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 (a) 125 MW (b) 60 MW (c) 54 MW (d) 35 MW

Answer: (b). Nuclei in 2 kg: N = 2/235×6.023×10²⁶ ≈ 5.13×10²⁴. Energy E = N×200×1.6×10⁻¹³ J ≈ 1.64×10¹⁴ J. Time t = 30×86400 = 2.59×10⁶ s. P = E/t ≈ 6.3×10⁷ W ≈ 60 MW. So the answer is (b) 60 MW.

In a reactor, 2 kg of 92U235 fuel is fully used up in 30 days. The… | JEE Main 2020 (2 Sep, Shift 1)

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Nuclear Fission & Fusion

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 $\times$ 1026 per kilo mole and 1 eV = 1.6 × 10–19 J. The power output of the reactor is close to

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Answer: (b)

Nuclei in 2 kg: $N = \dfrac{2}{235}\times6.023\times10^{26} \approx 5.13\times10^{24}$ .
Energy $E = N\times200\times1.6\times10^{-13}\text{ J} \approx 1.64\times10^{14}\text{ J}$ .
Time $t = 30\times86400 = 2.59\times10^{6}\text{ s}$ .
$P = \dfrac{E}{t} \approx 6.3\times10^{7}\text{ W} \approx 60\text{ MW}$ .
So the answer is (b) $60\text{ MW}$ .

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