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(B) The number of atoms in $1 \, g$ of $U^{235}$ is given by $N = \frac{N_A}{235} = \frac{6.023 	imes 10^{23}}{235} \approx 2.56 	imes 10^{21}$ atom

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$2.5 	imes 10^4 \, kWh$

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(B) The number of atoms in $1 \, g$ of $U^{235}$ is given by $N = \frac{N_A}{235} = \frac{6.023 	imes 10^{23}}{235} \approx 2.56 	imes 10^{21}$ atoms.Energy released per fission of one $U^{235}$ nucleus is approximately $200 \, MeV$.Total energy released $E = N 	imes 200 \, MeV = \frac{6.023 	imes 10^{23}}{235} 	imes 200 \, MeV \approx 5.12 	imes 10^{23} \, MeV$.Converting $MeV$ to Joules: $E = 5.12 	imes 10^{23} 	imes 1.6 	imes 10^{-13} \, J \approx 8.2 	imes 10^{10} \, J$.Converting Joules to $kWh$ $(1 \, kWh = 3.6 	imes 10^6 \, J)$:$E = \frac{8.2 	imes 10^{10}}{3.6 	imes 10^6} \, kWh \approx 2.27 	imes 10^4 \, kWh \approx 2.5 	imes 10^4 \, kWh$ (rounding to the nearest provided option).

What is the energy released by the fission of $1 \, g$ of $U^{235}$?

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