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JEE Main 2024 (9 Apr, Shift 2) · Nuclear Fission & Fusion

Nuclear Fission & Fusion question from JEE Main 2024 (9 Apr, Shift 2)

Q14JEE Main 2024 · 9 APR · SHIFT 2Difficult

Nuclear Fission & Fusion

The energy released in the fusion of 2 kg2\text{ kg} of hydrogen deep in the sun is EHE_{H} and the energy released in the fission of 2 kg2\text{ kg} of 235U{}^{235}\text{U} is EUE_{U}. The ratio EHEU\frac{E_{H}}{E_{U}} is approximately: (Consider the fusion reaction as 4 11H+2e−→24He+2v+6γ+26.7 MeV4\,{}^{1}_{1}H + 2\text{e}^{-} \rightarrow {}^{4}_{2}\text{He} + 2v + 6\gamma + 26.7\text{ MeV}, energy released in the fission reaction of 235U{}^{235}\text{U} is 200 MeV200\text{ MeV} per fission nucleus and NA=6.023×1023)\text{N}_{\text{A}} = 6.023 \times 10^{23})

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Answer: (a)

  1. Fusion: 26.7 MeV26.7\text{ MeV} per 4 H atoms (4 g per mole of reactions), so per gram =26.74NA=6.675 NA MeV= \dfrac{26.7}{4}N_A = 6.675\,N_A\text{ MeV}.
  2. Fission: 200 MeV200\text{ MeV} per U nucleus (235 g per mole), so per gram =200235NA=0.851 NA MeV= \dfrac{200}{235}N_A = 0.851\,N_A\text{ MeV}.
  3. EHEU=6.6750.851≈7.8\dfrac{E_H}{E_U} = \dfrac{6.675}{0.851} \approx 7.8, nearest option 7.627.62.
  4. So the answer is (a) 7.627.62.
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