According to my textbook, electrons which are emitted due to the photoelectric effect have kinetic energy given by: $$KE_{electron}=h\nu-h\nu_0$$ Where $\nu_0$ is the activation frequency of the metal.

My question is: when electrons are emitted in this way, is all of the photon's energy really converted to kinetic energy in the electron? Is there no loss of energy in this process?

If there is energy lost (which I suspect must be true), is it a sufficiently small amount so that we can say it is negligible for most purposes?

  • 1
    $\begingroup$ In your formula KE should actually be labelled KE(max). Electrons will be ejected with a range of kinetic energies from 0 to KE(max). $\endgroup$
    – ron
    Apr 15 '17 at 23:15
  • $\begingroup$ What exactly happens to the EMR when the photoelectric effect occurs? I would guess that the frequency of the radiation is reduced so that the energy gained by the electron is equal to the loss in photon energy, but what causes a different amount of energy transfer in different cases? $\endgroup$
    – Ramoose
    Apr 15 '17 at 23:43

$h\,\nu$ is photon's energy; $h\,\nu _0$ is the potential energy of the electron. All energy excess is converted in kinetic energy of the eletron - that's the left side of the equation.


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