Calculate the binding energy per nucleon of `._(20)^(40)Ca`. Given that mass of `._(20)^(40)Ca` nucleus `= 39.962589 u`, mass of proton `= 1.007825 u`. Mass of Neutron `= 1.008665 u` and `1 u` is equivalent to `931 MeV`.

Explanation

The graph showing the varitaiton of binding energy per nucleon with number of atomic nuclei s shown in below.

The nucleus `._(20)^(40)Ca` contains `20` protons and `20` neutrons.
Mass of `20` protons `=20xx1.007825`
`=20.1565 u`
Mass of `20` neutrons `=20xx1.008665`
`{:(=20.1733u),(=ul(40.3298u)):}`
Maa `._(20)^(40)Ca` nucleus `=39.962589 u`
Mass defect, `Delta m=0.367211u`
`B.E.-0.367211xx931=341.87 MeV`
`B.E.` per nucleon `=(341.87)/(40)=8.547MeV`