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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

3.125 × 1016

b

1000

c

931

d

2 × 108

answer is C.

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Detailed Solution

E = 200 MeV =32×10−12J
P=nEt;106=n×32×10−121n=10632×10−12=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