The energy released in the fusion of $$2 \text{ kg}$$ of hydrogen deep in the sun is $$E_H$$ and the energy released in the fission of $$2 \text{ kg}$$ of $$^{235}U$$ is $$E_U$$. The ratio $$\frac{E_H}{E_U}$$ is approximately: (Consider the fusion reaction as $$4 \mid H + 2e^- \to {}^4_2He + 2\nu + 6\gamma + 26.7 \text{ MeV}$$, energy released in the fission reaction of $$^{235}U$$ is $$200 \text{ MeV}$$ per fission nucleus and $$N_A = 6.023 \times 10^{23}$$)

Energy from fusion of 2 kg hydrogen:

The fusion reaction: $$4\,^1H + 2e^- \to \,^4_2He + 2\nu + 6\gamma + 26.7$$ MeV

This means 4 hydrogen atoms (combined mass $$\approx 4 \times 1 = 4$$ g/mol) produce 26.7 MeV per reaction.

Moles of H in 2 kg = $$\frac{2000}{1} = 2000$$ mol

Number of fusion reactions = $$\frac{2000}{4} = 500$$ mol

Total energy: $$E_H = 500 \times N_A \times 26.7$$ MeV

Energy from fission of 2 kg $$^{235}U$$:

Moles of $$^{235}U$$ = $$\frac{2000}{235}$$ mol

Total energy: $$E_U = \frac{2000}{235} \times N_A \times 200$$ MeV

Ratio:

$$\frac{E_H}{E_U} = \frac{500 \times 26.7}{\frac{2000}{235} \times 200} = \frac{13350}{\frac{400000}{235}} = \frac{13350 \times 235}{400000} = \frac{3137250}{400000} \approx 7.84$$

The closest option is 7.62. The slight difference comes from the exact atomic masses used.

The correct answer is Option A: 7.62.