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 ²³⁵ U is E_U. The ratio E_H/E_U is approximately: (Consider the fusion reaction as 4₁¹H+2 e⁻ → ₂⁴ He+2 v+6 γ+26.7 MeV, energy released in the fission reaction of ²³⁵ U is 200 MeV per fission nucleus and N_A= 6.023 × 10²³)

Solution

Physics sir · 3 steps

Count the reactions in 2 kg of each fuel, multiply by the energy per reaction, and N_A cancels.

1. Fusion side. 2 kg of hydrogen is 2000 g, which is 2000 N_A hydrogen atoms. Each fusion uses 4 of them, so there are 500 N_A fusions. Then E_H = 500 N_A × 26.7 = 13350 N_A MeV.
2. Fission side. 2 kg of ²³⁵U is 2000/235 = 8.51 mol, which is 8.51 N_A nuclei. Each gives 200 MeV, so E_U = 8.51 N_A × 200 = 1702 N_A MeV.

   E_H/E_U = (26.7/4)/(200/235) = (6.675 × 235)/200 ≈ 7.8

3. So. The ratio is about 7.8. Among the options, 7.62 is the closest, so the answer is (a).

The idea to keep. To compare fuels of equal mass, find the energy released per nucleon of fuel used. Fusion gives about 6.7 MeV per nucleon, fission only about 0.85 MeV, which is why fusion wins by roughly 8 times.

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