Spin-parity for excited O-15 states

[Warr]

Homework Statement

Use the shell model to find reasonable configurations for the first 5 excited states of O-15.

Homework Equations

The spin-parity and energy above the ground state for the first five excited states are:

ground state: (1/2)-

excited states:
5.18 MeV, (1/2)+
5.24 MeV, (5/2)+
6.18 MeV, (3/2)-
6.79 MeV, (3/2)+
6.86 MeV, (5/2)+

Parity of an unpaired nucleon is given by (-1)^l, with l_s = 0, l_p=1, etc.

The order of shells is: $$1s_{1/2}, 1p_{3/2}, 1p_{1/2}, 1d_{5/2}, 2s_{1/2}, 1d_{3/2}, 1f_{7/2}$$

The Attempt at a Solution

First off, O-15 has 8 protons and 7 neutrons, so the ground state config is given by

protons: $$(1s_\frac{1}{2})^2(1p_\frac{3}{2})^4(1p_\frac{1}{2})^2$$
neutrons: $$(1s_\frac{1}{2})^2(1p_\frac{3}{2})^4(1p_\frac{1}{2})^1$$

so for the neutrons we have $$(1p_\frac{1}{2})^{-1}$$. This has a spin of 1/2 and l_p = 1, so the parity is -1, hence the state (1/2)-, confirming the ground state.

For the first excited state, we need the parity to be (1/2)+. Here is where my first confusion is. Can the $$1p_{1/2}$$ in the protons be promoted? If so, how high can it be promoted. For example, let's say it is promoted to the next subshell, $$1d_{5/2}$$

protons: $$(1s_\frac{1}{2})^2(1p_\frac{3}{2})^4(1p_\frac{1}{2})^2$$
neutrons: $$(1s_\frac{1}{2})^2(1p_\frac{3}{2})^4(1d_\frac{5}{2})^1$$

Does this make the spin-parity (5/2)+? Or do we have to take something into account for the fact that there is now no neutrons in the $$1p_{1/2}$$ subshell (thus a gap between the protons in the $$1p_{3/2}$$ and the proton in the excited subshell $$1d_{5/2}$$), yet there are 2 in the $$1p_{1/2}$$ for the neutrons.

are other valid 'excitations' to have a neutron in the $$1s_{1/2}$$ or $$1p_{3/2}$$ be promoted to the $$1p_{1/2}$$ shell, thus creating a spin parity of (1/2)+ and (3/2)- respectively? If these are right, I think I have the first 3 states.

 

[malawi_glenn]

"Does this make the spin-parity (5/2)+? "

Yes, so you only look where you have particles and "holes", and then couple them. In this case you have only one particle outside closed shells and paired nucleons, so its quantum numbers decides the total spin and parity of the nucleus.

And yes, you my break a pair and move that nucleon to higher shell and couple the angular momenta and parities. But remember that it requires energy to break pairs (approx 1MeV).