To determine the power output of the nuclear reactor using U-235 as fuel, we need to follow these steps:
- Calculate the energy released per fission event in joules. The energy released per fission is given as 185 MeV. We convert this energy into joules using the conversion: \(1 \, \text{MeV} = 1.60218 \times 10^{-13} \, \text{J}\).
- Number of atoms of U-235 in 2 kg: The number of atoms is found using Avogadro's number and the molar mass of U-235, which is approximately 235 g/mol.
- Total energy released from 2 kg of U-235 is calculated by multiplying the energy per fission with the number of fission events.
- Calculate the power output: Power is energy per unit time. Use the total energy and the time period (30 days converted to seconds) to find power.
Let's perform these calculations:
- Energy per fission in joules: \(185 \, \text{MeV} \times 1.60218 \times 10^{-13} \, \text{J/MeV} = 2.963 \times 10^{-11} \, \text{J}\)
- The number of atoms in 2 kg of U-235: \(\frac{2 \times 10^3 \, \text{g}}{235 \, \text{g/mol}} \times 6.022 \times 10^{23} \, \text{atoms/mol} = 5.124 \times 10^{24} \, \text{atoms}\)
- Total energy released: \(5.124 \times 10^{24} \, \text{atoms} \times 2.963 \times 10^{-11} \, \text{J} = 1.517 \times 10^{14} \, \text{J}\)
- Time in seconds for 30 days: \(30 \times 24 \times 3600 = 2592000 \, \text{s}\)
- Power output: \(\frac{1.517 \times 10^{14} \, \text{J}}{2592000 \, \text{s}} = 5.85 \times 10^{7} \, \text{W} = 58.5 \, \text{MW}\)
Therefore, the power output of the reactor is 58.5 MW.
The correct answer is thus: 58.5 MW