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Kinetic energy of the photoelectrons

  1. Apr 6, 2016 #1
    1. The problem statement, all variables and given/known data
    A beam of ultraviolet light with wavelength of 200 nm is incident on a metal whose work function is 3.0 eV. Note that this metal is applied with +1.0 V with respect to the ground. Determine the largest kinetic energy of the photoelectrons generated in this process.

    2. Relevant equations
    Kmax = hf - phi

    Kmax = q*|V|

    3. The attempt at a solution

    I am not entirely sure what the statement about the +1.0 V implies and hence I simply used the formula
    Kmax = hf - phi . Substituting the appropriate values gives Kmax = 5.139*10^-19 J . Am I on the correct path ?
     
  2. jcsd
  3. Apr 6, 2016 #2

    DrClaude

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    Staff: Mentor

    You got the result in the absence of the applied potential. You must now modify the kinetic energy to take into account the potential.
     
  4. Apr 7, 2016 #3
    how do you take into account the potential?
     
  5. Apr 7, 2016 #4

    DrClaude

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    Staff: Mentor

    By calculating its effect on the released photoelectron.
     
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