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I've only a general grasp on how all this works, so it could be I'm asking this poorly or misunderstanding what happens. With that said:

The energy of EM radiation is a function of its frequency. With enough energy (a high enough frequency) I understand it's possible to knock an electron out of its orbit(al).

Let's say I had a neutral hunk of iron, and I want to try to knock out as many 3d6 electrons from it as I can. Beyond simply having a frequency with enough energy, would there be specific frequencies which were best able to knock out those electrons?

Or, asked differently, does anyone have a graph of the initial current produced when different frequencies of light hit some substance? If so, does it have local maxima and minima?

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The problem with using a chunk of iron is that in a metal the sharp orbital energies that you get in isolated atoms spread out to form energy bands. Typically when you shine light on an iron surface nothing will happen until the energy exceeds 4.5eV (275nm so that's in the near uv), at which point it will eject photoelectrons. Increasing the energy of the light will simply increase the kinetic energy of the emitted electrons. You'd have to get to soft X-ray energies before you see any sharp features. With X-rays you'll see absorption edges.

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  • $\begingroup$ Do you know if there would be specific frequencies of light which maximize energy in vs energy out? Or is the energy of the ejected electrons exactly equal to that of the incoming light? Maybe another way to say that would be: Are there energy losses involved that can be minimized, and does the frequency have an effect on those losses? (Not sure where that lost energy would be redirected. Maybe just left in the photon?) Also, I'd totally up-vote your answer, but I don't have the reputation to. OTL $\endgroup$
    – Casey
    Nov 4 '12 at 15:56
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    $\begingroup$ It takes an energy of 4.5eV to remove an electron from solid iron. If you shine a light of exactly 4.5eV (275nm) then the ejected electrons are essentially stationary. If you use light with an energy of 4.6eV then the photoelectrons have an energy of 0.1eV and so on. The energy of the photoelectron is always the light energy less 4.5eV. The energy goes into creating an electric field around the now positively charged iron surface. $\endgroup$ Nov 4 '12 at 16:02
  • $\begingroup$ Interesting. Does this imply that the all the light reflected in this example would revert to having a wavelength of 275nm (energy of 4.5eV)? Certain materials hit with ultraviolet and smaller wavelength radiation would glow then, wouldn't they? $\endgroup$
    – Casey
    Nov 4 '12 at 17:10
  • $\begingroup$ Wait, I'm not considering the energy of the bond that was broken. I think. Is all the photon's energy transferred? If so what happens to the photon? $\endgroup$
    – Casey
    Nov 4 '12 at 17:23
  • $\begingroup$ Light with a wavelength of more than 275nm is simply reflected unchanged. At wavelengths < 275nm some light doesn't produce photoelectrons and is reflected unchanged. The light that does produce photoelectrons is totally absorbed. The end result is that there is no change in the wavelength of the light reflected from the iron. $\endgroup$ Nov 4 '12 at 19:50

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