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

Options

A. `5xx10^(22)s^(-1)`
B. `5xx10^(16)s^(-1)`
C. `8xx10^(16)s^(-1)`
D. `20xx10^(16)s^(-1)`

Answer

A

Explanation

We know that, `E=Deltamc^(2)=0.5xx10^(-3)xx(3xx10^(8))^(2)`
`=4.5xx10^(13)`
`E=(4.5xx10^(13))/(3.6xx10^(6))=1.25xx10^(7)`kWh