If I place a proton and an electron close to each other, can they form an atom of hydrogen? Why or why not? Please explain the scenarios where this is not likely/unlikely.

Personally I doubt it, since the electron would have to spin around the proton to call it an atom.

Can it be well-explained by classical mechanics?

How does quantum mechanics explain this sort of a possible bonding? I mean,an electron and a proton can't collide with each other because they would have had infinite attraction between them (that is they would act like magnets), and we know that this does not happen.

Is there something in QM that defines a minimum distance of some sort that is relevant to the proton-electron forming hydrogen?

  • $\begingroup$ A proton by itself IS already a hydrogen atom - the electron doesn't matter in that regard at all. The proton is an atom that (also) is a positively charged ion, $\ce{H+}$ $\endgroup$ Oct 30, 2017 at 10:18
  • $\begingroup$ @StianYttervik That is quite wrong. $\endgroup$
    – Mithoron
    Oct 30, 2017 at 13:24
  • $\begingroup$ @Mithoron I might have been too generalizing, at the detriment of accuracy, in the second sentence, I'll admit that. BUT - even IUPAC recognizes, and calls a lone proton for "hydron" - cationic hydrogen ion. I won't admit defeat on that one, just yet... $\endgroup$ Oct 30, 2017 at 14:08
  • $\begingroup$ @Mithoron Elements are defined by the number of protons they have. Unless there is some crazy technicality thing that I'm missing, I am going to agree with Stian. $\endgroup$ Oct 30, 2017 at 17:26
  • $\begingroup$ @LordStryker Long time, no see; I could suspect this would draw you, though. BTW this question shouldn't be here, it's hardly about chemistry and rather broad/unclear. For a chemist proton may seem to be ionised hydrogen, but this here is rather physics and bare proton is no more hydrogen then single electron. $\endgroup$
    – Mithoron
    Oct 30, 2017 at 18:12

3 Answers 3


Both in classical and quantum mechanics you have to get rid of a (rather large) amount of energy, i.e. the difference in potential energy from zero (=infinite distance) to the bound state (= e.g. Bohr radius), also called the binding energy.

Nature has no way to do that without a third particle taking the excess energy with it. In quantum mechanics there is the additional problem with the spin, but also classically that doesn't work if you assume elastic interaction of proton and electron.

So the answer is no.

(Inelastic interaction requires a zoo of additional assumptions, which I let fall victim to Occams razor here. Inverse photodissociation is a possibility, if overall energy and momentum are conserved.)

  • $\begingroup$ Assuming initially stationary nucleus and electron and full conversion to kinetic energy, my back-of-the-envelope calculation gives 1.6 km/s for $\ce{H}$, which is a lot compared to 0.4 km/s for $\ce{O2}$ at r.t., but not completely outlandish? $\endgroup$
    – TAR86
    Oct 30, 2017 at 10:56
  • $\begingroup$ It actually doesn't matter how large the energy is, if you can't give it to a third particle, there is no way to reach a bound state. $\endgroup$
    – Karl
    Oct 30, 2017 at 20:52
  • 1
    $\begingroup$ The answer and question are rather old, but in QM this can happen via a process called radiative recombination in which a photon is formed which takes away the binding energy (this is actually how elements became neutral after the Big Bang). This is basically the reverse process of photoionization. $\endgroup$
    – Paul
    Sep 9, 2019 at 20:25
  • $\begingroup$ @Paul You are (obviously, any microscopic process is reversible) right. However the collision must occur in a way so the photon (the third particle) is emitted in exactly the right direction so the overall energy and momentum is conserved. This has a rather low probability. $\endgroup$
    – Karl
    Sep 9, 2019 at 21:54

If you put an electron and a proton together you get a hydrogen atom, assuming there's nothing else around it will stay a hydrogen atom.

The positive proton and the negatively charged electron will attract each other to form a stable atom. However, classical mechanics would predict that the electron would collide with the proton. That is not the case and to explain what is happening we need quantum mechanics. This also teaches us that electrons aren't spinning around the nucleus.

  • $\begingroup$ "this also teaches us that electrons aren't spinning around the nucleus". You mean that classical mechanics does not actually prove that an electron actually moves around the nucleus? But Bohr made an assumption that it is revolving? $\endgroup$
    – Karthik
    Oct 30, 2017 at 4:22
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    $\begingroup$ Yes, Bohr was wrong. There are pretty good explanations all over the internet why the electron does not fall into the nucleus, why Bohr was wrong and why the Bohr radius still works (it's the average distance between the electron and the core rather than an orbit). See for example here: chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/… $\endgroup$
    – DSVA
    Oct 30, 2017 at 4:43


This is exactly what happens when hydrogen is ionised in the gas phase and then returns to an unionised state.

When it is less likely to happen is when the proton has very high energy (eg. having been emitted by nuclear decay) such high energy protons are either absorbed by proton capture into another atomic nucleus or the lose their energy by inelastic collisions until they are able to capture an electron.

In the same way, most of the helium on earth was formed by alpha particles from nuclear decay capturing electrons.


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