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:
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.
Best Answer
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