21
$\begingroup$

I'm aware that the Q branch ($\Delta j = 0$) is usually not observed in rotational-vibrational spectra of diatomic molecules in the gas state, however, I've heard different things about when exactly this transition is forbidden.

I've sometimes been told that the transition is not allowed in general, for any diatomic molecule, but I've sometimes heard that this is only true if the electronic state has a $\Lambda \neq0$, so which is the correct selection rule here?

I'd love to have this clarified for me. If it is possible I would also like to know when is a Q branch observed in vibronic spectra of heteronuclear and homonuclear diatomics.

$\endgroup$
5
  • $\begingroup$ You are correct in that the electronic angular momentum $\Lambda \ne 0$ for a Q branch to be observed in a diatomic molecule, i.e. the molecule is not in a $\Sigma$ state. The rule is then $\Delta J =0; \Delta v = \pm1$. (Homo-nuclear diatomic molecules do not have an ir spectrum as $d\mu/dx = 0$, as the dipole is zero) $\endgroup$
    – porphyrin
    Sep 22, 2017 at 14:32
  • 1
    $\begingroup$ ahhh, I know homo nuclear diatomics don't have an IR spectrum, there was a typo in the question, thanks. Are you sure about the $ \Lambda$ selection rule? is a Q branch always permitted if $\Lambda \neq 0$? Is there some easy derivation of this or somewhere where I could find a derivation? I can't see where this comes from from a simple transition moment integral. $\endgroup$
    – Ignacio
    Sep 22, 2017 at 15:47
  • $\begingroup$ A photon carries one unit of angular momentum which is conserved. When the photon is destroyed on absorption, this ang mom must be transferred to the molecule in one way or another, in your case $\Lambda_i +1=\Lambda_f $ where i and f are initial and final states. This is the basis of the transition moment integral and so selection rules. $\endgroup$
    – porphyrin
    Sep 24, 2017 at 7:21
  • $\begingroup$ I understand that argument, however, $L$ is not really a good quantum number in diatomic molecules and $\Lambda$ is not the angular momentum, but the projection of the angular momentum in the direction of the internuclear axis, so I don't think that the fact that the total anguar momentum is conserved helps a lot. $\endgroup$
    – Ignacio
    Sep 25, 2017 at 15:12
  • $\begingroup$ In Hydrogen, for instance, although $\Delta l$ is constrained to be $\pm 1$, $\Delta m_l $, the z projection, can be zero, and this doesn't break angular momentum conservation. $\endgroup$
    – Ignacio
    Sep 25, 2017 at 15:26

0

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.