(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 proton from this nucleus.
(c ) Why are these energies different? Mass of `._(20)^(40)Ca =40.962278u`, mass of proton ` =1.007825 u`.

Explanation

`_(20)^(41)Ca` nucleus is formed after removing a neutron from `_(20)^(42)Ca`. The mass of `_(20)^(41)Ca` plus the mass of a free neutron =40.962278 +1.008665 =41.970943 u Difference between `_(20)^(41)Ca` plus the mass of a free neutron and the mass of `._(20)^(42)Ca` is `0.012321 u`, so the binding energy of the missing neuton `=(0.012321)(931.49)=11.4 MeV`
(b) When a proton is removed from `._(20)^(42)Ca`, the resulting nulceus is the potassium isotope .`_(19)^(41)K`. On a similar pattern as above, the binding energy for the missing proton can be calculated, results is `10.27 MeV`
( c) Neutron and proton have differnet energies becasue only attractive nuclear forces act on the neutron whereas the proton is acted upon by repulsive electric forces that decreases its binding energy