JEE Main 2025 Apr 2 Shift 2, Physics Q12: Mass-Energy and Nuclear Binding Energy

JEE Main2025Apr 2, Shift 2Physics

Energy released when two deuterons ( $_{1} H^{2}$ ) fuse to form a helium nucleus ( $_{2} He^{4}$ ) is :

(Given : Binding energy per nucleon of $_{1} H^{2} = 1.1$ MeV and binding energy per nucleon of $_{2} He^{4} = 7.0$ MeV)

Solution

Total binding energy of the two deuterons (each has $2$ nucleons): $2 \times (2 \times 1.1) = 4.4$ MeV.

Total binding energy of the helium-4 nucleus ( $4$ nucleons): $4 \times 7.0 = 28.0$ MeV.

Energy released in the fusion equals the increase in total binding energy:

$Q = 28.0 - 4.4 = 23.6$ MeV.

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