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Photoelectrons and Kmax

  1. Feb 6, 2009 #1
    1. The problem statement, all variables and given/known data

    A metal gives off photoelectrons with maximum kinetic energy Kmax at the rate of N per second when a beam of light of wavelength λ shines on it. How will Kmax and N change if the intensity (brightness) of light is doubled? Assume 100% efficiency.

    2. Relevant equations

    hc/wavelength=? then. P/?=

    h=planks constant
    c= speed of light
    p=power
    3. The attempt at a solution

    kinetic energy will remain the same
    n will double

    Im not sure how to do this problem
    if the brightness doubles, will the frequency double? or the wavelength?
    and im not sure how that would affect n per second

    so what im basically asking is. Is the intensity of light related to the frequency, and if so, in what way?
     
    Last edited: Feb 6, 2009
  2. jcsd
  3. Feb 7, 2009 #2

    Hootenanny

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    Light intensity is a measure of the [time-averaged] energy flux and has S.I. units of W.m-2. In other words, light intensity is a measure of the average amount of energy passing through the unit surface. In terms of photons, the intensity of light is directly proportional to the number of photons passing through the unit area per unit time.

    Perhaps this will help: http://www.newton.dep.anl.gov/askasci/phy05/phy05113.htm
     
    Last edited: Feb 7, 2009
  4. Feb 8, 2009 #3
    ok. so does that mean that the photons would double in number since the brightness doubles, and kmax would remain the same.
    is this right?
     
  5. Feb 9, 2009 #4

    Hootenanny

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    Sounds about right to me :approve:
     
  6. Feb 9, 2009 #5
    alright thanks :)
     
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