# Blackbody radiation and the quantization of energy?

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")?

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.
• 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. Dec 31, 2022 at 13:08
• 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. Dec 31, 2022 at 14:39