Ground State Assignment for nuclear shell model Why is the ground state assignment $0^{+}$ whenever the number of neutrons and protons are both even? 
 A: It is true that all even-even nuclei (hundreds of such isotopes have been measured) have spin-0 in the ground state. This is due to what is often called the pairing effect. Protons and neutrons are spin-½ particles, and they have a tendency to respectively pair up in proton-proton and neutron-neutron pairs so that their spin (and orbital) angular momentum adds to zero.
This is a pillar of the nuclear shell model which says that we can predict many properties of a nucleus by examining the "unpaired" nuclei. The spin and parity of odd-even and even-odd nuclei are generally determined by the "valence nucleon" (cf. the valence electron in the atomic shell model) that is left when all pairing has occurred. Odd-odd nuclei are not commonly found in nature, which we can describe to its tendency to convert the odd proton into a neutron (or vice versa) via $\beta$ decay to gain binding energy through the pairing force.
As to "why?", that is a larger question. This effect is an empirical observation in nuclear physics, and it is seen to be helpful in predicting how nuclei behave over a large range of isotopes. The "pairing term" is a part of the semi-empirical mass formula, which does a fairly good job of predicting nuclear properties, at least among heavier and stable nuclei.
The links to the nuclear shell model and the semi-empirical mass formula are probably good further reads.
