Two lighter nuclei combine to form a comparatively heavier nucleus by the relation given below: ${}_{1}^{2} \mathrm{X}+{}_{1}^{2} \mathrm{X}={}_{2}^{4} \mathrm{Y}$ The binding energies per nucleon ${}_{1}^{2} \mathrm{X}$ and ${}_{2}^{4} \mathrm{Y}$ are 1.1 MeV and 7.6 MeV respectively. The energy released in this process is ______ MeV.

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Modern Physics

Two lighter nuclei combine to form a comparatively heavier nucleus by the relation given below:

_{1}^{2} X +_{1}^{2} X =_{2}^{4} Y

The binding energies per nucleon

_{1}^{2} X

and

_{2}^{4} Y

are 1.1 MeV and 7.6 MeV respectively. The energy released in this process is ______ MeV.

Answer: 26

Solution

The energy released in a nuclear fusion reaction equals the difference between the binding energy of products and reactants.

Binding energy of product

_{2}^{4} Y

7.6 \times 4 = 30.4

MeV

Binding energy of reactants (two

_{1}^{2} X

nuclei) =

2 \times (1.1 \times 2) = 4.4

MeV

Energy released =

30.4 - 4.4 = 26

MeV

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Integer answer

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