Question

Obtain the binding energy of the nuclei `._26Fe^(56)` and `._83Bi^(209)` in units of MeV from the following data: `m(._26Fe^(56))=55.934939a.m.u.` , `m=(._83Bi^(209))=208.980388 a m u`. Which nucleus has greater binding energy per nucleon? Take `1a.m.u =931.5MeV`

Answer 1

Mass defect in `._26Fe^56`= 30 x 1.008665 + 26 x 1.007825 - 55.934939
=0.528461 amu
`therefore` Total binding energy =0.528461 x 931.5 MeV
=492.26 MeV
`therefore` B.E. per nucleon=`492.26/56`=8.760 MeV
Mass defect in `._83Bi^209` nucleus = 83 x 1.007825 + 126 x 1.008665 - 208.980388
=1.760872
Total B.E.=1.760872 x 931.5
=1640.26 MeV
`therefore` B.E. per nucleon =`1640.26/209`=7.848 MeV
`._26Fe^56` has greater B.E. per nucleon than `._83Bi^209`.