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If $200 \mathrm{MeV}$ of energy is released in the fission of one nucleus of ${ }_{92}^{236} \mathrm{U}$, the number of nuclei that must undergo fission to release an energy of $1000 \mathrm{~J}$ is
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The correct answer is:
$3.125 \times 10^{13}$
Energy released in the fission of one nucleus
$\begin{aligned}
& =200 \mathrm{MeV} \\
& =200 \times 1.6 \times 10^{-13} \mathrm{~J} \\
& =3.2 \times 10^{-11} \mathrm{~J}
\end{aligned}$
Energy to be released
$=1000 \mathrm{~J}$
$3.2 \times 10^{-11} \mathrm{~J}$ energy is released due to number of nucleus fissioned $=1$
$\therefore 1000 \mathrm{~J}$ energy is released due to number of nucleus fissioned
$\begin{aligned}
& =\frac{1 \times 1000}{3.2 \times 10^{-11}} \\
& =3.125 \times 10^{13} \text { nucleus }
\end{aligned}$
$\begin{aligned}
& =200 \mathrm{MeV} \\
& =200 \times 1.6 \times 10^{-13} \mathrm{~J} \\
& =3.2 \times 10^{-11} \mathrm{~J}
\end{aligned}$
Energy to be released
$=1000 \mathrm{~J}$
$3.2 \times 10^{-11} \mathrm{~J}$ energy is released due to number of nucleus fissioned $=1$
$\therefore 1000 \mathrm{~J}$ energy is released due to number of nucleus fissioned
$\begin{aligned}
& =\frac{1 \times 1000}{3.2 \times 10^{-11}} \\
& =3.125 \times 10^{13} \text { nucleus }
\end{aligned}$
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