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Physics
Physics
Problem
Consider the fusion reaction
2
H
+
3
H
→
4
H
e
+
n
^{2}\!\mathrm{H}+{}^{3}\!\mathrm{H} \rightarrow {}^{4}\!\mathrm{He}+n
2
H
+
3
H
→
4
He
+
n
. The atomic masses are:
2
H
=
2.014
102
u
^{2}\!\mathrm{H}=2.014\,102\,u
2
H
=
2.014
102
u
,
3
H
=
3.016
049
u
^{3}\!\mathrm{H}=3.016\,049\,u
3
H
=
3.016
049
u
,
4
H
e
=
4.002
603
u
^{4}\!\mathrm{He}=4.002\,603\,u
4
He
=
4.002
603
u
,
n
=
1.008
665
u
n=1.008\,665\,u
n
=
1.008
665
u
. Given
1
u
=
931.5
MeV
/
c
2
1\,u=931.5\,\text{MeV}/c^{2}
1
u
=
931.5
MeV
/
c
2
, the energy released per fusion reaction is closest to a) 20.2 MeV b) 17.6 MeV c) 14.1 MeV d) 11.3 MeV
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