Question

`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

Answer 1

Correct Answer - 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`.