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What is the maximum speed of the ejected electrons? good times!

  1. Apr 21, 2006 #1
    Hello everyone, i'm alittle confused on why this isn't right.
    THe problem asks:
    What is the maximum speed of the ejected electrons?

    They give you the following:
    With the help of others i found:

    If the work function for a certain metal is 1.5 eV, what is its stopping potential for electrons ejected from the metal when light of wavelength 337 nm shines on the metal?
    2.19 V

    Andrew Mason said it best:
     
    Last edited: Apr 21, 2006
  2. jcsd
  3. Apr 21, 2006 #2

    dav2008

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    eV is a measure of energy, not voltage, so in your equation [tex] V_s = h\nu/q_e - \phi[/tex] you are equating potential to potential plus energy.

    Also you set voltage equal to kinetic energy. Again, voltage is energy/charge.
     
    Last edited: Apr 21, 2006
  4. Apr 21, 2006 #3
    Your right, my units don't work out at all. I end up getting volts/kg hm...
    I don't see how this is going to work out, are you saying I should convert my 2.9 V into somthing else?
     
  5. Apr 21, 2006 #4

    dav2008

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    You have to realize what is going on.

    Once the electron has been shot out, it takes a potential of 2.19 Joules/coulomb(if you did your calculations correctly) to stop them.

    If it takes 2.19 Joules/Coulomb to stop the electron, what does that tell you about its initial kinetic energy?
     
  6. Apr 21, 2006 #5

    dav2008

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    I guess the key thing to realize is that all this is is conservation of energy. You don't need to memorize any formulas if you just understand the situation and write a conservation of energy statement for it.

    In the first case you have the energy of the photon converted into the work function energy plus the energy of the electron. Then you have the energy of the electron converted into electric potential (qV)
     
  7. Apr 21, 2006 #6
    Oooo! i got you so it should be:
    .5*mv^2 = (charge of an electron)(2.19 J/C)?
     
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