Timeline for Why spin-orbit coupling in TMDCs is strong?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Feb 22, 2019 at 10:35 | comment | added | wcc | and also, my comments about scalingare actually wrong...the r-scaling is $1/r^3$, not $1/r^4$ since $v \sim L/mr$. So you get your $Z^3$ from there and the last factor of $Z$ from the strength of the electric field $Ze/r^2$ | |
| Feb 22, 2019 at 10:16 | comment | added | wcc | really? I thought they would differ because of different principal quantum numbers. Anyway, focusing on length scale may be a little misleading since SO really comes from $B = v \times E$ and for hydrogenic atom it's written as $L/(mr^2) \times e/r^2$ but I guess the r-scaling could be different in a more complex situation | |
| Feb 22, 2019 at 8:56 | comment | added | user137289 | @IamAStudent Yes. But the relevant orbitals here are the valence orbitals, and those have approximately the same size for oxygen, sulfur, selenium and tellurium. | |
| Feb 22, 2019 at 5:37 | comment | added | KF Gauss | +1 this is the right answer. For the record, spin orbit coupling in these materials isn't even that strong, it's just much stronger than graphene or silicon. | |
| Feb 22, 2019 at 2:41 | comment | added | wcc | for hydrogenic atoms it scales as $Z^4$ (since Bohr radius decreases as $1/Z$ and SO depends on $\langle 1/r^4 \rangle$ | |
| Feb 12, 2018 at 8:51 | history | answered | user137289 | CC BY-SA 3.0 |