JEE Main 2024 Physics question
If three helium nuclei combine to form a carbon nucleus then the energy released in this reaction is ...... × 10 MeV. (Given 1 u = 931 MeV/c, atomic mass of helium = 4.002603 u)
Answer: 727
Solution
Δm = 3 × 4.002603 − 12 = 0.007809 u; E = 0.007809 × 931 = 7.27 MeV = 727 × 10⁻² MeV.