2 questions about the photo electric effect

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Supitha
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Graph of photoelectric current against collector plate potential for three light frequencies, with curves labelled v3 > v2 > v1 and stopping potentials -V03, -V02, -V01 on the horizontal axis, levelling off at saturation current.
Graph of photocurrent I against voltage V for two light intensities, I1 and I2, with a saturation plateau for I2 and a stopping voltage marked as -V0 on the horizontal axis.

  1. How can explain the difference of these red dots?
  2. Red line = Green line ?. How to explain it?
 
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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.
 
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Swamp Thing said:
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.
Thank you so much.
 
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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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