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KCET 2024 Physics Solved Paper
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© examsnet.com
Question : 50
Total: 60
Consider the nuclear fission reaction
0
1
n
+
92
235
U
⟶
56
144
B
a
+
36
89
K
r
+
3
0
1
n
Assuming all the kinetic energy is carried away by the fast neutrons only and total binding energies of
92
235
U
,
56
144
Ba
and
36
89
Kr
to be
1800
MeV
,
1200
MeV
and 780 MeV respectively, the average kinetic energy carried by each fast neutron is (in MeV )
200
180
67
60
Validate
Solution:
3
K
neutron
=
(
B
E
v
−
B
E
B
a
−
B
E
K
r
)
=
(
1800
−
1200
−
780
)
MeV
|
K
neutron
|
=
180
3
MeV
=
60
MeV
© examsnet.com
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