Two deuterium nuclei ²₁H combine to form one ⁴₂He nucleus. The binding energy per nucleon of ²₁H is 1.1 MeV and that of ⁴₂He is 7.2 MeV. What energy is released in the reaction?
Options

Question

Two deuterium nuclei ²₁H combine to form one ⁴₂He nucleus. The binding energy per nucleon of ²₁H is 1.1 MeV and that of ⁴₂He is 7.2 MeV. What energy is released in the reaction?

G1 Physics · Accepted Answer

Initial state contains two deuterium nuclei, giving a total of four nucleons. Initial total binding energy: BE(initial) = 4×1.1 = 4.4 MeV. Final helium nucleus also contains four nucleons. Final binding energy: BE(final) = 4×7.2 = 28.8 MeV. Energy released is the increase in total binding energy: Q = BE(final)-BE(initial) = 28.8-4.4 = 24.4 MeV. Correct option: B.