The photoelectric effect says that the energy of a photon, if greater than the work function of a metal will liberate an electron from it. The additional energy of an electron just gets converted into the kinetic energy of the electron.
Questions:
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Does this mean that a metal undergoing the photoelectric effect doesn't heat up due to it?
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One thing I notice is that when I place a slab of metal under the sun, it heats up. I think since the energy of the sunlight is lower than the work function of the metal, it just gets converted into heat? But then you also have radiation pressure. So does the energy get converted into the kinetic energy of the slab? (If the slab is placed in a vacuum and no other forces are acting on it). But if so, what is happening?
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To sum it all up, when a beam of light hits a slab of metal, where is the energy going? Does it heat up the metal? What about radiation pressure, does the energy get converted into the kinetic energy of the slab?
Best Answer
There's a continuum of effects happening. In some cases almost all the energy of the incoming photon is given to the KE of the electron, but that does not occur every time.
No. It means that in some (relatively small) fraction of interactions there is almost no energy transferred to the bulk metal, because the electron leaves with it. But other photons that arrive will leave most of the energy in the metal. Some arrivals will not eject an electron at all. The sum of these reactions result in heating the metal.
So some photons that arrive do not heat the metal, but others will.
Sunlight has photons with a spectrum of energies. You're correct that the low-energy photons will not drive the photoelectric effect. But not all of them contribute to heating. Some of them will simply be reflected.
Even in perfect conditions, almost no energy goes into the KE of the slab. Most will go into heating or reflected energy.