Calculate the binding energy of `""_(3)^(6)Li` assuming the mass of `""_(3)^(6)Li` atoms as 6.01512 amu.

Explanation

Mass defect is
`Delta = [Zm_(p) + (A-Z)m_(n) - M]`
`m_(p)` = mass of the proton = 1.007277 amu.
`m_(n)` = mass of the neutron = 1.00 8665 amu.
Mass of the bound `""_(3)^(6)Li` = 6.01512 amu.
Substituting this Eq.(1) .
[We know as per the symbol `""_(3)^(6)`Li, Lithium has 3 protons and three neutrons , since Z = 3,A = 6 and number of neutrons =A - Z = 6-3=3]
`therefore Delta m = [ 3 xx 1.007277 + 3(1.008665)]`
`Delta m = [ 3.02183 - 6.01512 = 0.032705` amu.
But the energy equivalent of 1 amu = 931 Me V .
`therefore` The binding energy which is equal to the energy equivalent of mass defect in amu is `E_(b) = Delta m xx 931` MeV
`= 0.032705 `xx` 931 = 30.44 MeV.