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200 MeV of energy may be obtained per fi...

`200 MeV` of energy may be obtained per fission of `U^235`. A reactor is generating `1000 kW` of power. The rate of nuclear fission in the reactor is.

A

`1000`

B

`2xx10^(8)`

C

`3.125xx10^(16)`

D

`931`

Text Solution

Verified by Experts

The correct Answer is:
C
Fission rate `= (1000xx 10^(3))/(200xx1.6xx10^(-3)) = 3.125xx10^(6)sec^(-1)`
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