Question

The energy released by fission from $2g$ of ${}_{92}^{235}U$ in kWh is (the energy released per fission is $200MeV$):

Answer 1

Number of atoms in $2g$ of ${}_{92}^{235}U=\frac{2}{235}\times N_A$ where $N_A$ is avogadro number

$=\frac{2}{235}\times 6.023\times {10}^{23}$

Energy per fission $=200MeV$

Therefore total energy $\left(K\right)=\frac{2}{235}\times 6.023\times {10}^{23}\times 200\times {10}^6\times 1.6\times {10}^{-19}J$

Total Energy in $kWh=\frac{k}{3600\times {10}^3}$$=4.556\times {10}^{4}kWh$