The binding energies per nucleon of ⁷₃Li and ⁴₂He are 5.60 MeV and 7.06 MeV, respectively. For the reaction ⁷₃Li + ¹₁H → 2(⁴₂He) + Q, find the energy Q released.
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Question

The binding energies per nucleon of ⁷₃Li and ⁴₂He are 5.60 MeV and 7.06 MeV, respectively. For the reaction ⁷₃Li + ¹₁H → 2(⁴₂He) + Q, find the energy Q released.

G1 Physics · Accepted Answer

Initial binding energy of ⁷₃Li is 7 × 5.60 = 39.20 MeV. The proton has no nuclear binding energy. Final binding energy of two ⁴₂He nuclei is 8 × 7.06 = 56.48 MeV. Therefore Q = 56.48 − 39.20 = 17.28 MeV ≈ 17.3 MeV. Correct answer: D.