The electrical power output of a large nuclear reactor facility is 900 MW. It has a 35.0\% efficiency in converting nuclear power to electrical power. (a) What is the thermal nuclear power output in megawatts? (b) How many \(^{235}\) U nuclei fission each second, assuming the average fission produces \(200 \mathrm{MeV}\) ? (c) What mass of \(^{235} \mathrm{U}\) is fissioned in 1 year of full-power operation?

First, we need to calculate the thermal nuclear power output. To do this, we can use the efficiency formula: Efficiency = (Useful output power / Input power) * 100 Rearrange the formula to solve for the input power (thermal nuclear power output): Input power = Useful output power / (Efficiency/100) Plugging in the given values: Input power = 900 MW / (35.0/100) Input power = 900 MW / 0.35 Input power ≈ 2571.43 MW Therefore, the thermal nuclear power output is approximately 2571.43 MW.

Now, we'll use the given average energy per fission (200 MeV) to calculate the rate of fission events. Remember, 1 MeV = 1.60218 × 10^{-13} Joules. First, we need to convert the thermal nuclear power output (in MW) to Joules per second: Thermal power output (J/s) = 2571.43 * 10^6 W = 2571.43 * 10^6 J/s Now, use the formula for energy released per fission event: Fission rate (nuclei/s) = Thermal power output (J/s) / (Average energy per fission * 1 MeV in Joules) Fission rate (nuclei/s) ≈ 2571.43 * 10^6 J/s / (200 * 1.60218 × 10^{-13} J) Fission rate (nuclei/s) ≈ 8.03708 × 10^19 nuclei/s The fission rate of \(^{235}\text{U}\) nuclei is approximately 8.03708 × 10^19 nuclei/s.

Finally, we'll calculate the mass of \(^{235}\text{U}\) fissioned in one year of full-power operation. Mass = (Number of particles * atomic mass) / Avogadro's number First, calculate the total number of \(^{235}\text{U}\) nuclei fissioned in one year: No. of nuclei fissioned in a year = Fission rate (nuclei/s) × Number of seconds in a year No. of nuclei fissioned in a year ≈ 8.03708 × 10^19 nuclei/s × 3.1536 × 10^7 s No. of nuclei fissioned in a year ≈ 2.5364 × 10^27 nuclei Now, plug the values in the mass formula, considering that the atomic mass of \(^{235}\text{U}\) is roughly 235 g/mol and Avogadro's number is 6.022 × 10^23 particles/mol: Mass = (2.5364 × 10^27 nuclei * 235 g/mol) / 6.022 × 10^23 particles/mol Mass ≈ 9.884 × 10^3 kg The mass of \(^{235}\text{U}\) fissioned in one year of full-power operation is approximately 9,884 kg.