Calculate the mass defect and the binding energy per nucleon of the `""_(47)^(108)Ag` nucleus.
[atomic mass of Ag = 107.905949]

Explanation

`""_(47)^(108)Ag` nucleus contains 47 protons & 61 neutrons.
Mass of the proton `m_(p)=1.00783u`
Mass of the neutron `m_(n)=1.00867u`
Mass of the Ag nucleus `m_(N)=107.905949u`
Mass of 47 protons = `47xxm_(p)=47xx1.00783`
`M_(p)=47.36801u`
Mass of 61 neutrons = `61xxm_(n)=61xx1.00867`
`M_(n)=61.52887u`
Mass of the nucleous = `(M_(p)+M_(n))`
= 47.36801 + 61.52887 = 108.89688
Mass defect = mass of the nucleous - mass of the nucleus
`A_(m)` = 108.89688 - 107.905949
`A_(m)` = 0.990931 M
Total B.E. = `0.990931xx931MeV`
= 922.556761 MeV.
B.E. per nucleon = `("B.E.")/("atomic no")=(922.556761)/108`
B.E. = 8.5 MeV.