Obtain the binding energy (in MeV) of a nitrogen nucleus $\left({ }_{7}^{14} \mathrm{~N}\right)$, given $m\left({ }_{7}^{14} \mathrm{~N}\right)=14.00307 \mathrm{u}$
Atomic mass of nitrogen $\left({ }_{7} \mathrm{~N}^{14}\right), m=14.00307 \mathrm{u}$
A nucleus of nitrogen ${ }_{7} \mathrm{~N}^{14}$ contains 7 protons and 7 neutrons.
Hence, the mass defect of this nucleus, Δm = 7mH + 7mn − m
Where,
Mass of a proton, mH = 1.007825 u
Mass of a neutron, mn= 1.008665 u
∴Δm = 7 × 1.007825 + 7 × 1.008665 − 14.00307
= 7.054775 + 7.06055 − 14.00307
= 0.11236 u
But 1 u = 931.5 MeV/c2
∴Δm = 0.11236 × 931.5 MeV/c2
Hence, the binding energy of the nucleus is given as:
Eb = Δmc2
Where,
c = Speed of light
$\therefore E_{b}=0.11236 \times 931.5\left(\frac{\mathrm{MeV}}{c^{2}}\right) \times c^{2}$
= 104.66334 MeV
Hence, the binding energy of a nitrogen nucleus is 104.66334 MeV.