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From what I understand, ultraviolet catastrophe means that at that time scientists believed that for energy to be higher the frequency should be higher. But frequency is higher in case of ultraviolet light but ultraviolet light has less heat.

Does this mean that even though ultraviolet light is having greater frequency it has less energy?

Also since violet color has most energy ultraviolet light has less energy than violet even though ultraviolet is more hotter than normal heat?

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  • $\begingroup$ Re, "ultraviolet light has less heat": Ten Watts of ultraviolet light has exactly as much "heat" as ten Watts of any other wavelength of electromagnetic radiation. When they say that ultraviolet light has "more energy" than longer wavelengths of light, they mean that each individual photon carries more energy. From those two facts, you can deduce that a ten Watt beam of ultraviolet light does not deliver as many photons per unit time as a ten Watt beam of infrared light. $\endgroup$ May 12, 2017 at 13:42
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    $\begingroup$ Re, "violet color has most energy." There is no "most energy," and "violet" and "ultra violet" are just arbitrary names. The electromagnetic spectrum is a continuum in which "violet" is the name for the highest energies/shortest wavelengths that are visible to the human eye. "Ultra violet" is the name for a very broad range of higher energies/shorter wavelengths that are not visible. And, "ultra violet" transitions into "X-rays" which are even more energetic/shorter wavelength. $\endgroup$ May 12, 2017 at 13:51
  • $\begingroup$ @SolomonSlow when you say "Ten Watts (J /s) of ultraviolet light has exactly as much "heat" as ten Watts of any other wavelength of electromagnetic radiation", was this a real-life observation? $\endgroup$
    – Lawrence
    Nov 5, 2022 at 2:15
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    $\begingroup$ @Meemaw, Do you mean, did I perform an experiment? No. I did not. I was commenting on the meaning of the word "Watt." Watts are the unit in which we measure the rate of energy transfer. "Ten Watts" specifies an amount of energy per unit time, and "heat" simply is a synonym for "energy" in the language of thermodynamics. "Ten Watts" means a certain amount of heat per second, and it means the same amount of heat per second no matter what mechanism or wavelength of light is transporting the heat. $\endgroup$ Nov 5, 2022 at 14:32
  • $\begingroup$ @SolomonSlow got it. $\endgroup$
    – Lawrence
    Nov 6, 2022 at 22:18

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The idea is related to thermal equilibrium of a black body. The classical theory predicts that this type of bodyy will emit light on all frequencies, with energy density increasing with frequency $B_\nu \sim\nu^2$.

And you can actually see this is not case (hence the catastrophe part), a furnace is a good example of a black body, the classical model will conclude that there'll be light with an arbitrarily high energy density coming out of the furnace, which is not observed in reality

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From what I understand, ultraviolet catstrophe means that at that time scientists believed that for energy to be higher the frequency should be higher. But frequency is higher in case of ultraviolet light but ultraviolet light has less heat.

This is a wrong understanding. More energy means also utimately more heat, once the electromagnetic frequencies are absorbed.

This the catatastropy:

black body

As the other answer explains, in classical electromagnetic theory the radiation (radiated power density) from a heated body could not be fitted. It resulted with a climbing frequency spectrum,( ~1/wavelength) whereas the quantized formula fitted the data with the Planck radiation formula. It was called a catastrophy because classically it went to a singularity, very large frequencies.

Does this mean that even though ultraviolet light is having greater frequency it has less energy?

No , the energy of a photon is h*nu, as discovered by Planck.

Also since violet color has most energy ultraviolet light has less energy than violet even though ultraviolet is more hotter than normal heat?

No .

Ultra violet, as the name says , is beyond violet in frequency, thus with higher energy.

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I think I have the answer.

An ideal black body at thermal equilibrium emits radiation in all frequency ranges. It emits more energy as the frequency increases. The higher the frequency the higher the energy. Since ultraviolet light has higher frequency it has more energy. By calculating the total amount of radiated energy, it can be shown that a blackbody would release an infinite amount of energy. The maximum wavelength emitted by a black body radiator is infinite.

Lets take an example of furnace here heat is provided to the furnace which is a blackbody, which means it absorbs all the radiation and reflects none. Now when we calculate the heat emitted we can say that the amount of heat emitted in the furnace is infinite. So since energy is infinite it violates the law of conservation of energy since input heat was not infinite but output emission was infinite. This is ultraviolet catastrophe.

I am politely asking whether my reasoning is correct?

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  • $\begingroup$ you have to qualify that this is what classical electromagnetism would predict. The data did not fit this Rayleigh Jeans law you are describing, but a curve that Planck's quantized contributions of frequency h*nu fit very well. $\endgroup$
    – anna v
    May 12, 2017 at 16:36
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Another way of explaining is this: The predictions suggested that an ideal black body at thermal equilibrium would emit energy proportional to the frequency squared. Now since an object if it is more hotter, it has more energy. since ultraviolet light has higher frequency it should be hotter(more energy) and an object burning with high heat should emit light at the ultraviolet end of the spectrum. This does not happen so this is ultraviolet catastrophe. Am I right?

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