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I want to define the spin of the following nuclei: $^{15}_{\phantom{1}7} \textrm{N}$, $^{27}_{12} \textrm{Mg}$ and $^{47}_{20} \textrm{Ca}$. I have a scheme for the niveaus of the energies, see below:

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Where left is for protons and right column for neutrons. I am not really sure how to work with such a pattern.

For $^{15}_{\phantom{1}7} \textrm{N}$ we have 7 protons and 8 neutrons. The neutrons completed two shells - so there is no contribute to the nucleus spin. But 1 proton is "left" in the niveau $1p_{1/2}$.

So the total spin is $1/2$?

$^{47}_{20} \textrm{Ca}$: 20 Protons (shells full; no contribution). But 7 Neutrons on the $1f_{7/2}$-Niveau. What's the total spin here?

And another question: What do to the numbers on the left represent? Principal quantum number $n$?

Thanks in advance.

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  • $\begingroup$ @Experts: would this be eligible as 'homework' question? $\endgroup$ – Effervescenza Naturale Nov 7 '16 at 10:41
  • $\begingroup$ Hi, is there still something that is not clear to you about your question, that I could help you with expanding/improving my answer? If I was able to fully solve your doubts, you could consider accepting my answer ;D $\endgroup$ – Effervescenza Naturale Nov 11 '16 at 12:08
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The scheme is: For a nucleus at the ground state, when the shells are full for one kind of nucleons, and

  • The other has the last shell with only one nucleon, the nucleon total angular momentum (in $1p_{1/2}$, this is 1/2) is the nucleus spin.
    The reasoning is that all the others give angular momentum $0$, so $| \vec 0 + \vec J_p | = | \vec J_p |$.

  • The other has the last shell with only one nucleon missing from being full, the missing nucleon total angular momentum (in $1 f_{7/2}$, this is 7/2) is the nucleus spin.
    The reasoning is that the full shell would give $0$. You subtract the angular momentum of the nucleon you are removing. $| \vec 0 - \vec J_n | = | \vec J_n |$.

You can make this more general considering that pairing favours $0$ angular momentum couples of nucleons. Given this, any even-even nucleus at ground state has spin $0$, any even-odd has spin equal to the lone nucleon angular momentum.

For "the number on the left", if I undestand which one you refer, yes, it is the principal quantum number (notice that the exact way you number the levels can vary, but they are just labels).

Edit: I see they explain this here on Wikipedia

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