2
$\begingroup$

There are spectroscopy selection rules marking some radiative electronic transitions as forbidden. For example, it is stated that single electron of H atom can not move from S2 to S1, or from P2 to S2 (and vice versa) with photon emission/absorption.

But is this true in the case of collisional (or, more generally, radiationless) transitions? Are there any selection rules and what they are if any?

I found the Wigner spin conservation rule. Are there any more rules?

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ I think what you are asking about is pressure broadening: en.wikipedia.org/wiki/Spectral_line#Pressure_broadening . From my experience in NMR, all transitions are allowed during relaxation induced by weak interactions with the "lattice." Not sure how this translates when looking at electronic spectroscopies. $\endgroup$ – Buck Thorn♦ Aug 13 '19 at 4:59
  • 1
    $\begingroup$ The photon carries a unit of intrinsic angular momentum which, with spin conservation, leads to selection rules of a change of 1 in angular momentum in optical transitions in the atom, so $s\to p,\;p\to d$ allowed but not $s\to s;\; p\to p;\,s\to d$ etc. In a collision, as generally the partner has no intrinsic angular momentum (excludes rotation) then there should be no other restrictions other than that to conserve spin. $\endgroup$ – porphyrin Aug 14 '19 at 7:47
  • $\begingroup$ @porphyrin It makes sense. And can the partner bring/take enough rotational angular momentum to cause changes like s→p etc? $\endgroup$ – Alexander Olikevich Aug 15 '19 at 17:02
  • 1
    $\begingroup$ I think that that is unlikely to influence electronic transitions, however, at high pressure 10 to 100 atm otherwise 'strictly forbidden' bands are observed in what have been called 'pressure induced' bands. This may be due to the mutual distortion of the charge distribution of the molecules because of interaction at close range, and electric dipole selection rules are broken. $\endgroup$ – porphyrin Aug 15 '19 at 18:17

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.