`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.

Options

A. 1000
B. `2xx10^(8)`
C. `3.125xx10^(8)`
D. 931

Answer

C

Explanation

`E=200 MeV`
`=200xx1.6xx10^(-19)xx10^(6)J`
`=200xx1.6xx10^(-13)J`
`P=1000 kW = 10^(6)` watt
decay rate `=(P)/(E )=(10^(6))/(200xx1.6xx10^(-13))`
`= 3.125xx10^(16)` per second