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In a reactor 2 kg of ₉₂U²³⁵ fuel is fully used up in 30 days. The energy released per fission is 200 MeV. Given that the Avogadro number N = 6.023 x 10²⁶ per kilo molecule and 1 eV = 1.6 x 10^-19 J, the power output of the reactor close to

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35 MW

60 MW

125 MW

54 MW

answer is B.

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Detailed Solution

Power $= \frac{E^{'}}{t} = \frac{m \cdot Na}{A} \cdot \frac{E}{t}$
$= \frac{2}{235} \times \frac{6.023 \times 10^{26} \times 200 \times 1.6 \times 10^{- 19} \times 10^{6}}{30 \times 24 \times 60 \times 60} = 60 MW$

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