Obtain the binding energy of the nuclei ⁵⁶₂₆Fe and ²⁰⁹₈₃BI in units of MeV from the following data:

m (⁵⁶₂₆Fe) = 55.934939 u m (²⁰⁹₈₃BI) = 208.980388 u

The number of neutron and Proton both in a ⁵⁶Fe₂₆ atom is respectively 30 and 26.

The mass defect Δm = m_p + m_n-m_Fe

And Binding energy is given by E_B = Δmc²

Mass of a proton = 1.007825 u

Mass of a neutron = 1.008665 u

In this case mass defect = Δm = (26 × 1.007825 + 30 × 1.008665-55.934939)u = 0.528461u

We know, 1 u = 931.5 MeV/c²

So, E = Δmc² = 0.528461 × 931.5(MeV/c²) × c² = 492.26MeV

So, Binding Energy per Nucleon = = = 8.79MeV

The number of neutron and Proton both in a ²⁰⁹Bi₈₃ atom is respectively 126 and 83.

The mass defect Δm = m_p + m_n-m_Bi

And Binding energy is given by E_B = Δmc²

Mass of a proton = 1.007825 u

Mass of a neutron = 1.008665 u

Mass Defect = Δm = 83 × 1.007825 + 126 × 1.008665-208.980388 = 1.760877 u

We know, 1 u = 931.5 MeV/c²

So, E = Δmc² = 1.760877 × 931.5 (Mev/c²) × c² = 1640.26 MeV

Average binding energy per nucleon = 1640.26/209 = 7.848 MeV