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Help: Photoelectric effect problem

  1. Apr 8, 2014 #1
    In a photoelectric cell, the stopping voltage is 2.00V. If the voltage applied across this cell is zero, what is the maximum speed of the electrons from the photoelectric surface?

    Confused.
     
  2. jcsd
  3. Apr 8, 2014 #2

    ZapperZ

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    First of all, tell us what you understand as the definition of "stopping voltage", and its significance in the photoelectric effect experiment.

    Zz.
     
  4. Apr 8, 2014 #3
    well i know that its used to increase the intensity of the negative plate to repel the electrons... the stopping voltage is the point at which the current stops because of that increased negative intensity...
    but i dont understand why theres a stopping voltage AND zero voltage applied... ??
     
  5. Apr 8, 2014 #4

    ZapperZ

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    Er... now it is my turn to not understand what you wrote. You do know that when you apply the stopping voltage, there IS an applied field that is being used to slow down and stop the photoelectrons from reaching the anode? The voltage whereby the most energetic electrons are just stopped (i.e. where the current first reaches zero) is the definition of the stopping potential.

    Now, see whether this jives with your understanding of what the stopping potential is, and see if you can figure out that if it takes this much potential to stop that electron, what was its kinetic energy that it started out with?

    Zz.
     
  6. Apr 8, 2014 #5
    ya i understand that part...
    so, if the applied voltage is zero, doesnt that make the kinetic energy zero? since KE=eV...? and if KE is zero... that makes the vmax zero too?
     
  7. Apr 8, 2014 #6
    im trying to understand what the question is asking me for? the KE when the Vstop is 2.00? or when the voltage is zero?
     
  8. Apr 8, 2014 #7
    The fact that the stopping voltage is 2.00 V does not mean that there is an applied voltage at all. It means that if there was one, it should provide 2.00 V to stop the electrons before reaching the anode.
     
  9. Apr 8, 2014 #8
    so what does the question mean when it asks "if the applied voltage is zero...?"
     
  10. Apr 8, 2014 #9
    Wouldn't really like to tell this one directly.

    Imagine your photocell which is in a circuit and tell me what objects are usually in a similar photocell circuit and why .
     
  11. Apr 8, 2014 #10
    so i should figure out what the ke is if the Vstop is 2.00V...then u use that information to solve for vmax?
     
  12. Apr 8, 2014 #11
    If you know the [itex] U_{stop} [/itex] you can figure out the [itex] E_{kin} [/itex] from that you know [itex] v_{max} [/itex].

    However, I recommend you to understand/revise the theory, too.
     
  13. Apr 8, 2014 #12
    im still not sure what voltage being zero has to do with this question?
    i came up with 2.65x10^7 m/s
    theory in physics is NOT my forte!! :S
    appreciate the help, big time!
     
  14. Apr 8, 2014 #13
    The proper English isn't mine...

    The incoming photons give their energy to the cathode. If these energy quantums are big enough ([itex] \phi ≤ [/itex] ) the cathode emit electrons which might (and usually) have kinetic energy. If you apply a voltage to this circuit you can speed up or slow down the emitted electrons.

    When you slow them down there will be a voltage which will not let the electrons to reach the anode. So at the end (when the electrons almost reached the anode) the electrons' kinetic energy will be zero. This voltage is the stopping voltage.

    So by knowing the stopping voltage you know the kinetic energy of the electrons which is constant if there isn't any applied voltage (and if λ is constant too).
     
  15. Apr 8, 2014 #14

    ZapperZ

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    OK, I think we have narrowed down a fundamental error in your understanding of the photoelectric effect.

    Most of the photoelectrons that come out of the metal already have an initial kinetic energy!. So they are already moving!

    This is why you apply a reverse voltage to stop them.

    Zz.
     
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