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## Homework Statement

The low-lying levels of C-13 are ground state, [tex]\frac{1^-}{2}[/tex]; 3,09 MeV, [tex]\frac{1^+}{2}[/tex]; 3,68 MeV, [tex]\frac{3^-}{2}[/tex]; 3,85 MeV, [tex]\frac{5^+}{2}[/tex]. Interpret these four states according to the shell model.

## Homework Equations

Negative parity --> [tex]\ell[/tex] = odd; the valence nucleon must occupy a level with the spectroscopic symbol p, f, h etc.

Positive parity --> [tex]\ell[/tex] = even; the valence nucleon must occupy a level with the spectr. symbol s, d, g etc.

## The Attempt at a Solution

In the ground state, the 7th neutron must be in the [tex]1p_{\frac{1}{2}}[/tex] level. All levels below are filled.

In the first excited state, that is, the [tex]\frac{1^+}{2}[/tex] state, I think the 7th neutron is excited from the [tex]1p_{\frac{1}{2}}[/tex] level to the [tex]2s_{\frac{1}{2}}[/tex] level. The remaining neutrons occupy the same levels as in the ground state.

In the [tex]\frac{3^-}{2}[/tex] state, one of the two pairs in the [tex]1p_{\frac{3}{2}}[/tex] level is broken, and a neutron is excited to the [tex]1p_{\frac{1}{2}}[/tex] level, where it forms a pair with the former valence neutron. The remaining neutrons occupy the same levels as in the ground state.

In the [tex]\frac{5^+}{2}[/tex] state, the single neutron in the [tex]1p_{\frac{1}{2}}[/tex] level is excited to the [tex]1d_{\frac{5}{2}}[/tex] level. The remaining neutrons occupy the same levels as in the ground state.

Is this correct? Do I seem to understand the shell-model somewhat?

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