Question

The most stable nucleus in terms of binding energy per nucleon is ⁵⁶Fe. If the atomic mass of ⁵⁶Fe is 55.9349 amu, calculate the binding energy per nucleon for ⁵⁶Fe. The mass of a hydrogen atom is 1.0078 amu, and the mass of a neutron is 1.0087 amu. (1 J = 1 kg･m²/s², 1 amu = 1.66 × 10⁻²⁷ kg)

Answer 1

Sol.

As Atomic Mass of 56Fe = m = 55.9349 amu

Speed of light = c = 3 × 10⁸ m / s

Mass of proton = mp = 1.0078 amu

Mass of neutron = mn = 1.0087 amu

Atomic Number of 56Fe = Z = 26

Mass Number of 56Fe = A = 56

Number of protons = Z = 26

Number of neutrons = A - Z = 56 - 26 = 30

Therefore , Mass defect = deltam

= 26 × mp + 30 × mn - m

= 26 × 1.0078 amu + 30 × 1.0087 amu - 55.9349 amu

= 0.5289 amu

= 0.5289 × 1.66 × 10^-27 Kg

= 0.8779 × 10^-27 Kg

Binding Energy per nucleon

= deltam × c² / A

= 0.8779 × 10^-27 Kg × ( 3 × 10⁸ m/s )² / 56

= 1.4109 × 10^-12 Kg m² s^-2  

= 1.4109 × 10^-12 J