Can Oscillating Charges Emit a Hydrogen-Like Spectrum Through Prism Dispersion?

[T.Roc]

hyrdo,

ok - photons(n energy) are created when an electron fall back to its' previous orbit after being bumped to a higher orbit by a photon of the same energy (n) - we are on the same page, and i didn't think you are an idiot.

the other lines in the spectrum (photons) are EACH created by subsequent changes in the possible orbits for the nucleus. every line you see is created by a different photon (frequency).

your idea seems to be that one oscillation is making all these photons? not so. if you want to look at this with wave forms, that is fine. the straight line between the emitter and receiver (90') is all that is required because for every angle above 90', there is a symmetrical angle below 90', and they cancel each other out.

even if this were not the case, the harmonics created by such a small range of frequency change is not enough to produce the wide range seen in the hydrogen spectrum (or others). ie. - you might get a "red" to "red-orange" change, but not orange to violet, and definitely not IR through visible and into UV.

TRoc

 

[Hydr0matic]

your idea seems to be that one oscillation is making all these photons?

Well that's the beauty of it .. By explaining discrete spectra as a result of atomic motion, you don't actually need to infer any energy levels at all. ... Of all the crazy statements you've ever heard, this takes the cake huh ? ... "We don't need atomic energy levels" .. wow, I even amaze my self.

Ironically, this isn't that far from our current model, is it? We started out with planetary orbits, then energy levels were introduced with fixed orbits. Now, there's a wavefunction for each electron describing the probability of finding it in a certain position.
Is it just me or is this another way of saying "- the electrons are somewhere within the atom" .. Were it not for the fact that each electron is associated with a certain energy level...

not so. if you want to look at this with wave forms, that is fine. the straight line between the emitter and receiver (90') is all that is required because for every angle above 90', there is a symmetrical angle below 90', and they cancel each other out.

Yes, I realize that. Of course, you need more than one oscillating atom to get a visible spectrum. Like, perhaps, a cloud of atoms oscillating in random directions.

even if this were not the case, the harmonics created by such a small range of frequency change

What makes you say it's small? The atoms are oscillating at very high speeds. And the amount of shifting is obviously very depent on the oscillator's speed.