cowboy_mortician said:
If opposites attract, why don't electrons crash into the nucleus. I've been told its because the electon is spinning so fast, it orbits, but I know this can't be so because then at absolute zero, the electron should then crash inward or even at a higher temperature. Or is it the "weak nuclear" forces? Please help!
As you have seen via all the responses to your question, this isn't a trivial thing to answer, at least it isn't trivial to answer if you want a simple picture or analogy. One can try explaining it by saying that an electron is a "wave" and that the smallest standing-wave configuration is the fundamental wavelength, very much like the Bohr-Sommerfeld quantization model. But this would be inaccurate also.
The second problem on why this is difficult to answer is because we have this picture of a well-defined "electron" zipping around in an atom, very much like a planetary model, etc. That perception in itself can be the root of the difficulty in trying to explain and answer this problem. When you ask why doesn't the electron "crash" or get arbitrary close to the nucleus, I could easily reply "But it does, but not in the way you think!" [I will explain this later]
H atom is one of the MOST well-known system in physics. In QM, it is one of the few exactly-solvable system (He atom is the other - the rest of the elements have not been solved exactly). When we ask how the H atom behaves, especially in terms of the dynamics of the electron within this atom, we have to first of all solve the Hamiltonian or the Schrodinger equation of the system. For most physics majors, this can be a painful exercise, because it is almost a requirement that one MUST know how to do this. When you solve this, you will find that the LOWEST possible state satisfying the Schrodinger equation with the appropriate H atom boundary conditions contains three quantum numbers n,l,m with values n=1, l=m=0. This state corresponds to the LOWEST energy possible given the H-atom configuration. There's nothing lower.
Now this derivation sort of answers your question on "why", i.e. it is because it is the lowest state possible with the given configuration and so, there's no other places for the electron occupy to get any closer. But unfortunately, it doesn't completely answers why. It merely says that based on QM, I can give you a description of the H atom that accurately match ALL of the experimental observations. QM doesn't allow us to go beyond that because such info are not imbedded within the QM description (or is this just a objectivist or solipsist view?).
However, the interesting thing here is that, knowing the solution for the lowest state, where is the electron? This would explain what I said earlier that the electron in the H atom does actually get very close to the nucleus, but not in ways you think. The "radial solution" of the H-atom, i.e. the part that describe the electron's position from the center of the atom, has a very interesting profile for the lowest state. The expectation value for |nlm>=|100> of the radial function actualy PEAKS at the origin,i.e. where the nucleus is. So the radial wavefunction seems to indicate that the highest probability of finding the electron is actually right at the nucleus! [Note: rR_nl is zero here].
The major caveat here that one needs to keep in mind is that this is based on a strictly non-classical view of what an electron is. The wavefuction does not describe a "well-defined" particle known as an electron, but rather it is describing the PROPERTIES of the electron that seem to be "smeared" in the space of the atom. It only based on this that we can accurately explain all the phonomena associated with the H atom.
If you wish to see the painful detail of such a derivation, you're welcome to look at the link below:
http://scienceworld.wolfram.com/physics/HydrogenAtom.html
It will give you an idea why I mentioned in the beginning that answering such a question isn't as trivial as it appears.
Zz.
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