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Finding the work function using stop potential

  1. Feb 20, 2006 #1
    Apologies, but I have another photoelectron question. In this one I get a result that seems reasonable, however it does not show up on my list of work functions that I have available to choose from. The problem is:

    A photoelectric-effect experiment finds a stopping potential of 1.93V when light of 200nm is used to illuminate the cathode. What is the cathode made of?

    I initially thought this would be an easy problem as Vstop = (hf - Eo) / e

    I have all the values except Eo which is what I am trying to solve for.

    Vstop = 1.93 V
    h = 4.14x10^-15 eV
    f = 3x10^8/200x10^-9 = 1.5x10^15
    e = 1.6x10^-19

    So, I plug all of the values into the equation and solve for Eo

    1.93 = ((4.14x10^-15)(1.5x10^15)-Eo)/1.6x10^-19

    The resulting value of Eo was 6.21
    This seems valid, except the list I have to choose from is:

    Potassium: 2.3
    Sodium: 2.75
    Aluminum: 4.28
    Tungsten: 4.55
    Copper: 4.65
    Iron: 4.7
    Gold: 5.1

    Likely I am having the same problem as my previous question and not seeing a simple unit difference. Anyhow, I do appreciate your help.
     
  2. jcsd
  3. Feb 20, 2006 #2

    Tom Mattson

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    That is not Planck's constant in SI units. In fact h is not expressed in eV in any system of units.
     
  4. Feb 20, 2006 #3
    Tom is right and wrong. The problem IS units, but the Planck's constant being used by Calmeir is in eV-s, which is correct. The problem is that he is already working in eV, and so needs to drop the "1.6x10^-19" at the end of his equation.

    I reworked it and got e=4.28 eV.

    -Dan
     
  5. Feb 21, 2006 #4

    Tom Mattson

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    No, Tom is right and right! :biggrin:

    While the numerical value of h that he used corresponds to h in eV-s, he did report the units in eV, which is wrong.

    That's one fix, but I was hinting at a different one. Since he put all of his other quantities in SI units I was trying to get him to put h in SI units as well. He would have obtained E0 in Joules and then convert to eV. Changing h to eV-s and leaving the electron charge as e is certainly more direct though...
     
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