Question

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

Answer 1

Atomic mass of $(_7N^{14})$ nitrogen, m = 14.00307 u
A nucleus of $(_7N^{14})$ nitrogen contains 7 protons and 7 neutrons.
Hence, the mass defect of this nucleus, $∆m = 7mH + 7mn − m$
Where,
Mass of a proton,$m_H$ = 1.007825 u
Mass of a neutron, $m_n$= 1.008665 u
$∆m$ = 7 × 1.007825 + 7 × 1.008665 − 14.00307
$∆m$= 7.054775 + 7.06055 − 14.00307
$∆m$ = 0.11236 u
But 1 u = 931.5 $\frac{MeV}{c^2}$
∆m = 0.11236 × 931.5 $\frac{MeV}{c^2}$
Hence, the binding energy of the nucleus is given as:
$E_b = ∆mc^2$
Where, c = Speed of light
$E_b$ = 0.11236 × 931.5
$E_b$ = 104.66334 MeV
Hence, the binding energy of a nitrogen nucleus is 104.66334 MeV.