# Two Questions on the nuclear spin

To those who are familiar with nuclear physics, I have two questions of understanding:

1. Why does it happen that nuclei have a spin greater than 1 (e.g. I=8 at 90Nb)?

2. How can we infer the parity of a nucleus?

My ideas:

1): I have the nuclear shell model in mind. If the levels for protons and neutrons are filled one after the other. In the case of "even-even" nuclei, all levels are fully occupied, so the total spin adds up to 0. If we now have an odd number of protons and neutrons, then according to this model one level would only be half full. If the spins are parallel, this adds up to I=1. But how do I get I=8 (or other integer values)?

To 2): My idea about "even-even" nuclei: The total spin is 0, so the spin wave function is antisymmetric. Since according to the Pauli principle the total wave function must be antisymmetric, it follows that the local wave function is symmetric, i.e. the parity is positive.

Is this argumentation correct?

What about "odd-odd"-nuclei or "even-odd"-nuclei? How can we conclude parity here?

I would be very grateful, if someone could help me. Sorry if my text is not understandable in some places, English is not my mother tongue.

## 2 Answers

Note that a the total spin of a nucleus includes both the intrinsic spins and the orbital angular momenta of the nucleons. For a simpler example, compare oxygen-16 (which, like all stable even-even nuclei, has ground state spin-parity $$0^+$$) with oxygen-17, which has $$J^P=5/2^+$$. A simple shell-model-ish description would put the ninth neutron in oxygen-17 in an empty orbital around an oxygen-16 core. That ninth orbital must have even parity, since O-16 and O-17 have the same parity, which means that the orbital angular momentum of the ninth neutron is even. But it can't be in an $$s$$-wave, $$\ell=0$$ orbital, because then there is not enough angular momentum to reach $$J=5/2$$. Apparently the ninth neutron in oxygen-17 has orbital angular momentum $$\ell=2$$, and its spin and orbital angular momenta are aligned.

Of course, the wave function for an oxygen nucleus is much more complicated than a core and single orbital. But this should give you an idea about how the quantum numbers have to work out.

$$^{90}$$Nb has 41 protons and 49 neutrons. In the non-deformed single particle shell model according to Wikipedia nuclear shell article (I don't have my textbookds with me), the single particle state for both proton 41 and neutron 49 is $$g\frac{9}{2}$$. That means the total contribution (particle spin plus orbital angular momentum) has a quantum number, $$j=\frac{9}{2}$$. The $$g$$ is the orbital state and has quantum number $$\ell = 4$$. That means the parity of each single particle will be positive (+). Therefore, the total ground state parity is predicted to be +.

The total angular momentum quantum number will be between the sum and the difference of the individual particle $$j$$ so $$|\frac{9}{2}-\frac{9}{2}|\le \mathcal{I} \le\frac{9}{2}+\frac{9}{2} \\ 0 \le \mathcal{I} \le 9 .$$ Consequently, 8+ is totally consistent with the single particle shell model due to the orbital angular momentum contribution of each particle.

To summarize

1. You must look at the total angular momentum quantum number of the extra particle(s) and use the q.n. combination rule to find possible total q.n.

2. You must look at the $$\ell$$ of each extra (odd) particle to determine its parity from $$\left(-1\right)^{\ell}$$. Standard spectroscopic notation of $$s,~p, ~d,~f, ~g, ...$$ has $$\ell= 0,1,2,3,4,...$$ Final parity is the product of the odd particle parities.