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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

`2 xx 10^8`

C

`3.125 xx 10^16`

D

931

Text Solution

Verified by Experts

The correct Answer is:
C
( c) Power `= 1000 kW = 10^6 J//s`
Rate of nuclear fission `= (10^6)/(200 xx 1.6 xx 10^-13)`
=`3.125 xx 10^16`.
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