- #1

Warr

- 120

- 0

## 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: [tex]1s_{1/2}, 1p_{3/2}, 1p_{1/2}, 1d_{5/2}, 2s_{1/2}, 1d_{3/2}, 1f_{7/2}[/tex]

## The Attempt at a Solution

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

protons: [tex](1s_\frac{1}{2})^2(1p_\frac{3}{2})^4(1p_\frac{1}{2})^2[/tex]

neutrons: [tex](1s_\frac{1}{2})^2(1p_\frac{3}{2})^4(1p_\frac{1}{2})^1[/tex]

so for the neutrons we have [tex](1p_\frac{1}{2})^{-1}[/tex]. 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 [tex]1p_{1/2}[/tex] 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, [tex]1d_{5/2}[/tex]

protons: [tex](1s_\frac{1}{2})^2(1p_\frac{3}{2})^4(1p_\frac{1}{2})^2[/tex]

neutrons: [tex](1s_\frac{1}{2})^2(1p_\frac{3}{2})^4(1d_\frac{5}{2})^1[/tex]

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 [tex]1p_{1/2}[/tex] subshell (thus a gap between the protons in the [tex]1p_{3/2}[/tex] and the proton in the excited subshell [tex]1d_{5/2}[/tex]), yet there are 2 in the [tex]1p_{1/2}[/tex] for the neutrons.

are other valid 'excitations' to have a neutron in the [tex]1s_{1/2}[/tex] or [tex]1p_{3/2}[/tex] be promoted to the [tex]1p_{1/2}[/tex] 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.