Here is an extract from my Physics textbook. According to classical physics, each of these standing waves carries energy, and as their frequency increases, so does their energy . I don't understand how this is true . I think that the classical wave theory says that the energy is proportional to the intensity while on the other hand it is the Quantum theory which predicts : E = [itex]h[/itex] f So, is there an error in the book regarding this issue ?
See http://www.egglescliffe.org.uk/physics/astronomy/blackbody/bbody.html on how the standing wave analysis of standing waves in a cavity led to an incorrect result of the radiation emmitted by a blackbody. Yes, in classical physics way back then it was known that the energy of an electromagnetic wave increases with frequency. http://galileo.phys.virginia.edu/classes/252/black_body_radiation.html gives some more history on the subject.
I don't think you get my point. I want to know how is it possible that classical physics which was formulated 400 years ago predicted what is now known to be true . We now know that the energy of a photon of E.M wave is equal to E = h f In other words it is dependent upon frequency. From classical physics, i think we all know that for a wave, its energy is propotional to its intensity. What exactly did classical physics say ? (frequency or intensity)
It is not the energy of radiation which increases with frequency. It is the photon energy - the step size of possible energies. This quantization is not classical - you do not get it in classical physics.
I read this link http://www.egglescliffe.org.uk/physi...ody/bbody.html I don't really understand how the introduction of quantized (special and discrete) frequency of oscillation of electrons solved the Ultra Violet Catastrophe According to classical physics, the energy emitted by the black body would increase to infinity as the wavelength decreases. (how is this problem solved by quantized energy levels) Also, I am facing another problem : If the wavelength decreases ie frequency increases(to infinity); according to quantum theory, wouldn't the energy of each photon tend to infinity as well ? and hence give the exact same result as the classical theory ?
It's not either-or. A classical wave has energy proportional to its frequency and also proportional to the square of its amplitude (sometimes called intensity).
I agree with Vanadium. In particular, for the example of standing waves on a string, the frequency is quantised, so it is highly analogous to quantum physics. But what they didn't know back then was that light is only emitted in quanta. That is a more recent development (along with the rest of quantum mechanics).
@hms.tech: In classical physics, radiation is not quantized. A black body could emit x-rays (or any other frequency) without the requirement of a minimal amount of energy. As every frequency should receive the same intensity and the frequency range is not limited, the total radiation should be infinite. Quantum mechanics solves this issue with the photon energy: The minimal energy required to emit radiation of a specific frequency. For x-rays, that energy threshold is so high that materials at room temperature do not reach it (or so extremely rare that it does not matter), and the used frequency range is finite. It does, but the number of photons drops quicker.
Yeah, the way I think of it, there is a bunch of molecules flying around, also with rotational degrees of freedom. And for a molecule to go from one state to the other, it can absorb or emit a photon. So to get a high frequency photon, you would need a molecule to go from a very high energy state to a much lower energy state. So this requires a molecule in a high energy state, but this doesn't happen very often, therefore high energy photons do not happen very often. Of course, this is a very qualitative, hand-wavey description, but I think it is roughly a good idea of what is going on.
According to Wikipedia : (on Ultraviolet Catastrophe) Max Planck solved the problem by postulating that electromagnetic energy did not follow the classical description, but could only be emitted in discrete packets of energy proportional to the frequency, as given by Planck's law. This has the effect of reducing the number of possible excited modes with a given frequency in the cavity described above, and thus the average energy at those frequencies Can someone explain the bolded part of the extract from this article
The probability of a very high frequency mode with very high energy (according to quantum theory) getting a big fraction of the available energy, at the expense of the huge number of less greedy modes, is very low. Hence the non-occurrence of the u-v catastrophe. [How hand-waving can you get?] The word you want is 'emboldened', by the way!