Obtain the binding energy of the nuclei `._26^56Fe` and `._83^209Bi` in units of MeV from the following data .
`m_H`=1.007825 u, `m_n`=1.008665 u
`m(._26^56Fe) =55.934939 u " " m(._83^209Bi)` = 208.980388 u
Which nucleus has greater binding energy per nucleon ?

Explanation

Mass defect in `""_(26)Fe^(56)=30xx1.008665+26xx1.007825-55.934939=0.528461` amu
`therefore` Total binding energy `=0.528461 xx931.5 MeV =492.26 MeV`
`therefore` B.E. per nucleon `=(492.26)/(56)=8.790 MeV`
Mass defect in `""_(83)Bi^(209)` nucleus `=83xx1.007825 +126xx1.008665-208.980388=1.7660872`
Total B.E `=1.760872xx931.5 =1640.26` MeV
`therefore` B.E. per nucleon `=(1640.26)/(209)=7.848` MeV
`""_(26)Fe^(56)` has greater B.E per nucleon than `""_(83)Bi^(209)`