I am studying the Shell Model.
If a nucleus has an odd number of nucleons, there is either an unpaired proton or an unpaired neutron. In this case, the unpaired proton or neutron determines the ground state spin of the nucleus.
However, for an odd-odd nucleus, there is both an unpaired proton and an unpaired neutron. In this case the ground state spin of the nucleus is then given by coupling the spins of these unpaired nucleons.
After learning about addition of angular momentum, none of these options make sense to me: Why do we not always couple the spin of all nucleons?
Best Answer
One can start building a nucleus by adding protons and neutrons.
Normally These will always fill the lowest available level. Thus the first two protons fill level zero,
the next six protons fill level one, and so on.
We see that for the first two numbers we get 2 (level 0 full) and 8 (levels 0 and 1 full), in accord with experimental observations.
We next include l.s. interaction.
Now the system is described by the j, mj instead of l, ml and ms as in the atomic case.
For every j there are 2j +1 different states from different values of mj .
Due to the spin-orbit interaction, the energies of states of the same level but with different j will no longer be identical.
THE SPIN STATES-
The shell model also predicts or explains with some success other properties of nuclei, in particular, spin and parity of the ground state.
Let us take 17 O 8 as an example: ( written as A O Z )
Its nucleus has eight protons filling the three first proton "shells", eight neutrons filling the three first neutron "shells", and one extra neutron.
All protons in a complete proton shell have total angular momentum zero since their spin angular momenta cancel each other.
The same is true for neutrons.
All protons in the same level (n-value) have the same parity (either +1 or −1), and since the parity of a pair of particles is the product of their parities, an even number of protons from the same level (n) will have +1 parity.
Thus the total angular momentum of the eight protons and the first eight neutrons is zero, and their total parity is +1.
This means that the spin (i.e. angular momentum) of the nucleus, as well as its parity, are fully determined by that of the ninth neutron.
This one is in the first (i.e. lowest energy) state of the 4th shell, which is a d-shell (l = 2), this gives the nucleus an overall parity of +1.
This 4th d-shell has a j = 5/2, thus the nucleus of 17 O 8 is expected to have positive parity and total angular momentum 5/2. which is corroborated by experiments.
The order of the first few terms are listed as follows:
1s, 1p3⁄2, 1p1⁄2, 1d5⁄2, 2s, 1d3⁄2...
one must add the assumption that due to the relation between strong nuclear force and angular momentum nucleons with the same n tend to form pairs of opposite spin angular momenta.
Therefore, a nucleus with an even number of protons and
an even number of neutrons has 0 spin and positive parity.
In the case of a nucleus with an odd number of protons and an odd number of neutrons, one must consider the total angular momentum and parity of both the last neutron and the last proton.
The nucleus parity will be a product of their individual parity,
while the nucleus spin will be one of the possible results of the sum of their angular
momenta (with other possible results being excited states of the nucleus).
For nucleon pairs, however, it is often energetically favorable to be at high angular momentum, even if its energy level for a single nucleon would be higher.