JEE Main 2026 Physics question
The binding energy per nucleon of Bi is MeV. [Take m(Bi) = 208.980388 u, = 1.007825 u, = 1.008665 u, 1u = 931 MeV/c²]
Answer: (B)
Solution
Δm = 83(1.007825) + 126(1.008665) − 208.980388 = 1.760877 u. BE = 1.760877 × 931 = 1639.4 MeV; BE/nucleon = 1639.4/209 ≈ 7.84 MeV.