Does the relative angular momentum of the nucleons contribute to the spin of the nuclei? Imagine we have a nucleus. We do talk about spins of nuclei. Is this spin the total angular momentum of the nucleus, that is, spins and relative angular momentum of all of the nucleons?
If so, why? If not so, why?
 A: To a first approximation we can describe the nucleus with a wavefunction that ignores the electrons and also ignores the fact the nucleons are made up from quarks. Actually solving the Schrodinger equation for a many body strongly interacting system like a nucleus is impossible, but we expect there will be a ground state and excited states just as there are for the electrons in atoms. These states will have an associated spin, just as the states of atoms have an associated spin.
So the spin of the nucleus is the spin of the eigenfunction that describes whatever energy state it's in. In other words it's a property of the nucleus as a whole.
Having said this, the overall nuclear spin will be related in a basically simple way to the spins of its individual nucleons, just as the overall spin of an atom is related to the spins of the individual electrons. For example we expect nuclei with an even number of nucleons to have zero spin in their lowest energy state, and nuclei with an odd number of nucleons will have spin half as their lowest energy state.
As with atoms, there will be exceptions to the lowest spin being the lowest energy state. For example the ground state of $^{55}$Fe has a spin of $3/2$ and the ground state of $^{59}$Co has a spin of $7/2$. To explain these higher spins you need to consider the wavefunction of the nucleus as a whole.
