Q. 13.3 Obtain the binding energy( in MeV ) of a nitrogen nucleus , given m
mn = 1.00866 u
mp= 1.00727 u
Atomic mass of Nitrogen m= 14.00307 u
Mass defect m=7mn+7mp - m
m=71.00866+71.00727 - 14.00307
m=0.10844
Now 1u is equivalent to 931.5 MeV
Eb=0.10844931.5
Eb=101.01186 MeV
Therefore binding energy of a Nitrogen nucleus is 101.01186 MeV.