In a reactor, \(2\, kg\) of \({ }_{92} U ^{235}\) fuel is fully used up in \(30\) days. The energy released per fission is \(200\, MeV.\) Given that the Avogadro number, \(N =6.023 \times 10^{26}\) per kilo mole and \(1\, eV =1.6 \times 10^{-19}\, J .\) The power output of the eactor is close to\(.....MW\)

\(=\frac{2 \times 6.023 \times 10^{26}}{235}\)

energy from one atom is \(200 \times 10^{6}\) e.v. hence total energy from \(2 kg\) uranium

\(=\frac{2 \times 6.023 \times 10^{26}}{235} \times 200 \times 10^{6} \times 1.6 \times 10^{-19} J\)

\(2 kg\) uranium is used in \(30\) days hence this energy is recieved in 30 days hence energy recived per second or power is

Power \(=\frac{2 \times 6.023 \times 10^{26} \times 200 \times 10^{6} \times 1.6 \times 10^{-19}}{235 \times 30 \times 24 \times 3600}\)

Power \(=63.2 \times 10^{6}\) watt or \(63.2\) Mega Watt