Looking at the spin of 7N7 and 9F10

[bluestar]

I have worked out the total spin for the elements up to Neon however I hit a snag with Nitrogen and Fluorine.

The shell sequence of nitrogen 7N7 is 1s2, 2s2, 2p3 for both the proton and neutron and the total angular spin is 1

Likewise, Fluorine 9F10 is 1s2, 2s2, 2p5 for the proton and 1s2, 2s2, 2p6 for the neutron and the total angular spin is ½

In the nitrogen isotope it would appear that the 3 protons and neutrons are not paired and occupy separate states in the 2p shell. Thus the total orbital angular momentum would be 3 and the total spin would be 3/2 for the protons and additional for the neutrons suggesting the total angular momentum of 9 which is wrong.

So do 2 of the protons in different states in the 2p shell pair up their orbital and spin angular momentum? Likewise do 2 of the neutrons in the 2p shell pair up their orbital and spin angular momentum? If so, this would leave yield an orbital momentum of 1 for the proton and 1 for the neutron; likewise, the spin of the two separate nucleons sum up to 1. This would then yield a total angular momentum of 2 + 1 = 3 with is also wrong.

The only way I can see to achieve a total angular momentum of 1 is if the spin of the proton and neutron were both negative, which I considered wrong because unpaired particle spin were always positive. So can an unpaired particle have a negative spin?

With respect to 9F10 there are 2 sets of paired protons plus 1 separate in the 2p shell and all neutrons are paired. This would produce an orbital angular momentum of 1 and a spin of ½ for a total of 3/2; however JANICE indicates this isotope has only ½ spin suggesting the proton spin was -1/2.

Can anyone shed any light on the total spin of these two isotopes?

 

[PeterDonis]

Are you looking at the nuclear shells or the electron shells? They're not the same. What you're giving look like electron shell configurations.

Check out the Wikipedia page: http://en.wikipedia.org/wiki/Nuclear_shell_model" . According to the nuclear shell model as described there (see the graph on the right in the "predicted magic numbers" section), 7N7 would have 1s(1/2) - 2, 1p(3/2) - 4, 1p(1/2) - 1 for protons, and the same for neutrons. The total spin would be 1 (1/2 each for the unpaired proton and neutron).

9F10 would have 1s(1/2) - 2, 1p(3/2) - 4, 1p(1/2) - 2, 1d(5/2) - 1 for protons, and the same except 1d(5/2) - 2 for neutrons. The total spin would be 1/2 for the unpaired proton.

(Pedantic note: these are, of course, the ground states of these nuclei. This actually does make a difference for the 7N7 case, because there is also a spin-0 state possible for that nucleus, with the same shell configuration, where the unpaired proton and neutron have opposite spins; however, this state has higher energy than the spin-1 state where the unpaired proton and neutron spins are aligned. I've been unable to find a reference online with a good simple explanation of why that is, but I suspect it's because the magnetic moments of the proton and neutron are of opposite signs--the proton is about +2.8 and the neutron is about -2, in the standard units for magnetic moments of elementary particles--which means that when their spins are aligned, their magnetic moments attract each other--opposite magnetic poles attract just as opposite charges do--so that the nucleus as a whole is a little bit more tightly bound--lower total energy--whereas when the spins are opposite, the magnetic moments repel each other and the nucleus as a whole is a little bit less tightly bound--higher total energy.)