Answer:
The power output of the reactor is approximately 63.387 MW
Step-by-step explanation:
The energy released per one fission reaction = 200 MeV
The Avogadro's number,
= 6.023 × 10²³
The mass of fuel used = 2 kg
The mass of one mole of uranium 235, U²³⁵ = 235 grams = 0.235 kg
Therefore, the number of moles of U²³⁵ in 2 kg = 2 kg·mol⁻¹/(0.235 kg) = 400/47 moles ≈ 8.51 moles
The number of uranium atoms in 400/47 moles of U²³⁵ = (400/47) × 6.023 × 10²³ ≈ 5.126 × 10²⁴ atoms of U²³⁵
The energy released per fission of an atom = 200 MeV = 200 × 10⁶ eV × 1.602177 × 10⁻¹⁹ J/(eV) ≈ 3.204354 × 10⁻¹¹ J
The energy, 'E', released by the 5.126 × 10²⁴ atoms of U²³⁵ in the 2 kg of U²³⁵ is given as follows;
E = 3.204354 × 10⁻¹¹ J/atom × 5.126 × 10²⁴ atoms = 1.643 × 10¹⁴ Joules
The power, P = Energy/Time
The time = 30 days = 30 × 24 × 60 × 60 = 2,592,000 seconds
∴ P = (1.643 × 10¹⁴ Joules)/(2,592,000 s) = 63.3873457 megawatts ≈ 63.387 MW
The power output of the reactor, P ≈ 63.387 MW