Nuclear Spin

  1. jcsd
  2. mfb

    Staff: Mentor

    The value is the sum from the individual nucleons: their angular momentum and their spin.
    Both protons and neutrons have energy levels similar to the electrons, but their energy structure is more complicated. Pairs always cancel out, but you can have a single proton or neutron (or both) leading to the total spin. If you know the structure of the energy levels, you can predict the state they are in, and therefore the total spin.
     
  3. Vanadium 50

    Vanadium 50 18,137
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    I agree with what you say except for "pairs always cancel out". To pick a nucleus at random, Vanadium-50 is spin 6+. Lots of aligned pairs.
     
  4. mfb

    Staff: Mentor

    Vanadium 50 is odd/odd, you have two unpaired nucleons there. That is the "or both" case I mentioned.
     
  5. Vanadium 50

    Vanadium 50 18,137
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    That gets you to spin-1. It's 6+.

    Pierce, to get back to your question, the answer is "yes", but it's not simple. The short answer is that you have shells like you do in atoms, but because of the fact that the nuclear interaction is more complicated than electromagnetism, the order in which the shells fill moves around quite a bit, and you can end up with bizarre situations like 12 of 50 nuclei all spinning in the same direction.
     
  6. Is this kind of stuff studied in QCD?
     
  7. mfb

    Staff: Mentor

    Doesn't that come from orbital angular momentum? Okay, that's interesting. That would probably require weird energy levels.

    @pierce15: QCD is more relevant "inside" hadrons. Nuclear physics uses effective models as a full QCD analysis gets too complex for large nuclei.
     
  8. Vanadium 50

    Vanadium 50 18,137
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    As mfb said, it's too hard to study a system like V-50 in QCD. It has 150 valence quarks.
     
  9. So do are all the protons and neutrons (or individual quarks) in a nucleus described by orbitals?
     
  10. Vanadium 50

    Vanadium 50 18,137
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    Yes, the nuclei are in orbitals, but predicting these orbitals is harder than it is in chemistry.
     
  11. Does the Schrodinger equation then describe the system? If so, what would the potential be for ionized deuterium?
     
  12. ChrisVer

    ChrisVer 2,327
    Gold Member

    It's not "hard" to study QCD for nuclear physics... I'd better say it's meaningless... QCD stops working perturbatively at the nucleus range [energies]. So your results are not predicting at all...

    Yes the Schrod. equation can describe the system [because the nucleons are not relativistic]. You can look for the potential... it depends on what interactions you allow... eg some standard potential well for the nuclear force, maybe spin-orbit coupling, spin-spin coupling etc... The nuclear potentials in general can be very difficult to be determined, they can contain many terms, some coming from "theoretical" background, others coming straightforward from experiments, and then other experiments are needed to determine your parameters.
     
  13. Vanadium 50

    Vanadium 50 18,137
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    That would be true if the only way to do a calculation was perturbation theory. But it's not. There's also the lattice.
     
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