Let us first find the binding energy of

56

‌

26

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

Fe nucleus E = Δ mc2 = Δm 931.5 MeV
= 0.528461 (931.5) MeV = 492.26 MeV

B.E

nucleon

of

56

‌

26

Fe =

492.26

56

MeV = 8.79 MeV
(b) Now binding energy of

209

‌

83

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

Bi nucleus ⇒ Δ mc2
⇒ Δm (931.5 MeV) ⇒ 1.760877 × 931.5 MeV ⇒ 1640.3 MeV

B.E

nucleon

of

209

‌

83

Bi =

1640.3

209

MeV = 7.85 MeV
So,

56

‌

26

Fe is much more stable than

209

‌

83

Bi, due to more binding energy per nucleon.