Just why electrons don't fall into nucleus?

Question

I think I finally got it, after hearing 1000 times.

Long story short: in ground state and some similar,
electron's wavefunction is spherical, with a nucleus being at the center, which is exactly what can be considered a "collision".

The difference is that there're no such quantum numbers onboard both an electron and nucleus that would allow them to, say, annihilate, like in a case with positron.

Therefore, they're basically ignoring each other; except for EM interaction, of course.

And the shapes of orbitals? Well, they depend on just a kinetic energy of an electron.

..So, yeah :) question is, how incorrect is that.

This may help - Why doesn't an electron ever hit (and stick on) a proton? — mmesser314, Aug 02 '23 at 01:02
Electron capture (with neutrino emission to carry away the extra 1/2 spin) is a real thing, but neutrons have more energy than a ground state hydrogen atom, not less, so I wouldn't call it "crashing". — g s, Aug 02 '23 at 03:15
@gs that's interesting; "wouldn't" call it misquotation either. — Victor Novak, Aug 02 '23 at 13:48

score 4 · Answer 1 · answered Aug 02 '23 at 02:38

4

There’s a lot of confusion here, so let me try to get to the heart of the matter the way I do with my general chemistry students. The key to seeing why electrons don’t cascade into the nucleus is not that they have to orbit in some particular shape, nor that they have to have any particular numbers describe the way they move. It is simply that the governing equation called the Heisenberg uncertainty principle requires that we cannot ever know the position and momentum of an electron simultaneously (or exactly at all). If the electron crashed into the nucleus, we would know that the electron has no momentum. This would violate the uncertainty principle. Thus, we are forced to make the rather bizarre proclamation that the electron is not a little ball, but actually is smeared out over a large amount of space all at the same time. The orbitals that chemists sometimes talk about electrons “occupying” might have accidentally confused you. The orbital is not a place the electron rattles around in; it is a description of what the electron is really like! And the shape of the electron orbitals certainly does not need to be spherical. For instance, consider a 2p orbital. So when we say the electron is in a 2p state, that means that the electron really is smeared out in a sort of weird dumbbell shape, and that it is likely to be on two opposite sides of the nucleus without ever actually being in between these locations! If this all seems weird, it is! Quantum mechanics is deeply weird and it will take a long time before you are comfortable and familiar with these ideas.

answered Aug 02 '23 at 02:38

Matt Hanson

2,550

2

It is a traditional way to preset the argument. However, it contains two questionable points: 1) Heisenberg relations are not postulates (principles) of QM but a consequence of the principles, and their validity refers to the statistical interpretation of wavefunctions. Therefore, the applicability to an individual system requires some justification; 2) the intuition built in this way is in contradiction with the experiment of the double slit with single electrons, where one sees that the "wave" is built as the consequence of many individual point-like events. – GiorgioP-DoomsdayClockIsAt-90 Aug 02 '23 at 05:20
I agree that in a more rigorous presentation we would not begin from Heisenberg, but show that it is a consequence of the general uncertainty principle arising from non-commutativity of incompatible observables. But this is for freshmen ;) 2. I’m not sure I agree that this builds faulty intuition. Unless we perform some sort of perturbation measurement, it is simply the case that the wavefunction describing the electron is spread over space. The only time you need to appeal to localized wavefunction collapse like you mentioned is for measurement events, which does not apply here.

Matt Hanson

Aug 02 '23 at 12:27

@MattHanson oh, yeah? And how is your non-commutativity doing when it comes to “occupying” near a positron, that I specifically mentioned? – Victor Novak Aug 02 '23 at 12:54

I’m not exactly sure where your hostility is coming from, but annihilation is a completely different subject. You can’t deal with that in the context of regular QM, which is where this discussion came from. Instead you need to use quantum field theory, as it does not constrain particle number. It just so happens that at the energy scales we deal with there is no feasible way to annihilate a proton with an electron. Instead you just make atoms. But positrons, being different than protons, actually can readily annihilate with an electron to form anti parallel photons. – Matt Hanson Aug 02 '23 at 13:55

Sorry, didn't mean to confuse you by not placing a qft tag. I actually used to place it near my past questions exactly for the reason you obviously mean here, but only for some other "teacher" to remove it, and never give an answer from the point of more interesting framework. "It just so happens". – Victor Novak Aug 02 '23 at 21:36

score 2 · Accepted Answer · answered Aug 03 '23 at 06:39

If I understand correctly you are already happy with the idea that the electron is delocalised over the region around the nucleus, and that the uncertainty principle prevents it from shrinking down to a point. In that case your key question is related to your statement:

The difference is that there're no such quantum numbers onboard both an electron and nucleus that would allow them to, say, annihilate, like in a case with positron.

That is you are really asking why even though the electron overlaps with the proton it does not interact with the proton to form some new particle.

If this is the case then your statement is basically correct but with one exception - an electron can react with a proton to form a neutron. This reaction is energetically unfavourable so it does not happen in most atoms. However there are a few atoms where the conversion of the proton to a neutron increases the nuclear binding energy and this supplies enough energy for the reaction to happen. This process is called electron capture.

Yeah, I mean, would've gladly discussed here more details of that process, but ran out of concepts I'm confident enough in. Thank you! — Victor Novak, Aug 03 '23 at 08:20
I think you captured (no pun intended) the part of this question that I failed to address quite nicely! — Matt Hanson, Aug 07 '23 at 13:26

Just why electrons don't fall into nucleus?

2 Answers2