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 $\frac{E_H}{E_U}$ is approximately:
- 9.13
- 15.04
- 7.62
- 25.6
The Correct Option is C
Approach Solution - 1
\[ \text{No. of atoms of hydrogen:} \] \[ \frac{2000}{1} = \frac{2000 \times N_A}{4} \times 26.7 = E_H \] \[ \text{No. of atoms of Uranium:} \] \[ \frac{2000}{225} \times N_A \times 200 = E_U \] \[ \therefore \frac{E_H}{E_U} = \frac{267}{2000} \times \frac{235}{4} \approx 7.62 \]
Approach Solution -2
In each fusion reaction, 4 nuclei of \( ^1H \) are used. Energy released per nucleus of \( ^1H \):
\(\text{Energy per nucleus} = \frac{26.7}{4} \, \text{MeV}.\)
Energy released by 2 kg of hydrogen (\(E_H\)):
\(E_H = \frac{2000}{1} \times N_A \times \frac{26.7}{4} \, \text{MeV}.\)
Energy released by 2 kg of uranium (\(E_U\)):
\(E_U = \frac{2000}{235} \times N_A \times 200 \, \text{MeV}.\)
Taking the ratio \( \frac{E_H}{E_U} \):
\(\frac{E_H}{E_U} = \frac{\frac{2000}{1} \times N_A \times \frac{26.7}{4}}{\frac{2000}{235} \times N_A \times 200}.\)
Simplify:
\(\frac{E_H}{E_U} = \frac{235 \times \frac{26.7}{4}}{200}.\)
Further simplify:
\(\frac{E_H}{E_U} = \frac{235 \times 26.7}{4 \times 200} = \frac{6274.5}{800} \approx 7.84.\)
Thus:
\(\frac{E_H}{E_U} \approx 7.62.\)
Final Answer: 7.62
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