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1

Electrons and nuclei both have spin. A spinning charged particle has a magnetic dipole moment. When a magnetic dipole is in a magnetic field, it experiences a force. This oversimplified description gives some brief intuition on the origin of the fine and hyperfine splittings. Fine structure is due to the interaction of the electron's spin with the ...


1

The most probable radius is found by maximising probability per unit radius, whereas $|\psi^2|$ gives us probability per unit volume. To find the conversion factor from one to the other, we need to ask how much volume is there per unit radius near a radius $r$? The answer to this is $4 \pi r^2$.


4

The short story is that the Hyperphysics article you link to is using classical and semiclassical heuristics to justify the numbers it present. As I'm sure you're aware, the hydrogen atom cannot be described in any rigorous detail using classical mechanics, and instead requires quantum mechanics for any appropriate treatment, particularly where the ...


3

In a comment elsewhere you write that you're interested in understanding how quantum-mechanical theory describes the radiation that a hydrogen atom does and does not emit. In your question you ask about another answer that suggests some significance to the electron having zero total momentum; I think that's a feature of the coordinate system choice rather ...


-2

The answers posted so far repeat the common fallacy that Maxwell's Equations do not apply to the hydrogen atom. They may not work for the Bohr atom, but they certainly explain everything the hydrogen atom does in terms of its emission and absorption of radiation. In the Schroedinger equation there is a charge density, and for the eigenfunctions of the ...


15

The existence of hydrogen atoms is enough to demonstrate that the electrons don't emit radiation. If they did, that energy would have to come from somewhere. The only place it could come from would be a reduction of orbital radius until the electron finally reaches the nucleus. If you accept that electrodynamics applies, then you have to accept that atoms ...


4

In addition to the answers already given, which answer the question pretty-well, I'll say that, historically, this exact question was the one which puzzled Niels Bohr enough to inspire him to advance his famous theoretical-explanation for the several observed frequencies of the radiations emitted from hydrogen-atoms ... in general, the fact that electrons in ...


9

Because of its wave nature, the electron in its ground state is actually smeared symmetrically about the proton (ignoring spin-spin effects), and spherically symmetric charge distributions do not radiate (there's no special direction). Accelerated charges do not always radiate em radiation. See also How to find the magnetic field due to a revolving electron ...


29

You have your "prove" in the wrong place. The way to prove that ground-state electrons in hydrogen atoms don't emit radiation is the following: Construct a sample of ground-state neutral hydrogen atoms. Place this sample near a detector which is sensitive to the sort of EM radiation you expect. Die of old age waiting for a signal, because ground-state ...


8

I believe some of the answer in the links are correct, others are less obvious and might even be confusing. I am not gonna repeat the arguments there, but to stress the following idea. You cannot demonstrate that using classical electrodynamics. The theory as is does not apply to quantum objects and thus it was modified. The equations are the same, they are ...



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