Calculate the binding energy per nucleon of 20Ca40 . Given that mass of 20Ca40 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.

Q: Calculate the binding energy per nucleon of 20Ca40 . Given that mass of 20Ca40 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 ={Zm_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.