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Find the photoelectric work function for this metal

  1. May 22, 2013 #1
    1. The problem statement, all variables and given/known data

    The graph in fig shows the stopping potential as a function of the frequency of the incident light falling on the metal surface.

    Find the photoelectric work function for this metal

    2. Relevant equations

    [itex]V_0 =\frac{hf}{e}-\frac{\phi}{e}[/itex]

    3. The attempt at a solution

    From the graph, fth =1.25×10 Hz.
    I am completely lost.
    The solution gives 4.8ev
    is it h times threshold?
    how do you find h?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

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  3. May 22, 2013 #2

    BruceW

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    Homework Helper

    yep. the work function is just the threshold frequency times h. Although, I don't think the threshold frequency is 1.25*10 Hz... shouldn't it be 10 to the power of something? I can't tell because the graph is a bit hard to read. and how to find h? Um. It is just a constant. It should be given to you in the question, or you can look it up online.
     
  4. May 22, 2013 #3
    h is the gradient of the straight line, you need 'e' to convert stopping potential into energy in joules
     
  5. May 22, 2013 #4
    The solutions said fth =1.25×10^15 Hz.
    h = planks constant = 6.26x10^-36
    if you multiply you get 8.2825 i have to assume thats in jewels
    so i convert that to ev
    8.2825 / 1.6 x 10^-19 = 5.17ev

    but the answer is 4.8ev
     
  6. May 22, 2013 #5
    I think i got it.
    h= (e)(slope) = (1.60×10−16 C)(3.8×10−15 V ⋅ s) = 6.1×10^−34 J ⋅ s
    (6.1×10^−34 Js)(1.25x10^15) = 7.75x10^-19
    then convert to ev
    7.75x10^-19 / 1.6x10^-19 = 4.84 x10^-19
     
  7. May 22, 2013 #6

    BruceW

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    Homework Helper

    ah, good work. I see, you got h from the graph. nice, I didn't think of that.
     
  8. May 22, 2013 #7
    Yes, i got h from the graph. Thank you.
     
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