Question about photoelectric effect equation

  1. If I write photoelectric effect equation to nhv = 1/2mv2 + W with n = 1,
    I am puzzled whether the n is larger than 1, such as 2 or 3, for which I will get the two- or three- photon absorption.
    Could you give me some comments on these question?

    Best regards.
    sandf
     
  2. jcsd
  3. n represents the number of photons ejected from the metal. According to the frequency of the light wave & its intensity, we can obtain the no. of photons.

    E=nhv
     
  4. Dear lawmaker,
    Thanks for your reply.
    But I cannot understand what you mean.
    In terms of your opinion, may I create the energy conservation equation
    n1hv1 = 1/2mv2+W+n2hv2 ?

    I understand the photoelectric effect that this phenomenon cannot occur when the energy of one photon is not large enough, i.e. hv < W. But I think nhv could be larger than W, i.e. nhv > W, thus the effect will occur.

    What is wrong with my understanding?

    Best regards.
    sandf
     
  5. Bandersnatch

    Bandersnatch 1,288
    Gold Member

    The explanation of the photoelectric effect that Einstein proposed used the assumption that at most one photoelectron can be ejected from the source by a single photon. It implies n=1.
    Therefore, you cannot use n>1 to calculate the energy of a single photoelectron.
    The equation should look like this:

    [itex]h\nu=E_k+W[/itex]

    If you have a beam containing n quanta of light(photons) at sufficient frequency for Ek to be >0, then they'll eject n photoelectrons from the source.

    [itex]nh\nu=n(E_k+W)[/itex]
     
  6. Dear Bandersnatch,
    Thanks for your help.
    Is it also assumed that only a single photon can be absorbed?
    Otherwise, could I have the equation.
    [itex]nh\nu=E_k+W[/itex], which also means the emisstion of a single photon.

    Best
    sandf
     
  7. Bandersnatch

    Bandersnatch 1,288
    Gold Member

    Yes, that's what the quantified approach means. The beam is not a continuous wave, but a collection of discrete quanta(photons) that hit the electrons in the metal surface and eject them.
    One photon hits one electron. If its energy is too low, the electron does not get ejected and the energy is wasted on heating up the bulk of the material instead.

    Also, it's not
    but the emission of an electron by a photon. The electron is often called "photoelectron" in this context, but it's still just your regular electron.
    Perhaps you got confused by lawmaker's post. There are no photons ejected. It's the photons that do the ejecting, so to speak. hv is the energy of the photon, while Ek can be only defined for a massive particle - here, the electron.

    So your equation is wrong, as it presupposes multiple photons (n) pooling their energy(hv) to eject one electron. That's exactly contrary to what the whole shebang about light quanta was about.
     
  8. Dear Bandersnatch,
    Great thanks for your help.
    Best regards.
    sandf
     
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