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Problem
For the fusion reaction
1
2
H
+
1
3
H
→
2
4
He
+
0
1
n
^{2}_{1}\text{H}+^{3}_{1}\text{H}\;\rightarrow\;^{4}_{2}\text{He}+^{1}_{0}\text{n}
1
2
H
+
1
3
H
→
2
4
He
+
0
1
n
, calculate the energy released per reaction event in MeV. Atomic masses:
1
2
H
=
2.0141
u
^{2}_{1}\text{H}=2.0141\,\text{u}
1
2
H
=
2.0141
u
1
3
H
=
3.0160
u
^{3}_{1}\text{H}=3.0160\,\text{u}
1
3
H
=
3.0160
u
2
4
He
=
4.0026
u
^{4}_{2}\text{He}=4.0026\,\text{u}
2
4
He
=
4.0026
u
0
1
n
=
1.0087
u
^{1}_{0}\text{n}=1.0087\,\text{u}
0
1
n
=
1.0087
u
Take
1
u
=
931.5
MeV
/
c
2
1\,\text{u}=931.5\,\text{MeV}/c^{2}
1
u
=
931.5
MeV
/
c
2
. a)
14.0
MeV
14.0\,\text{MeV}
14.0
MeV
b)
1.76
MeV
1.76\,\text{MeV}
1.76
MeV
c)
17.6
MeV
17.6\,\text{MeV}
17.6
MeV
d)
10.4
MeV
10.4\,\text{MeV}
10.4
MeV
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