Question

Obtain the binding energy (in MeV) of a nitrogen nucleus. Refer image for question and equation.

Answer 1

we know,
mass of proton , $m_p=1.00783u$
mass of neutron, $m_n=1.00867u$

also we know, in $^{14}_7N$, there are 7 protons and 7 neutrons.

so, mass defect , $\Delta{m}=(7m_p+7m_n)-m$
where m is given mass .

= 7 × 1.00783u + 7 × 1.00867u - 14.00307u

= 0.11243u

hence, binding energy of nitrogen nucleus is given by E = ∆m × 931MeV
= 0.11243 × 931 MeV
= 104.67233 MeV