If the energy spectrum is continuous, a blackbody would radiate shorter wavelengths with higher intensity with no upper limit (the "ultraviolette catastrophe") for any temperature.

How can the correct behaviour of a blackbody be explained by quantization of energy? I know the emittance from the blackbody occurs due to recieve of energy to the molecules, and these molecules emit waves which can be in any mode.

  1. I have read that according to none-quantum theory, each mode of such a standing wave is formed with equal probability. Is that correct (according to none-quantum theory, then)?
  2. Does each mode of such a standing wave contribute to the intensity of a certain wavelength?
  3. If the energy was not quantized, would the molecules "collect" energy (i.e. heat from the surroundings) until it has enough to relase another wave with a higher mode (and this go on and on leading to the "UV catastrophe")?

1 Answer 1


The spectrum is continuous, but the energy (for each frequency of the spectrum) is emitted in discrete chunks. That is the energy stored in each mode is $n\hbar\omega$ rather than $\propto |E|^2$, which gives very different results when substituted into the Boltzmann distribution. The rest is math.

  • $\begingroup$ Energy difference is $1/e$ between two modes. Actually there is no energy levels, it is the power which we calculated and power is constant for all modes. $\endgroup$ Dec 31, 2022 at 13:08
  • $\begingroup$ @NeilLibertine what you say is incorrect. $\endgroup$
    – Roger V.
    Dec 31, 2022 at 13:16
  • $\begingroup$ Use Baye's theorem for probability, so $nE_n=mE_m$. So probability of number or particles at lower energy is higher and vice-versa. $\endgroup$ Dec 31, 2022 at 14:39

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