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113. Energy released in the fission of a single uranium nucleus is $200 \mathrm{MeV}$. Then the number of fissions per second to produce $5 \mathrm{~mW}$ power is
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The correct answer is:
$1.56 \times 10^8$
Number of fissions per second,
$\begin{aligned} & \mathrm{n}=\frac{5 \times 10^{-3}}{200 \times 10^6 \times 1.6 \times 10^{-19}} \\ & =\frac{50}{32} \times 10^8=1.56 \times 10^8 \mathrm{~m} / \mathrm{s}\end{aligned}$
$\begin{aligned} & \mathrm{n}=\frac{5 \times 10^{-3}}{200 \times 10^6 \times 1.6 \times 10^{-19}} \\ & =\frac{50}{32} \times 10^8=1.56 \times 10^8 \mathrm{~m} / \mathrm{s}\end{aligned}$
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