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Calculate power output of \({ }_{92}^{235} \mathrm{U}\) reactor, if it takes 30 days to use up \(2 \mathrm{~kg}\) of fuel, and if each fission gives \(185 \mathrm{~MeV}\) of useable energy. Avogadro's number \(=6 \times 10^{23} / \mathrm{mol}\) ?
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Verified Answer
The correct answer is:
\(58.3 \mathrm{~MW}\)
Mass of \({ }_{92}^{235} \mathrm{U}\) per second in the reactor is
\(m=\frac{2 \times 10^3}{30 \times 24 \times 60 \times 60}=7.72 \times 10^{-4} \mathrm{~g} / \mathrm{sec}\)
\(\therefore\) Number of fissions reaction per second
\(\begin{aligned}
& =\frac{6 \times 10^{23}}{235} \times m=\frac{6 \times 10^{23} \times 7.72 \times 10^{-4}}{235} \\
& =1.97 \times 10^{18} / \mathrm{sec}
\end{aligned}\)
Power of nuclear reactor \(=1.97 \times 10^{18} \times 185 \mathrm{MeV} / \mathrm{s}\) \(=1.97 \times 10^{18} \times 185 \times 10^6 \times 1.6 \times 10^{-19} \mathrm{~J} / \mathrm{s}\). \(=58.3 \mathrm{~MW}\).
\(m=\frac{2 \times 10^3}{30 \times 24 \times 60 \times 60}=7.72 \times 10^{-4} \mathrm{~g} / \mathrm{sec}\)
\(\therefore\) Number of fissions reaction per second
\(\begin{aligned}
& =\frac{6 \times 10^{23}}{235} \times m=\frac{6 \times 10^{23} \times 7.72 \times 10^{-4}}{235} \\
& =1.97 \times 10^{18} / \mathrm{sec}
\end{aligned}\)
Power of nuclear reactor \(=1.97 \times 10^{18} \times 185 \mathrm{MeV} / \mathrm{s}\) \(=1.97 \times 10^{18} \times 185 \times 10^6 \times 1.6 \times 10^{-19} \mathrm{~J} / \mathrm{s}\). \(=58.3 \mathrm{~MW}\).
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