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Work function

  1. Feb 13, 2005 #1


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    A light source of wavelength lambda illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 1.00 eV. A second light source with half the wavelength of the first ejects photoelectrons with a maximum kinetic energy of 4.00 eV. What is the work functin of the metal?

    I know I can use the equation for maximum kinetic energy defined by K=hf - phi where K is the maximum kinetic energy for emmitted electrons, h Planck's constant and f is frequency. Since we don't have frequency, I thought of rewriting the equation as K=hc/lambda - phi but now I don't know how to incorporate both light sources and to actually to solve for phi. Do I have to write two separate linear equations and solve them or is there something esle I'm supposed to be doing? Help please? :confused:
  2. jcsd
  3. Feb 13, 2005 #2
    You have the right idea. Two seperate equations and solve them. Go with it.
  4. Feb 13, 2005 #3


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    Staff: Mentor

    Bingo! :smile:

    Any time you have two unknown quantities, you should suspect that you'll have to set up two equations for them, and solve them together.
  5. Feb 13, 2005 #4


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    Haha I just reread the question. I think I was having another blonde moment just a little while ago. Thanks for your help guys :smile:
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