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`

Explanation

We know that 1u `=931.5 MeV//c^(2)`
For `""_(20)Fe^(56)`,
Atomic number , Z=26
Atomic mass, A = 56
Number of neutrons, N=A-Z = 56-26=30
Binding energy (B.E.) = `Deltam xx c^(2)`
B.E. of `""_(26)Fe^(56) = (Zm_(H) + Nm_(n)) -m(""_(26Fe^(56))c^(2))`
`=492.26` MeV
B.E. per nucleon `=(492.26)/A = 492.26/56 = 8.79 MeV`
For `""_(53)Bi^(209)`
Atomic number, Z = 83
Atomic mass, A = 209
B.E. of `""_(83)Be^(209) =(83 xx (1.007825 + 126 xx (1.008665) -208.980388)u xx 931.5 MeV//u`
`=1.760877 xx 931.5` MeV
=1640.1 MeV
B.E. per nucleus `=(1640.1)/209 = 7.84 MeV`
`Fe^(56)` has greater binding energy per nucleon.