JEE Main 2024 Physics question
The atomic mass of ₆C¹² is 12.000000 u and that of ₆C¹³ is 13.003354 u. The required energy to remove a neutron from ₆C¹³, if mass of neutron is 1.008665 u, will be :
Answer: (D)
Solution
₆C¹³ → ₆C¹² + ₀n¹. Δm = 12.000000 + 1.008665 − 13.003354 = 0.005311 u, so ΔE = 0.005311 × 931.5 MeV ≈ 4.95 MeV.