Spin-parity for excited O-15 states

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SUMMARY

The discussion focuses on determining the spin-parity configurations for the first five excited states of O-15 using the shell model. The ground state is established as (1/2)-, with excited states at 5.18 MeV (1/2+), 5.24 MeV (5/2+), 6.18 MeV (3/2-), 6.79 MeV (3/2+), and 6.86 MeV (5/2+). The parity of an unpaired nucleon is calculated using the formula (-1)^l, and the order of shells is specified as 1s_{1/2}, 1p_{3/2}, 1p_{1/2}, 1d_{5/2}, 2s_{1/2}, 1d_{3/2}, 1f_{7/2}. The discussion emphasizes the importance of considering particle-hole configurations and the energy required to break pairs when promoting nucleons to higher shells.

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  • Understanding of the shell model in nuclear physics
  • Familiarity with quantum numbers and their significance in nuclear states
  • Knowledge of nucleon configurations in O-15
  • Basic principles of energy levels and parity in quantum mechanics
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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.
 
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"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).
 

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