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Photoelectric effect and work functions

  1. Jun 2, 2005 #1
    what can be said about the work functions of two metals when the threshold wavelength in the photelectric effect increases.

    I'm having trouble finding and equation to describe this relationship.

    ANyone have any ideas?
  2. jcsd
  3. Jun 2, 2005 #2
    Well, the work function, [itex]\phi = hf_{0}[/itex], where [itex]f_{0}[/itex] is the threshold frequency. Do you know the relationship between frequency and wavelength?
  4. Jun 2, 2005 #3
    Threshold wavelength is inversely proportional to the work function of the metal .
    Of course the threshold wavelength is decided by the metal itself.

    For the critical case the work function is [itex]\phi[/itex] is related to the wavelength as:

    [itex] \phi [/itex][itex]= \frac{hc}{ \lamda }[/itex]
  5. Jun 2, 2005 #4
    You spelt lambda wrong, that's why it won't come out.
  6. Jun 3, 2005 #5
    If the photon energy exceeds the work function, the excess energy appears as the kinetic energy of the electron.
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