JEE MAIN - Physics (2024 - 9th April Evening Shift - No. 11)
Explanation
To determine the ratio $$\frac{E_H}{E_U}$$, let's first consider the energy released in the fusion of hydrogen and the energy released in the fission of $$^{235}U$$.
For the fusion reaction:
The given reaction is:
$$4_1^1H + 2 \mathrm{e}^{-} \rightarrow {}_2^4 \mathrm{He} + 2 \mathrm{\nu} + 6 \gamma + 26.7 \, \mathrm{MeV}$$
This reaction shows that 4 hydrogen atoms and 2 electrons produce 1 helium atom, 2 neutrinos, and 6 gamma photons while releasing 26.7 MeV of energy.
The mass of 1 mole of $$H_1^1$$ is approximately 1 gram, so 2 kg of hydrogen is equal to:
$$2 \times 10^3 \, \mathrm{g}$$
The number of moles of hydrogen in 2 kg is:
$$\mathrm{N}_\text{moles} = \frac{2 \times 10^3}{1} = 2 \times 10^3 \, \mathrm{moles}$$
The number of hydrogen atoms in 2 kg is:
$$\mathrm{N}_\text{atoms} = N_A \times 2 \times 10^3 = 6.023 \times 10^{23} \times 2 \times 10^3 = 1.2046 \times 10^{27} \, \mathrm{atoms}$$
Since 4 hydrogen atoms release 26.7 MeV, the total energy released ($$E_H$$) is:
$$E_H = \frac{1.2046 \times 10^{27}}{4} \times 26.7 \, \mathrm{MeV}$$
$$E_H = 3.0115 \times 10^{26} \times 26.7 \, \mathrm{MeV}$$
$$E_H \approx 8.04 \times 10^{27} \, \mathrm{MeV}$$
For the fission reaction:
The energy released per fission of one $$^{235}U$$ nucleus is 200 MeV.
The mass of 1 mole of $$^{235}U$$ is approximately 235 grams, so 2 kg of $${}^{235}U$$ is equal to:
$$2 \times 10^3 \, \mathrm{g}$$
The number of moles of $$^{235}U$$ in 2 kg is:
$$\mathrm{N}_\text{moles} = \frac{2 \times 10^3}{235} \approx 8.51 \, \mathrm{moles}$$
The number of $$^{235}U$$ nuclei is:
$$\mathrm{N}_\text{nuclei} = N_A \times 8.51 \approx 6.023 \times 10^{23} \times 8.51 \approx 5.12 \times 10^{24} \, \mathrm{nuclei}$$
The total energy released ($$E_U$$) is:
$$E_U = 200 \, \mathrm{MeV} \times 5.12 \times 10^{24} \approx 1.024 \times 10^{27} \, \mathrm{MeV}$$
Finally, the ratio $$\frac{E_H}{E_U}$$ is:
$$\frac{E_H}{E_U} \approx \frac{8.04 \times 10^{27}}{1.024 \times 10^{27}} \approx 7.85$$
Therefore, the ratio is closest to option A: 7.62.
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