Atomic Spin: Origin in Nucleus & Impact of Electrons

[RandallB]

Is the spin of an atom come from (or based on) the Nucleus (i.e. net of Neutrons and Protons)?
With the spin of the Electrons not really noticed or counted though a sense of being ‘lost’ or masked by being in an orbital?

 

[dextercioby]

It all depends on the "pairings".You know the electrons build shells and subshells according to Pauli's Principle & Hund's rules.So the total spin of the electronic configuration depens of the # of unpaired electrons in the last shell.

However,we know that both the nuclear particles (the neutron & the proton) have spin.There's a nice model,called "shell model" which builds the nucleus (and the energetic configuration of the component particles) just in the same way as the electronic configuration is built.Nucleons are spin 1/2 fermions,too...

I'm sure u can read more in a nuclear physics treatise...

Daniel.

 

[Gokul43201]

Is the spin of an atom come from (or based on) the Nucleus (i.e. net of Neutrons and Protons)?
With the spin of the Electrons not really noticed or counted though a sense of being ‘lost’ or masked by being in an orbital?

Actually, the electron spin is over 3 orders of magnitude greater than a proton spin. Thus, magnetism can be explained fairly accurately by dealing only with electron spins.

When you suggest that the electron spin is "lost", what kind of measurement are you talking about ? The only thing that comes to mind is NMR, where the nuclear spin precession is of primary interest (and is roughly in the RF range, while the Larmor frequency for electrons is much higher - 3 orders, so in the microwave range).

 

[RandallB]

Actually, the electron spin is over 3 orders of magnitude greater than a proton spin.

How does "order of magnitude" get involved? I though all leptons had 1/2 spin not '3 orders'? bigger than a proton

When you suggest that the electron spin is "lost", what kind of measurement are you talking about ?

I mean lost to the measurement of the spin of the atom.
SO - Can you say that spin one direction while the electron was on one side of the nucleus for 50% of the time would be a spin in the opposite direction while it was on the other side for 50% of its “orbit”. So the net spin from the electron would be “lost” or canceled from viewing the entire atom.
Is this the correct way to look at it?
Thus the spin of the atom comes from the nucleus alone (P’s N’s etc.) with the electrons not really contributing to this one measure (spin) of the Atom.

Example: Boron B10 spin = 3 when Boron B11 spin =1.5 (or 3/2)
Difference due only to the extra neutron and nucleus configuration?
I assume configuration is involved as the nutron by it self could only contribute a change of 1/2 not 3/2.
Or could electron spin (and/or its orbital ‘shape’) be involved in this parameter as well.

Also:

Thus, magnetism can be explained fairly accurately by dealing only with electron spins.

I thought magnetism was dependent on the electron orbital shapes as defined by QM. Not the “quasi rotation” of the electron (Spin). Are both involved in magnetism?

 

[dextercioby]

How does "order of magnitude" get involved? I though all leptons had 1/2 spin not '3 orders'? bigger than a proton

He didn't mean spin operator eigenvalue,but spin magnetic moment...

Thus the spin of the atom comes from the nucleus alone (P’s N’s etc.) with the electrons not really contributing to this one measure (spin) of the Atom.

You have been told already that electrons,too carry spin & spin magnetic moment and this "contribution" that u mention,for a computation,required knowledge of the electronic configunation of electrons in (sub)shells.

Also: I thought magnetism was dependent on the electron orbital shapes as defined by QM. Not the “quasi rotation” of the electron (Spin). Are both involved in magnetism?

Magnetism depends on the magnetic moment.So how about searching & documenting on this issue...?

Daniel.

 

[Gokul43201]

SO - Can you say that spin one direction while the electron was on one side of the nucleus for 50% of the time would be a spin in the opposite direction while it was on the other side for 50% of its “orbit”. So the net spin from the electron would be “lost” or canceled from viewing the entire atom.
Is this the correct way to look at it?

No it is completely incorrect.

First of all, you are confusing the orbital angular momentum with that of spin. Secondly, you have a wrong picture of orbital angular momentum (even in terms of the classical analogue).

And please do not continue to hold the misconception that when the spin (first understand what you mean by 'spin') of an atom is measured (do you know how this is measured ?), only the nuclear contribution dominates. It is most definitely the other way round.

 

[RandallB]

No it is completely incorrect.

First of all, you are confusing the orbital angular momentum with that of spin. Secondly, you have a wrong picture of orbital angular momentum (even in terms of the classical analogue).

And please do not continue to hold the misconception that when the spin (first understand what you mean by 'spin') of an atom is measured (do you know how this is measured ?), only the nuclear contribution dominates. It is most definitely the other way round.

Misconception?? How can a question include a misconception?
The only conception I have is I don’t know. And you’ve left me thinking you may be under the misconception that you have provided an answer.

I’ll try to clarify the question for you.
In an advanced periodic table you can find the Nuclear spin by Isotope of an atom.
(Yes I’m trusting that someone has ‘measured’ that. No I don’t know how they did. If you want to share the how, ok that’d be cool)
Iron (Fe-56) has spin of “0” Zero (also F-54 & F-58)
& Boron B10 spin = 3 vs. Isotope Boron B11 spin = 3/2
So I’m interpreting this to mean the even number of neutron changes to contribute a (+½ -½) spin for the net 0 change where F-57 has a ½ spin change for one neutron change.
Following this reasoning I’d have to guess that adding a neutron could contribute a –1/2 AND convince an existing neutron to flip from +1/2 to –1/2 in order to get to the net 3/2 change.

But this interpretation seemed incomplete without accounting for the spins of all the individual electrons.

On the issue of considering the electron spins, if I understand you correctly I can ignore the individual spin of the electrons. BUT I should also ignore the spin from the nucleus (Protons & Neutrons) as “it’s the other way around” in that the “orbital angular momentum” of the electrons dominates the spin of an atom!
Since it dominates, does that mean this electron net momentum is somehow changing with the changing # of neutrons in an isotope change ??

Frankly this makes less sense than when I was just trying to figure out if the spin of the electrons was included in the spin of an atom at all.
I’d have thought that the “orbital” stuff was only involved in magnetic field generation.

 

[Gokul43201]

Misconception?? How can a question include a misconception?
The only conception I have is I don’t know. And you’ve left me thinking you may be under the misconception that you have provided an answer.

I'm sorry if I misrepresented your effort. I apologize. I got the impression that you had read somewhere, or had heard from someone, something that gave you this incorrect picture. Let me try again.

I’ll try to clarify the question for you.
In an advanced periodic table you can find the Nuclear spin by Isotope of an atom.

...

But this interpretation seemed incomplete without accounting for the spins of all the individual electrons.

The numbers you talk of (and you've actually said this above) are Nuclear Spins, so they deal only with the nucleus. There is no attempt to incorporate the electron spins here. So, this is not talking about "atomic spins" and so the role of electrons doesn't come into the picture at all.

The electron angular momentum is the sum of the spin and orbital contributions. The first of these is always positive (spin +1/2), and the second can either be positive or negative, making the material paramagnetic or diamagnetic depending on the sign of the total angular momentum. In any case, the magnetic moment from the electrons is much larger than that from the nucleus, so the (magnetic) susceptibility of an element can be calculated fairly accurately neglecting the nuclear contributions. The proof of the order of magnitudes comes from an NMR type measurement, where the Larmor frequency for electrons is measured to be much higher than that for the nucleus. This part, of course, has nothing to do with the previous discussion of the total nuclear spin.

 

[ZapperZ]

Misconception?? How can a question include a misconception?
The only conception I have is I don’t know. And you’ve left me thinking you may be under the misconception that you have provided an answer.

I’ll try to clarify the question for you.
In an advanced periodic table you can find the Nuclear spin by Isotope of an atom.
(Yes I’m trusting that someone has ‘measured’ that. No I don’t know how they did. If you want to share the how, ok that’d be cool)
Iron (Fe-56) has spin of “0” Zero (also F-54 & F-58)
& Boron B10 spin = 3 vs. Isotope Boron B11 spin = 3/2
So I’m interpreting this to mean the even number of neutron changes to contribute a (+½ -½) spin for the net 0 change where F-57 has a ½ spin change for one neutron change.
Following this reasoning I’d have to guess that adding a neutron could contribute a –1/2 AND convince an existing neutron to flip from +1/2 to –1/2 in order to get to the net 3/2 change.

But this interpretation seemed incomplete without accounting for the spins of all the individual electrons.

This is incorrect. If you include the electronic spin, then this is the net spin of the whole atom. It is then pointless to call the earlier one NUCLEAR SPIN, isn't it?

The nucleus has its own spin, the electrons have their own spins and orbital angular momentum. You can select which ones you want to "tickle" because each of them have different Lamour frequency when they are all in an identical external magnetic field. You can tune an external RF frequency to select just the frequency corresponding to the precession of the nuclear spin about this external field. This means that you are at RESONANCE to the nuclear spin frequency. Thus, the name "Nuclear Magnetic Resonance". NMRs are done in the presence of the whole atom, and in the presence of the atom in a bulk material. There is no problem whatsoever in probing just the nucleus spin in this scenario.

Hence, the quoted nuclear spin IS really the nuclear spin, without having the influence of the rest of the electrons.

Zz.

 

[RandallB]

...

So for a short version
atom spin as shown on a periodic table = from nuclius (protons & nutrons);
ignore electron "spin" and "orbital spin" both

Atom Magnetic "spin" (or whatever) = from electron "orbital stuff" ignore individual electron spins or at least very small if at all.

Individual Electron spins = not involved (or very small) in the above, but having one keeps an electron happy within its self.

That generally makes things simpler, we like simple.

 

[Gokul43201]

So for a short version
atom spin as shown on a periodic table = from nuclius (protons & nutrons);
ignore electron "spin" and "orbital spin" both

Atom Magnetic "spin" (or whatever) = from electron "orbital stuff" ignore individual electron spins or at least very small if at all.

Individual Electron spins = not involved (or very small) in the above, but having one keeps an electron happy within its self.

That generally makes things simpler, we like simple.

Nearly there ! A few points to address.

The periodic table will never call it atom spin. It is only the nuclear spin, and nothing more.

Electron (intrinsic) spins are not at all ignored. The total electron angular momentum is the sum of the spin and the orbital terms. Both are involved, and neither is ignored. In an isolated hydrogen atom, the spin contribution to the total magnetic moment is a factor of 2 (roughly) greater than the orbital contribution.