Question

the nuclear shell model, orbitals are filled in the order 1s1/2,1p3/2,1p1/2,1d5/2,2s1/2,1d3/2, etc.
(a) What is responsible for the splitting between the p3/2 and p1/2 orbitals?
(b) In the model, 16O (Z = 8) is a good closed-shell nucleus and has spin and parity Jπ = 0+ . What are the predicted Jπ values for 15O and 17O?
(c) For odd-odd nuclei a range of Jπ values is allowed. What are the allowed values for 18F (Z = 9)?
(d) For even-even nuclei (e.g. for 18O) Jπ is always 0+ . How is this observation explained?

Answer 1

(A) The success of the Shell Model hinges on the fact that it predicts magic number nuclei. Magic number nuclei are when the number of protons or neutrons (or both) are 2, 8, 20, 28, 50, 82, 126, the nucleus shows unusual stability. To explain this, a spin-orbit correction term is added to the potential in the shell model. This term takes into account the interaction of the nucleon's spin with its motion.

Just like in atomic physics the total angular momentum is given by the combination of spin and orbital angular momentum.

The total angular momentum J has the quantum number j, which can take the values

since, s can either be 1/2 or -1/2.

If we calculate the expectation value of the spin-orbit interaction, we get two different energies, corresponding to two different energy levels.
I'll do the math for one of them.

Substituting for

And this is why the energy splits when l > 0, and s = -1/2 or +1/2. Of course, in my attempt to focus only on the actual splitting and not exactly how the Hamiltonian or the expectation values are arrived at, I have not written everything.

(B) For the first one,
     
    
Hence, the J? value for the ground state is j = tot. ang. momentum and parity = (-1)^l
So,

Similarly for the other one,
    
    
So,

(C) For odd-odd nuclei we use Brennan-Berstein rules. They empirically take into account the nucleon-nucleon interaction between the outermost proton and nucleon. Using them, for 18F, we get,

Here, using rule number 2,

Here,

(D) Since, all of the nucleons are paired with each other, the total spin cancels out and since each pair has even parity the nuclear spin of such a nucleus is always J? = 0+