Two deuterons (²₁H) fuse to form a helium nucleus (⁴₂He). The binding energy per nucleon of deuterium is 1.1 MeV and that of helium-4 is 7.0 MeV. The energy released is:
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Question

Two deuterons (²₁H) fuse to form a helium nucleus (⁴₂He). The binding energy per nucleon of deuterium is 1.1 MeV and that of helium-4 is 7.0 MeV. The energy released is:

G1 Physics · Accepted Answer

Binding energy of one deuteron: BE(D) = 2×1.1 = 2.2 MeV. There are two deuterons, so total initial binding energy is BE(initial) = 2×2.2 = 4.4 MeV. For helium-4: BE(final) = 4×7.0 = 28.0 MeV. Energy released is the increase in total binding energy: Q = BE(final) - BE(initial) = 28.0 - 4.4 = 23.6 MeV. Correct option: B.