Say a H atom is ionised and then it captures a free electron at a later time, do the atomic orbitals then have to go through a transitory phase to accommodate the electron before they form the correct wave function?

  • $\begingroup$ I don't understand the question. A ionized H-atom has no electron, therefore there are no orbitals at all. $\endgroup$
    – ACuriousMind
    Nov 26, 2014 at 11:15
  • $\begingroup$ that's my question are orbitals a permanent property of the nucleus or do they spontaneously appear when electrons are some set distance from a nucleus. might be a stupid question. $\endgroup$
    – Peter
    Nov 26, 2014 at 11:17
  • $\begingroup$ I think this is a deeper question than it appears - rephrasing: "when an electron is capture by a proton does it immediately assume a stable orbital?". I think the answer is "no" because there is initial uncertainty in the energy (Heisenberg) so you can't "know" that the electron is in the orbital.until some time has passed. Suggesting that people answering include that consideration... $\endgroup$
    – Floris
    Nov 26, 2014 at 11:45
  • 1
    $\begingroup$ @Floris - I agree with the spirit of your comment, but answer yes, but note also that the uncertainty principle means that if the state formed only lives for a very short time then the energies of photons released will be broad due to lifetime broadening - I agree with you about the physics of what is happening, but think it goes straight into the orbital, but the orbital energy is not well defined if it is only there for a very short time. --- Sorry don't want to have an argument over language- I can see it from your point of view, does what I have written make sense? $\endgroup$
    – tom
    Nov 26, 2014 at 11:55
  • $\begingroup$ @tom - yes it does $\endgroup$
    – Floris
    Nov 26, 2014 at 12:40

3 Answers 3


Atomic orbitals are states of electrons. If there's no electron, then all orbitals are empty, since there's no electron state occupied. Whether a space of states that is fully unoccupied "exists" is an irrelevant matter of definition.


Adding an electron to a bare $H^+$ ion is an important process in astrophysics.

It is also a relatively low probability process because energy must be released by photon emission (or some other process) to enable the electron to drop in energy into an Atomic Orbital.

I agree with ACuriousMind that whether the orbital exists or not when unoccupied is a matter of definition.

I would like to add that there is not any

transitory phase

An electrons drops into unoccupied orbital level and a photon is released.

This process is known as radiative recombination. A description is given here.

Edit after good comment from Floris. If the electron only remains in the atomic orbital for a very short time after it is captured then energy/lifetime uncertainty broadening means that the energy of the atomic orbital will be a bit uncertain and a somewhat broad range of photon energies would be observed. If, on the other hand, the state formed is stable on a longer time scale the photon released would have a narrower energy range and the energy of the atomic orbital would be better defined. (Final comment - the energy of the photon will be the energy of the electron intially plus the energy release in formation of the hydrogen atom - if the free electrons which are captured by the ions have all have close to zero kinetic energy - or all have the same energy above zero then we might expect the photons emitted to give a spectrum of lines reflecting the energy lost by the electron as it is captured)


Atomic orbitals are the states of the mutual motion! Few people know that the nucleus moves around the atomic center of mass and produces a "positive charge cloud" or "orbital" strongly depending on the state $\psi_n$. If you consider a positronium, both "clouds" coincide in size. When we speak of an orbital, it is normally the relative distance who is an "orbital".

  • $\begingroup$ It is a bad idea to talk about "motion" of particles on quantum mechanical scales. Motion is a classical concept. $\endgroup$
    – ACuriousMind
    Nov 26, 2014 at 12:56
  • $\begingroup$ Yes, I wanted first to write "a wave motion", but the I decided that it would be too complicated to visualize ;-) $\endgroup$ Nov 26, 2014 at 13:07
  • $\begingroup$ @ACuriousMind: By the way, have you heard of quasi-classical description? $\endgroup$ Nov 26, 2014 at 13:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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