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Determine Planck's Constant from gradient

  • 21-09-2012 07:45PM
    #1
    Smythe
    Registered Users, Registered Users 2 Posts: 434 ✭✭


    I have a graph entitled "Photoelectric effect: stopping voltage as a function of light frequency"

    The y-axis is the Stopping Voltage (V), the x-axis the Frequency (Hz).

    How do I determine Planck's Constant (h) from the gradient?

    Thanks!


Comments

  • #2
    krd
    Banned (with Prison Access) Posts: 3,449 ✭✭✭krd


    Is this the graph?

    g0003.gif


    It's your homework, isn't it?


  • #3
    Smythe
    Registered Users, Registered Users 2 Posts: 434 ✭✭Smythe


    Yes. That's the graph.


  • #4
    krd
    Banned (with Prison Access) Posts: 3,449 ✭✭✭krd


    Smythe wrote: »
    Yes. That's the graph.


    There's more here.

    http://www.physics.usyd.edu.au/teach_res/qm2/qm0003.htm


    I've got a headache and a brain fog. But in terms of determining Planck's constant from the gradient, differentiation may work.

    If E = hf, then dE/df = h

    If the stopping voltage is directly proportional to the energy then

    dVs/df = h

    To put it another way, dy/dx = h


    I could be wrong but I have a headache and a brain fog.


    And as it turns out I was incorrect and had made a silly boo boo. See below for the correct answer.


  • #5
    bnt
    Registered Users, Registered Users 2 Posts: 13,228 ✭✭✭✭bnt


    Well, the formula is in the standard linear form y = mx +c, where m is the gradient and c is the intercept on the y-axis. You can read those off the graph, where m = Δy/Δx. The m value equates to (h/e) from the formula.

    In its pure form, fascism is the sum total of all irrational reactions of the average human character.

    ― Wilhelm Reich



  • #6
    krd
    Banned (with Prison Access) Posts: 3,449 ✭✭✭krd


    bnt wrote: »
    Well, the formula is in the standard linear form y = mx +c, where m is the gradient and c is the intercept on the y-axis. You can read those off the graph, where m = Δy/Δx. The m value equates to (h/e) from the formula.

    I believe you are correct.

    And yes, I know where I went wrong.


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