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Atoms and Nuclei Part 3

Section: Physics

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Question : 45 of 98

Marks: +1, -0

The energy released in the fusion of

2\,\text{kg}

of hydrogen deep in the sun is

E_H

and the energy released in the fission of

2\,\text{kg}

{}^{235}\mathrm{U}

E_U

\frac{E_H}{E_U}

is approximately : (Consider the fusion reaction as

4\,{}^{1}_{1}\mathrm{H} + 2\,\mathrm{e}^{-} \rightarrow {}^{4}_{2}\mathrm{He} + 2 v + 6\gamma + 26.7\,\text{MeV}

, energy released in the fission reaction of

{}^{235}\mathrm{U}

200\,\text{MeV}

per fission nucleus and

N_A = 6.023 \times 10^{23}

[9 Apr 2024 Shift 2]

9.13
15.04
7.62
25.6

Solution: 👈: Video Solution

In each fusion reaction,

4\,{}^{1}_{1}\mathrm{H}

nucleus are used.

Energy released per Nuclei of

{}^{1}_{1}\mathrm{H} = \frac{26.7}{4}\,\text{MeV}

\therefore

Energy released by

2\,\text{kg}

hydrogen

(E_H)

= \frac{2000}{1} \times N_A \times \frac{26.7}{4}\,\text{MeV}

\therefore

Energy released by

2\,\text{kg}

Vranium

(E_V)

= \frac{2000}{235} \times N_A \times 200\,\text{MeV}

So,

\frac{E_H}{E_V} = 235 \times \frac{26.7}{4 \times 200} = 7.84

\therefore

Approximately close to 7.62

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