The energy released in the fusion of 2 kg of hydrogen deep in the sun is and the energy released in the fission of 2 kg of is . The ratio is approximately

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Solution

Q.
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 ${}^{235}U$ is $E_{U}$ . The ratio $\frac{E_{H}}{E_{U}}$ is approximately

(Consider the fusion reaction as $4 {}_{1}^{1}H + 2 e^{-} \to {}_{2}^{4}H e + 2 v + 6 γ + 26.7$ MeV, energy released in the fission reaction of ${}^{235}U$ is 200 MeV per fission nucleus and $N_{A} = 6.023 \times 10^{23}$ ) [2024]

1 9.13

2 15.04

3 7.62

4 25.6

Ans.
(3)

In each fusion reaction, $4 {}_{1}^{1}H$ nucleus is used.

  Energy released per Nuclei of ${}_{1}^{1}H = \frac{26.7}{4}$ MeV

   $∴$ Energy released by 2 kg hydrogen,

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

   $∴$ Energy released by 2 Kg Uranium,

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

$So, \frac{E_{H}}{E_{U}} = 235 \times \frac{26.7}{4 \times 200} = 7.84$

$∴$   Approximately close to 7.62

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