# Nuclear spin of Fluorine 19

1. Feb 4, 2015

### Clear Mind

I was trying to calculate the $^{19}_{9}F_{10}$ nuclear spin using the nuclear shell method. From what i know the nuclear spin for a nucleous with odd $A$ is the total angular momentum $J$ of the stand-alone nucleon:
P: $(1d_{5 \backslash 2})^1$ (stand-alone nucleon)
N: $(1d_{5 \backslash 2})^2$
which in this case is a proton with $J=\frac{5}{2}$ so it leads to a nuclear spin of $I=\frac{5}{2}$. But this isn't what i found in lecture, from the pdg (http://ie.lbl.gov/toi/nuclide.asp?iZA=90019) $^{19}_{9}F_{10}$ result to have $I=\frac{1}{2}$.
I missed something?

2. Feb 4, 2015

### ChrisVer

You have 9 protons and 10 neutrons. Well no need to care about the neutrons, since they will pile up into pairs together (even number). From the protons you will have 1 extra remaining unpaired. You will fill until [itex]1d_{5/2} [itex]. So your nucleus spin will be determined by the unpaired proton, the proton has spin 1/2 so 1/2 is the spin of your nucleus...
Sometimes things are mixed with J, S in nuclear physics, calling both of them "spins". The total angular momentum is indeed 5/2, the spin is 1/2.

Last edited: Feb 4, 2015
3. Feb 4, 2015

### Clear Mind

My apologies but i'm not understanfding what you mean, for example take the $^{43}_{21}Sc_{22}$. It have 22 neutron (so we don't need to care about), and 21 proton which are filled until $1f_{7 \backslash 2}$. There's only one proton unpaired so nuclear spin $I$ should be $\frac{1}{2}$ (equal to the spin of the proton) but is $I=\frac{7}{2}$ (always from pdg http://ie.lbl.gov/toi/nuclide.asp?iZA=210043)

4. Feb 4, 2015