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Two deuterium nuclei fuse according to
PHYSICS · Single correct · CBSE · Physics · Nuclei · Nuclei
Problem
Two deuterium nuclei fuse according to
1
2
D
+
1
2
D
→
2
3
H
e
+
n
^2_1\mathrm D + ^2_1\mathrm D \rightarrow ^3_2\mathrm{He} + n
1
2
D
+
1
2
D
→
2
3
He
+
n
. Given atomic masses
1
2
D
=
2.014102
u
^2_1\mathrm D = 2.014102\,\text{u}
1
2
D
=
2.014102
u
,
2
3
H
e
=
3.016029
u
^3_2\mathrm{He} = 3.016029\,\text{u}
2
3
He
=
3.016029
u
,
n
=
1.008665
u
n = 1.008665\,\text{u}
n
=
1.008665
u
, and
1
u
=
931
MeV
/
c
2
1\,\text{u}=931\,\text{MeV}/c^2
1
u
=
931
MeV
/
c
2
, calculate the energy released per fusion reaction. (a)
6.54
MeV
6.54\,\text{MeV}
6.54
MeV
(b)
2.27
MeV
2.27\,\text{MeV}
2.27
MeV
(c)
3.27
MeV
3.27\,\text{MeV}
3.27
MeV
(d)
3.02
MeV
3.02\,\text{MeV}
3.02
MeV
Related theory
Mass-Energy Equivalence
Nuclear Fusion
Nuclear Reactions
Nuclear physics
Atomic Structure
Nuclear fusion (stars as energy sources)
Nuclei
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