Obtain the binding energy (in MeV) of a nitrogen nucleus
Question:

Obtain the binding energy (in MeV) of a nitrogen nucleus $\left({ }_{7}^{14} \mathrm{~N}\right)$, given $m\left({ }_{7}^{14} \mathrm{~N}\right)=14.00307 \mathrm{u}$

Solution:

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, Δ= 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

Δ= 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.