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On bombardment of U^235 by slow neutrons...

On bombardment of `U^235` by slow neutrons, `200 MeV` energy is released. If the power output of atomic reactor is `1.6 MW`, then the rate of fission will be

A

`5xx10^(10)`

B

`5xx10^(12)`

C

`5xx10^(4)`

D

`5xx10^(16)`

Text Solution

Verified by Experts

The correct Answer is:
D
Output power `1.6MW=1.6xx10^(6)J//s`
Energy released/fission =200MeV
`=200xx1.6xx10^(-13)J`
`:.` Nuclear of fission/sec
`=(1.6xx10^(6))/(200xx1.6xx19^(-13))=5xx10^(16)`
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