Electrons can move through potential barriers by tunneling. Atomic/molecular orbitals are separated by energy differences. Therefore I was wondering if an electron can tunnel from one orbital to another with higher energy, say from a 2s to a 2p, or if it is limited to spatial barriers.

Clarification: It is possible to tunnel to a lower potential (as in scanning tunneling microscopy, that provides a voltage (potential difference)). It is also possible to tunnel to inside the potential barrier (see this question). If at the other side of the barrier there is a state of medium energy (higher than the previous but lower than the barrier), the wavefunction (and therefore the probability of finding the particle) cannot drop immediately to zero, because the wavefunction must be continuous. So, even if by a small probability, the particle must be able to tunnel to another state of higher energy. The question is if this applies only to spatial barriers (as the space between the tip and the sample in STM) or to different states in general, as in atomic orbitals.

  • 1
    $\begingroup$ This is rather unclear - does not make sense. In tunnelling particle doesn't get energy. $\endgroup$ – Mithoron Jul 11 '20 at 21:05
  • $\begingroup$ Remember that all electrons in a molecule are indistinguishable from each other, so there's no practical meaning to the concept of an electron being confined to a specific orbital. What you're really considering is the transition between two electronic states of a whole molecule, for example between a ground and excited state. The barriers to different types of transitions are different for different cases (eg "spin-forbidden" or "symmetry-disallowed"). Since spectroscopy makes use of these transitions, a book on that subject is a good place to start. $\endgroup$ – Andrew Jul 14 '20 at 11:58

There are no spatial barriers, much like there are no sky-high blue walls along all meridians and parallels, even though the map of the world might have convinced you otherwise.

There is, however, the energy barrier. You can't skip that by tunneling. The tunneling is about getting to the state of equal energy when getting there seems to be topologically impossible. Getting to a state of other energy requires energy. There is no going around that.

So it goes.

  • $\begingroup$ Hey, what about tunneling to a lower energy state? Isn't that what happens in a tunneling microscopy (because of the applied voltage)? $\endgroup$ – peruca3d Jul 12 '20 at 13:53
  • $\begingroup$ Sure, that's possible all right. $\endgroup$ – Ivan Neretin Jul 12 '20 at 14:22
  • $\begingroup$ By looking at a "tunneling diagram", I see that by the continuity of wavefunction requisite, the probability of finding the particle inside the barrier is non zero (also these answers). If at the other side of the barrier there is a higher energy region (significantly less than the barrier but still higher than the other side), the probability cannot vanish to zero immediately as well. Could this then provide a really small but possible way to tunnel to a higher energy state? $\endgroup$ – peruca3d Jul 12 '20 at 15:31
  • $\begingroup$ Let's say there is a really small but possible way for a particle to act as if it where in a higher-energy state for a short while. That's pretty much what tunneling is about. To reside in that state permanently, though, would violate the conservation of energy and hence is impossible. $\endgroup$ – Ivan Neretin Jul 12 '20 at 19:54

Yes, it is possible to for an electron to move between orbitals, provided it can gain or radiate away the energy involved. The energy involved is often in the range of visible light or ultraviolet radiation, which means that UV/vis spectroscopy typically deals in electron transitions. More precisely, the transitions are between electronic states, which may involve "movement" of more than one electron.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.