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Planck's Constant and the Work Function

  1. Aug 11, 2009 #1
    I have recently graphed my Planck's Constant data from my Planck's Constant kit on Excel. The R squared value is great but there are a few things that confuse me.

    1. I presume that the Equation of V = f(h/e) - W/e can be used with y = mx + c

    If so, then I presume that my equation of y = 3e-15x - 1.2163 from excel can solve the equation:

    c = -1.2163 = The work function (I find this a bit strange though because the cathode is covered in cesium which is meant to be 2.13 eV)



    V = y = backing voltage

    f(h/e) = gradient = m

    Is this right? Or do are my presumptions wrong?

    Also is the letter for frequency f or v? Because there seems to be some variation whenever I attempt to browse sites.
     
    Last edited: Aug 11, 2009
  2. jcsd
  3. Aug 11, 2009 #2

    Redbelly98

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    That's the work function for pure cesium. Your cathode could have some cesium oxide or other impurity that lowers the work function.

    Note quite. This should be
    f(h/e) = mx
    The variable in f(h/e) would equate to x, and the constants would equate to the slope m.

    Both are used, though it's actually the Greek letter nu, not v.
    f is a more general or universal symbol, referring to the frequency of anything that oscillates sinusoidally. ν is used strictly for the frequency of electromagnetic waves -- but the more general f can be used for those as well.
     
  4. Aug 11, 2009 #3
    I also calculated my percentage error to be 25.4%. Is this alright or should I consider re-doing it again?
     
  5. Aug 12, 2009 #4
    How did you calculate that.
     
  6. Aug 12, 2009 #5
    To calculate the experimental Planck’s constant:

    V = f(h/e) – W

    =3.132×〖10〗^(-15) x - 1.22
    This correlates to the equation
    y=mx+c


    ∴m=3.132×〖10〗^(-15) and ∴c=- 1.22

    According to theory, the gradient = h/e

    ∴h/(1.6×〖10〗^(-19) )=3.132×〖10〗^(-15)
    ∴h=3.132×〖10〗^(-15)×1.600×〖10〗^(-19)
    ∴h=5.011×〖10〗^(-34) J

    Experimental Uncertainty:
    Percentage Error=(| Experimental Value-Accepted Value |)/(Accepted Value)×100%
    =(| 5.011×〖10〗^(-34)-6.626×〖10〗^(-34) |)/(6.626×〖10〗^(-34) )×100%
    =0.244×100%
    =24.4%
     
  7. Aug 12, 2009 #6

    Redbelly98

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    Your math is correct.

    I do wonder how many significant figures are appropriate for your 5.011e-34 value. Did you make the best-fit line by hand, or did you use a calculator or computer to get it?
     
  8. Aug 13, 2009 #7
    I used the computer to get the R squared value and the equation. Microsoft Excel 2007.
     
  9. Aug 13, 2009 #8

    Redbelly98

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    Looks good then. By the way, errors are usually reported to no more than 2 significant figures.
     
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