Expression for nuclear binding energy: 

i. Consider a nuclide \(^A_ZX\) that contains Z protons and (A – Z) neutrons. Suppose the mass of the nuclide is m. The mass of proton is m_p and that of neutron is mn.

ii. Total mass = (A – Z)m_n + Zm_p + Zm_e …..(1)

Δm = [(A – Z)m_n + Zm_p + Zm_e] – m

= [(A – Z)m_n + Z(m_p + m_e] – m

= [(A – Z)m_n + Zm_H] – m …..(2)

Where (m_p + m_e) = m_H = mass of H atom.

Thus, (Δm) = [Zm_p + (A – Z)m_n] – m

Where Z = atomic number 

A = mass number 

(A – Z) = neutron number

m_p and m_n = masses of proton and neutron, respectively 

m = mass of nuclide

iii. The mass defect, Δm is related to binding energy of nucleus by Einstein’s equation, ΔE = Δm × c²

Where, ΔE = Binding energy, Δm = mass defect. 

iv. Nuclear energy is measured in million electro volt (MeV).

v. The total binding energy is then given by, B.E. = Δm (u) × 931.4 

Where 1.00 u = 931.4 MeV 

B.E. = 931.4 [Zm_H + (A – Z)m_n – m] ……(3)

Total binding energy of nucleus containing A nucleons is the B.E. 

vi. The binding energy per nucleon is then given by, 

\(\bar{B}\) = B.E./A