>>13398694>Photons hit the wall and disappear. What happened to them?If a photon hits a bound electron, the electron will absorb its energy and jump to a higher energy level. This will happen with the highest probability if the wavelength of a photon is very close to one of the absorption lines of the material. Wall is a type of material that has a very dense forest of absorption lines in the visible spectrum and since absorption lines have a natural width + the thermal energy of the wall increases their width even further (through Doppler broadening) thus there is a very high chance a visible spectrum photon is going to become absorbed. After a delay, an electron can then drop to its original level (emitting the photon of the same wavelength), or it can cascade down (emitting several photons of different wavelengths), or it can drop an energy level non-radiatively and transfer its energy to vibrational energy of the molecule (contributing to the heat of the wall) without emitting a photon at all. Those photons that were radiated back are likely going to get absorbed by a different molecule which means all photons are eventually going to be converted to wall's thermal energy. A more energetic photon could kick the electron out and make a free electron (the photoelectric effect).
If a photon hits a free electron it cannot absorb it and can only scatter it. It can scatter it elastically (through Thompson or Rayleigh scattering) or it can scatter inelastically (through Compton scattering). Elastic scattering will just deflect the photon away (eventually, it will find an atom or a molecule and get absorbed depending on its wavelength) while inelastic scattering will take away photon's energy which will change its wavelength which can make it more likely or less likely of getting absorbed.
The resulting effect is that the energy that visible photons had are converted into thermal energy of the wall and get radiated back into the environment as black body radiation.
