Re: Why is Energy Quantized?
- From: Bjoern Feuerbacher <feuerbac@xxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Wed, 04 May 2005 17:09:00 +0200
ica wrote:
Bjoern Feuerbacher wrote:
[snip]
3. What atomic principle prevented Ultraviolet Catastrophe... or is it independent of the atom?
It is indeed independent of the atom. It is prevented by the quantization of energy in electromagnetic waves: you can only generate a wave with at least one photon in it. But the higher the frequency of the wave, the more energy you need to create such a photon ( E = h f ). Hence the higher the frequency, the lower the probability that such a wave is generated.
Bye, Bjoern
How can it be independent of the atom.
I just explained it.
It needs the atom to produce the photon to produce the electromagnetic wave.
Not necessarily. Already existing photons can also scatter off each other and thereby change their energies. So you can get a wave with high energy plus a wave with low energy out of two waves of medium energy, in a sense.
Additionally, if there are charged particles (ions or electrons) around, inverse Compton scattering can happen, which also increases the energy of a photon.
You said the higher is the frequency of the wave, the more enegy must the photon have.
Yes. E=hf.
Since the photons came from the electron in the orbitals.
Again, not necessarily.
How come the atoms can't produce high energy electrons or high frequency wave (or photon). Isn't it the frequency of the photon is the difference between certain energy levels of the electrons?
Yes. And that difference isn't *that* big. At most 100 keV, I would say.
Such photons are indeed emitted now and then - but if the temperature
of the source (i.e. the velocity of the ions etc.) isn't high enough,
such photons will soon undergo Compton scattering and loose most of their energy.
In Iron for example. Know what is the frequency of the higher level of the atom and the lowest level.
Don't know.
Can't the difference be equal to ultraviolet but only topping in the blue.
Yes, the difference is probably in the UV range. So what?
I begin to suspect that you don't know what "ultraviolet catastrophe" actually means!
When iron produce red light. Which of the 2 orbital differences produced red.
Don't know.
If you are talking about blackbody radiation, the initial photons probably do not correspond to red color. Only after a lot of scattering events, when the system is in thermal equilibrium, you'll have mainly "red" photons.
But you said other colors are also produced. How come it peak at red.
That's for a given temperature simply the wavelength/frequency
where the probability to have a photon of that frequency is highest.
Thermal equilibrium. Planck curve. Do the math. There is nothing mysterious or special about the color red.
Bye, Bjoern .
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