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The Photelectric Effect

  1. Feb 20, 2006 #1
    Alright, I am having some trouble with the following problem, because my solution I arrive at seems to be way too large. The problem is:

    What is the maximum photelectron speed if potassium is illuminated with light of 250 nm?

    From that I have

    λ = 250 nm
    f = c/λ = 3x10^8 / 250x10^-9 = 1.2x10^15
    Eo of potassium = 2.3 eV

    From there I tried getting the velocity by obtaining the kinetic energy.

    Eelec = hf = (4.14x10^-15)(1.2x10^15) = 4.968
    Kmax = Eelec - Eo = 4.968 - 2.3 = 2.668

    V = √2K/m = √2(2.668)/(9.11x10^-31) = 2.42x10^15 m/s

    So my calculated answer is 2.42x10^15 m/s which is much faster than the speed of light, so I know it must be wrong. I can't seem to figure out another way to arrive at an answer though. Any help is much appreciated.
     
  2. jcsd
  3. Feb 20, 2006 #2
    Your units don't match. You either need to convert the electrons energy to joules or it's mass to eV/c^2 (that's about 511keV/c^2).
    Doing a bit of dimensional analysis is usually useful if you get bizarre results from a calculation.
     
  4. Feb 20, 2006 #3
    Ok, so taking the K of 2.668 eV and converting to joules I got 4.3x10^-19, then plugging those results into the my final equation I ended up with 9.7x10^5 m/s which sounds much more reasonable, but I just want to double check that I did that portion correctly. Was that the result you ended up with?
     
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