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How to calculate stopping potential

  1. Sep 24, 2009 #1
    1. The problem statement, all variables and given/known data
    A photoelectic experiment was performed at two different wavelengths of light: 400 nm and 600 nm. The stopping potential ("push back" voltage) at 400 nm s 0.9V. What is the stopping potential at 600 nm


    2. Relevant equations
    eV = h(v - v0)



    3. The attempt at a solution
    eV/h = v - v0
    v0 = v - eV/h
    v0 = 400 *10^-9 - (1.602*10^-19)(0.9)/(6.6*10^-34)
    v0 = -2.18*10^14

    V = h(v - v0)/e
    V = (6.6*10^-34)(600*10^-9 - -2.18*10^14)/(1.602*10^-19)
    V = 0.9

    But this can't be right, since that is the stopping potential for a wavelength of 400nm. The stopping potential should be higher.
     
  2. jcsd
  3. Sep 24, 2009 #2

    kuruman

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    I am not sure I understand the symbols in your "relevant equation". Could you please identify them one by one? Thanks.
     
  4. Sep 24, 2009 #3
    e is the charge of an electron.
    V is the stopping potential
    h is plancks constant
    v is the frequency
    v0 is also a frequency
     
  5. Sep 24, 2009 #4

    kuruman

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    I understand now. To begin with, what you call vo is related to the work function and cannot be negative. Your problem is that you confused wavelength with frequency. When you are given 400 nm and 600 nm, these are wavelengths not frequencies. You need to learn to interpret the given numbers in terms of your equations and not just plug in without second thought.
     
  6. Sep 24, 2009 #5
    That makes sense, but when I tried it out using frequency instead of wavelength, I get a huge number for v0. Maybe that's ok, but then when I try to use that to solve for V, I get a negative number.
     
  7. Sep 24, 2009 #6

    kuruman

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    Show what you did and what you got. Maybe I can find something wrong.
     
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