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C
HAPTER
30
Quick Quizzes
1.
(c). The total energy released was
( )
( )
39
30
10 ton
4.
01
0
Jton
1.
21
1
4
J
E
=×
×
. The mass
equivalent of this quantity of energy is
( )
14
3
2
2
8
0 J
31
0 kg ~ 1 g
3.
ms
E
m
c
−
×
==
= ×
×
2.
(a). This reaction fails to conserve charge and cannot occur.
3.
(b). This reaction fails to conserve charge and cannot occur.
4.
(c), (e). The proton and the electron each have spin
1
2
s
=
. The two possible resultant spins
after decay are 1 (spins aligned) or 0 (spins antialigned). Neither equal the spin of a
neutron,
1
2
s
=
, so spin angular momentum is not conserved. The are no leptons present
before the proposed decay and one lepton (the electron) present after decay. Thus, the
decay also fails to conserve lepton number.
441
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Problem Solutions
30.1
1
235
141
92
1
0
56
36
0
n
U
Ba + Kr +
3n
+→
To conserve both charge and the total number of nucleons, it is seen that this reaction
must yield
3 neutrons
in addition to the other fission products.
30.2
The energy released in the reaction
is
12
3
5
9
8
1
3
5
1
40
52
0
Zr + Te +
()
98
135
22
UZ
r
2
n
Qm
cm
mm m
⎡⎤
=∆
=
−
−
−
⎣⎦
T
e
c
( )
( )
235.043 924 u
2 1.008 665u
97.912 0 u
134.908 7 u
931.5 M eV u
=−−
−
=
192
MeV
30.3
The energy released in the reaction
is
1
88
136
1
38
54
0
Sr + Xe + 12 n
US
r
11
n
=
−
−
−
X
e
c
( )
( )
11 1.008 665 u
87.905 618 u
135.907 215u
=− −
−
=
126
30.4
Three different fission reactions are possible:
1
90
144
1
09
23
8 5
4
0
nUS
rX
e
2
+→+ +
n
Xe
1
143
1
4
0
e
3
n
3
5 9
0 1
4
2
1
4
0
e
4
n
142
30.5
(a) With a specific gravity of 4.00, the density of soil is
3
4000 kg m
ρ
=
. Thus, the mass
of the top 1.00 m of soil is
( )
2
27
3
kg
1 m
4000
1.00 m
43 560 f
t1
.
62
10 kg
m3
.
281 ft
mV
⎛⎞
==
= ×
⎢⎥
⎜⎟
⎝⎠
442
CHAPTER 30
At a rate of 1 part per million, the mass of uranium in this soil is then
7
U
66
1.62
10 kg
10
m
m
×
==
=
16.2 kg
(b) Since 0.720% of naturally occurring uranium is
235
, the mass of
in the soil of
part (a) is
U
U
( ) ( )( )
33
U
U
7.20
0.117 kg
mm
−−
=×
=
=
117 g
30.6
At 40.0% efficiency, the useful energy obtained per fission event is
( )
( )
13
11
0.400 200 M eV event 1.60
J M eV
1.28
J event
event
E
=
×
The number of fission events required each day is then
( )
( )
94
24
1.00
10 J s 8.64
10 s d
6.75
eventsd
t
N
E
−
××
℘⋅
×
Each fission event consumes one
atom. The mass of this number of
atoms is
U
U
atom
mN
m
=
( ) ( )
( )
27
events d
235.044 u
1.66
kg u
−
⎡
⎤
×
=
⎣
⎦
2.63 kg d
In contrast, a coalburning steam plant producing the same electrical power uses more
than
6
61
0
k
gd
×
of coal.
30.7
The mass of
in 1.0kg of fuel is 0.017 kg, and the number of
nuclei is
U
U
()
( )
22
0.017 kg
4.35
235.044
u1
.66
m
N
m
−
×
At 208 MeV per fission event and 20% efficiency, the useful energy available from this
number of fission events is
( )
( )
( )
events
208 M eV event 1.60
J M ev
0.20
2.
91
0 J
E
−
⎡⎤
×
=
×
⎣⎦
443
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From
, the distance the ship can travel on this 1.0kg of fuel is
drag
Work F
s E
=⋅
=
11
6
5
2.
91
0 J
0 m
1.
01
0 N
E
s
F
×
==
=
×
=
3
0 km
×
×
(or about 1800 miles)
30.8
The total energy released was
( )
( )
39
20
10 ton
4.
0
Jton
8.
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This homework help was uploaded on 03/21/2008 for the course PY 211 / 212 taught by Professor Chilton during the Spring '08 term at N.C. State.
 Spring '08
 Chilton

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