Nuclear Shell Model - Spin-parity of excited states

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SUMMARY

The discussion focuses on the spin-parity values of the excited states of _{6}Cl^{13}, specifically 1/2-, 1/2+, 3/2-, and 5/2+. The initial configuration of the ground state is identified as (1s_{1/2})^2(1p_{3/2})^4(1p_{1/2})^1. The configurations leading to the first excited state with spin-parity 1/2+ include (1s_{1/2})^2(1p_{3/2})^3(1p_{1/2})^1(1d_{5/2})^1 and (1s_{1/2})^2(1p_{3/2})^4(1p_{1/2})^0(1d_{5/2})^0(2s_{1/2})^1. The discussion reveals the complexity of determining configurations for subsequent excited states.

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  • Knowledge of nuclear configurations and excitations
  • Basic grasp of quantum numbers and their significance
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Manman
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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:

<br /> (1s_\frac{1}{2})^2(1p_\frac{3}{2})^4(1p_\frac{1}{2 })^1<br />

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

<br /> (1s_\frac{1}{2})^2(1p_\frac{3}{2})^3(1p_\frac{1}{2 })^1(1d_\frac{5}{2})^1<br />

or

<br /> (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<br />

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.
 
Last edited:
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no luck so far, anyone?
 

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