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Work function and wavelength

  1. Mar 30, 2007 #1

    kingwinner

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    1) The maximum kinetic energy of photoelectrons is 2.8eV. When the wavelength of the light is increased by 50%, the maximum energy decreases to 1.1 eV. What are the "work function" of the cathode and the "initial wavelength"?
    K.E.initial=K.E.max=2.8eV=4.48x10^-19J
    lambda.final=1.5*lambda.initial
    Final energy = 1.1 eV (<---is this purely kinetic energy? why?)

    K.E.max = hc/(lambda.initial) - work function
    But now I have 2 unknowns: work function and lambda.initial, what can I do to solve for both?

    I am stuck here...

    Does anyone have any idea or insight? Any help is greatly appreciated.
     
    kingwinner, Mar 30, 2007
  2. jcsd
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  3. Mar 30, 2007 #2

    kingwinner

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    Can someone please help me? I am sure that there are a lot of genius here...
     
    kingwinner, Mar 30, 2007
  4. Mar 30, 2007 #3

    Dick

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    Make that:

    K.E.max.initial = hc/(lambda.initial) - work function

    Then how about

    K.E.max.final = hc/(lambda.final) - work function

    as a second equation?
     
    Dick, Mar 30, 2007
  5. Mar 30, 2007 #4

    kingwinner

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    Thanks for your help! I got it!
     
    kingwinner, Mar 30, 2007
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