(a) Find the energy needed to remove a neutron from the nucleus of the calcium isotopes `._(20)^(42)Ca`
(b) Find the energy needed to remove a proton from this nucleus
(c ) Why are these energies different ?
Atomic masses of `._(20)^(41)Ca` and `._(20)^(42)Ca` are `40.962278 u` and `41.958622 u` respectively.

Explanation

(a) Removing a neutron from `._(20)^(42)Ca` leaves `._(20)^(41)Ca`. The mass of `._(20)^(41)Ca` plus the mass of a free neutron is:
`40.962278 u +1.008665 u =41.970943 u`
The difference between this mass and the mass of `._(20)^(42)Ca` is `0.012321 u` , so the binding energy of the mission neutron is:
`(0.012321 u) (931.49 MeV//u)=11.48 MeV`
(b) Removing a proton from `._(20)^(42)Ca` leaves the potassium isotope `._(19)^(41)K`. A similar calculation gives a binding energy of `10.27 MeV` for the missing proton.
(c ) The neutron was acted upon only by attractive nuclear whereas the proton was also acted upon by repulsive electric forces that decrease its binding energy.