Advertisement
If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)
Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie
Hi there,
There is an issue with role permissions that is being worked on at the moment.
If you are having trouble with access or permissions on regional forums please post here to get access: https://www.boards.ie/discussion/2058365403/you-do-not-have-permission-for-that#latest

Determine Planck's Constant from gradient

  • 21-09-2012 6:45pm
    #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: 3,455 ✭✭✭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: 3,455 ✭✭✭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: 13,077 ✭✭✭✭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.

    You are the type of what the age is searching for, and what it is afraid it has found. I am so glad that you have never done anything, never carved a statue, or painted a picture, or produced anything outside of yourself! Life has been your art. You have set yourself to music. Your days are your sonnets.

    ―Oscar Wilde predicting Social Media, in The Picture of Dorian Gray



  • #6
    krd
    Banned (with Prison Access) Posts: 3,455 ✭✭✭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.


  • Advertisement
Advertisement