why does light incident on a material heat up the material?
The electrons in the metal absorb the light energy (or some of it), and this energy is distributed among the electrons/atoms of the metal.
I can only give a partial answer. Essentially, all substances absorb some photons without re-emitting (refracting) them. The energy thus gained is expressed as motion, which on that scale is heat.
edit: Ha! I should have known the bearded wonder would beat me to it. Don't you ever rest?
Danger - do you mean subtances does gain heat energy when light is incident on them?
Yes. Whether or not it's measurable depends upon the energy of the light as well as the characteristics of the material. A mirror will absorb an infinitesimal amount from an LED, but I'm sure you'll agree that black vinyl car seats can soak up an uncomfortable amount from the sun.
Very interesting. What happens there exactly, I mean how is the photon's energy transferred to kinetic energy of atoms?
What about non-metals? What for example if I take my magnifying glass and burn dark paper with it?
Or what happens in the microwave oven, when water is heated up?
I thought that the atoms get a momentum kick when absorbing the photon. Although it's not clear to me what happens exactly when the atom gets a kick. In my imagination the electron gets a kick and the electron pulls the nucleus with it due to Coulomb force such that the whole atom is dragged. What do you think?
If the photon, like you say, is not reemitted again, what happens with the energy? Does the atom stay in an excited state? How is the photon's energy completely transferred to kinetic energy of the atom?
I'm afraid that I have to turn this over to an atomic physicist for that kind of detail; I don't know all that much about it myself. There are resonance frequencies involved (how microwave photons heat water), excitation/relaxation cycles, and whatnot. Eventually a fair bit of the light is reradiated as infrared. Sorry I can't be of more help.
Well, as for how the energy of a photon can be transformed into the kinetic energy of an atom, just think of the atom as a collection of charged particles and the photon as an oscillating EM wave. If the photon "hits" the atom and is absorbed by it, the atom's nucleus will then respond to the applied field by moving in such a way that it cancels out that field, effectively destroying the photon. At the same time it gains kinetic energy. Likewise, if an atom is forced to oscillate due to collisions with another one, a photon can be created.
As for what happens to the electrons in the excited state, you're right: they won't necessarily stay there. What happens is that some electrons will fall back down to lower energy states, emitting another photon. However, it is likely that another atom in the material (a non-excited one) will be able to absorb this photon, putting one of its electrons in an excited state. So, if you inject some photons into a material, the atoms of that material will basically continue to "trade" that energy back and forth, unless some of it manages to escape through the surface. This energy is accounted for when we talk about the heat of a system.
Great explanation, Manchot. Thanks for clearing that up.
My question is why we can consider the atoms as charged particles, since atoms are neutral. Or is it when you get close enough to the atom and you see the positive charge of the nucleus and the negative one of the electrons, i.e. you see an electric dipole moment.
Again, there's something missing here. When you ask how light heats up a MATERIAL, you asking about a solid object and not "isolated atoms". Please keep in mind that when atoms conglomerate into a solid, in many cases, their "individuality" is lost as far as the property of the material is concerned. This is because the valence shell of each of these atoms are modified, sometime severely, when they are in close proximity with other atoms in forming a solid. This is the origin of the continuos bands that we get in metals, insulators, and semiconductors.
At the SIMPLEST and naive case, one can then picture a solid as a chain of + ions with - charge in between, so you get something like
.... + - + - + - + - + .....
This is roughly a highly simplified 1D lattice chain of dipoles. One can then calculate the VIBRATIONAL mode of such a thing. This is where one gets the optical and acoustic modes for the different oscillation. It is the primitive idea of PHONONS.
Now, why am I wasting your time with this? This optical mode is what dictates the optical property of the solid, NOT the "electron transition/absorption" of the individual atom (at least not within the normal visible range). If the phonon mode is available, the EM radiation will be able to oscillate or vibrate the chain and the solid acquires vibrational energy, i.e. HEAT! This makes the material absorbs that particular frequency, causing it to not be completely transparent to that EM radiation, or even be opaque.
Refer to Kittel, or Ashcroft and Mermin texts for a more in-depth discussion on this. In any case, please remember that most of the property of the material you deal with, their properties that you are familiar with are NOT due to the property of the individual, isolate atom of that material. The property of a solid seldom depends on that. It is why Solid State Physics/Condensed matter physics is not the same as atomic/molecular physics. We could study the same based element, but we're studying them under very different characteristics and conditions.
Even better... thanks, ZZ.
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