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Q.

200 MeV of energy may be obtained per fission of U235. A reactor is generating 1000 kW of power. The rate of nuclear fission in the reactor is

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a

1000

b

2×108

c

3.125×1016

d

931

answer is C.

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Detailed Solution

Power =1000kW=106J/s  Rate of nuclear fission =106200×1.6×10−13=3.125×1016.
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200 MeV of energy may be obtained per fission of U235. A reactor is generating 1000 kW of power. The rate of nuclear fission in the reactor is