Photoelectric effect question

No, it's independent of the intensity.

 

What about an intensity high enough to increase the temperature of the material? The work function is a function of temperature right? Of course i'm nitpicking a little bit but i'm interested :) Of course the material will have to absorb that wavelength.

Edit bad grammar!: I think the proof behind the photoelectric effect was the fact that by increasing the intensity of the of the light impinging on a material there was no change in the work function. Only by changing the wavelength and thus the energy of the light hitting the material changed whether or not electrons were knocked out of the material. Pretty sure that is the basis for the photoelectric affect and the discovery of quantized energy levels. I'm sure someone can correct me if i'm wrong.

 

Without the sufficient frequency (small enough wavelength) the photons will be just reflected and not absorbed. There are only two possibilities, one photon of sufficient frequency (hence sufficient energy opposing the work function) is absorbed to knock one electron off the metal, or the photon being knocked off (reflected from the metal).

Higher intensity means that there are more photons in the stream of light, but without the sufficient frequency, they are all going to be reflected, and not increasing the temperature of the metal because no energy is absorbed by the metal to be converted to heat energy.

(I have sufficient knowledge in photoelectric effect, but I'm not very advanced at thermodynamics, so my answer might be wrong).

 

But that's only if electrons are the only thing absorbing the photons. What about the lattice, ie phonons, absorbing the photons? Isn't that why the absorption of a material vs wavelength isn't a flat drop off at the work function? (which you'd expect if the photoelectric effect was the only thing happening).