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B 2 questions about the photo electric effect

  1. Jun 17, 2016 #1
    22906991854_266de9cdfe_o.png image741.jpg
    1. How can explain the difference of these red dots?
    2. Red line = Green line ?. How to explain it?
     
  2. jcsd
  3. Jun 17, 2016 #2
    Even when the applied voltage is zero, there is a kind of "virtual" voltage that pulls the electrons back into the metal. This is called the "work function", I believe.
    So at low photon frequencies, there is a smaller chance that an electron will end up with enough energy to leave the metal (because total energy = photon energy + random thermal energy). For high photon energies, there is more chance of escaping i.e. more chance that electron's original thermal energy plus photon energy will carry it over the work function threshold.
     
  4. Jun 17, 2016 #3
    Thank you so much.
     
  5. Jun 17, 2016 #4
    What's happenning in second one?
     
  6. Jun 17, 2016 #5
    No. The second pic
     
  7. Jun 17, 2016 #6
    I'm guessing here, but the second figure probably shows two different intensities at the same wavelength. The "stopping potential" is the same for both curves because the photon energy is the same, but once you allow some current to flow then the more intense light means more photons per second which means more electrons per second.

    Going back to fig. 1, all curves involve the same intensity (same rate of photons per second) so they all saturate at the same number of electrons per second. If you increase the applied voltage to a huge value, you can still pull only those electrons that are liberated by photons, so "pulling harder" won't make much difference -- so the curves saturate at some point.
     
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