Assuming that 200 MeV of energy is released per fission,the mass of ²³⁵U that undergoes fission to operate a 250-MW reactor for 1 y is approximately A)100 kg B)200 kg C)400 kg D)500 kg E)1 Mg

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The energy released per fission event is 200 MeV, which is equivalent to $200 \times 10^6$ eV. Since $1 \text{ eV} = 1.602 \times 10^{-19} \text{ Joules}$, the energy per fission in Joules is $200 \times 10^6 \times 1.602 \times 10^{-19} = 3.204 \times 10^{-11} \text{ Joules}$.For a 250-MW (megawatt) reactor operating for 1 year, the total energy output in Joules is calculated as follows:$$250 \times 10^6 \text{ Watts} \times (365 \times 24 \times 3600) \text{ seconds} = 250 \times 10^6 \times 31,536,000 = 7.884 \times 10^{15} \text{ Joules}$$To find the mass of $^{235}U$ that undergoes fission, we divide the total energy output by the energy released per fission:$$\frac{7.884 \times 10^{15}}{3.204 \times 10^{-11}} = 2.46 \times 10^{26} \text{ fissions}$$The mass of $^{235}U$ per fission can be estimated using its molar mass (235 g/mol) and Avogadro's number ($6.022 \times 10^{23}$ atoms/mol):$$\frac{235}{6.022 \times 10^{23}} = 3.90 \times 10^{-25} \text{ kg/fission}$$Thus, the total mass of $^{235}U$ that undergoes fission is:$$2.46 \times 10^{26} \times 3.90 \times 10^{-25} = 9.59 \times 10^1 \text{ kg} \approx 100 \text{ kg}$$Therefore, approximately 100 kg of $^{235}U$ undergoes fission to operate a 250-MW reactor for 1 year.

Assuming That 200 MeV of Energy Is Released Per Fission,the