Why is spin-orbit splitting larger in heavier atoms?

Spin-orbit interaction is the interaction of an electron's spin with the magnetic moment generated by the orbital motion of the electron. Or, from the point of view of an electron, it is the interaction between electron's spin and orbital motion of the nucleus. And, for different atoms, the spin-orbital splitting of energy levels was found to be different. And the trend is that for heavier atoms the splitting is larger. I do not completely see why is that so qualitatively... Let's consider only the valence electrons. The heavier the atom, the more electrons it has and the more electrons are there in between the valence electrons and the nucleus, and the more efficient the screening effect. The last, as I understand, means that the Coulomb interaction between the valence electrons and the nucleus is largely reduced. And reduction is larger for the heavier elements. But larger spin-splitting should mean stronger interaction, right? So why is the spin-orbit splitting larger in heavier atoms?

• In heavier atoms i.e. atoms with bigger nuclear charges, spin-orbit interactions are frequently as large as or larger than spin-spin interactions or orbit-orbit interactions. In this situation, each orbital angular momentum l tends to combine with the corresponding individual spin angular momentum s originating an individual total angular momentum j. These then couple up to form the total angular momentum J – drvrm Aug 12 '16 at 11:19
• The spin-orbit splitting is partly due to the effect you describe, but it's also partly a special-relativistic effect. – Ben Crowell Aug 10 '17 at 17:50
• Are you sure it's even true that the splitting is bigger for the valence electrons in heavier atoms? The energy scale of the fine structure is $m\alpha^4$, so that the splitting in a hydrogenlike atom goes like $Z^4$. That's why the splitting is so much bigger for the inner electrons of a heavy atom. – Ben Crowell Aug 10 '17 at 17:53

Some insight into this question may be gained from considering calculations of the spin orbit splitting $\Delta$ by F. Herman et al., Phys. Rev. Letts. $\textbf{11}$, 541 (1963), which I have plotted below. There we see a saw-tooth dependence of $\Delta$ with atomic number $Z$, with $\Delta = 0$ for a single valence electron in an outer shell (alkali metals) to a maximum for an electron in a full shell (inert gases).

This suggests that it is only the core electrons that screen the nucleus, with little or no screening of a valence electron by other electrons in the outer shell. For a given row of the periodic table, the number of screening core electrons will stay the same as $Z$ increases, so the local electric field seen by the electron will indeed increase with the size of the atom and hence give rise to a stronger spin-orbit splitting.

Large text incoming, but none of this is unnecessary.

Someone here said very misleading things:

-Imagine you are an electron orbiting a nucleus. You are moving so fast that it appears the nucleus is orbiting you.

-> in the system of the electron, the core seems to be orbiting the electron. this has nothing to do with relativistic effects! the relativistic effects come into play if you consider that the frame in which the electron rests is not an inertial frame (there is always a force acting on the electron, since it is not flying on a straight line with constant velocity) to double down, it was said:

-it is due to the larger positive charge of the core orbiting the electron.

->This is verrrrry oversimplified. The atom itself is neutral. In the system of the electron, the rest of the atom has +1 charge (all other electrons + the core), just as in any other atom.

The quantum mechanical calculation gives a factor of H_SO ~ Z/r^3 for, and this is important, a "hydrogen-atom-like" atom. It comes from the assumption that there is one single electron orbiting one positive core with the charge +Z. WELL NOW we can all see, that this is not the case in a neutral atom!

what we can see though is the dependency of 1/r^3, meaning, just as someone suggested, it depends on the radius (and then he got downvoted, while being perfectly right).

Of course the radius itself (in this hydrogen like assumption) depends on the charge of the core, because the bigger the charge, the nearer the electron will be to the core. But the reasonable interpretation of the orbital momentum, with l=rxp, suggests NO dependency AT ALL of the core charge at first. It simply comes from p being dependent of the radius, and the radius being dependent of the charge. So the reasonable assumption would be to suggest that it depends on the radius, rather than the charge.

So in the end we will end up with a term H_SO ~ Z^4 BUT AGAIN, this is the assumption for a hydrogen-like atom with a core having more positive charge than the electron (meaning we would need to neglect all other electrons for this assumption)

So why is the effect still bigger for bigger atoms? For one, when we talk about the spin orbit effect we also consider valence electrons usually. these obviously are very close to the core with very high positive charge. But usually we are interested in effects around the fermi-level. I would suggest, although im not perfectly sure since I cannot find an elaborate explanation anywhere, that this is due to the distribution of the electrons in heavier elements. The 1s electrons are very close to the core for example, so the 2p electrons will see a "shielded" core potential. but if we go to heavier elements and compare for example 3p and 3d orbitals, then we see they have huge overlaps, so the are of comparable size. Thus this shielding is not as pronounced anymore (sure, the 1s electrons still shield the core as well as before, but those are only 2 charges, for elements as heavy as iron, those are rather insignificant. the rest of the positive charges has to be shielded as well, which isnt as pronounced as in the case of the lighter elements anymore). Here are the sizes of the orbitals: https://www.chem.fsu.edu/chemlab/chm1046course/orbitals.html

not only are the 3p and 3d orbitals of comparable size. consider also, that d shells take 5*2=10 electrons, which orbit the atom all on equal distances. same principle for the p-shells which take 6 electrons.

TLDR: to simply argue its just dependent on Z is pretty misleading, because the 'typical H_SO formula' is for the case of a hydrogen like atom.

And again, im unsure about the last part, these are my own thoughts.

But after having written all this, ive read Martin V's answer who suggests the same, so I guess its solid then.

Imagine you are an electron orbiting a nucleus. From your perspective it appears the nucleus is orbiting you. Since the nucleus is circling you it is creating a ring of current. A ring of current produces a magnetic field. So you are an electron within a magnetic field. As such, this magnetic field depends on the charge of the nucleus you're orbiting. So larger atoms have a higher nuclear charge so consequently this atomic magnetic field felt by the electron increases with atomic number/nuclear charge. Since spin orbit coupling is just a change in angular momentum as felt by as electron (due to spin angular momentum coupling with orbital angular momentum), this coupling is dependent on magnetic field, since spinning electrons produce a magnetic field with which to interact with the nucleus' magnetic field. And since larger atoms produce larger atomic magnetic fields relativistically, the spin orbit coupling is amplified for larger atoms. As has been pointed out, simply having a larger nuclear charge does not directly increase the electric field felt by the electron of interest. This depends also on electronic shielding, so the orbital occupied by our electron and the other electrons between it and the nucleus. However generally, larger atoms have larger nuclear charge felt by unpaired (typically valence) electrons so hence a larger magnetic field is felt and a larger spin orbit effect is observed. Smaller atoms have too small a nuclear charge to observe spin orbit coupling, although it is definitely present where unpaired electrons exist.

I think it has less to do with coulombic interactions between the nucleus and electron and more to do with the radius of the "orbit" of the electron. The movement of the electron generates a magnetic field which interacts with its spin and this field is affected by the path of the electron.

• I think the OP's description of the physics is actually somewhat more accurate than the description given here, and this also doesn't seem to provide an answer. – Ben Crowell Aug 10 '17 at 17:55