(a) Find the energy needed to remove a neutron from the nucleus of the calcium isotope `._(20)^(42)Ca` . (b) Find the energy needed to remove a prtron from the nucleus. (c ) Why are these energies different?
Given: Mass of `._(20)^(40)Ca =40.962278 u`., mass of neutron .`=1.007825 u`,
Mass of `._(20)^(40)Ca =40.974599 u`., mass of protron .`=1.007825 u`.

Explanation

(a) .`_(20)^(41)Ca` nucleus is fromed after removing a neutron from `._(20)^(42)Ca`. The mass of `._(20)^(41)Ca` plus the mass of a free neutron equals `40.962278 u`. Mass of neutron `=40.962278 u+ 1.008665 u=41.970943 u` Difference between `._(20)^(42)Ca` plus the mass of a free neutron and the mass of `._(20)^(42)Ca` is `0.012321 u`. S0, the binding energy of the missing neutron is `(0.012321 u) (931.49 MeV/u)=11.5 MeV`.
(b) When a proton is removed from `._(20)^(42)Ca` ,the resulting nucleus is the postassium isotpoe `._(19)^(41)K` . On a similar pattern as above, the binding energy for the missing proton can be calculated, result is `10.27 MeV`.
(c ) Neutron and proton have different energies because only attractive nuclear forces act on the neutron whereas the proton was also acted upon by repulisve electric force that decreases its binding energy.