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Calculate vmax and voltage from X-ray distribution function

  1. Apr 18, 2017 #1

    VH1

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    1. The problem statement, all variables and given/known data
    The X-ray intensity distribution function for an X-ray lamp is
    given on the figure. What is the maximum velocity of the electrons? What is the potential difference under which the lamp is operating? Figure: http://imgur.com/01kCBc8
    01kCBc8.png

    2. Relevant equations
    Kmax = 0.5m(vmax)^2
    Kmax = hf - work function
    λ = c/f
    c = speed of light = 3 x 10^8 m/s
    Peak wavelength (from the figure) = 3.5 x 10^-9 m
    h = Planck's constant = 6.63 x 10^-34
    m = mass of electron = 9.11 x 10^-31

    3. The attempt at a solution
    Doubt I'm starting at the right place.
    0.5mv^2 = hf (work function??)
    0.5mv^2 = h(c/λ)
    v = sqrt [2h(c/λ)/m]
    v = 1.1 x 10^7 m/s
    Given answer is 1.5 x 10^7 m/s

    As for the potential difference, I have no idea how to calculate it.
     
    Last edited by a moderator: Apr 18, 2017
  2. jcsd
  3. Apr 18, 2017 #2

    gneill

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    Staff: Mentor

    Hi VH1,

    Welcome to Physics Forums!

    X-Ray production isn't quite the same thing as the photoelectric effect. In this case electrons are stopped by a target and shed their energy in the form of photons.

    Looking at the intensity distribution, which wavelength on the curve represents the highest energy photons (not the total intensity of emitted photons, but the highest photon energy)?

    For the potential difference, how much energy does a charge gain by "falling through" a given potential difference? What's the formula?
     
  4. Apr 18, 2017 #3

    VH1

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    Thanks gneill for the warm welcome and help - I managed to work out both questions with that.

    Cheers
     
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