(i) \\\\(_{26}Fe^{56}\\\\) nucleus contains 26 protons .Number of neutrons \\\\(=(56-26) = 30 \\\\) neutronsNow,Mass of 26 protons\\\\( = 26\\\\) X \\\\(1.007825 = 26.20345 u\\\\)Mass of 30 neutrons\\\\( = 30\\\\) X \\\\(1.008665= 30.25995 u\\\\)Total mass of 56 nucleons\\\\(= 56.46340 u\\\\)Mass of \\\\(_{26}Fe{56}\\\\) nucleus\\\\(= 55.934939 u\\\\)Therefore,Mass defect, \\\\(\ riangle m\\\\) \\\\(= 56.46340-55.934939 = 0.528461 u\\\\)Total Binding Energy\\\\(= 0.528461\\\\) X \\\\(931.5 Mev = 492.26 MeV\\\\)Average binding energy per nucleon\\\\(= \\frac{492.26}{56}=8.790 MeV\\\\) (ii) \\\\(_{83}Bi^{209}\\\\) nucleus contains 83 protonsNumber of neutrons\\\\(=209-83=126 neutrons\\\\)Now,Mass of 83 protons\\\\(= 83\\\\) X \\\\(1.007825= 83.649475 amu\\\\)Mass of 126 neutrons\\\\( = 126 X 1.008665= 127.091790 amu\\\\)Therefore, total mass of nucleons\\\\(= 83.649475+127.091790=210.741260amu\\\\)Given, mass of nucleus\\\\(=208.980388 amu\\\\)Now, mass defect \\\\(\ riangle m\\\\) \\\\(=210.741260-208.980388=1.760872\\\\)Total binding energy\\\\(=1.760872\\\\) X \\\\(931.5 = 1640.26 MeV\\\\)Therefore, average binding energy per nucleon = \\\\(\\frac{1640.26}{209}=7.848 Mev\\\\)