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Vacuum Photocell help

  1. May 19, 2008 #1
    Hello. This is actually my first post in these forums and I've never really thought of posing my questions online but I've decided to give it a try.

    Well about the Vacuum Photocell. I know what the setup looks like and all, but there's one thing I don't get. I'm going to say everything I know in here, so please feel free to correct me: There's a plate and a ring in the photocell surrounded by vacuum. Now if the plate is connected to the -ve terminal of a generator, it will be the -vely charged cathode while the ring becomes relatively positively charged. Supposing the UV radiation is of enough energy to extract the electrons from the plate, these electrons which leave the plate will be absorbed by the +vely charged ring and a complete circuit is achieved. What if the polarities are reversed though? What happens? The plate becomes +vely charged while the ring is now -vely charged. So now the radiation will extract electrons from this +ve plate which will be attracted/absorbed again by the plate when they leave it? Moreover, I don't really get the "cut-off potential" concept. What is really going on in the photocell, in other words? Have I misunderstood the whole concept?

    Thanks in advance
  2. jcsd
  3. May 19, 2008 #2


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    Welcome to PF.

    You've got it about right. Reversing the polarity will make the cell much less sensitive, but it should still work.

    The electrons near the surface are held in a potential well which defines the photoelectric 'work function'. To free the electron requires a certain minimum energy, called the work function ( I think ). Any energy left over will contribute to the electrons kinetic energy outside the material.

    It's explained here


  4. May 19, 2008 #3
    Well yea I know about the work function and Kinetic energy left and all.. but my question is when the polarities are reversed, the electrons leaving the plate would go where? How is the cut-off potential reached? Can you define it for me please?
  5. May 19, 2008 #4


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    When the polarity is reversed, you are trying release electrons from a postively charged electrode , so the effective work function is greater.

    A released electron will fall back to the anode unless it has enough energy to escape.

    I don't know what 'cut-off' voltage you mean.
  6. May 19, 2008 #5


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    If you reverse the power source, like Mentz says, "will make the cell much less sensitive". The plate on which the light falls on has a larger area which will give it a greater chance of being struck by an incoming photon. As to 'cut-off' could you be meaning saturation?
  7. May 20, 2008 #6
    I'm not sure but perhaps it's also called the "stopping potential" ? Also, when polarities are reversed, wouldn't it be a good thing if the electron comes back to the anode? The polarities are reversed so the flow of the current should also be opposite to what it was, meaning into that +ve plate. The ring would reject the electron since it repels it anyway, no ?
  8. May 20, 2008 #7


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    Karimz is right.
    If the electron can't leave the anode, no current will flow. In fact I don't think it will work at all unless light falls on the cathode.
  9. May 20, 2008 #8
    Well at some point an electron does leave the anode but it's attracted back to it when it leaves. But the idea that the polarities were reversed makes the anode a cathode now since it's role would be to receive electrons not lose them. So when it does lose them and attracts them again that's considered as a current flow no ?

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