The energy released in the fusion of 2 kg of hydrogen deep in the sun is E(H) and the energy released in the fission of 2 kg of ²³⁵U is E(U). The ratio E(H)/E(U) is approximately :
(Consider the fusion reaction as 4 ¹₁H + 2e⁻ → ⁴₂He + 2ν + 6γ + 26.7 MeV, energy released in the fission reaction of ²³⁵U is 200 MeV per fission nucleus and N(A) = 6.023 × 10²³)

Question

The energy released in the fusion of 2 kg of hydrogen deep in the sun is E(H) and the energy released in the fission of 2 kg of ²³⁵U is E(U). The ratio E(H)/E(U) is approximately :

(Consider the fusion reaction as 4 ¹₁H + 2e⁻ → ⁴₂He + 2ν + 6γ + 26.7 MeV, energy released in the fission reaction of ²³⁵U is 200 MeV per fission nucleus and N(A) = 6.023 × 10²³)

G1 Physics · Accepted Answer

In one fusion reaction, four hydrogen nuclei are consumed. Energy released per hydrogen nucleus: = 26.7/4 MeV For 2 kg of hydrogen: E(H) = (2000/1)N(A) × (26.7/4) MeV For 2 kg of ²³⁵U: E(U) = (2000/235)N(A) × 200 MeV Therefore: E(H)/E(U) = [26.7 × 235]/[4 × 200] ≈ 7.84 The closest listed value is 7.62. Hence, Option C is the keyed answer.