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The binding energy per nucleon of ²⁰⁹₈₃Bi is ______ MeV. [Take m(²⁰⁹₈₃Bi)=208.980388 u, mₚ=1.007825 u, mₙ=1.008665 u, 1 u=931 MeV/c²]

Asked in JEE Main 2nd April 2nd Shift 2026 · Mass defect and binding energy

Answer: (2) 7.84

Step-by-step solution

Z=83, N=126.

Nucleon mass =83×1.007825+126×1.008665=210.741265 u

Δ m=1.760877 u, B=1639.4 MeV

B/A=(1639.4)/(209)=7.84 MeV

Why the other options are wrong

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