# What happens to the absorbed light energy?

When light comes across with a solid material, some of it is reflected, some of it passes through and some of it is absorbed. I understand the reflection and passing through, but I don't understand what happens when light is absorbed.

Suppose that there is an enclosed box. Its material is thick enough so that no light can escape out. We put a Led diode in it whose internal resistance is zero (there is no loss as heat energy, all energy dissipated is given out as light). What are the possible steady state scenarios that can happen?

Why and how some wavelengths of sun light heats up black objects while some other wave lengths do not?

Why and how some wavelengths of light pulls the outer most electrons of atoms to higher energy levels while other wavelengths have no such effect? (I understand that energy of a photon must be enough to take electron to higher energy level, but why doesn't it take the electron temporarily a littler higher in the current energy level and then make it return back? I understand that an electron cannot stay between two energy levels, because those are unstable regions; but why doesn't electrons stay there even if it is as short as an instant?)

Certainly vinas is correct. The absorbed energy is converted to heat energy.

The scenario you mention with the LED is very close to the blackbody problem known as the "ultraviolet catastrophe." There is a Wikipedia article about it here. What happens in the situation you described is that the light proof box gets hotter. It will increase in heat until the heat leaving the box due to conduction, convection, and radiative effects balance the energy emitted by the LED. Given the typical power required by an LED, the temperature gain would be minimal.

For all practical purposes, all wavelengths of sunlight will heat an object of any color. The amount of heat produced by light incident upon an object depends on the material properties. Some wavelengths will be reflected, some absorbed, and some transmitted.

As you point out, the associated energy of the photon is important in determining what sort of interaction will take place. A photon of a certain threshhold energy will be able to free a bound electron from an atom causing the photoelectric effect. A photon of lesser energy can still free less strongly bound electrons causing the Compton effect. In this case, the photon continues on with less energy. If the energy is high enough, a photon can cause pair production and create an electron and a positron.

The case you are wondering about, where a photon transfers energy but does not release an electron do occur. Depending on whether the photon interacts with the electrons or the entire atom the process is called Thompson or Raleigh scattering respectivly. In the first case, the electron simply returns to the lower energy level and re-emits a photon of the same energy; the result is effectively an elastic scattering event. In the second case, the atom re-emits a photon of slightly less energy in essentially the original direction. The drop in energy of the photons is due to transfer of momentum because of recoil effects.

• While it is not strictly true, it is a good rule of thumb that photons interact with objects of dimension similar to their wavelength. Because of this, it is unlikely that a photon with energy too low to raise an electron's energy level to a partial state would interact with the electron in the first place. Hence the Raleigh scattering. Commented Aug 30, 2011 at 13:32
• Does that mean, if two exactly similar objects( made from same material, shape and size) are kept together in same environment, with only difference that one is painted white, and another black. The black object would be warmer than the white object since black absorbs all energy and white reflects all? Commented Sep 3, 2015 at 4:59

it usually gets converted to heat... unless it bounces off or goes right through.

That's why black things get hotter than white things.