(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.

(a) Find the energy needed to remove a neutron from the nucleus of the

1.	(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.
Answer» Solution :(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.

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