  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  B
297 W  C
297 MW  D
27MW  Solution

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

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