Answer to Question #120865 in Atomic and Nuclear Physics for Dheeraj

Question

200MeV of energy obtained in the fission of one neucleus of 235^U. A reactor is generating 1000KW power. The rate of nuclear fission in the reactor is?

Answer 1

The rate is the number of reactions per second. So we should divide the power by the energy in the fission of one nucleus. We should express the power in W and energy in J (1eV $\approx 1.60\cdot10^{-19}\,\mathrm{J}$ )

$N = \dfrac{P}{E} = \dfrac{10^6\,\mathrm{W}}{2\cdot10^8\cdot (1.60\cdot10^{-19}\,\mathrm{J})} = 3.13\cdot10^{16}$ fissions per second.