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Homework Help: Find current induced in the photoelectric effect

  1. Apr 28, 2016 #1
    1. The problem statement, all variables and given/known data
    A metal surface had dimensions of 1.5mm x 2mm. The intensity of light incident on surface is 4.5×10^-6 W/m^2. On average one electron is emitted for every 300 photons incident on the surface. Determine the initial current leaving the metal.
    ●wavelength of light incident is 420nm
    ●work function is 3.4×10^-19

    2. Relevant equations

    3. The attempt at a solution
    I know the intensity is related to the number of electrons that are emitted but not how, that's the only way I can think of aproaching this problem. Is there another way?
  2. jcsd
  3. Apr 28, 2016 #2
    I dont think so..

    Intensity is energy per unit time per unit area or power per unit area.
    You can relate intensity to the number of photons incident per second just by replacing Power in the Formula of Intensity
    by (Energy of 1 photon * n) where n is the no. of photons incident per second. Calculate for n.

    It is given that 300 incident photons emit one electron, try calculating the current from this.

    P. S current is charge emitted per unit time.
  4. Apr 29, 2016 #3

    rude man

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    First compute the number of photons per unit time incident on the metal.
    The rest is a gimme.
  5. Apr 29, 2016 #4
    Ok so,
    I=P/A becomes, P=IA=1.35×10^-11 J/s
    But the power = Ep *(n/t) , {Ep= energy of photon=work function=3.4×10^-19J}
    n/t ≈ 39700000 {n = number of photons}
    Now I divide this number by three hundred to get the number of eldctrons per time and then multiply by the charge of the electron to get the value of the current = 2.1 ×10^-14 A or 21fA.
    So tiny did I do this right?

    By the way the units for the work function was J
  6. Apr 29, 2016 #5
    Are you sure?
    What is the energy of a photon dependant on?
  7. Apr 30, 2016 #6
    hf, oh so I use this instead of the work function?
  8. Apr 30, 2016 #7
  9. Apr 30, 2016 #8

    rude man

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    The work function is what determines the minimum photon energy to get photoelectric emission, but the way the problem is stated its effect is included in the wording. You get 1 electron per 300 photons regardless of the work function. If the problem had given just the incident power and frequency (assuming monochromatic light) then you would have had to have taken the work function into account, which varies from metal to metal.
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