1
$\begingroup$

When we have to choose, given $n$, $\ell$, where to put an electron, when there are several $m$ orbitals, with the same energy, is there a concrete quantum mechanic rule or some superselection rule to choose? Example: 1s²2s²2p³, is the last electron $m=+1$, or $m=-1$? If it were 1s²2s²2p⁴, same question...How to choose? In some places I find a rule of "maximum $J=L+S$", and sometimes I have seen however the place the electrons from minimum $m$ to maximum $m$. I suspect this is conventional in the last case, unless there is some interaction that breaks down the degeneration in energy. Are the schemes $LS$ or $jj$ coupling related to this?

$\endgroup$
  • $\begingroup$ It might be useful to read about Hund’s rule hyperphysics.phy-astr.gsu.edu/hbase/Atomic/Hund.html#c1 $\endgroup$ – ZeroTheHero Oct 28 '17 at 17:53
  • $\begingroup$ Yes, I read and I knot that rule. I teach it, but sometimes, when I think about quantum mechanics, I feel this (semiempirical?) law is not clear when we have half-filled or half-fulled orbitals tht are degenerated in energy, I have doubts about how to place and where electrons, or is there free choice with degenerated energy? That is very interesing, since it is a little bit like relativity ... $\endgroup$ – riemannium Oct 28 '17 at 18:21
  • $\begingroup$ Related: physics.stackexchange.com/q/11042/2451 , physics.stackexchange.com/q/255465/2451 and links therein. $\endgroup$ – Qmechanic Oct 28 '17 at 19:27
  • $\begingroup$ Indeed an interesting question. I take that you can place the electron indifferently in the available degenerated orbitals. From a merely energetic point of view this is natural. However there should be a deeper significance . It should else be important to distinguish in which orbital an electron is located. At the end, with all quantistic indeterminacy, I shall ask/know if the electron is about x, or y, or z....... perhaps this is due to the fact that we can always rotate the atom to get the right j. Symmetry $\endgroup$ – Alchimista Oct 28 '17 at 22:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.