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Determining Planck's Constant With Photoelectric Effect

  1. Mar 5, 2015 #1
    1. The problem statement, all variables and given/known data
    I'm trying to determine Planck's Constant through an experiment with the Photoelectric Effect, however, the equations I'm given and the data I've collected are not getting me to the 6.63e-34 that I need to be at. I'm graphing Stopping Potential (V) vs 1/λ and then using the slope to find Planck's Constant. The experiment is designed so that stopping potential equals the work function. I don't know whether the trendline's slope that Excel plotted for me is wrong because of outlier data or whether my manipulation of equations is wrong.

    2. Relevant equations
    E = hc/λ
    e(V + Φ) = hc/λ
    e(V + Φ) = hc/ λ --> 2eV = hc/λ if stopping potential = work function --> h = 2Veλ/c
    V = hc/eλ (what's given on assignment) --> h = Veλ/c
    3. The attempt at a solution
    Using h = 2Veλ/c, where Vλ is the slope of the line, I get an average value for Planck's Constant of 7.76e-34. Using h = Veλ/c, I get an average value of Planck's Constant of 3.88e-34. Also, if I disregard my stopping potential reading for λ = 470 nm, I get a value for Planck's constant that is probably the most accurate, 5.72e-34.
     
  2. jcsd
  3. Mar 5, 2015 #2
    Calcualate freq. By given wavelengths. Plot a graph between freq. And stopping potential then slop of the graph will give you h/e multiply it with charge it will give value of planck constant.
     
  4. Mar 5, 2015 #3
    If I multiply that by 2 it gives me the most accurate result so far, but the assignment specifically asks for a graph of V vs 1/λ.
     
  5. Mar 5, 2015 #4
    Then plot graph between V and 1/λ but the slop will give you hc/e.
     
  6. Mar 5, 2015 #5
    So my problem is most likely outlier data that is skewing the slope of the line then.
     
  7. Mar 5, 2015 #6
    Ye
    Yeah try plotting graph by urself by adjusting the slope to get required value..
     
  8. Mar 6, 2015 #7

    BvU

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    Show your data. Typically you have only a few frequencies of spectral lines to work with. (Or did you use LEDs instead of a mercury lamp?)
    Low-frequency light may be polluted with some of the other frequencies to give a tail out to high lambda. Just inserting a factor of 2 out of nowhere isn't physics.

    Finding a much lower h isn't uncommon (see here )
     
  9. Mar 6, 2015 #8

    BvU

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    Oh, and: Hello MA, welcome to PF :smile: !
     
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