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AIIMS 2013 Physics Question Paper for free online practice with solutions
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© examsnet.com
Question : 3
Total: 60
When a slow neutron is captured by a
232
92
U
nucleus, a fission energy releasing 200 MeV. If power of nuclear reactor is 100 W then rate of nuclear fission is
3.6
×
10
6
s
−
1
3.1
×
10
12
s
−
1
1.8
×
10
4
s
−
1
4.1
×
10
6
s
−
1
Validate
Solution:
Number of fission per second =
total
power
energy
fission
Here, total power = 100 W
energy/fission = 200 MeV =
200
×
10
6
×
1.6
×
10
−
19
J
=
3.2
×
10
−
11
J
.
∴ fission rate =
100
3.2
×
10
−
11
=
3.1
×
10
12
s
−
1
© examsnet.com
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