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 $^{235}U$ is $E_U$. The ratio $\frac{E_H}{E_U}$ is approximately:

• A. 9.13
• B. 15.04
• C. 7.62
• D. 25.6

Answer 1

The Correct Option is C

\[ \text{No. of atoms of hydrogen:} \] \[ \frac{2000}{1} = \frac{2000 \times N_A}{4} \times 26.7 = E_H \] \[ \text{No. of atoms of Uranium:} \] \[ \frac{2000}{225} \times N_A \times 200 = E_U \] \[ \therefore \frac{E_H}{E_U} = \frac{267}{2000} \times \frac{235}{4} \approx 7.62 \] 

Answer 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