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Homework Help: Photoeletric threshold

  1. Apr 14, 2008 #1
    1. The problem statement, all variables and given/known data
    The photoelectric threshold wavelength of a tungsten surface is 272nm

    A) what is the threshold frequency of this tungsten
    b) what is the work function in electonVolts of this tungsten
    C) Calculate the maximum kinetic energy in eV of the electrons ejected from this tungsten by ultraviolet surface.

    2. Relevant equations
    A) f= W/h
    B) ????
    C) Kmax = hf-W

    3. The attempt at a solution

    i have no idea how to implement these equations, im only given the wavelegnth can some one point me in the correct direction and also if any one know a formula for part B of this problem
  2. jcsd
  3. Apr 14, 2008 #2
    a) for any wave phenomenon: wavelength = wave speed * frequency
    b) I'm not sure what you mean by a "work function".
    c) should be easy if you have a)
  4. Apr 14, 2008 #3


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    For part a, kamerling is right. You can calculate the frequency of the photon directly from its wavelength.

    For part b, the work function (W) is a term specific to the photoelectric effect. W has units of energy and represents the amount of "work" required to completely liberate an electron from the metal (in this case tungsten). Since you know the frequency of a photon that *will* accomplish this, you know its energy, therefore you know the work function.
  5. Apr 14, 2008 #4
    wave speed = wavelength*frequency
  6. Apr 14, 2008 #5


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    yeah...good catch.
  7. Apr 14, 2008 #6
    (Sorry about the long-winded intro, but you said you had some trouble manipulating the equations; I myself am not good at remembering equations so I find it helpful to keep the main (applicable) concepts in mind and more often than not, that helps me to come to terms with a problem much more easily than digging out seemingly meaningless equations from a book.)

    The key thing to remember with the Photoelectric Effect is that one photon gives all of its energy to one electron.

    The energy of the photon is given by:

    [tex]E = hf = \frac{hc}{\lambda}[/tex]

    Considering conservation of energy for the emission:

    [tex]E = W + K[/tex]

    Where, W is the work done in liberating the electron from the surface and K is the kinetic energy it leaves the surface of the metal with.

    The work function (a property of the metal in question) is defined as being the very least energy (of an incoming photon) that will stimulate electron emission; thus, the emitted electron has a kinetic energy of approximately zero. Such a photon is said to be at a threshold frequency, [tex]f_{0}[/tex] (or wavelength [tex]\lambda_{0}[/tex]) and hence it's energy is given by:

    [tex]E_{0} = hf_{0} = \frac{hc}{\lambda_{0}}[/tex]

    But, since the kinetic energy of the electron here is zero this energy is equal to the work function:

    [tex]W = hf_{0} = \frac{hc}{\lambda_{0}}[/tex]

    Putting this into the conservation of energy equation above:

    [tex]\frac{hc}{\lambda} = \frac{hc}{\lambda_{0}} + K[/tex]

    [tex]K = \frac{hc}{\lambda} - \frac{hc}{\lambda_{0}}[/tex]

    Now, for part (a) you are right; the equation you are using will give you the work function of the metal in units of Joules.

    For part (b) all you need to know is the value of the electron volt. The electron volt is defined as the energy gained/lost by an electron when it is accelerated/decelerated by a potential of 1 volt:

    [tex]1eV = 1.6 x 10^{-19}J.[/tex]

    So, in order to get the work function of the metal in eV, all you need to do is multiply your answer from part (a) by that number.

    You are also correct about part (c); your equation is the same as my expression for K.
    Last edited: Apr 14, 2008
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