# Momentum-Parity, ground state quantum no even-even odd-odd

1. Feb 15, 2015

### binbagsss

Notation: $J^p$ - $J$ the total angular momentum, $p$ the parity = $+$ or $-.$

Ok so I'm given a diagram of energy levels, all have a ground state of $0+$, except one which has a ground state of $1+$.
I'm askeed to indentify which set of energy levels belong to which nucleus, the nuclei are all even-even except Fe : $A=18, Z=9$,
So the answer is that $1+$ ground state is only possible from Fe, and that all even-even nuclei must have a ground state of $0+$.

Questions:
- I don't understand why ANY even-even nuclei has $J^p=0^+$. I know that a filled shell has $J=0$So this is fine for double-magic nucleus, but not for a magic, or general even-even nucleus.

- I don't understand the addition of angular momenta given by my text book :
Because both neutrons and protons have one nucleon in the $1d^{5/2}$ level, the addition of angular momenta is: $J=5/2+5/2=5,4,3,2,1$*

What I don't understand is which angular momenta you are supposed to be adding so here there's only one extra proton/neutron, but in general, for a odd-odd nuclei say you had 3 extra protons in a energy level, say it has $J=J_{p}$and 1 extra neutron in a energy level, say J$=J_{n}$, then how should the addition work? Would you add $J_{p}$ and $J_{n}$ in accord to * ? how would you account for the fact there are 3 extra protons but only one extra neutron?

- In terms of addition of angular momentum, how do you see it sums to $0$for an even-even nuclei. I see that as magic numbers are even, there will be an even number of surplus nucleons in a energy level for non double-magic numbers. Does this somehow differ the addition of $J$ compared to odd-odd so that 0 is possible?

Thanks any help really appreciated.

2. Feb 15, 2015

### Staff: Mentor

The strong force is not a nice 1/r^2 potential, while you still get some shell-like structure all different orbitals within those shells have different energies. As a result, they are always filled in pairs (spin up/down), so every pair of protons (and neutrons) will cancel each other in terms of spin contribution.
At most, you have one proton and one neutron to consider for the ground-state.

3. Feb 27, 2015

### binbagsss

So it is assumed that $l=0$? (The question does not say.)

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