200 MeV of energy may be obtained per fission of U235. A reactor is generating 1000 kW of power. The rate of nuclear fission in the reactor is
Answer & explanation
Correct answer: option 3
Power = 1000 kW = $10^6$ J/s
Rate of nuclear fission $=\frac{10^6}{200 \times 1.6 \times 10^{-13}}$ = 3.125 × 1016