In a reactor, $2\ \text{kg}$ of ${}_{92}\text{U}^{235}$ fuel is fully used up in $30$ days. The energy released per fission is $200\ \text{MeV}$. Given that the Avogadro number, $N = 6.023 \times 10^{26}$ per kilo mole and $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$. The power output of the reactor is close to:
Answer: (B) $60\ \text{MW}$
Number of nuclei in $2$ kg: $N = \dfrac{2}{235}\times 6.023\times10^{26} = 5.13\times10^{24}$.
Energy per fission $= 200\times10^{6}\times1.6\times10^{-19} = 3.2\times10^{-11}$ J, so total energy $= 5.13\times10^{24}\times3.2\times10^{-11} = 1.64\times10^{14}$ J.
Time $= 30\times 86400 = 2.59\times10^{6}$ s.
$$P = \frac{1.64\times10^{14}}{2.59\times10^{6}} \approx 6.3\times10^{7}\ \text{W} \approx 60\ \text{MW}$$
Solution by Sreeraj P, M.Sc Physics