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Question about the Photoelectric Effect

  1. Oct 16, 2007 #1
    I'm not sure if I'm posting this in the right forum, so I apologize in advance if I'm not. I just have a question about the photoelectric effect. As I understand it, the photoelectric effect is where light/photons can knock electrons off a metal surface. I think that the intensity of the light gives the electrons more energy and the light frequency can knock more electrons loose. Anyway, my question is this: where do those electrons come from? I understand that most metal elements give up electrons easily, but if electrons are coming off the nucleus of, say, an atom of copper, doesn't this mean that the entire chemical structure of the copper is being changed?????? Could someone please resolve this question????
  2. jcsd
  3. Oct 16, 2007 #2
    In metals valence electrons of atoms are not strongly bound to their nuclei. They can freely move between nuclei. This is why metals have exceptionally good electric and heat conductivity. The bonding between atoms is provided by this diffuse electron "jelly". So, if you knock a few electrons out of the metal, it wouldn't change its bonding structure at all. In this respect metals are quite different from dielectrics where valence electrons are well-localized on bonds, and removing even a single electron may lead to a permanent damage in the crystal lattice (defect).

  4. Oct 16, 2007 #3
    But why? It just seems to me that the number of electrons would have to remain the same on each atom regardless of whether or not the electrons are moving between different atoms in order to maintain the identity of the element. Does this mean that some of the atoms of the metal will not have the same amount of valence electrons than if if was a single atom of the metal by itself?

    Okay, wait a second, I think I'm getting this now, I forget about the fact that protons define the identity of the element, not electrons! Even though valence electrons are missing (or there are more valence electrons than usual) in ions, they still maintain their identity as a certain element because the amount of protons is what really defines the element.

    Okay here's how I understand it now: electrons in a metal are darting chaotically between each atom of metal, causing the metal give up electrons easily, but still keeping it bonded together at the same time by a nasty jumble of electrons. A photon comes along with the required amount of energy to knock the electron loose. Since there are still many electrons in that disgustingly chaotic sea of electrons, the atoms of metal remain bonded while still maintaining their elemental structure.

    Am I right in saying that? But still would this mean that their are lots of positively charged ions in metal? If so, would this change the properties of the metal at all??????
  5. Oct 16, 2007 #4


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    There are a couple of mistakes in your description of the photoelectric effect.

    1. the "intensity" only affects the NUMBER of electrons, not the overall energy distribution. If you increase the intensity, the highest energy electrons do not change. You just get more electrons.

    2. "light frequency" is the one responsible for getting electrons with different energy.

    The electrons in a typical photoelectric effect come from the conduction band of the metal. These are the same electrons that are responsible for electrical conduction. None of these come from the "atom" - you only do that if you use energetic light such as x-ray and do core-level photoemission.

  6. Oct 16, 2007 #5
    Yes, that's about right.

    I don't quite understand this question. Yes there are positive ions in metal (nucleus + core electrons). They are immobile. When few valence (or conduction band) electrons are removed from the metal, the change in the average density of the electron sea is negligible. So, there is no visible change in the forces that hold positive ions together. Nothing happens to the metal, except it acquires some positive charge.

  7. Oct 16, 2007 #6
    Thanks for correcting me on that Zapper, I did get that a little mixed up didn't I?
  8. Oct 16, 2007 #7
    Thank you Eugene, I think I'm understanding this now. In response to the comment about the question that you didn't understand, what I meant was this: if light is constantly knocking electrons off the metal, wouldn't that mean that somewhere in the metal, there are atoms that are missing valence electrons because of this effect, therefore making them positively charged ions?
  9. Oct 16, 2007 #8


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    You are forgetting that when we do a photoelectric experiment, the cathode (metal) is grounded.

    If not, you'll get a charging effect that will raise the work function as the metal becomes positively charged and after a short while, no more electrons will be given off with that frequency of light.

  10. Oct 16, 2007 #9


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    Well, yes and no. Ions in metals do not have their "own" valence electrons. A metal is essentially a more or less immobile (negleting temperature for the moment) lattice of ions
    surrounded by a "sea" of electrons that move around more or less freely , i.e. they are not bound to specific sites. This is why metals are good conductors. It is also why you can understand many basic properties of metals by considering the electrons (i.e. the "sea") alone,
  11. Oct 17, 2007 #10
    Okay, so the valence electrons are not bonded to any particular place, and when light knocks them out, it doesn't really do anything to the structure of the metal except give it a positive charge since the negatively charged electrons aren't in the metal anymore?
    And it doesn't really matter if there are less electrons than before?
  12. Oct 17, 2007 #11


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    Oy vey!

    Don't you know what "grounded" means?

  13. Oct 17, 2007 #12


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    Yes, provided that the conductor is insulated so that replacement electrons cannot flow in. In this case, it becomes more and more difficult to eject further electrons because of the electrostatic attraction between the remaining electrons and the positively-charged conductor.

    If the conductor is grounded (some parts of the English-speaking world say "earthed" instead :wink:), then new electrons flow in to replace the ones that are ejected.
  14. Oct 17, 2007 #13
    jtbell, thanks so much, I think you just answered my question.

    Okay, so if the conductor is grounded, electrons come off due to the photoelectric effect, but other electrons come in to take their place? I'm not sure I really understand in what context you guys are using the term 'grounded'. Do you mean that the metal is simply uninsulated or that there is literally another object touching it? Sorry if I'm being difficult.
  15. Oct 17, 2007 #14
    That's correct.

  16. Oct 18, 2007 #15


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    By the photoelectric effect a space charge ("cloud of electrons") builds up near the conductor. If you turned off the "light" the charges would get pulled back into the conductor.

    What you do instead in the experiment is apply an external electric field (a voltage source) between your conductor (minus) and some other plate (plus). This field pulls off the space charge to the positive pole of your voltage source. The voltage source (acting as a "pump") pushes electrons from its negative pole into your conductor again, so the conductor remains neutral. The current you get in this circuit depends on the rate of space charge that is generated by the light quanta.

    Generally, connecting a conductor to some specified pole of your voltage source is referred to as grounding. This comes from electrical engineering where the current "back to the power plant" goes through the earth, so there need only be one cable to the house where you live (not two => more economical). In the PE experiment you need not actually connect one pole to the earth (although this might be necessary to reduce disturbances, don't know).

    Edit: I am not so sure about that one-cable-powerplant stuff anymore. Don't bet on it.
    Last edited: Oct 18, 2007
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