If the energy released per fission of a ${ }_{92}^{235} \mathrm{U}$ nucleus is 200 Me V . the energy…
If the energy released per fission of a ${ }_{92}^{235} \mathrm{U}$ nucleus is 200 Me V . the energy released in the fission of 0.1 kg of ${ }_{92}^{235} \mathrm{U}$ in kilowatt - hour is.
$22.8 \times 10^5$
$22.8 \times 10^7$
$11.4 \times 10^5$
$850 \times 10^{10}$
Solution
Energy released per fission per atom, $\mathrm{E}=200 \mathrm{MeV}$ Number of atoms in 0.1 kg of ${ }_{92}^{235} \mathrm{U}$ is
$\mathrm{N}=\left(\frac{\mathrm{m}}{\mathrm{~m}}\right) \mathrm{N}_{\mathrm{A}}=\frac{0.1 \times 6.023 \times 10^{23}}{235 \times 10^{-3}}=25.63 \times 10^{22}$
$\therefore \quad$ Energy released per fission is
$\begin{aligned}
& \text { E/fission }=\text { NE } \\
& =\frac{25.63 \times 10^{22} \times 200 \times 10^6 \times 1.6 \times 10^{-19}}{3.6 \times 10^6} \\
& =22.8 \times 10^5 \mathrm{kwh}
\end{aligned}$