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.