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

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Modern Physics

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

E_{U}

. The ratio

\frac{E _{H}}{E _{U}}

is approximately :
(Consider the fusion reaction as

4_{1}^{1} H + 2 e^{-} \to_{2}^{4} He + 2 ν + 6 γ + 26.7 MeV

, energy released in the fission reaction of

^{235} U

is 200 MeV per fission nucleus and

N_{A} = 6.023 \times 1 0^{23}

)

Answer: (C)

Each fusion event consumes 4 hydrogen-1 nuclei and releases 26.7 MeV, so energy released per hydrogen nucleus is

\frac{26.7}{4}

MeV.

For 2 kg (

2000

g) of hydrogen, number of atoms

= \frac{2000}{1} N_{A}

, so

E_{H} = \frac{2000}{1} N_{A} \cdot \frac{26.7}{4} MeV

For 2 kg of

^{235}

U, number of atoms

= \frac{2000}{235} N_{A}

, each fission releasing 200 MeV:

E_{U} = \frac{2000}{235} N_{A} \cdot 200 MeV

Ratio:

\frac{E _{H}}{E _{U}} = \frac{235 \times 26.7}{4 \times 200} = \frac{6274.5}{800} \approx 7.84

This is closest to option (3),

7.62

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Single correct option