Question

Obtain the binding energy (in MeV) of a nitrogen nucleus, given =14.00307 u

Answer 1

Atomic mass of nitrogen, m = 14.00307 u

A nucleus of nitrogen contains 7 protons and 7 neutrons.

Hence, the mass defect of this nucleus, Δm = 7m_H + 7m_n − m

Where,

Mass of a proton, m_H = 1.007825 u

Mass of a neutron, m_n= 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:

E_b = Δmc²

Where,

c = Speed of light

∴E_b = 0.11236 × 931.5

= 104.66334 MeV

Hence, the binding energy of a nitrogen nucleus is 104.66334 MeV.