The value of binding energy per nucleon of 20 40 Ca nucleus is Given: Mass of 20 40 Ca nucleus = 39.962589 u Mass of proton = 1.007825 u Mass of neutron = 1.008665 u and 1 u = 931 MeV C - 2

Step-by-step solution

The nucleus 20 40 Ca contains 20 protons and 20 neutrons. Mass of 20 protons = 20 × 1.007825 = 20.1565 u Mass of 20 neutrons = 20 × 1.008665 = 20.1733 u Total mass = 40.3298 u ___________ Mass 20 40 Ca nucleus = 39.962589 u Mass defect, Δ m = 0.367211 u Binding energy = 0.367211 × 931 = 341.87 MeV Binding energy per nucleon = 341.87 40 = 8.547 MeV