Question

Obtain the binding energy (in MeV) of a nitrogen nucleus `(""_7^14"N")`, given `"m"(""_7^14"N")` = 14.00307 u.

Answer 1

Atomic mass of nitrogen `(""_7^14"N")`, m = 14.00307 u

A nucleus of nitrogen `""_7^14"N"` 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 xx 931.5(("MeV")/"c"^2) xx "c"^2`

= 104.66334 MeV

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