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In the nuclear fission of one nucleus of $U^{255}$ the energy released is $188 \mathrm{MeV}$. The energy released in the nuclear fission of $235 \mathrm{~g}$ of $\mathrm{U}^{235}$ is nearly (Avogadro number $=6.02 \times 10^{23} \mathrm{~mol}^{-1}$ )
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The correct answer is:
$18.11 \times 10^{12} \mathrm{~J}$
Energy released, is given by
$\begin{aligned}
& \Delta \mathrm{E}=188 \times 1.6 \times 10^{-13} \times \frac{235}{235} \times 6.02 \times 10^{23} \\
& =18.11 \times 10^{12} \mathrm{~J}
\end{aligned}$
$\begin{aligned}
& \Delta \mathrm{E}=188 \times 1.6 \times 10^{-13} \times \frac{235}{235} \times 6.02 \times 10^{23} \\
& =18.11 \times 10^{12} \mathrm{~J}
\end{aligned}$
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