1.

Obtain the binding energy of the nuclei `._26Fe^(56)` and `._83Bi^(209)` in units of MeV form 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» (i) `._(26)F^(56)` nucleus contains 26 protons and `(56-26)=30` neutrons
Mass of 26 protons `=26xx1.007825=26.20345 a.m.u`
Mass of 30 neutrons `=30xx1.008665=30.25995 a.m.u`
Total mass of 56 nuleons `=56.46340 a.m.u`.
mass of `._26Fe^(56)` nucleus `=55.934939`
`:.` Mass defect `Deltam=56.46340-55.934939=0.528461 a.m.u.`
Total binding energy `=0.528461xx931.5MeV=-492.26 MeV`
Av. B.E. per nucleon=492.26/56=8.790MeV
(ii) `._83Bi^(209)` nucleus 83 protons and `(209-83)=126` neutrons.
Mass of 83 protons =`83xx1.007825=83.649475 a.m.u.`
Mass of 126 neutrons `=126xx1.008665=127.091790 a.m.u.`
Total mass of nucleons `=210.741260 a.m.u`.
Mass of `._83Bi^(209)` nucleus `=208.980388 a.m.u`.
Mass defect, `Deltam=210.741260-208.980388=1.760872`
Total B.E. `=1.760872xx931.5MeV=1640.26 MeV.`
Av. B.E. per nucleon =`1640.26/209=7.848 MeV`
Hence `._26F^(56)` has greater B.E. per nucleon than `._83Bi^(209)`


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