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What determines if the photoelectric effect occurs?

  1. Aug 8, 2011 #1
    What determines if the photoelectric effect occurs?

    There are several aspects on this topic that I don't really understand such as:

    - What is the relationship between the frequency of the incident photon, threshold frequency and the ejection of electrons?

    - What is the relationship between the energy of the incident photon, the work function and the ejection of electrons?

    - What is the relationship between kinetic energy of the ejected electrons, the energy of the incident photon and the work function?

    - What is the relationship between the intensity of the incident light and the average kinetic energy of the ejected electrons?

    - What is the relationship between the intensity of the incident light and the number of ejected electrons?

    Thanks
     
  2. jcsd
  3. Aug 8, 2011 #2
    Forgot to include my work so far, please assist me if this is correct so far, thanks

    1)The energy of a photon is a function of the frequency of the photon

    Energy = (Planck's constant * Speed of Light) / wavelength = Planck's Constant * Frequency

    If energy is large enough it will knock an electron loose

    2) Any energy in excess of what is needed to knock an electron loose (Work Function) goes to KE of the electron

    4) Not related the number of photons per second has no relationship to the KE of the ejected electrons

    5) The more high energy photons you have the more electrons you knock loose
     
  4. Aug 8, 2011 #3
    You have posted down various doubts over the photoelectric effect ,I will try to be as concise as possible.

    P.F.E. occurs when light of enough/over frequency* is absorbed by the electron of the photo-cathode end[often cesium] (note intensity has no direct correlation with individual photon energy),as a result of this electron is ejected but this happens when K.E is minimum.
    Which can be found by the following equation:
    EKmax = hf- w [w is the work function)
    At min. K.E = 0
    so hf = work function
    What triggers this,I am not quite sure about but if I vaguely recall since Light acts both as a wave and particle,it has an E-M field which has oscillation,it's these oscillation which excite/kick the electron[Don't take this to be valid for I maybe wrong].
    Also you may have not come across this yet but stopping potential is achieved when there's zero current between the circuit this is achieved by changing the polarity.
    Just to remind you again intensity has no affect over individual K.E of electrons however if the no. of photons is increasing so does the no. ejected electrons which results in a greater current value between the circuit.
    I hope this answers your doubts.
    P.S: I think this thread should be under classic sub-section ? :s
    -ibysaiyan
     
    Last edited: Aug 8, 2011
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