JEE MAIN - Physics (2024 - 9th April Evening Shift - No. 11)

The energy released in the fusion of $$2 \mathrm{~kg}$$ of hydrogen deep in the sun is $$E_H$$ and the energy released in the fission of $$2 \mathrm{~kg}$$ of $${ }^{235} \mathrm{U}$$ is $$E_U$$. The ratio $$\frac{E_H}{E_U}$$ is approximately: (Consider the fusion reaction as $$4_1^1H+2 \mathrm{e}^{-} \rightarrow{ }_2^4 \mathrm{He}+2 v+6 \gamma+26.7 \mathrm{~MeV}$$, energy released in the fission reaction of $${ }^{235} \mathrm{U}$$ is $$200 \mathrm{~MeV}$$ per fission nucleus and $$\mathrm{N}_{\mathrm{A}}= 6.023 \times 10^{23})$$
7.62
25.6
9.13
15.04

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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