Nuclear Shell Model - Spin-parity of excited states

[Manman]

Homework Statement

The ground state of $$_{6}Cl^{13}$$ has spin-parity 1/2- and the next three excited states have values of 1/2+, 3/2- and 5/2+. Explain these values in terms of the shell model.

The Attempt at a Solution

The problem is that i don't know what is being asked of me...i initially tried to go through the excitations and work out which configurations fit each one, but lots do!

In more detail, I know that initially there is 1 extra p in $$1P_{1/2}$$

The initial configuration is:

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

A configuration that could give a spin-parity of 1/2+ for the first excitation:

$$(1s_\frac{1}{2})^2(1p_\frac{3}{2})^3(1p_\frac{1}{2 })^1(1d_\frac{5}{2})^1$$

or

$$(1s_\frac{1}{2})^2(1p_\frac{3}{2})^4(1p_\frac{1}{2 })^0(1d_\frac{5}{2})^0(2s_\frac{1}{2})^1$$

Now, if this is correct then we can follow on from this in 2 directions, and each one of these splits into many options for the 2nd excited state. And we do not getting any definite configuration for any of the excited states.

Am i going about this correctly? Maybe the question is asking something entirely different.

 

[Manman]

no luck so far, anyone?