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When it comes to atoms electrons can't fall into the nucleus, which besides the off hand uncertainty explanation, I'm not sure which force prevents them from falling into the nucleus. I thought I heard it was the strong nuclear force-but is there a similar force preventing protons and electrons, outside of atoms, from falling into each other? I figure there must be otherwise the force will become infinite… I'm lost.

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it's due to the uncertainty principle. SE shows that the ground state isn't completely localized as it would be if the electron behaved classically and fell towards the center of the potential well. –  user18764 Apr 14 '14 at 22:50
What force holds it out? –  user24082 Apr 14 '14 at 22:50
Possible duplicates: physics.stackexchange.com/q/20003/2451 , physics.stackexchange.com/q/9415/2451 and links therein. –  Qmechanic Apr 14 '14 at 22:54
The question isn't "What force holds it out?" the question is "What force holds it in?", because quantum states tend to spread out. And the answer is that the electromagnetic interaction (mostly electrostatic) is what holds it in. –  dmckee Apr 14 '14 at 23:16

3 Answers 3

up vote 5 down vote accepted

A very tempting mental model of an atom, reinforced by many illustrations in books, has protons and neutrons as "large" spheres in the nucleus and electrons as "small" spheres somewhere near the nucleus. If you assume that all of these particles are made of some "stuff" that has roughly the same density (which is the case for everyday solid and liquid matter, to within an order of magnitude), then this big-ball small-ball picture of nucleons and electrons also makes sense.

But that's not how size works in quantum mechanics.

In quantum mechanics, every object has a wavelength, and the wavelength sets a fundamental limit on how well you can answer the question "where is this object?" If I have an electron in the $2p$ orbital around a nucleus, and you ask me where the electron is, I'll tell you your answer: it's in the $2p$ orbital around the nucleus. Unlike a planet, which always occupies a particular place in its orbit, the wavefunction doesn't contain any information about "where" an electron "is" in its orbital. There are many questions you can answer using the wavefunction, but that's not one of them. It's much closer to reality to think of the electron as spread out over the entire volume described by the wavefunction (though this starts to get into thorny territory about "interpreting quantum mechanics").

Of course if you have a particle with a very well-defined momentum, you won't be able to localize it over many wavelengths. But that's not the case for electrons bound in an atom: the wavefunction for a bound state is quite different from a long sine wave.

The hydrogen atom has well-described electron orbitals which are different for different values of orbital angular momentum. For the $s$ orbitals, with $\ell=0$, the wavefunction actually has a maximum at $r=0$: the electron is actually more likely to be found inside the nucleus than in any other comparable volume! From the electron's perspective this isn't very much overlap at all, since the volume of the nucleus is something like $10^{-15}$ the volume of a sphere with the Bohr radius, but it's there. The inner $s$-shell electrons of heavier atoms have larger overlap with their nuclei, since (a) the nuclei are bigger, and (b) the electric attraction is stronger, reducing the effective "Bohr radius" for those atoms. This sort of overlap is important for weak nuclear decays by electron capture.

The wavefunctions for orbitals with higher angular momentum --- the $p$ orbitals with $\ell=1$, the $d$ orbitals with $\ell=2$, etc. --- do vanish at $r=0$. (I believe that the general wavefunction for hydrogen has a factor of $r^\ell$ out front of the angular term, but I haven't looked at it for a while.) So these electrons are even less likely to be found inside the nucleus than the $s$ electrons. However even here the overlap between the electron and the nucleus isn't quite zero; knowing this overlap is important in some precision experiments, such as electric dipole moment searches using pear-shaped nuclei.

So the short answer is the atomic electrons may be held away from their nuclei by angular momentum; electrons without orbital angular momentum do overlap with their nuclei, but the overlap is small because the electrons are relatively cold, which makes them great big. This overlap does have observable effects on the behavior of atoms.

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+1 for a clear explanation of the issues - including the fact that weak nuclear decays are caused by electrons "finding themselves in the nucleus". –  Floris Apr 18 '14 at 21:29

I have always thought that electrons can and do spend some (tiny) fraction of their time in the nucleus, depending on the orbital they occupy - the same quantum mechanics that says "they must remain in orbit" does in fact allow for orbitals that, while strictly speaking having zero probability at r=0, have a very small probability at a radius comparable to the radius of the nucleus (which is itself not a point). I have always understood this to be the basis of beta capture decay in which a proton in a (larger) nucleus captures an electron, and turns into a neutron.

This is the mechanism by which Rb-83 turns to Kr-83, for example.

You can see some beautiful images of orbitals in an earlier answer given to this same question - and one of the comments states exactly the same thing: there is a non-zero probability of finding electrons in the nucleus.

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So firstly, it is not the strong nuclear force that keeps electrons in fixed orbits around the nucleus.

The strong nuclear force, that is, the Quantum Chromodyanmic interactions, hold the protons and neutrons together in the nucleus, as well as holding the quarks and/ or antiquarks together in other hadrons and mesons. These interactions do not come into play here.

It is the very nature of Quantum Mechanics which keeps the electrons in fixed orbits around the nucleus, and does not allow the electron to 'fall in' on top of the proton, in a Hydrogen atom say.

So firstly, we know from the Bohr model, that the energy levels of electrons orbiting atoms are discrete energy levels. That is, it is a quantum system.

In order for this to be the case, it turns out that the radius of the electron orbiting the proton must also be in descretised, namely in multiples of the Bohr radius

$$ r_0^2 = \frac{1}{4 \pi \varepsilon_0}\frac{\hbar^2}{e^2 m_e} $$

You can see a derivation of this done out here

This was a big problem in turn-of-the-century physics, "Why is the atom stable?", because Classical Electrodynamics would predict that indeed the electron would fall into the centre of the atom. This was part of what led to the creation of Quantum Mechanics. in the early 1900s.

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So if have just a proton and an electron next to each other, they won't fall onto each other either, will they? I mean there's no angular momentum in the Hamiltonian now, but I feel like a similar argument must hold. –  user24082 Apr 14 '14 at 22:54
Though actually, having just posted an answer, I somehow recall reading that indeed there is a non-zero probability of finding a 1s electron of the Hydrogen atom at the centre where the proton is. I think it may have been in Heisenberg - The Physical Principles of The Quantum Theory. I'll have to look it up when I get home... –  Flint72 Apr 14 '14 at 22:54
@Anthony well, you can't properly consider this Quantum Mechanical system without spin angular-momentum. The electron and proton are both spin-1/2 fermions, and a proper treatment of their dynamics requires taking this into account. –  Flint72 Apr 14 '14 at 22:56
I see. But the same argument holds for just one proton, and one electron? –  user24082 Apr 14 '14 at 22:58
Not orbiting, if I just release an electron and a proton from rest! –  user24082 Apr 14 '14 at 23:29

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