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Kinetic energy of the emitted electrons

  1. Aug 12, 2012 #1
    1. The problem statement, all variables and given/known data
    When the frequency of light incident on a metallic plate is doubled, the K.E. of the emitted photoelectrons will be?

    2. Relevant equations
    The answer should be either doubled or increased but more than double of the previous kinetic energy


    3. The attempt at a solution
    λ= h/ √2mK.E.

    therefore, 1/v= h/c*√2mK.E.
     
  2. jcsd
  3. Aug 13, 2012 #2

    AGNuke

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    Gold Member

    K.E. of photoelectron = (KE of incident photon) - (Work Function of Metal).

    Since you doubled the energy of incident photon, it will no doubt increase the KE of photoelectron, but the extra energy will be purely dedicated in doing so, unlike before doubling the energy, in which some of the energy was required to free electron from the metal.

    So, the new KE will be more than double the initial KE.
     
  4. Aug 13, 2012 #3
    Let's use this equation and say that this is the starting point.

    Now, you have doubled the incident photon's energy, which is one term on the right side. However, the second term, the work function, is still the same. This is the crux of the problem.

    K.E. of photoelectron 1 = (KE of incident photon 1) - (Work Function of Metal).

    K.E. of photoelectron 2 = (KE of incident photon 2) - (Work Function of Metal).

    Now, rewriting this energy in terms of the energy of incident photon,

    K.E. of photoelectron 2 = 2*(KE of incident photon 1) - (Work Function of Metal).


    Difference of the two energies = KE of incident photon 1

    Question: How did I get this relation? Simple enough, but just saves me some typing


    So, the energy of the second photoelectron EXCEEDS that of the earlier one by a magnitude equal to the KE of incident photon 1.

    BUT, KE of incident Photon 1 is MORE than the KE of photoelectron 1 (by first equation.)

    So, yes, answer is more than twice of the initial photoelectron.
    Likewise, the work function has to be paid off only once, for the electron to surface. It has got nothing to do with the incident energy of the photon.
     
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