1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Photo-electric Effect

  1. Jan 21, 2007 #1
    1. The problem statement, all variables and given/known data


    It's question 2.

    2. Relevant equations

    Work = U/t
    E = hf
    f = c/lambda

    3. The attempt at a solution

    f = c/lambda = (3.00x10^8 m/s)/(530x10^-9 m) = 5.66x10^14 Hz

    E = hf = (6.626x10^-34 Js)(5.66x10^14Hz) = 3.75x10^-19 J/photon

    Work = 2.00x10^-3 J/s ----> in one second U = 2.00x10^-3 J

    #photons = U / E = 5.33 x 10^15 photons

    #photo-electrons produced = (5.33 x 10^15)/(10^5) = 5.33x10^10 photo-electrons

    Q = (5.33x10^10)(1.60x10^-19 C) = 8.54x10^-9 C

    I = 8.54x10^-9 C/s

    The thing is I did not use the work function of cesium to solve this. Hence I am unsure if this is right... Any help would be appreciated.

  2. jcsd
  3. Jan 21, 2007 #2


    User Avatar
    Homework Helper

    btw, when you say "work", I think you mean "Watts" or more precisely "power".
    your method looks good to me.
    work function only comes in when you want to know the energy of the photo-electron..but you are only concerned with the amount of them (ie. current) the most you could do is to check that the incoming photon energy can indeed overcome the work function: [tex]E_\text{photon} > \phi[/tex]. I don't think there is any problem with that
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Photo-electric Effect
  1. Photo electric effect (Replies: 1)

  2. Photo electric effect (Replies: 5)