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Frequency of EMR

  1. Oct 7, 2005 #1
    Can some please review this question and tell me if the working looks correct? Please advise of any mistakes i may have made. Thanks

    Q: UV radiation (wavelength = 250 nm) falls on a metal target, and electrons are liberated. If the maximum kinetic energy of these electrons is 1.00*10^-19 J, what is the lowest frequency EMR that will initiate a photocurrent on this target?

    A: E = Ø + K = h*(c/λ)

    Therfore Ø = ((h*c)/250*10^-9) - 1*10^-19
    Which = 6.96*10^-19 J
    = h*f

    Therefore f = 1.05*10^15 Hz

    Is this what i am supposed to do? Thanks for any help! :smile:
  2. jcsd
  3. Oct 7, 2005 #2


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    Your final answer is correct.

    I find it peculiar that you group " Ø + K " together
    - it's NOT the total Energy " E " of any configuration!
    Rather, it's the NEGATIVE Energy change of the electron
    (- E_initial + E_final) ... so it does equal E_photon,initial.

    Most of us learn conservation laws as
    "final = initial + change" or "change = final - initial" ,
    once in a while as "in - out = change".
    Is there a word for this "negative change"?
  4. Oct 9, 2005 #3
    This is what my demonstrator told me to use?
  5. Oct 9, 2005 #4


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    Do you see how it comes from E conservation?
    Ø is just the negative of the electron's initial PE, etc.
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