Obtain the binding energy of the nuclei `._(26)^(56)Fe` and `._(83)^(209)Bi`in units of MeV from the following data : `m (._(26)^(56)Fe)= 55.934939 u, m (._(83)^(209)Bi)=208.980388 u`

Explanation

(i) `._(26)F^(56)` nucleus contains 26 protons and (56 - 26) = 30 neutrons
Mass of 26 proton `= 26xx1.007825`
`= 26.26345 a.m.u`
Mass of 30 neutrons `= 30xx1.008665`
`= 30.25995 amu`
Total mass of 56 nucleons
`= 56.46340 a.m.u`
Mass of `._(26)F^(56)` Nucleus
`= 55.934939 a.m.u`
Mass defect `Delta m = 56.46340-55.934939`
`= 0.528461 a.m.u`
Total binding energy `= 0.528461xx931.5 MeV`
`= 492.26 MeV`
Average B.E per nucleon `= (492.26)/(56)`
`= 8.790 MeV`.
(ii) `._(83)Bi^(209)` nucleus contains 83 protons and (209-83) = 126 neutrons
Mass of 83 protons `= 83xx1.007825`
`= 83.649475 a.m.u`
Mass of 126 Neutrons `= 126xx1.008665`
`= 127.09170 a.m.u`
Total mass of nucleons `= 210.741260 a.m.u`
Mass of `._(83)Bi^(209)` nucleus = 208.986388 a.m.u
Mass defect `Delta m = 210.741260-208.980388`
= 1.760872
Total B.E `= 1.760872xx931.5 MeV`
= 1640.26 MeV
Average B.E per nucleon `= (1640.26)/(209)`
= 7.848 MeV
Hence `._(26)Fe^(56)` has greater B.E per nucleon than `._(83)Bi^(209)`