Answer :

The number of neutron and Proton both in a ^{56}Fe_{26} atom is respectively 30 and 26.

The mass defect Δm = m_{p} + m_{n}-m_{Fe}

And Binding energy is given by E_{B} = Δmc^{2}

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/c^{2}

So, E = Δmc^{2} = 0.528461 × 931.5(MeV/c^{2}) × c^{2} = 492.26MeV

So, Binding Energy per Nucleon = = = 8.79MeV

The number of neutron and Proton both in a ^{209}Bi_{83} atom is respectively 126 and 83.

The mass defect Δm = m_{p} + m_{n}-m_{Bi}

And Binding energy is given by E_{B} = Δmc^{2}

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/c^{2}

So, E = Δmc^{2} = 1.760877 × 931.5 (Mev/c^{2}) × c^{2} = 1640.26 MeV

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

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