Please Register or Log in to view the hidden image! I have a number of concerns about this figure (ultraviolet catastrophe): 1) I know that higher frequency means higher energy. But does higher light intensity means higher energy? (is "light intensity" related to "energy" at all?) 2) The "expected curves" is such that the intensity keeps increases as wavelength decreases (i.e. the 2 curves labelled as classical theory). WHY is this the expected way? I don't understand. 3) The "actual curves" shows that as wavelength decreases, intensity doesn't increase indefinitely, but will eventually decreasse. And this suggests the energy of an atom is quantized. WHY? I mean, how does this fact contribute to the thinking that the energy is quantized? 4) The curve at the bottom is red and the one at the top is blue, is there a reason for doing it this way? (is it realted to temperature?) Can someone explain? Thanks a lot! Please Register or Log in to view the hidden image!
I am certainly no expert, but I think you are confusing intensity and energy (frequency). The intensity of light has to do with the number of photons... the enery of each individual photon is determined by its wavelength.
Kingwinner, With no reference to quantum mechanics I interpret the curves as indicating that the peak intensity is simply the wave length where the matter is best able to absorb energy. This ability tails off on both sides of the peak. Intensity is certainly related to energy but the curves could be misleading. The total volume of photons must certainly reflect the "total energy present". However, increasing wavelength as is assumed modernly means a decrease in frequency which is a measure of photons capacity to store individual levels of energy. The photo electric effect is impervious to intensity but is observed significantly affected by wavelength, meaning that the electrons boiling off the substance are only amplified when there is a wavelength match. If there were 100 electrons available to boil off and some wave length was determined to be the most efective length then one photon would boil off one electron. Multiplying the number of photons, increases the intensity as an increase the number of electrons measured in the output current, up to a maximum of 100 photons. After that excesses in photons would have no effect. When observing accelerated electrons there are at least two remarkable observations. Without more, the electron can reach only near the speed of light but the energy of the electron ia not confined by the velocity alone.Some of the accelerating energy that is not absorbed as velocity increases is absorbed and sometimes referred to as mass increases. However, as the electron increases velocity the ability of the electron to transfer all absorbed energy into velocity decreases. Instead the electron merely loads the excess as a burden of "false mass". We must, however, recognize that the accelerated electron is also radiating throughout the acceleration process. Apparently there is a preferred wavelength for the various segments of radiant energy with peaks that correlate, somewhat. Geistkiesel
The label "intensity" on the y axis of that graph is misleading. In fact, it is a graph of intensity per unit wavelenght vs wavelength. Higher temperature of the black body leads to a higher overall intensity. According to the Stefan-Boltzmann law, the total intensity is proportional to the fourth power of the black body temperature. That total intensity is equal to the area under the graph you have given. The classical arguments which lead to that conclusion are quite complex. They depend on thermodynamic considerations and electromagnetic ones. I can't really explain in detail in a short space. If you're really interested, I suggest you get a good advanced textbook which will take you through the arguments which lead to the classical prediction. According to thermodynamics, the available energy of the oscillators in equilibrium with the radiation is partitioned equally among the allowed quantised modes of oscillation. Modes with smaller wavelengths require higher energies to become excited. At very small wavelengths, there simply is not enough energy to excite the relevant modes (i.e. bump the atomic energies to higher levels). On the other hand, in the classical picture, without quantised energies, if you wait long enough you can excite any frequency oscillation you like. The blue one is hotter than the red one.
