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Determination of planck's constant

  1. Dec 3, 2007 #1
    I have a graph of voltage verse frequency.
    Given that the charge of an electron = 1.6 x 10^-19 I need to obtain an estimate of the value planck's constant. What equation do I need to use? Do I find the gradient of the graph?
     
  2. jcsd
  3. Dec 3, 2007 #2

    robphy

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    What is the experiment that led to these measurements (this graph)?
    [I have a guess... but you should really describe it.]
     
  4. Dec 3, 2007 #3
    its the stopping voltage versus frequency for a metal surface.
    The only part of the graph shown is from 5.4 to 6.8 f(X 10^14) and .26 to .85 volts. The graph is linear.
     
  5. Dec 3, 2007 #4

    rock.freak667

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    the work done in stopping the electron is equal to the kinetic energy lost...
    that is,
    [itex]eV_s=\frac{1}{2}mv^2[/tex]

    so from [itex]hf=hf_0+\frac{1}{2}mv^2[/itex]=>[itex]hf=hf_0+eV_s[/itex]

    rearrange and make it into the form you want...
     
  6. Dec 3, 2007 #5
    How can I solve for h when h is on both sides of the equation?
     
  7. Dec 3, 2007 #6

    rock.freak667

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    [itex]eV_s=hf-hf_0[/itex]
    [itex]V_s=\frac{h}{e}f-\frac{h}{e}f_0[/itex]

    Y=MX+C, can you see what M and C are?
     
  8. Dec 3, 2007 #7
    I can't see what they are.
    The values are
    m = .421428
    c = -2.2757 x 10^14
     
  9. Dec 4, 2007 #8

    rock.freak667

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    Well from [itex]V_s=\frac{h}{e}f-\frac{h}{e}f_0[/itex]

    if a graph of [itex]V_s[/itex] and f is drawn, then the gradient,m. would be equal to the fraction: [itex]\frac{h}{e}[/itex] and so you know what e is, find h
     
  10. Dec 4, 2007 #9
    I tryed this but but solving for h in m=h/e does not give planck's constant.
     
  11. Dec 4, 2007 #10
    Are you being careful with your units when you are calculating your gradient????

    Remember that your frequency scale will probably be (x 10^14) Hz

    You should get a much smaller gradient.
     
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