Hello everybody,
I have a question about the spin-orbit coupling ?
Lanthanides have unpaired electrons in their f orbitals and transition metals have unpaired electrons in their d orbitals however lanthanid ions have the strong angular momentum and strong magnetic anisotropy caused by effective spin-orbit coupling. My question is why the spin-orbit coupling in f orbitals is bigger than d orbitals?
All the best,
Spin-orbit can be calculated and exits for all the elements of periodic table, even in the hydrogen you have a splitting between the states 2s1/2 and 2p1/2 at a frequency of 1054 MHz (by the way, not fully splained its strength for Bohr, Dirac or even QED is difficult to do it ) known as Lamb shift. But usually we only mentioned the f elements because this is a relativistic effect which depends of the atomic number very strongly, Z4.
The problem is that the orbits can cancel each other and therefore its Zeeman energy would be zero. Thus it is important also to know the total angular momentum j of the orbital. But in any case, this effect is much higher for the f elements than for the d elements in general, although if you choose special transitions you can find important ones for d-electrons provided the L orbits which surrrounds them enhance or decrease the whole angular momentum under a magnetic field force.
Thank you for notes!
Actually I did not get the reason why SO coupling is much higher in f orbitals than d orbitals ? any phisical reason behind ?
from the mechanich quantoum point of view it is said that SO caoupling is higher for f orbitals because they have bigger quantoum nomber (L).
I thought the Lamb shift, the splitting between 2S1/2 and 2P1/2, came from vacuum fluctuation, not spin-orbit coupling (which would be fine structure when incorporated with relativistic mass and Darwin term)...
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