Q: Calculate the binding energy per nucleon of _{20}Ca^{40} . Given that mass of _{20}Ca^{40} nucleus = 39.962589 u, mass of proton = 1.007825 u. mass of Neutron = 1.007825 u. mass of neutron = 1.008665 u and 1 u is equivalent of 931 MeV.

Sol: A = 40, Z = 20, A-Z = 20

Δ m ={Z m_{P} + (A -Z) m_{n}} – M_{n}

= {(20 x 1.007825 + (20 × 1.008665)} – 39.962589

= 40.329800 – 39.962589; Δ m = 0.367211

Binding energy per nucleon = (Δm × 931)/A

= (0.367211 × 931 )/40

= 8.547 MeV.