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Photoelectric effect and kinetic energy

  1. Oct 5, 2008 #1


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    1. The problem statement, all variables and given/known data
    Light of a wavelength 2000 Å falls on an aluminum surface. In aluminum, 4.2 eV are required to move an electron. What is the kinetic energy of

    (a) the fastest, and
    (b) the slowest emitted photoelectrons?

    2. Relevant equations

    [tex]K_{{\rm{max}}} = eV_0 [/tex]

    3. The attempt at a solution
    [tex]K_{{\rm{max}}} = eV_0 = - {\rm{1}}{\rm{.602}} \times 10^{ - 19} {\rm{C}} \times 4.2{\rm{eV}} \times {\rm{1}}{\rm{.602}} \times 10^{ - 19} \frac{{\rm{J}}}{{{\rm{eV}}}} = - 1.078 \times 10^{ - 37} {\rm{JC}}[/tex]
    I doubt I'm doing this right, by using the required energy as the stopping potential. The units don't work out. But there's nothing in the book, except the formula I listed, that tells me how to do such a problem. The book gives me the impression that the stopping potential is experimentally derived. They give a graph for sodium, with points and a trendline, but it doesn't tell me what it is for aluminum (unless they buried it in an appendix). How do I do this problem?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Oct 6, 2008 #2
    It looks like you need the work function:

    [tex] K_{electron} = E_{photon} - W_{metal} [/tex]

    where [tex] W_{metal} [/tex] is the minimum energy needed to free an electron from some particular metal.

    The joules-electronvolt conversion would be helpful as well:

    1 eV = 1.6022 X [tex] 10^{-19} [/tex] J
  4. Oct 7, 2008 #3
    wow..i have this same exact problem.
  5. Oct 7, 2008 #4


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    @ buffordboy; the Work function is [itex]eV_0=4.2 eV[/itex]

    @tony; why have you multiplied by the fundamental charge? you are given an energy (4.2eV) not a voltage; to convert it to Joules you only need to multiply it by [itex]1.602x10^{-19} \frac{J}{eV}[/itex]

    also, [itex]eV_0[/itex] gives you the work function not the maximum kinetic energy of the photo emitted electrons. The energy of these electrons is coming from the light that is bombarding the aluminum, not from a voltage [itex]V_o[/itex]. What is the maximum amount of energy that an electron will get from 2000 angstrom light? How much of that energy will be leftover kinetic energy after it escapes the aluminum?
  6. Oct 7, 2008 #5
    Yes, you are right. My quoted equation will yield the maximum kinetic energy.

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