 – Astrophysics and Cosmology
13-20. Equation 13-33:
3H 2
8 G
c
3
8
1/ H
2
G
3
c
8
1.5 1010 y 3.15 107 s / y
8.02 10
2
6.67 10
11
27
kg / m3
Nm 2 / kg 2
(This is about 5 hydrogen atoms/m3 !)
13-21. Present size
1010 c y
Sp
1
T
1
with T
T
Sp
2.7 K
2.7 1010 c yK
(a) 2000 years ago, S = Sp
(b) 106 years ago, S = Sp
(c) 10 seconds after the Big Bang, S
2.7 1010 c yK / 109 K 2.7 10 9 S p
2.7 1010 c yK / 5 109 K 5.4 10
(d) 1 second after the Big Bang, S
10
Sp
25c y
5c y
(e) 10-6 seconds after the Big Bang,
S
13-22.
2.7 1010 c yK / 5 1012 K 5.4 10
proton
3
pl
10
1.67 10
10
osmium
5.5 10 8 kg
m pl
Planck time
15
3
27
kg
m
3
35
m
13
Sp
0.005c y
6.4 10 4 AU
5.5 1097 kg / m3
3
1.67 1018 kg / m3
3
2.45 104 kg / m3
13-23. Wien’s law (Equation 3-11):
max
2.898mm K
T
2.898mm K
2.728K
1.062mm
(this is in the microwave region of the EM spectrum)
13-24. Muon rest energy
208me
106MeV / c2 . The universe cooled to this energy (average)
at about 10-3s (see Figure 13-34). 2.728K corresponds to average energy = 10-3 eV.
Therefore, m
10 3 eV 1.6 10
c2
19
J / eV
337
1.8 10
39
kg
Chapter 13 – Astrophysics and Cosmology
13-25.
0
M / 4/3
r 3 t0
r t
R t r t0
t
(Equation 13-37) \
M
0
0
4/3
R3 t
r3 t
M / 4/3
r t /R t
M
R3 t
3
4/3
r2 t
t
13-26.
B
If Hubble’s law applies in A, then
vBA
FBC
vCA
HrCA .
From mechanics,
FBA
vBC
C
FCA
HrBA ,
vBA
vCA
H rBA
rCA
HrBC
and Hubble’s lab applies in C, as well, and
A (Milky Way)
by extension in all other galaxies.
13-27. At a distance r from the Sun the magnitude of the gravitational force acting on a dust
GM m
where m (4 / 3) a 3 . The force acting on
r2
the particle due to the Sun’s radiation pressure at r is given by: (See Equation RP-9.)
article of radius a is: F
grav
Frad
a 2 Prad
a2
U
3
where
a2 is the cross sectional area of the particle and U
is the energy density of solar radiation at r. U is given by: (See Equation 3-6.)
U
4
R
c
4 L
c 4 r2
Therefore, Frad
a2
1 4L
3 4 r 2c
The minimum value of a is obtained from the condition that Fgrav
338
Frad :
Chapter 13 – Astrophysics and Cosmology
(Problem 13-27 continued)
GM m
1 4L
a2
2
r
3 4 r 2c
GM (4 / 3) a 3
1 4L
a2
2
r
3 4 r 2c
Simplifying this expression yields:
L
4 cGM
a
3.84 1026 W
4 (3.00 108 m/s)(6.67 10 11 N m2 / kg 2 )(1.99 1030 kg)(5500 kg/m3 )
a 1.40 10 7 m or 1.40 10 5 cm
Note that (i) a is very small and (ii) the magnitude of a is independent of r.
a
13-28. (Equation 13-31) Robs
M / 4/3
Z
Remit 1 Z
r3
0
Substituting for r0 in the
0
0
M / 4/3
Z / 1 Z
0
or
13-29. (a) H available for fusion = M
(b) Lifetime of H fuel =
r 1 Z
M / 4/3
r03
equation:
r 1 Z
3
r0
3
M / 4/3
Z
0
1 Z
r3 1 Z
3
3
0.75 0.13 2.0 1030 kg 0.75 0.13 2.0 1029 kg
2.0 1029 kg
6.00 101 kg / s
3.3 1017 s
3.3 1017 s / 3.15 107 s / y
1.03 1010 y
(c) Start being concerned in 1.03 1010 y 0.46 1010 y
5.7 109 y
13-30. SN1987A is the Large Magellanic cloud, which is 170,000c•y away; therefore
(a) supernova occurred 170,000 years BP.
339
Chapter 13 – Astrophysics and Cosmology
(Problem 13-30 continued)
(b) E
K
mo c 2
mo c 2
1
109 eV ,
v2
c2
mo2
9.28 108 eV
109
9.38 108
9.38 108
v2
1 2
c
0.875c
or v
Therefore, the distance protons have traveled in 170,000y
=v
170, 000 y 149, 000c y. No, they are not here yet.
1.99 1030 kg.
13-31. M
(a) When first formed, mass of H = 0.7 M , m 1H
number of H atoms
0.7 M
1.007825u 1.66 10
He; 4 1H
(b) If all H
produced =
4
27
1.007825u 1.66 10
kg / u
27
8.33 1056
He 26.72eV . The number of He atoms
8.33 106
.
4
Total energy produced =
8.33 106
4
5.56 1057 MeV
26.72MeV
8.89 1044 J
(c) 23% of max possible = 0.23 8.89 1044 J
0.23 8.89 1044
L
tL
13-32. (a) F
Gm1m2 / r 2
v2 / r
ac m2
5.53 1017 s 1.7 1010 y
3.85 1026W
L
v 2 / r m2
Gm1 / r 2 and orbital frequency f
v/2 r
Substituting for f and noting that the period T
or, T 2
kg / u, thus
1/ f , 4
4 2 r 3 / Gm1, which is Kepler’s third law.
(b) Rearranging Kepler’s third law in part (a),
340
2
f2
Gm1 / r 3
Chapter 13 – Astrophysics and Cosmology
(Problem 13-32 continued)
mE
2 3
moon
4
r
/ GT
4
2
6.67 10
11
2
3.84 108 m
3
Nm2 / kg 2 27.3d 8.64 104 s / d
2
6.02 1024 kg
(c) T
2
11
2
T
12d
2 2
2
1.97 107
r2
2
2
12 24 3600
r2
6.06 10 6 / s 2
2
v2
r
G
m1m2
and
r
2 3
m2
r
G
from the graph v1
200km / s and v2
3.3 1010 m and, similarly, r2
100km / s
1.6 1010 m
4.9 1010 m
Assuming circular orbits,
m1
1.48 1022 kg
2
m1m2
, then
m1 m2
1
5.44 103 s 1.5h
24
3
6.46d 3.1 10 s / d
, and
1/ 2
3
6.02 10
4
200 103 m / s
6.06 10 6 / s
r1
11
Gm1m2
or m1
r2
r
r
r1 1, v2
r
6.67 10
m2 : reduced mass
m1m2
m1 m2
r1
6.67 106
2
6.67 10
(b) For m1
(c) v1
1/ 2
4
(d) mcomb
13-33. (a) T
rsh3
GmE
m1v12
r1
6.63 1030 kg and m2
m2v22
and m1
r2
1.37 1031 kg
341
r1v22
m2 Substituting yields,
r2v12
Chapter 13 – Astrophysics and Cosmology
13-34. E
1 2
mv
2
or GM m / r
GmM / r
1 2
mv
2
mv 2
1 GM m
2 r
E
FG
GM m / r 2
mv 2 / r
1
GM m / r
2
GM m
r
1
2
GM m
r
13-35.
dV
20km / s
106 c y
H
universe V
Current average density = 1H atom / m3
4 3
R
3
V
dV
4 R 2 dR
dR
The current expansion rate at R is:
v
dV
HR
20km / s
1010 c y
6
10 c y
dR
20 107 m / s 3.16 107 s / y
4 R 2 dR
1010
4
7.07 1074 m3
106 c y
Current volume V
2
20 104 km / s
20 107 m / s
106 y
106 y
9.45 1015 m / c y
2
20 107 m / s 3.16 107 s / y
106 y
106 y
# of H atoms
to be added
106 c y
4
3
"new" H atoms =
1010
3
8.4 1077 m3
7.07 1074 atoms / 106 c y
8.4 1077 m3
342
0.001 "new" H atoms / m3 106 c y ; no
Chapter 13 – Astrophysics and Cosmology
13-36. (a) Equation 8-12: vrms
3RT / M is used to compute vrms vs T for each gas ® = gas
constant.
M
Gas
3
( 10 kg )
vrms (m/s) at T = :
3R / M
50K
200K
500K
750K
1000K
H2O
18
37.2
263
526
832
1020
1180
CO2
44
23.8
168
337
532
652
753
O2
32
27.9
197
395
624
764
883
CH4
16
39.5
279
558
883
1080
1250
H2
2
111.6
789
1580
2500
3060
3530
He
4
78.9
558
1770
1770
2160
2500
The escape velocities vsc
2 gR
2GM / R , where the planet masses M and
radii R, are given in table below.
Planet
Earth
Venus
Mercury
Jupiter
Neptune
Mars
vesc (km/s)
11.2
10.3
4.5
60.2
23.4
5.1
vesc/6 (m/s)
1870
1720
750
10,000
3900
850
On the graph of vrms vs T the vesc/6 points are shown for each planet.
343
Chapter 13 – Astrophysics and Cosmology
(Problem 13-36 continued)
2GM / R
(b) vesc
2GM Pl / RPl
vPl
M Pl / RPl
M E / RE
vPl
vE
2GM E / RE
vE
M E / RE
M E / RE
vPl
vE
M Pl / M E
RPl / RE
(c) All six gases will still be in Jupiter’s atmosphere and Netune’s atmosphere, because
vesc for these is so high. H2 will be gone from Earth; H2 and probably He will be gone
from Venus; H2 and He are gone from Mars. Only CO2 and probably O2 remain in
Mercury’s atmosphere.
13-37. (a) α Centauri d in pc
d
1AU
sin 0.742 "
(b) Procyon d
Earth's orbit radius (in AU)
sin p
2.78 105 pc
1AU
sin 0.0286 "
9.06 105 c y
7.21 105 pc
2.35 106 c y
13-38. Earth is currently in thermal equilibrium with surface temperature
T 4 and I
Earth radiates as a blackbody I
is f
102 L then f
0.338 1.36 103 102
away.
vrms
4
459W / m2 . The solar constant
1.36 103W / m 2 currently, so Earth absorbs 459/1360 = 0.338 of incident solar
energy. When L
I
300
300K. Assuming
However,
3RT / M
102 f . If the Earth remains in equilibrium.
T 4 or T
the
vrms
3 8.31 949
18 10 3
solution to problem 14-26).
994 K
for
676 C sufficient to boil the oceans
H2 O
molecules
1146m / s 1.15km / s.
at
994K
is
The vesc = 11.2km/s (see
Because vrms ≈ 0.1 vesc, the H2O will remain in the
atmosphere.
344
Chapter 13 – Astrophysics and Cosmology
13-39. (a)
a
n = grains/cm3
a
total scattering area =
dust grain
R
which is
R
dx
photons
N/mrs
N
R 2 na 2 dx
R2 a 2 n dx
a2
R2 n dx of the
total area = fraction scattered = dN/N
d
dN
N
N0
n R 2 dx or N
N 0e
n R2d
0
From those photons that scatter at x = 0 (N0), those that have not scattered again after
traveling some distance x = L is N L
N0e
n R2 L
. The average value of L (= d0) is
given by:
L
d0
(b) I
0
dN L
dL
dL
dN L
dL
dL
0
I 0e
d / d0
1
n R2
Note:
near the Sun d0
1
n
n R 2 N 0e
n R2 L
R 10 5 cm
3000c y
3000c y 9.45 1017 cm / c y
(c)
dN L
dL
10
n 1.1 10
2
5
12
/ cm3
2 gm / cm3
grains
mgrains
3
cm of space
mass in 300c y
2
4
3
10
5
3
1.1 10
9.41 10 27 gm / cm3
M
0.0012
12
9.41 10
27
9.45 1017 cm / c y
0.1%M
345
/ cm3
gm / cm3
3
300
Chapter 13 – Astrophysics and Cosmology
13-40. 56 11H
14 142 He
56
26
Fe 2
2e
14 4.002603 m56Fe
14 m4He
2
2e
2.04MeV / c 2
55.939395u
56.036442u
14 26.72 MeV
2.04MeV
Net energy difference (release) =
90.40MeV
466.5MeV
56
2 26
Fe
2m
56
Fe
112
48
Cd
4
4e
4
2 55.939395u
m
Net energy required = 2m56Fe
13-41. (a) dt
4.08MeV / c 2
4e
112
111.902762u
Cd
0.023972u 4.08MeV
m112Cd
18.25MeV
1.024 104 2G 2 M dM
hc 4
hc 4
1.024 1024
dM
dt
rearranging, the mass rate of change is
2
G2M
Clearly, the larger the mass M, the lower the rate at which the black hole loses mass.
(b) t
t
(c) t
1.024 104
2
6.62 10
2
2.0 1030
2
hc 4
3.35 1044 s
1.024 104
1.06 1037 y far larger than the present age of the universe.
4
6.67 10
6.63 10
t
11
34
11
2
2.0 1030 1012
3.00 108
4
3.35 1068 s 1.06 1061 y
346
2