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10¹⁴ fissions per second are taking place in a nuclear reactor having efficiency 40%. The energy released per fission is 250 MeV. The power output of the reactor is

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400 W

800 W

200 W

1600 W

answer is D.

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Detailed Solution

$P = \frac{40}{100} (\frac{N}{t}) E$
$P = \frac{40}{100} \times 10^{14} \times 250 \times 10^{6} \times 1.6 \times 10^{- 19}$
$= 1600 W$

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