200 MeV energy is released when one nucleus of U235 undergoes fission. Find the number of fission per second required for producing a power of 1 megawatt.

Q: 200 MeV energy is released when one nucleus of U²³⁵ undergoes fission. Find the number of fission per second required for producing a power of 1 megawatt.

Sol: Energy released = 200 MeV

= 200 × 10⁶ × 1.6 × 10^-19 = 3.2 × 10^-11 J

P = 1 mega watt = 10⁶ watts.

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

n = 10⁶/(3.2 × 10^-11 ) = 3.125 × 10¹⁶ fission.