Chapter 30
Nuclear Energy and
Elementary Particles
Problem Solutions
30.6
At 40.0% efficiency, the useful energy obtained per fission event is
Eevent = 0.400( 200 M eV event) ( 1.60 × 10−13 J M eV ) = 1.28 × 10−11 J event
The number of fission events required each day is then
9
4
P⋅t ( 1.00× 10 J s)( 8.64 × 10 s d )
N =
=
= 6.75× 1024 events d
−11
Eevent
1.28× 10 J event
Each fission event consumes one
235
U atom. The mass of this number of
235
U atoms is
m = N m atom
= ( 6.75× 1024 events d ) ⎡⎣( 235.044u ) ( 1.66 × 10−27 kg u ) ⎤⎦ = 2.63 kg d
In contrast, a coal-burning steam plant producing the same electrical power uses more
than 6× 106 kg d of coal.
t 1800 miles)
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CHAPTER 30
30.8
The estimated weight of the naturally occurring uranium is
⎞
1N
⎛ 2000 lbs⎞ ⎛
12
w = ( 1.0 × 109 tons) ⎜
⎟ = 8.9× 10 N
⎟⎜
⎝ 1 ton ⎠ ⎝ 0.224 8 lbs⎠
m=
and its mass is
w 8.9× 1012 N
=
= 9.1× 1011 kg
g 9.80 m s2
The total number of uranium nuclei contained in this mass of uranium is
N total =
9.1× 1011 kg
m
=
= 2.3× 1036
−27
average m ass ofuranium atom ( 238.03 u ) ( 1.66 × 10 kg u )
Of this total, 0.720% is the fissionable 235 U isotope (see percentage abundance in
Appendix B). Assuming all will fission, releasing 208 M eV per event (see statement of
Problem 1), the total energy potentially available is
E = ( 208 M eV ) N 235 = ( 208 M eV ) ( 0.720 × 10−2 ) N total
= ( 208 M eV ) ( 0.720× 10−2 )( 2.3× 1036 )( 1.60× 10−13 J M eV ) = 5.5× 1023 J
At a rate of P = 7.0 × 1012 J s , the time that this energy could supply the world’s energy
needs is
∆t=
30.9
1 yr
E
5.5× 1023 J ⎛
⎞
3
=
⎜
⎟ = 2.5× 10 yr
12
7
P 7.0× 10 J s⎝ 3.156× 10 s⎠
The total energy required for one year is
E = ( 2000 kW h m onth ) ( 3.60 × 106 J kW h ) ( 12.0 m onths) = 8.64× 1010 J
The number of fission events needed will be
N =
E
Eevent
=
8.64 × 1010 J
= 2.60× 1021
−13
( 208 M eV ) ( 1.60× 10 J M eV )
and the mass of this number of
235
U atoms is
m = N m atom = ( 2.60 × 1021 ) ⎡⎣( 235.044 u ) ( 1.66× 10−27 kg u ) ⎤⎦
= 1.01× 10−3 kg = 1.01 g
Nuclear Energy and Elementary Particles
30.13
The energy released in the reaction 12 H + 13H → 24H e + 10n is
Q = ( ∆m ) c2 = ⎡⎣ m 2 H + m 3 H − m 4 H e − m n ⎤⎦ c2
= [ 2.014 102u + 3.016 049u − 4.002 602u − 1.008 665u ] ( 931.5 M eV u )
= 17.6 M eV ( 1.60× 10−13 J M eV ) = 2.81× 10−12 J
The total energy required for the year is
E = ( 2000 kW h m onth )( 12.0 m onths yr) ( 3.60× 106 J kW h ) = 8.64× 1010 J yr
so the number of fusion events needed for the year is
N =
8.64 × 1010 J yr
E
=
= 3.07 × 1022 events yr
Q 2.81× 10−12 J event
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