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Homework Help: Einstein's photoelectric equation

  1. Jul 18, 2007 #1
    1. The problem statement, all variables and given/known data

    1. How can Planck's constant and the work function be determined from the following graph?

    http://img507.imageshack.us/img507/9076/untitledgo5.jpg [Broken]

    2. Calculate the frequencies corresponding to the cut-off wavelengths for filters 15 and 29 in table 1. Take speed of light as 3*10^8 m/s.

    http://img515.imageshack.us/img515/2240/untitled2jh5.jpg [Broken]

    3. The attempt at a solution

    I know the following equations, but can't seem to peice it all together

    KE = hf - W

    KE = eVs (where Vs is called the stopping potential, the potential for which the photocurrent reaches a value equal to zero)

    and Vs = (h/e)f - (W/e)
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jul 18, 2007 #2
    A hint: the equation of a line which intersects the x axis at A and has slope m is

    y = m*(x-A)

    (Think about why... the case y=0 )
  4. Jul 18, 2007 #3


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    Homework Helper

    1. You can see from your final equation that the gradient of the graph will be h/e and the y-intercept of the graph (which is not shown in the picture) will be (- W/e). Your final equation is a linear function of the frequency of the incident light.

    2. The filters transmits colours with longer wavelengths up to the minimum wavelengths stated. Just calculate the frequencies (cut-off ...) for these minimum wavelengths from

    [tex]c = \lambda f[/tex]

    This means that these filters will transmit light of lower frequencies up to these calculated ones. Since c is constant in the equation above one need to choose a larger f with a smaller [tex]\lambda[/tex] to obtain the same product. That is a short wavelength causes the peaks to pass a point quicker (larger frequency) than a longer wavelength when they both move at the same speed.
  5. Jul 18, 2007 #4
    In regards to q1, gradient will be h/e. Therefore h = gradient*e and work function = y-intercept*-e?

    Sorry, I'm slightly confused.
  6. Jul 19, 2007 #5


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    Homework Helper

    Correct. The first graph plots the stopping potential for the emitted electrons as a function of the frequency of the incident photons on the material. It therefore displays the relationship between Vs and f. That is exactly what your final equation gives - the relationship between Vs and f. Both h/e and W/e are constants in this equation. So this is a linear relationship between Vs and f.
  7. Jul 19, 2007 #6
    Gotcha, now in regards to q2.
    I get the explanation behind c = lambda*f (i.e. as wavelenth gets smaller, frequency gets larger and vice versa), but how do I actually find the numerical values of the frequencies?
  8. Jul 19, 2007 #7


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    Homework Helper

    The minimum wavelengths are given in the table so you just calculate the maximum frequencies with

    [tex]c = \lambda f[/tex]
  9. Jul 19, 2007 #8
    You forgot some equations.

    [tex]c = \lambda f[/tex]

    [tex]E = h f[/tex]

    Where h -> Plank's Cosnt, f -> frequency, and c -> speed of light, E -> energy, lambda -> wavelength
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