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Use the atomic masses given below to find the energy released
PHYSICS · Subjective · CBSE · Physics · Nuclei · Nuclei
Problem
Use the atomic masses given below to find the energy released, in MeV, in the fission reaction
235
U
+
n
→
141
B
a
+
92
K
r
+
3
n
^{235}\mathrm{U}+n\rightarrow{}^{141}\mathrm{Ba}+{}^{92}\mathrm{Kr}+3n
235
U
+
n
→
141
Ba
+
92
Kr
+
3
n
Atomic masses:
235
U
=
235.0439
u
^{235}\mathrm{U}=235.0439\,\text{u}
235
U
=
235.0439
u
, neutron
=
1.00866
u
=1.00866\,\text{u}
=
1.00866
u
,
141
B
a
=
140.9144
u
^{141}\mathrm{Ba}=140.9144\,\text{u}
141
Ba
=
140.9144
u
,
92
K
r
=
91.9261
u
^{92}\mathrm{Kr}=91.9261\,\text{u}
92
Kr
=
91.9261
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
.)
Related theory
Nuclear Fission
Nuclear Reactions
Mass-Energy Equivalence
Nuclear physics
Mass defect and binding energy
Mass-energy relation
Nuclei
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