What is the power output of a „U reactor if it takes 30 days to use up 2kg of fuel Given that energy release per fission is 200MeV and Avogadro's number = 6,023 x 10°
per kilomole

The power output of the reactor is approximately 63.387 MW

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

answer:

yes

explanation: