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Difference in energy between potential and light in photoelectric effect

  1. Dec 26, 2012 #1

    I have a quick question I'm not quite understanding.

    If I shine some light on a surface, metal or something, and the energy of the incoming wave is larger than the energy-barrier for the electron on the surface, an electron will be emitted.
    But my question is, if the incoming photon has the energy of, lets say, 1 ev, and the energy barrier is 0.5 ev, then the light will have twice the amount of energy it needs to emit the electron from the surface. But what happens with the rest then ?

    If the barrier is 1.1 ev, nothing will happen, and the photon will just pass through or diffract, right ?
    But in this case, will the electron absorb 0.5 ev from the photon, and the re-emit another photon with a changed wavelength that is equal to the left over energy of 0.5 ev ?

    Thanks in advance.

  2. jcsd
  3. Dec 26, 2012 #2
    The electron will leave the atom with 0.5 ev of kinectic energy and this electron now become a free electron and this kinectic energy will become heat like the electron moveing in metal.
  4. Dec 26, 2012 #3
    So the electron just get the energy from the photon, minus the work function of the electron, in kinetic energy ?
  5. Dec 26, 2012 #4
    the left energy will become the kinectic energy of electron
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