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

Solution

Physics sir · 3 steps

Count the nuclei, multiply by energy per fission, then divide by the time in seconds.

1. Count the nuclei. 2 kg of ²³⁵U is 2/235 kilomole. With N = 6.023 × 10²⁶ per kilomole, the number of nuclei is 2/235 × 6.023 × 10²⁶ = 5.13 × 10²⁴.
2. Energy per fission. 200 MeV = 200 × 10⁶ × 1.6 × 10⁻¹⁹ J = 3.2 × 10⁻¹¹ J. So the total energy is 5.13 × 10²⁴ × 3.2 × 10⁻¹¹ = 1.64 × 10¹⁴ J.

   P = E/t = (1.64 × 10¹⁴ J)/(30 × 86400 s) = (1.64 × 10¹⁴)/(2.592 × 10⁶) W

3. So. P ≈ 6.3 × 10⁷ W = 63 MW. The closest option is 60 MW, option (b).

The idea to keep. Reactor power is (number of nuclei) × (energy per fission) ÷ (time in seconds). Watch the units: kilomole with N per kilomole, MeV to joules, days to seconds.

Answer from the NTA final answer key. Solution by PrepWiser.