If three helium nuclei combine to form a carbon nucleus then the energy released in this reaction i s ______ × 10⁻² MeV. (Given 1 u = 931 MeV/c², atomic mass of helium = 4.002603 u)

Question

G1 Physics · Accepted Answer

Reaction: 3(⁴₂He) → ¹²₆C + γ-rays Mass defect: Δm = 3m(He) - m(C) = 3 × 4.002603 - 12 = 0.007809 u Energy released: E = 931 × Δm MeV = 931 × 0.007809 ≈ 7.27 MeV = 727 × 10⁻² MeV Hence, answer = 727.