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Photoelectric exp - calculate intensity of incident light

  1. Oct 3, 2014 #1
    1. The problem statement, all variables and given/known data
    In an photoelectric effect experiment , a photoelectric current of 100µA is obtained when lights of 550nm is incident on metal cathode of surface area 1.0cm^2 .... calculate the intensity of the incident light. The ans is 2.26w/m^2

    2. Relevant equations


    3. The attempt at a solution

    intensity= power /area
    intensity= (voltage x current) /area
    intensity= ( (hc/λ) x I ) / area
    = (6.63x10^-34)(3x10^8)(100x10^-6) / (550x10^-9)(1.0x10^-4) = 3.61x10^-19 (W/m^2 )

    why am i wrong?
     
  2. jcsd
  3. Oct 4, 2014 #2

    ehild

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    How is the photoelectric current related to the number of electrons ejected? What is the relation between the number of electrons ejected and the number of the incident photons?
    If you check the unit of your result, it is not W/m2.
     
  4. Oct 4, 2014 #3
    can you please tell me what's the unit of my 3.61x10^-19 should be? i am confused now.
     
  5. Oct 4, 2014 #4

    ehild

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    The problem is not with the units. If you check the unit of your result and it is not what it should be then your solution is wrong. Your solution is wrong, as its dimension is not energy/(time area).
    Read the first two sentences of my previous post.

    ehild
     
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