Calculate the binding energy per nucleon for the nuclei 8636

Calculate the binding energy per nucleon for the nuclei 86_36 K r and 180_73 Ta. Do your results confirm what is shown in the binding energy diagram, that for A greater than 62 the binding energy per nucleon decreases as A increases.

Solution

Binding energy per nucleon De = Dmc², Where De= Binding energy per nucleon, Dm= Mass defect, c=Speed of light

Mass defect is the difference between the mass of a nucleus and the sum of the masses of the nucleons of which it is composed.

Mass of ⁸⁶₃₆Kr = 86 amu

Combined mass of proton and neutron = 36protons*(1.00728 amu/proton)+50neutrons*(1.00867 amu/neutron) = 86.69558amu

So mass defect Dm = 86.69558amu -86amu = 0.69558amu = 1.155*10^-27kg , Since (1 amu = 1.6606 x 10^-27 kg)

So binding energy pernucleon De = 10.395*10^-11 J/nucleon

Similarly for ¹⁸⁰₇₃Ta, Mass defect Dm = 73protons*(1.00728 amu/proton)+107neutrons*(1.00867 amu/neutron)-1801mu=181.45913amu -180amu = 1.45913amu = 2.423*10^-27kg

So binding energy pernucleon De = 21.807*10^-11 J/nucleon