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I had originally thought that electrons can be moved from a lower energy level to a higher energy level with an applied voltage. Thus, putting an applied voltage on a given material will allow a smaller light frequency to emit electrons, but this turns out to be false.

Where in my reasoning did I go wrong? Also, if this isn't true, how is it possible that not all emitted electrons have the same kinetic energy?

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  • $\begingroup$ what do you mean by "will allow a smaller light frequency to emit electrons"? Light does not emit electron... $\endgroup$ – ZeroTheHero Jan 27 at 15:35
  • $\begingroup$ @ZeroTheHero it's about photoelectric effect $\endgroup$ – Goldname Jan 28 at 23:20
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The voltage is able to create an electric field. This represents a change in potential over some distance ($V/m$). The effect of an electric field therefore depends not only on the strength, but on the distance over which it can act.

While the field might provide a force to charges (like free electrons), it will have almost no effect on an atom. The change in the field over the size of the atom is so small that the electron cloud orientation is not affected. So the field can't energize the individual atoms.

how is it possible that not all emitted electrons have the same kinetic energy?

The ejection is a messy process. Imagine having a device that can deliver a very precise "kick" to a ball. You place a ball on it and it will always kick the ball up exactly 2 meters. Now you take the device into a ball pit. Sometimes you get the maximum KE into a single ball and it goes up 2m into the air. Sometimes it goes a bit sideways and doesn't reach 2m. Sometimes the ball hits another and just shuffles some balls around.

What you can say is that you never see any ball go higher than 2m. The same thing into the photoelectric experiment. By turning up the voltage, it's like squashing down how high the balls can go. The idea is to find the maximum extent, not the proportion of ejections that reach that extent.

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  • $\begingroup$ Does this mean that those electrons with highest KE were the least "tightly bound"? $\endgroup$ – Goldname Jan 25 at 19:09
  • $\begingroup$ At some level, that statement is correct. But the variation doesn't tell you that others were more tightly bound. $\endgroup$ – BowlOfRed Jan 25 at 19:22
  • $\begingroup$ Your first two paragraphs confuse me. It seems like you are saying that because atoms are small they can't experience large fields. $\endgroup$ – Aaron Stevens Jan 26 at 1:49
  • $\begingroup$ I'm saying that to ionize an atom directly with a field, the strength would have to be enormous. The fields used in the photoelectric apparatus are inadequate to do so. $\endgroup$ – BowlOfRed Jan 26 at 3:30
  • $\begingroup$ Yes I agree with that. But that's not because electric fields have units of $V/m$ and that they are small. Also the $V/m$ unit doesn't mean the strength of the field depends on the distance over which it acts. It just means the strength of the field depends on how the potential varies over some distance. $\endgroup$ – Aaron Stevens Jan 26 at 12:35
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Electrons and atoms belong to the quantum mechanical framework. Voltage is a collective manifestation of an enormous number of electrons and belongs to the classical electromagnetic theory.

The band theory of solids is an intermediate, quantum mechanical model , that can describe the collective behavior of electrons in solids.

For special situations atoms may be ionized , emitting an electron, see the answer to a relevant question here , but this happens when accelerated free electrons (or semi free in the conduction band of metals) can transfer energy to an electron in an orbital by scattering, and eject it.

The photoelectric effect does not involve electron-electron scattering but photon electron scattering, the photon transferring energy to the ejected electron. The introduction of a macroscopic voltage cannot change the atomic structure and consequently the band structure of the energy levels of the electrons at the surface.

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