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Find minimum velocity of emitted electron

  1. Aug 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Photoelectric experiment show that about 5eV of energy are required to remove an electron from platinum, if light of 150nm wavelength is used, what is the velocity of the emitted electron?


    2. Relevant equations
    de broglie's equation, E=hv


    3. The attempt at a solution
    Tried by using the de Broglie's equation, to find the velocity but wasn't successful.
     
  2. jcsd
  3. Aug 6, 2012 #2
    5eV is energy, and 150 nm can also be converted to the energy of the photon. The diff between these two gives you the energy of the electron.
     
  4. Aug 6, 2012 #3
    The energy of the electron, i got is 77eV, is it correct?
    But I am suppose to find the velocity of the electron?

     
  5. Aug 6, 2012 #4
    The excess energy becomes kinetic energy.
     
  6. Aug 7, 2012 #5
    The answer what i got after considering 77eV as the kinetic energy, was not the original one, the answer should be 1.058m/s
    I mentioned in my previous comments all the possible values that I managed to find, kindly check them and let me know if i have made a mistake, even the concept and its application seems to be right to me as per the hints received.
     
  7. Aug 10, 2012 #6
    Hi ill try to explain it.
    Convert the workfunction ( 5ev) into energy and then solve
    E=hv for v ( frequency) youll get the threshold frequency ( we call it x) that is the frequency required. then
    put the value in the following equation
    K.E.=h(v-x) v is the frequency of the light and then solve it for velocity and youll get the answer.
     
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