Question

What is the minimum photon energy required to remove the least bound neutron of ${}_{20}^{40}Ca$ and ${}_{18}^{40}Ar$. The necessary atomic masses (in u) are given below:
$M{(}^{40}Ca)=39.062591u$
$M{(}^{39}Ca)=38.970719u$
$M{(}^{40}Ca)=39.962383u$
$M{(}^{39}Ar)=38.964314u$
$m_n=1.008665u$

Answer 1

Total mass (neutron plus product nucleus) after the removal is expected to be more than the initial nucleus mass.
This mass difference when expressed in energy units (MeV) is the binding energy of the least bound neutron.
The mass after break-up $<{(}_{30}^{39}Ca)=38.970719u$
$m_n=1.08665u$
Total mass $=39.979384u$
The mass difference is, $\Delta m=\left[M{(}^{39}Ca\right)+m_n]-m{(}^{40}Ca)$
$=(39.979384-39.962591)=0.016793u$
$=\left(0.016793\right)\times \left(931.5\right)MeV=15.64MeV$

So, the binding energy of least bound neutron is $15.64MeV$.
Similarly, for ${}_{20}^{39}Ar$,
$\Delta m=\left[M{(}_{20}^{39}Ar\right)+m_n]-M{(}_{18}^{40}Ar)]$
$=(38.964314+1.008665)-39.962383u$
$=0.010596u=\left(0.010596\right)\times \left(931.5\right)MeV$
$=9.870MeV$

The inert elements are relatively unreactive because their outer shells of electrons are full. Large energies are involved in gaining or losing electrons which is therefore unlikely. An analogous behaviour takes place in the nucleus. Experimental evidence does indicate the existence of 'closed nuclear shell' when the number of protons or neutrons is $2,8,20,28,50,82$ or $126$ although the concepts of individual nucleon 'orbits' and filled shells inside the nucleus is hard to believe.
The above mentioned number of protons or neutron are called magic numbers. The elements with magic number of protons have unusually high number of stable isotope. For example, aluminium with 13 protons has just one stable isotope ${}^{27}Al$, but tin with $Z=50$ (a magic number) has 10 stable isotopes ranging from $N=62$ to $N=74$, whereas neighboring indium with $Z=49$ and antimony with $Z=51$ have only two.
Thus, the magic numbers are associated with extra binding energies, implying higher stability.
Thus, in the given example it takes about 6 MeV more energy to remove a paired neutron from a pair that completes a closed nuclear shell (a magic number) than from a pair that does not complete a shell.