`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. 2 MeV
C. `3.125 xx 10^16`
D. 931

Answer

C

Explanation

( c) Power `= 100 kW = 10^6 J//s`
Rate of nuclear fission
=`(10^6)/(200 xx 1.6 xx 10^-13) = 3.125 xx 10^16`.