QJEEM24Phd3a1: JEE Main 2024 Physics - Class 12 JEEM24Phd3a1 Solution: The energy released in the fusion of of hydrogen deep in the…

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.