Let us first find the binding energy of $\,{}^{56}_{26}\mathrm{Fe}$ No. of protons in Fe = Z = 26 Mass of protons = 26 × 1.007825 u = 26.203450 u No. of neutrons in Fe, n = A – Z = 56 – 26 = 30 Mass of neutrons = 30 × 1.008665 u = 30.259950 u Total theoretical mass of nucleus = 26.203450 u + 30.259950 u = 56.463400 u Actual mass of Fe nucleus 55.934939 u Mass defect Δm = Total mass – Actual mass = 0.528461 u B.E. of $\,{}^{56}_{26}\mathrm{Fe}$ nucleus E = Δ $mc^2$ = Δm 931.5 MeV = 0.528461 (931.5) MeV = 492.26 MeV $\frac{\mathrm{B.E.}}{\mathrm{nucleon}}$ of $\,{}^{56}_{26}\mathrm{Fe}$ = $\frac{492.26}{56}$ MeV = 8.79 MeV (b) Now binding energy of $\,{}^{209}_{83}\mathrm{Bi}$ No. of protons in Bi = Z = 83 No. of neutrons in Bi ⇒ n = A – Z = 209 – 83 = 126 Mass of protons = 83 × 1.007825 u = 83.649475 u Mass of neutrons = 126 × 1.008665 u = 127.091790 u Total theoretical mass of nucleus = 210.741265 u Actual mass of Bi nucleus = 208.980388 u Mass defect, Δm = 210.741260 – 208.980388 = 1.760877 u B.E. of $\,{}^{209}_{83}\mathrm{Bi}$ nucleus ⇒ Δ $mc^2$ ⇒ Δm (931.5 MeV) ⇒ 1.760877 × 931.5 MeV ⇒ 1640.3 MeV $\frac{\mathrm{B.E.}}{\mathrm{nucleon}}$ of $\,{}^{209}_{83}\mathrm{Bi}$ = $\frac{1640.3}{209}$ MeV = 7.85 MeV So, $\,{}^{56}_{26}\mathrm{Fe}$ is much more stable than $\,{}^{209}_{83}\mathrm{Bi}$, due to more binding energy per nucleon.