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

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(B) Total energy required $E = 	ext{Power} 	imes 	ext{Time} = (3.70 	imes 10^7 	ext{ J/s}) 	imes (144 	imes 10^4 	ext{ s}) = 5.328 	imes 10^{

Vedclass · Accepted Answer

$0.705$

Vedclass · Answer

(B) Total energy required $E = 	ext{Power} 	imes 	ext{Time} = (3.70 	imes 10^7 	ext{ J/s}) 	imes (144 	imes 10^4 	ext{ s}) = 5.328 	imes 10^{13} 	ext{ J}$.Energy per fission $\epsilon = 185 	ext{ MeV} = 185 	imes 10^6 	imes 1.6 	imes 10^{-19} 	ext{ J} = 2.96 	imes 10^{-11} 	ext{ J}$.Number of fissions required $N = \frac{E}{\epsilon} = \frac{5.328 	imes 10^{13}}{2.96 	imes 10^{-11}} = 1.8 	imes 10^{24} 	ext{ atoms}$.Mass of fuel $m = \frac{N 	imes M}{N_A} = \frac{1.8 	imes 10^{24} 	imes 235}{6 	imes 10^{23}} = 705 	ext{ g} = 0.705 	ext{ kg}$.

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