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Momentum-Parity, ground state quantum no even-even odd-odd

  1. Feb 15, 2015 #1
    Notation: ##J^p## - ##J## the total angular momentum, ##p## the parity = ##+## or ## -. ##

    Ok so I'm given a diagram of energy levels, all have a ground state of ##0+##, except one which has a ground state of ##1+##.
    I'm askeed to indentify which set of energy levels belong to which nucleus, the nuclei are all even-even except Fe : ##A=18, Z=9##,
    So the answer is that ## 1+ ## ground state is only possible from Fe, and that all even-even nuclei must have a ground state of ##0+##.

    Questions:
    - I don't understand why ANY even-even nuclei has ##J^p=0^+##. I know that a filled shell has ##J=0 ##So this is fine for double-magic nucleus, but not for a magic, or general even-even nucleus.

    - I don't understand the addition of angular momenta given by my text book :
    Because both neutrons and protons have one nucleon in the ##1d^{5/2}## level, the addition of angular momenta is: ##J=5/2+5/2=5,4,3,2,1##*

    What I don't understand is which angular momenta you are supposed to be adding so here there's only one extra proton/neutron, but in general, for a odd-odd nuclei say you had 3 extra protons in a energy level, say it has ##J=J_{p} ##and 1 extra neutron in a energy level, say J##=J_{n} ##, then how should the addition work? Would you add ##J_{p}## and ##J_{n} ## in accord to * ? how would you account for the fact there are 3 extra protons but only one extra neutron?

    - In terms of addition of angular momentum, how do you see it sums to ##0 ##for an even-even nuclei. I see that as magic numbers are even, there will be an even number of surplus nucleons in a energy level for non double-magic numbers. Does this somehow differ the addition of ##J## compared to odd-odd so that 0 is possible?

    Thanks any help really appreciated.
     
  2. jcsd
  3. Feb 15, 2015 #2

    mfb

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    Staff: Mentor

    The strong force is not a nice 1/r^2 potential, while you still get some shell-like structure all different orbitals within those shells have different energies. As a result, they are always filled in pairs (spin up/down), so every pair of protons (and neutrons) will cancel each other in terms of spin contribution.
    At most, you have one proton and one neutron to consider for the ground-state.
     
  4. Feb 27, 2015 #3
    So it is assumed that ##l=0##? (The question does not say.)
     
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