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In a reactor, 2 kg of 92U²³⁵ fuel is fully used up in 30 days. The energy released per fission is 200 MeV. Given that the Avogadro number, N = 6.023 × 10²⁶ per kilo mole and 1 eV = 1.6 × 10⁻¹⁹ J. The power output of the reactor is close to:

1. 125 MW
2. 60 MW
3. 35 MW
4. 54 MW

Correct answer: 60 MW

Solution

Atoms = (2000 g / 235) * 6.023e23 = 5.13e24. Energy = 5.13e24 * 200e6 * 1.6e-19 = 1.64e14 J. Time = 30*86400 = 2.59e6 s. Power = 1.64e14 / 2.59e6 = 6.3e7 W ~ 63 MW, closest to 60 MW.

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