On bombarding `U^(235)` by slow neutron , 200 Me V energy is released . If the power output of atomic reactor is 1.6 MW , then the rate of fission will be

Options

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

Answer

D

Explanation

Energy released per fission of uranium `=200xx10^(6)xx1.6xx10^(-19)J` Power output `=1.6xx10^(6)W`.
`therefore` Number of fission/s
`=(1.6xx10^(6))/(200xx10^(6)xx1.6xx10^(-19))=5xx10^(16)//s``.
This is the rate of fission.