Is there a formula for determining the total spin of a nucleus? I know protons and neutrons have 1/2 spin and I know even number A nuclei have 0 spin. But I don’t know how the spin is determined for values larger than 1/2. For example: 55Fe spin 3/2 56Co spin 4 57Co spin 7/2 60Co spin 5 I didn’t find anything in the PF library. Other book reference in Inter-Library queue. Any online reference would be helpful.
Do you know what the Stern-Gerlach experiment is? That's one way, in principle, that nuclear spins are measured. (Even though it's rarely used in practice, it's conceptually how one might do it). Not all even A nuclei have 0 spin. There are many exceptions, starting with in fact the lightest even-A nucleus, the deuteron, which has spin 1.
The nucleons have orbital angular momentum, just as the electrons do. I can't remember off the top of my head but you need to find the orbital ang momentum, l, of the unpaired nucleons, now: [tex]L = l_{1} + l_{2}... [/tex] [tex]S = s_{1} + s_{2}...[/tex] [tex]J = L + S [/tex] where, due to Pauli exclusion priciple : [tex]Parity = (-1)^{J} [/tex] PEP defines the allowed values of J as Parity = 1 for Bosons; -1 for ferminons but check it, I'm not 100%. Hope that helps
The nuclear shell model (Bohr-Mottelson) can be used to calculate most nuclear spins. A good Nuclear Physics text should discuss it. It is probably best described in an older text, because newer texts have gone on to other things.
Yes the Stern-Gerlach experiment was a rather interesting experiment where in applying a longitudinal inhomogeneous magnetic field resulted in the spectra splitting at the nuclear level. Even though I did read several pieces to understand their technique I don’t recall calculations for determining the full range of nuclei spin; although, I will revisit the material. Thanks for the information that not all ‘even’ A type nuclei have 0 spin. This is good information that I had not see yet and hope to pursue it further. *** The formulas [tex] L = l_{1} + l_{2}... [/tex] [tex] S = s_{1} + s_{2}... [/tex] [tex] J = L + S [/tex] were the first things I had considered but I could not achieve the spin numbers for the higher values 55Fe spin 3/2 56Co spin 4 57Co spin 7/2 60Co spin 5 An interesting note is the continuation of the L and S sequence. I had first seen these formulas where the total L or S was equal to just 2 subterms. I took that to mean the single outer most unpaired proton and the single outer most unpaired neutron. If this is not consistent with your understanding I would appreciate your comments. *** I also had considered the “rules” for shell filling where for the 0f shell there are 14 available positions. The first 7 positions would be filled by single nucleons (proton or neutron) and then at the 8 position the nucleon would couple up with the “1st position” with opposing spin directions. Thus the remaining 6 uncoupled nucleons determined the spin for that family of nucleons. However, that sequence didn’t seem to work out either. *** Thanks for the information on “nuclear shell model (Bohr-Mottelson)”. I will start looking for references on this model to see if any calculations are presented to determine nuclear spin. *** In May of last year member “neu”, in the following link, indicated the limiting range of spin nuclear spin was https://www.physicsforums.com/showthread.php?t=56506 [tex]\vert j_{1} - j_{2}\vert[/tex] to [tex] j_{1} + j_{2}[/tex] In the same note ‘malawi_glenn’ followed with this comment: “Also you must search for every unpaired nucleon in shells, and add their spins according to the formula given by neu. And then multiply each unpaired nucleon parity (parity is given by quantum number l) Then also the shell model just gives you the possible quantum numbers (you do not know witch one is the ground state if you get three possible solutions for a given configuration) the experiment gives you the outcome of this. “ *** I understood that nucleons filled shells from the bottom up, thus the 0s shell is filled before the 0p. Likewise, the 0p is filled before the 0d and so on up the sequence of energy levels for each respective shell. If this be true then all shells below the last nucleon would be filled. So, at most, there would only be 2 nucleons to consider, the last unpaired proton and the last unpaired neutron. So I am unsure if malawi_glenn is referring to more than 2 nucleons; likewise, I am unclear how parity figures into the spin computation. Just like anyone,… I just need a little clarity. Appreciate the tips!
The comment about ‘even’ A nuclei having spin =0 was extracted from the third paragraph at the Hyperphysics site: http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/nspin.html The comment was taken out of context because in the original text they are referring to the iron nuclei. “For example, in the nuclear data table for iron below, all the even A nuclides have spin I=0 since there are even numbers of both neutrons and protons.” My comment should have read for all even protons and neutrons the total spin would equal to 0. I would be curious to learn if there are any exceptions to this ‘rule’.
i think you are confused with spin of a particle and nuclear spin spin(Proton and neutron ) s =1/2 (always) nuclear spin = 1/2,5/2,7/2 the reason being it is the total nuclear spin means I= j = L+s the given example will tell you what is the difference between nuclear spin and spin of nucleon oxygen O(z=8,n=9,A=17) what you except its spin ,i think =1/2( you are right) but it is not nuclear spin because nuclear spin is I=L+s in our case the last nucleon is in d (5/2) shell (easily seen from nuclear shell model spectrum) I =L+s =2+1/2 =5/2 which is its nuclear spin so in this way you can find out the nuclear spin of any state ok bye
I think (sushil rathi) your rule only aplicable for only O.It can not be applicable for Cl (z=17,n=18,A=35). its nuclear spin is 3/2