If $200 \mathrm{MeV}$ of energy is released in the fission of one nucleus of ${ }_{92}^{235} \mathrm{U}$,…
If $200 \mathrm{MeV}$ of energy is released in the fission of one nucleus of ${ }_{92}^{235} \mathrm{U}$, then the number of nuclei that must undergo fission to release an energy of $1000 \mathrm{~J}$ is
$3.125 \times 10^{13}$
$6.25 \times 10^{13}$
$12.5 \times 10^{13}$
$3.125 \times 10^{14}$
Solution
$\begin{array}{llll}\text {Number of nuclei } =\frac{E_{\text {total }}}{E_1}=\frac{1000 \mathrm{~J}}{200 \mathrm{MeV}} \\ =\frac{1000 \mathrm{~J}}{200 \times 10^6 \times 1.6 \times 10^{-19} \mathrm{~J}} \\ \Rightarrow N =\frac{10^{14}}{3.2} \\ \Rightarrow N =3.125 \times 10^{13}\end{array}$