What is the effect of a magnetic field on an atom's magnetic moment?

[Malamala]

Hello! I want to make sure I understand (mainly qualitatively) what happens to an atom in a magnetic field. Assume we have an atom with an even number of protons and electrons. This means that all proton (electrons) are paired up, except for one of them (I am not totally sure if this pairing is always the case, especially for mid-shell nuclei, but let's assume this for now). If the electron has a total angular momentum of $j_e$ and proton of $j_p$, they have a magnetic moment of $\mu_e = g_e\mu_B j_e$ and $\mu_p=g_p\mu_Nj_p$, with $g_e$ and $g_p$ the effective g-factor of the electron and proton. Ignoring higher order effects, in a magnetic field these 2 magnetic moment would interact with the magnetic field and assuming we have very low temperature, they would align (or anti align) with the magnetic field. Is this it, or does something else happens in terms of the behavior of the 2 magnetic moments? Can I treat them separately i.e. if for example I need polarized nuclei, would the electron magnetic interaction have a major effect?

 

[Charles Link]

Besides aligning with the applied magnetic field, an additional effect with the nuclear spins is that energy splittings occur, e.g between the different $m_z$ states with $\Delta E=-\vec{\mu}_N \cdot \vec{B}= -g m_z \mu_N B$, so that resonant absorption can be observed at r-f frequencies, when a large static magnetic field is present. These magnetic resonance effects have been studied in great detail, and are used in magnetic resonant imaging.

 

[Malamala]

Besides aligning with the applied magnetic field, an additional effect with the nuclear spins is that energy splittings occur, e.g between the different $m_z$ states with $\Delta E=-\vec{\mu}_N \cdot \vec{B}= -g m_z \mu_N B$, so that resonant absorption can be observed at r-f frequencies, when a large static magnetic field is present. These magnetic resonance effects have been studied in great detail, and are used in magnetic resonant imaging.

Thank you for this! So shouldn't the exactly same effect exist for the electron (energy splitting between the 2 different states)? So overall there would be a 4 level splitting of the whole atom (2 from nucleus times 2 from electron)?

 

[Charles Link]

Thank you for this! So shouldn't the exactly same effect exist for the electron (energy splitting between the 2 different states)? So overall there would be a 4 level splitting of the whole atom (2 from nucleus times 2 from electron)?

The same type of effect , i.e. energy splittings from a static magnetic field, can be seen for the electron. The (visible) atomic spectral lines have splittings, known as the Zeeman effect. (The excited electron energy level can be split into multiple (closely spaced) levels, and the lower electron energy level involved in the atomic transition can also be split, with the result that the atomic spectral line, that may be a single line without the magnetic field, can show up as several closely spaced lines in the presence of a magnetic field).
There is also a resonant absorption that can occur between adjacent levels for the electron. I believe these are usually in the microwave region, and the effect is EPR=electron paramagnetic resonance.
The splittings from the nucleus are much, much smaller than those from the electron. For the electron the observed splittings of the spectral lines is commonly known as fine structure in the spectrum, where for the nucleus, it is referred to as hyperfine structure. In general, you need a very high resolution spectrometer to observe the hyperfine structure. Otherwise, the splittings are so small that the spectral line(s) simply looks like a single line.

 

[DrClaude]

Usually, the nuclear and electronic angular momenta do not align independently with the external field. One has to consider the total angular momentum F.
https://en.wikipedia.org/wiki/Hyperfine_structure

 

[Malamala]

Usually, the nuclear and electronic angular momenta do not align independently with the external field. One has to consider the total angular momentum F.
https://en.wikipedia.org/wiki/Hyperfine_structure

Thank you for you reply! So you mean that it is the overall atomic magnetic moment that aligns with the magnetic field? Is this because of coupling between the magnetic moments of the nucleus and electrons? It can be the case that the atomic moment is aligned but the 2 individual moment of nucleus and electron are not? Also, if you want to make sure you align the nucleus, you would need an atom with zero electronic angular momentum? So in this case the nuclear angular momentum would come from a neutron, not a proton.

 

[Charles Link]

Just a comment on getting the magnetic moment of the nucleus to align with an applied static magnetic field: Because the energy splittings $\Delta E=-\vec{\mu_N} \cdot \vec{B}$ are so small, typically it requires low temperatures and/or a very strong magnetic field, or thermal effects will keep the populations of the various nuclear spin states nearly equal.
There are nuclear magnetic resonance experiments where they first get the nuclear spins to align with a static magnetic field in the z direction, and then by applying a transverse r-f magnetic field at the resonant frequency, the nuclear spins can be made to change their direction in unison. See also https://en.wikipedia.org/wiki/Nuclear_magnetic_resonance#:~:text=Nuclear magnetic resonance (NMR) is,magnetic field at the nucleus.

 

[DrClaude]

So you mean that it is the overall atomic magnetic moment that aligns with the magnetic field? Is this because of coupling between the magnetic moments of the nucleus and electrons? It can be the case that the atomic moment is aligned but the 2 individual moment of nucleus and electron are not? Also, if you want to make sure you align the nucleus, you would need an atom with zero electronic angular momentum?

It is a question of the relative strength of the coupling between the momenta compared to the coupling with the external field. In strong magnetic fields, the moments will couple with the nuclear and electronic magnetic moments "before"(*) they couple with each other, so you can have independent alignment.
[(*): in a perturbation theory sense]

So in this case the nuclear angular momentum would come from a neutron, not a proton.

I don't know of any case where this would be true. You can consider the nucleus as a whole with a given spin $I$.

 

[Malamala]

I don't know of any case where this would be true. You can consider the nucleus as a whole with a given spin $I$.

What I meant is, if we have a nucleus with 2n protons and 2m+1 neutrons, we would have 2(m+n) pairs with an overall angular momentum of zero (as this gives the minimum energy) so the spin (and hence the magnetic moment) of the nucleus would come from the unpaired neutron alone. Moreover, the electrons would also pair up to give an overall angular momentum of zero. So in the end won't the whole angular momentum of the nucleus be equal to that of the unpaired nucleon?

 

[DrClaude]

What I meant is, if we have a nucleus with 2n protons and 2m+1 neutrons, we would have 2(m+n) pairs with an overall angular momentum of zero (as this gives the minimum energy) so the spin (and hence the magnetic moment) of the nucleus would come from the unpaired neutron alone.

I don't know much nuclear physics, but by that logic wouldn't nuclei only have spin 0 or spin 1/2? How would you explain ²H having spin 1 or ¹³³Cs having spin 7/2?

Moreover, the electrons would also pair up to give an overall angular momentum of zero. So in the end won't the whole angular momentum of the nucleus be equal to that of the unpaired nucleon?

The electrons do not pair up the way you think. For example, oxygen has 8 electrons but its ground state is ³P₂, so L = 1, S = 1, and J = 2.

 

[Vanadium 50]

or 133Cs having spin 7/2?

Or ⁵⁰V having spin 6?

 

[PeterDonis]

if we have a nucleus with 2n protons and 2m+1 neutrons, we would have 2(m+n) pairs with an overall angular momentum of zero (as this gives the minimum energy)

Not necessarily. The proton and neutron magnetic moments interact and have opposite sign, and there are also isospin symmetry and parity to consider, so proton-neutron pairs with opposite spin is not always the lowest energy state. For example, the deuterium nucleus is spin 1 in its ground state, not spin 0.

 

[Malamala]

I don't know much nuclear physics, but by that logic wouldn't nuclei only have spin 0 or spin 1/2? How would you explain ²H having spin 1 or ¹³³Cs having spin 7/2?The electrons do not pair up the way you think. For example, oxygen has 8 electrons but its ground state is ³P₂, so L = 1, S = 1, and J = 2.

I am sorry I didn't phrase my question well. I didn't mean that always under these conditions, we would have these quantum numbers. I was asking basically, if the angular momentum of the electrons is 0 and the angular momentum of the nucleus is given by only one neutron and it is hence 1/2, will the whole interaction of the atom with the magnetic field be given just by the interaction of the magnetic field with nucleus? Such an example would be $^{199}Hg$ (probably there are others too).

 

[Malamala]

Or ⁵⁰V having spin 6?

I am not sure why would V have a certain intuitive value of the spin. Odd-odd nuclei are well known to have weird values of spins. My argument was about even-even (in which case you would have spin zero) or even-odd (which sometimes have spin 1/2, but not always, especially mid-shell). As far as I know there is no easy consistent way to assign the spin value to an odd-odd nucleus without doing some actual calculations. So I don't see it as surprising for V to have spin 6 I guess.

 

[PeterDonis]

if the angular momentum of the electrons is 0

Which it won't always be, even if there are an even number of electrons.

and the angular momentum of the nucleus is given by only one neutron and it is hence 1/2

Which it won't always be.

will the whole interaction of the atom with the magnetic field be given just by the interaction of the magnetic field with nucleus?

For a small nucleus, this might be a reasonable approximation. For a large nucleus, I'm not so sure.

 

[Malamala]

For a small nucleus, this might be a reasonable approximation. For a large nucleus, I'm not so sure.

How about for the example I gave: $^{199}$Hg? That is for sure not a small nucleus, but it fulfills the conditions I mentioned. Is the whole atom magnetic moment given just by the nucleus in that case?

 

[Charles Link]

How about for the example I gave: $^{199}$Hg? That is for sure not a small nucleus, but it fulfills the conditions I mentioned. Is the whole atom magnetic moment given just by the nucleus in that case?

I think I can answer that one. Hg is a very good atom for demonstrating the Zeeman effect with the splitting of the green spectral line into about 9 lines if I remember correctly. It clearly has an electronic magnetic moment. Edit: See http://instructor.physics.lsa.umich.edu/adv-labs/Zeeman/zeeman experiment.pdf
It might be worth mentioning, that these are excited states. Perhaps the ground state has zero electronic angular momentum, but I think that is also unlikely. For the noble gases with filled orbitals/shells that would be the case, but not for Hg.
One thing that perhaps is also of relevance is whether we are discussing $J$ or $m_J$. The $m_J$ may be zero for the ground state, but I think the various $m_J$'s might be degenerate, i.e. same energy, in the absence of an external magnetic field. Perhaps someone else can elaborate more on this.

 

[DrClaude]

the angular momentum of the nucleus is given by only one neutron and it is hence 1/2

You really have got to get that picture out of your mind. You have to consider the nucleus as a whole. The nuclear spin of ¹⁹⁹Hg doesn't come from "only one neutron."

will the whole interaction of the atom with the magnetic field be given just by the interaction of the magnetic field with nucleus?

No. The electrons will interact with the magnetic field, as @Charles Link explained.

 

[Charles Link]

One additional comment: The Zeeman effect experiment for Hg is performed by putting a Hg lamp, i.e. an electrical discharge=arc type Hg lamp between the poles of a magnet, which supplies the magnetic field. In any case, the electrons of Hg even in the ground state are likely to play a role in how the atom responds to a magnetic field.
For something the OP may find of interest, they do use some of the noble gases in NMR studies as a useful probe. The filled electronic shells may be a useful property here. One additional note is these are isotopes they are using here, and not necessarily the most common one.
See https://www.tandfonline.com/doi/abs/10.1080/02678292.2019.1630489?journalCode=tlct20#:~:text=NMR active noble gases, 21,stem exclusively from environmental effects.

 

[Vanadium 50]

Perhaps the ground state has zero electronic angular momentum, but I think that is also unlikely.

Please don't guess. Mercury has electronic structure [Hg]4f14 5d10 6s2. L=0 and S=0, so J=0.

You really have got to get that picture out of your mind. You have to consider the nucleus as a whole. The nuclear spin of 199Hg doesn't come from "only one neutron."

Exactly. If that were the case, the nucleus (and atom) would have the same magnetic moment as the neutron. That's not what we see.