Nuclear shell model spin and parity?

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SUMMARY

The shell model spin and parity of the isotope _{38}^{89}Sr is determined primarily by the presence of unpaired nucleons. In this case, all six protons occupy the completely filled level 1f_{5/2}, while one neutron occupies the level 1g_{7/2}. Consequently, the spin parity of _{38}^{89}Sr is \frac{7}{2}^{+}, as the unpaired neutron dictates the overall spin and parity of the nucleus.

PREREQUISITES
  • Understanding of nuclear shell model concepts
  • Familiarity with nucleon configurations and their implications
  • Knowledge of spin and parity notation in nuclear physics
  • Basic principles of pairing in nucleons
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  • Research the significance of unpaired nucleons in determining nuclear properties
  • Learn about spin and parity calculations for various isotopes
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Homework Statement


What is the shell model spin and parity of _{38}^{89}Sr?

2. The attempt at a solution

If i fill the levels as we usually do,

Protons will end up in the level 1f_{5/2} with 6protons,ie the level is completely filled

If the neutron number is considered,1g_{7/2} with 1neutron is the final level.

Then what is the spin parity?
\frac{5}{2}^{+} or \frac{7}{2}^{+}?
 
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In a book i saw tha spin parity of oxygen is found by considering proton number only.I want to confirm if it is so even when the proton shell is completely filled(ie, there's no unpaired proton).or do i need to consider neutron number?
 
You have to consider any unpaired nucleon (protons or neutrons).
In this case the protons are all paired up and you have 1 unpaired neutron.
The unpaired nucleons determine these properties of the nucleus.
 

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