Q 12-13-033JEE MainJEE Main 2026 (2 Apr, Shift 2)Medium
The binding energy per nucleon of $^{209}_{83}Bi$ is ______ MeV.
[Take $m(^{209}_{83}Bi)=208.980388$ u, $m_p=1.007825$ u, $m_n=1.008665$ u, $1$ u $=931$ MeV/$c^2$]
Answer: (B) $7.84$
The nucleus has $83$ protons and $209-83=126$ neutrons.
Mass of separate nucleons $=83\times1.007825+126\times1.008665=83.649475+127.091790=210.741265$ u
Mass defect $\Delta m=210.741265-208.980388=1.760877$ u
Binding energy $=1.760877\times931\approx1639.4$ MeV
Per nucleon: $\dfrac{1639.4}{209}\approx7.84$ MeV
Solution by Sreeraj P, M.Sc Physics