If electrons and protons had the same mass, would they still be in a stable orbit around their barycenter, or would they eventually collide? Similarly, a positronium(or protonium) only lasts extremely briefly until it annihilates, would a similar situation happen where the electrons would join the protons and neutrons in the nucleus(this wouldn't violate Pauli exclusion principle as they are different fermions).

If this were the case, would Chemistry still be possible in anyway resembling the current form?

Would electrons/protons still have energy states?

  • $\begingroup$ More or less, they have... (105.6582+1776.82)/2= 941.24 MeV, and proton mass is 938.27 MeV. You could do some funny barycenters playing with muons and taus instead of electrons. (PS: note that the first average are yukawa-higgs masses, the second mass is QCD mass; completely unrelated. Funny coincidence) $\endgroup$
    – arivero
    Oct 19, 2014 at 2:08
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    $\begingroup$ Usually hypothetical questions don't do very well here but I like this one. $\endgroup$ Oct 19, 2014 at 2:22
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    $\begingroup$ @BrandonEnright, this kind of hypothetical question should (ought) to do well here. Let's see if it does. $\endgroup$ Oct 19, 2014 at 2:26
  • $\begingroup$ @AlfredCentauri Seconded. I could almost answer this question by analogy with positronium, but lack the knowledge of the electroweak interaction to answer about the proposed system's stability. I suspect there are many readers in exactly my position with this level of knowledge (or similar, i.e. lacking key pieces) who would greatly benefit from a good answer. $\endgroup$ Oct 19, 2014 at 4:39
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    $\begingroup$ Robert N. Cahn wrote a beautiful article along this line of inquery: radford.edu/~brockway/18_parameters.pdf $\endgroup$
    – CuriousOne
    Oct 19, 2014 at 5:23

1 Answer 1


This is a hypothetical question, since electrons are elementary particles and protons are composite.

The solutions of the potential problem would give stable orbitals with smaller average radii. Here is a Bohr model solution for the muonic hydrogen, where the muon is 200 times heavier than the elecron. The energies become KeV instead of eV. To go to the mass of the proton another factor of 10 enters.

One could imagine a chemistry coming out of such atoms, except then one should imagine also a new electroweak sector, to reduce the probability of instabilities.

a) The stability of atoms comes because electron capture is very rare, due to the small probabilities of overlap with the nucleus and consequent creation of a neutron and a neutrino. The proximity and the large energies involved would increase these probabilities. Have a look at exotic atoms.

b)excited states of the "hydrogen" atom might have enough energy and overlap with proton to turn into a neutron + neutrino and the cosmological models would have to be redone.

c) as @CuriousOne has pointed out in a comment, muon catalyzed fusion has shown another instability with respect to normal solid state matter:

If a muon replaces one of the electrons in a hydrogen molecule, the nuclei are consequently drawn 207 times closer together than in a normal molecule. When the nuclei are this close together, the probability of nuclear fusion is greatly increased, to the point where a significant number of fusion events can happen at room temperature.

Once the compositeness of protons ( three quarks) enters the program a completely new standard model would be needed as a number of reactions would not go by energy balances alone , example :

neutron decay graph

The Feynman diagram for beta decay of a neutron into a proton, electron, and electron antineutrino via an intermediate heavy W boson

All in all my opinion is that no stable solid state would result from such a hypothesis.

  • $\begingroup$ I think one could add, that even if a stable hydrogen state could form (by casting the standard model physics aside), the example of muon-catalyzed fusion shows that instead of making stable molecules, one would probably have lots of fusion reactions in addition to electron capture. $\endgroup$
    – CuriousOne
    Oct 19, 2014 at 10:52
  • $\begingroup$ @CuriousOne good point, I will add it. $\endgroup$
    – anna v
    Oct 19, 2014 at 11:24

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