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I was not able to recall my memories, so:

What is the formula that states the frequency of electrons revolving around nucleus is equal to the frequency of light (or photon) emitted (or radiated)?

(I am of course talking of Maxwell's theory; in reality, we know that this is not true.)

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    $\begingroup$ It is true for Rydberg atoms, and close enough to true in quantum systems to allow Bohr to find the spectrum. There is no formula for this--- it is just obvious that a time varying source with a given frequency produces light with the same period. $\endgroup$
    – Ron Maimon
    Jun 26, 2012 at 3:34

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The rule is that if you have a classical source with frequency f, the outgoing radiation is a superposition of frequencies f,2f,3f, etc, according to the Fourier decomposition of the source frequency.

If you write Maxwell's equation in Lorentz gauge:

$$ \partial_\mu \partial_\mu A = J $$

And assume that J is periodic, by Fourier tranform:

$$ k^2 A(k,\omega) = J(k,\omega) $$

So that the fourier transform of A is supported on the same frequencies as J. This is obvious--- a periodic source gives rise to a periodic wave with the same period.

The rule is correct quantum mechanically too in the correspondence limit: the emission between level n and n-k is at a frequency which is k times the inverse classical orbital frequency at level n. You can see how Borh used this to derive the quantization rule here: Bohr Model of the Hydrogen Atom - Energy Levels of the Hydrogen Atom .

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This answer is triggered by Ron Maimon's comment: the Rydberg formula gives the energy (and hence frequency) of light emitted as a result of transitions between electron energy levels of a hydrogen-like Bohr atom.

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    $\begingroup$ It's true, but it's really the other way around--- the frequency is equal to the orbital frequency because it has to be so classically. This is how Bohr derived the spectrum condition. $\endgroup$
    – Ron Maimon
    Jun 26, 2012 at 5:10
  • $\begingroup$ @RonMaimon, I think it's a transition between electron energy levels that results in emission of a photon (or conversely absorption of a photon that causes a transition between electron energy levels), with the energy of the radiation equal to the difference between the levels, not the absolute energy of a particular level. Ref the Rydberg formula section of the Wikipedia article on the Bohr model. I guess a limiting case would be complete ejection of an electron, where the photon energy would in fact be the electron energy level. Are you saying something different? $\endgroup$
    – Art Brown
    Jun 26, 2012 at 5:47
  • $\begingroup$ All I was saying is that the frequency of the emitted photon is equal to the classical orbit frequency if the orbit is at large N so that it is semiclassical. I also was saying that historically, this was what Bohr used to derive the level-spacing formula, so that in the historical development, the fact that the emitted photon frequency is equal to the orbital frequency is prior to the quantum mechanics. Remember that the classical orbital frequency is not the quantum frequency (the energy), it is the difference in energy between "adjacent" levels. $\endgroup$
    – Ron Maimon
    Jun 26, 2012 at 6:05
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    $\begingroup$ @RonMaimon: As I review the Rydberg formula write-up, I am reminded that the formula was originally wholly empirical and not explainable by classical physics, which predicted atoms were unstable. Bohr's triumph was a new model (an ad hoc first step towards a quantum mechanical model) which correctly calculated the Rydberg constant. I'm sure you're aware of all this; I'm just having trouble fitting your comments into this picture. Basically, my impression is that classical physics was helpless in this regard. $\endgroup$
    – Art Brown
    Jun 26, 2012 at 6:23
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    $\begingroup$ @RonMaimon: Thanks, I'll take a look. Sorry for the bother... $\endgroup$
    – Art Brown
    Jun 27, 2012 at 6:24

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