Q: 200 MeV energy is released when one nucleus of U^{235} undergoes fission. Find the number of fission per second required for producing a power of 1 megawatt.

Sol: Energy released = 200 MeV

= 200 × 10^{6} × 1.6 × 10^{-19} = 3.2 × 10^{-11} J

P = 1 mega watt = 10^{6} watts.

No. of fission per second (n) = (Total energy)/(Energy per fission)

n = 10^{6}/(3.2 × 10^{-11} ) = 3.125 × 10^{16} fission.