Question

36) a) What is the binding energy in kJ/mol nucleons for antimony-121? ___ kJ/mol nucleons
^{The required masses (g/mol) are 1/1 H= 1.00783 ; 1/0 n= 1.00867 ; 121/51 Sb= 120.90380}

b) What is the binding energy in kJ/mol Cl for chlorine-35? ___ kJ/mol
^{The required masses (g/mol) are 1/1 H= 1.00783 ;1/0 n= 1.00867 ; 35/17 Cl= 34.96885}

Answer 1

a)

Mass of Sb = 120.90380 gm

Mass of 51×H(proton)= 51 ×1.00783 gm = 51.39933 gm

Mass of 70× n = 70 ×1.00867 gm = 70.6069 gm

Difference in mass = (51.39933 gm+ 70.6069 gm) - 120.90380 gm Diff. in mass =1.10243 gm

Diff. In mass = 1.10243 ×10^-3 Kg

For Binding energy,

Binding energy = ∆m ×c^2

B.E =(1.10243 ×10^-3 ) * (3*10^8)^2

B. E = 9.92 ×10^13 J/mol

Binding energy = 9.92×10^10 KJ/mol.

b)

mass of Cl-35=34.96885 gm

mass of 17×H(proton)= 17×1.00783 gm = 17.13311gm

mass of 18× n = 18 ×1.00867 gm = 18.15606 gm

Diff. in mass = (17.13311 gm+ 18.15606 gm) - 34.96885 gm

Diff. in mass =0.32032 gm

Diff. in mass = 0.32032×10^-3 Kg

Binding energy = ∆m * c^2

B. E =(0.32032×10^-3 ) * (3*10^8)^2

B. E = 2.88288×10^13 J/mol

Binding energy= 2.88288×10^10 KJ/mol