# Why does Fluorine-19 have a nuclear spin of 1/2?

According to the nuclear shell model, $$^{19}F$$ has one unpaired proton in the $$6$$-fold degenerate $$1d_{5/2}$$ state, which means the orbital angular momentum is $$l = 2$$ and the total angular momentum is $$j = l+ \frac12 = 5/2$$. If we follow the rule highlighted by Lubos Motl that nuclear states with higher $$m_j$$ are filled first, then the one unpaired proton goes in the $$m_j = 5/2$$ state predicting a nuclear spin of $$5/2$$. However, we know that $$^{19}F$$ has a nuclear spin of $$1/2$$.

Why is that? Is the selection rule claimed by Lubos Motl simply wrong? If so, what is the appropriate way to fill the states and calculate nuclear spin? For example, why does $$^{23}Na$$ have a nuclear spin of $$3/2$$?

• I always thought that selection "rules" were better called selection "suggestions" - in multi-nucleon, like multi-electron, systems there are many factors at play. The 5/2 state is only 200keV above the 1/2 ground state for F19. Commented Jan 14, 2019 at 16:21
• Is the selection rule claimed by Lubos Motl simply wrong? Yes, it's simply wrong. I left a comment on his answer.
– user4552
Commented Jan 14, 2019 at 20:31

Here is a paper on the nuclear structure of this nucleus. It's not spherical, it's deformed. Basically the very first thing you should do when trying to guess the ground-state configuration of a nucleus is to try to find out whether or not it's spherical. If it's deformed, then you're not going to get the right answers by looking at a diagram of energy levels at spherical deformation. For a diagram showing the relevant energy levels in deformed nuclei of this region, see this paper, p. 30. The $$d5/2$$ shell splits up with deformation into states with definite values of $$j_z$$ along the axis of symmetry. For a prolate nucleus, the lowest $$j_z$$ drops the fastest. When you couple this odd particle to the rotational degrees of freedom, you get a rotational band in which the lowest spin equals the single particle's value os $$j_z$$.