# Spin and Parity of $^{17}_8 O$ excited states

$^{17}_8 O$ quoted here has a spin of 5/2 and parity of +1 for the ground state, I agree with this, the unpaired neutron is in the $1d_{1/2}$ state so l = 2, spin = 5/2.

Now I want to figure out the spin an parity of the first two excited states, but how do I know which proton and which neutron move up.

I am thinking that the first excited state is where the unpaired neutron moves to the $2s_{1/2}$ state, so that l = 2, and spin = 1/2, thus the parity remains 1.

But how do I figure out the configuration for the second excited state.

• On the same LBL site there is a full set of level drawings. Now all I have to do is recall how to read them. Here's the one for A=17 (PDF link). Feb 21, 2013 at 0:32
• For any future visitors, the $2s_{1/2}$ is the next excited state only by some authors (e.g. dommelen.net/quantum2/style_a/ntsm.html)... Other authors might have the $1d_{5/2}$ as above the $1d_{3/2}$ which would predict a different first excited state (namely $J^P={\frac32}^−$). This shows the deficiencies in the nuclear shell model. I believe the different orderings are due to to the strength of the spin-orbit coupling used. May 12, 2023 at 8:56

A better reference for determining these nuclear states from experiment is NuDat at Brookhaven. Figuring out the interaction energy of a neutron in any particular energy level with a $^{16}O$ core is extremely difficult. There is some model intuition that can help you, as you used for the first excited state. In this case, the second excited state $1/2^-$ must be more complicated. I would guess that it results from a neutron from the closed core joining the extra neutron in the d-shell, leaving a hole in the p-shell that gives the quantum numbers here. As you can see, there's a huge extra energy needed to make this configuration.