JEE Main 2024 Physics question
The energy released in the fusion of 2 kg of hydrogen deep in the sun is and the energy released in the fission of 2 kg of ²³⁵U is . The ratio / is approximately : (Consider the fusion reaction as 4¹₁H + 2e⁻ → He + 2ν + 6γ + 26.7 MeV, energy released in the fission reaction of ²³⁵U is 200 MeV per fission nucleus and = 6.023 × 10²³)
Answer: (C)
Solution
Each fusion reaction uses 4 hydrogen nuclei, so the energy per hydrogen nucleus is 26.7/4 MeV. = (2000/.7/4) MeV and
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