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Class 12 NEET Physics Atoms and Nuclei Part 1
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© examsnet.com
Question : 11
Total: 100
In the reaction
1
2
H
+
1
3
H
→
2
4
H
e
+
0
1
n
,
ifthe binding energies of
1
2
H
,
1
3
H
and
2
4
H
e
are respectively
a
,
b
and
c
(in
M
e
V
), then the energy (in MeV) released in this reaction is
(2005)
a
+
b
+
c
a
+
b
−
c
c
−
a
−
b
c
+
a
−
b
.
Validate
Solution:
Energy released,
E
=
(
Δ
m
)
×
931
M
e
V
Δ
m
=
mass of product
−
mass of reactant
Δ
m
=
c
−
a
−
b
E
=
(
Δ
m
)
×
931
or,
E
=
(
c
−
a
−
b
)
.
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