Physics for Scientists and Engineers with Modern Physics 6th Edition – Complete Solutions Manual (All Chapters Included)

SOLUTION MANUAL
All Chapters Included


For

PHYSICS For SCIENTIST And ENGINEERS

6TH EDITION

, Chapter 45 Solutions

*45.1 m = (mn + M U) – (M Zr + MTe + 3mn)

m = (1.008 665 u + 235.043 924 u) – (97.912 0 u + 134.908 7 u + 3(1.008 665 u))

m = 0.205 89 u = 3.418  10 –28 kg so Q = mc 2 = 3.076  10 –11 J = 192 MeV



1 235
45.2 Three different fission reactions are possible: n U 90Sr  144
Xe  2 1n 144
0 92 38 54 0 54
Xe

1 235
n U 90Sr  143
Xe  3 1n 143
1
n 235
U 90Sr  142
Xe  4 1n 142
0 92 38 54 0 54
Xe 0 92 38 54 0 54
Xe



1 232 233
45.3 n Th 233Th 233
Pa  e    Pa 233U  e  
0 90 90 91 91 92



1 238 239 239 239 239
45.4 n U U Np  e   Np Pu  e  
0 92 92 93 93 94




45.5 (a) Q  mc2  mn  M U235  MBa141  M Kr92  3mnc2

m  1.008 665  235.043 924  140.913 9  91.897 3  3  1.008 665u  0.215 39 u
Q  0.215 39 u931.5 MeV u  201 MeV
m 0.215 39 u
(b) f   9.13  104 
0.0913%
mi 236.052 59 u



45.6 If the electrical power output of 1000 MW is 40.0% of the power derived from fission
reactions, the power output of the fission process is
1000 MW  J  8.64  104 
 2.50  109   2.16  1014 J/d
0.400  s  d 

 J  1 fission   1 eV  24 1
The number of fissions per day is 2.16  1014  6.74  10 d
 d  200  106 eV   1.60  1019 J

This also is the number of 235U nuclei used, so the mass of 235U used per day is

6.74  1024 nuclei 235 g/mol 
 2.63  103 g/d =
  2.63 kg/d
 d   6.02  1023 nuclei/mol 
In contrast, a coal-burning steam plant producing the same electrical power uses more than
6  106 kg/d of coal.

, Chapter 45 Solutions 593


45.7 The available energy to do work is 0.200 times the energy content of the fuel.

 0.0340 235
U 1000 g  1 mol 6.02  1023  (208)(1.60  1013 J)
1.00 kg fuel  1 kg  235 g  
 fuel  mol  fission 


2.90  10 J0.200  5.80  10
12 11

J  1.00  105 N  d
d = 5.80  106 m = 5.80 Mm

Goal Solution
Suppose enriched uranium containing 3.40% of the fissionable isotope 23592U is used as fuel for a ship. The
water exerts an average frictional drag of 1.00  105 N on the ship. How far can the ship travel per
kilogram of fuel? Assume that the energy released per fission event is 208 MeV and that the ship's
engine has an efficiency of 20.0%.

G: Nuclear fission is much more efficient for converting mass to energy than burning fossil fuels.
However, without knowing the rate of diesel fuel consumption for a comparable ship, it is difficult
to estimate the nuclear fuel rate. It seems plausible that a ship could cross the Atlantic ocean with
only a few kilograms of nuclear fuel, so a reasonable range of uranium fuel consumption might be
10 km / kg to 10 000 km/kg.

O: The fuel consumption rate can be found from the energy released by the nuclear fuel and the work
required to push the ship through the water.
A: One kg of enriched uranium contains 3. 40% 235
U so m  1000 g0.0340  34.0 g
92 235

In terms of number of nuclei, this is equivalent to
 1 
N 235  34.0 g   23 22
 6.02  10 atoms / mol  8.71  10 nuclei
 235 g / mol

If all these nuclei fission, the thermal energy released is equal to

 
8.71  1022 nuclei 208 MeV 
 
1.602  1019 J eV  2.90  1012 J
 nucleus
work output fd cos
Now, for the engine, efficiency  or e
heat input Qh
So the distance the ship can travel per kilogram of uranium fuel is

d
eQh
 
0.200 2.90  1012 J   5.80  10 6 m
5
f cos0 1.00  10 N

L: The ship can travel 5 800 km/kg of uranium fuel, which is on the high end of our prediction range.
The distance between New York and Paris is 5 851 km, so this ship could cross the Atlantic ocean on
just one kilogram of uranium fuel.




© 2000 by Harcourt, Inc. All rights reserved.

, 594 Chapter 45 Solutions

1 3
 3V  A 4r 2
45.8 (a) For a sphere: V  4 r 3 and r so  
3
4.84V – 1/ 3
 4  V ()r 3


3 A 6l2
(b) For a cube: V and  V1 3 so  
l l 6V – 1/ 3
V l 3




3 V
1/3 A 2a 2
 8a2 
(c) For a parallelepiped: V = 2a and a so   6.30V – 1/ 3
 2 V 2a 3



(d) Therefore, the sphere has the least leakage and the parallelepiped has the greatest leakage
for a given volume.




 
45.9 
mass of 235U available  0.007 109 metric tons  10 6 g  7  1012 g
 1 metric ton
 
 7  1012 g   nuclei
number of nuclei ~ 6.02  1023  1.8  1034 nuclei
 235 g mol  mol 
 

The energy available from fission (at 208 MeV/event) is


E ~ 1.8  1034 events 208 MeV / event1.60  10 13

J / MeV  6.0  1023 J

E 6.0  1023 J  1 yr 
This would last for a time of t
P
~
7.0  1012 J s

 8.6  10
10

s 
3.16  107 s
~ 3000 yr




60.0 s
45.10 In one minute there are = 5.00  10 4 fissions.
1.20 ms

So the rate increases by a factor of (1.000 25)50000 = 2.68  105




45.11 P = 10.0 MW = 1.00  107 J/s

If each decay delivers 1.00 MeV = 1.60  10–13 J, then the number of decays/s = 6.25  10 19 Bq