U^{235} nuclear reactor generates energy at a | vedclass

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(B) Total energy required $E = P 	imes t = (3.70 	imes 10^7 	ext{ J/s}) 	imes (144 	imes 10^4 	ext{ s}) = 5.328 	imes 10^{13} 	ext{ J}$.Energy

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$0.705$

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(B) Total energy required $E = P 	imes t = (3.70 	imes 10^7 	ext{ J/s}) 	imes (144 	imes 10^4 	ext{ s}) = 5.328 	imes 10^{13} 	ext{ J}$.Energy released per fission $E_f = 185 	ext{ MeV} = 185 	imes 10^6 	imes 1.6 	imes 10^{-19} 	ext{ J} = 2.96 	imes 10^{-11} 	ext{ J}$.Number of fissions $N = E / E_f = (5.328 	imes 10^{13}) / (2.96 	imes 10^{-11}) = 1.8 	imes 10^{24} 	ext{ atoms}$.Mass of fuel $m = (N / N_A) 	imes M = (1.8 	imes 10^{24} / 6 	imes 10^{23}) 	imes 235 	ext{ g} = 0.3 	imes 235 	ext{ g} = 70.5 	ext{ g} = 0.0705 	ext{ kg}$.Re-evaluating the calculation: $N = 1.8 	imes 10^{24}$ atoms. Mass $= (1.8 	imes 10^{24} / 6 	imes 10^{23}) 	imes 235 = 3 	imes 235 = 705 	ext{ g} = 0.705 	ext{ kg}$.

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