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Question

Mass of $$_{92}{U}^{235}$$ required to produce electrical energy in a nuclear reactor for 30 days is $$23.5  kg$$. If 40% of the energy released by the fission is converted into electrical energy, determine the power generated by the nuclear reactor. (Assume that $$200  MeV$$ energy is released per fission of $$_{92}{U}^{235}$$ )


A
390 MW
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B
297 W
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C
297 MW
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D
27MW
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Solution

The correct option is C 297 MW
$$ n\left(\frac{235}{92} U\right)=\frac{23.5 \times 10^{3}}{235} \times 6.023 \times 10^{23} \text { atoms } $$
 $$ \begin{array}{l} \text { Energy released by fission }=200 \mathrm{MeV} \text { per atom } \\ \text { Total energy released }=0.1 \times 10^{3} \times 6.023 \times 10^{23} \times 200 \times 10^{6} \times 1.6 \times 10^{-19} \mathrm{~J} \\ =6.023 \times 2 \times 1.6 \times 10^{14} \mathrm{~J} \end{array} $$
 $$ \begin{array}{l} \text { Electricity produced }=\frac{40}{100} \times 6.023 \times 2 \times 1.6 \times 10^{14} \mathrm{~J} \\ \text { Power }=\frac{0.4 \times 6.023 \times 2 \times 1.6 \times 10^{14}}{30 \times 24 \times 60 \times 60} \\ =296.29 \times 10^{6} \mathrm{~W} \end{array} $$

Physics
NCERT
Standard XII

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