On fission,a U^{235} nucleus releases 3 times | vedclass

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(C) The power output of the reactor is $P = 1 \, GW = 10^9 \, J/s$.The useful energy required in half an hour $(t = 1800 \, s)$ is $E_{useful} = P 	i

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$500$

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(C) The power output of the reactor is $P = 1 \, GW = 10^9 \, J/s$.The useful energy required in half an hour $(t = 1800 \, s)$ is $E_{useful} = P 	imes t = 10^9 	imes 1800 = 1.8 	imes 10^{12} \, J$.Since only $4.2 \%$ of the total fission energy is converted to useful energy,the total energy required from fission is $E_{total} = \frac{E_{useful}}{0.042} = \frac{1.8 	imes 10^{12}}{0.042} \approx 4.286 	imes 10^{13} \, J$.The number of $U^{235}$ nuclei required is $N = \frac{E_{total}}{3 	imes 10^{-11} \, J/nucleus} = \frac{4.286 	imes 10^{13}}{3 	imes 10^{-11}} \approx 1.428 	imes 10^{24} \, nuclei$.The number of moles is $n = \frac{N}{N_A} = \frac{1.428 	imes 10^{24}}{6.023 	imes 10^{23}} \approx 2.37 \, moles$.The mass of $U^{235}$ consumed is $m = n 	imes M = 2.37 	imes 235 \approx 557 \, g$.Comparing this with the given options,the closest value is $500 \, g$.

On fission,a $U^{235}$ nucleus releases $3 \times 10^{-11} \, J$ of energy. In a $1 \, GW$ nuclear reactor,$4.2 \%$ of this energy is converted to useful energy. The $U^{235}$ consumed (in grams) in half an hour is closest to (Avogadro number $N_A = 6.023 \times 10^{23}$)

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