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Photoelectric effect question

  1. Feb 5, 2012 #1
    1. The problem statement, all variables and given/known data
    What is the maximum kinetic energy (in eV) of a photoelectron emitted when a surface, whose work function is 5eV, is illuminated by photons whose wavelength is 400 nm
    1. -1.89
    2. 1.89
    3. 0
    4. 3.1

    2. Relevant equations
    hc/(lambda)=work function + KE max

    3. The attempt at a solution
    After plugging numbers in the formula above, I get KE max=-1.89 eV, but I guess I must have done something wrong, since energy cannot be negative.
  2. jcsd
  3. Feb 5, 2012 #2


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    Gold Member

    looks like your calculation is correct. Think about the physical significance.
  4. Feb 6, 2012 #3
    thank you, i've figured it out :)
    since lambda > lambda0, the photoelectric effect cannot occur, then KEmax = 0
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