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Finding velocity

  1. Jul 31, 2011 #1
    1. The problem statement, all variables and given/known data
    If the energy of H-atom in the ground state is -E, the velocity of the photo-electron emitted when a photon having energy EP strikes a stationary Li2+ ion in the ground state is given by:
    (a)[itex]v=\sqrt{\frac{2(E_P-E)}{m}}[/itex]
    (b)[itex]v=\sqrt{\frac{2(E_P+9E)}{m}}[/itex]
    (c)[itex]v=\sqrt{\frac{2(E_P-9E)}{m}}[/itex]
    (d)[itex]v=\sqrt{\frac{2(E_P-3E)}{m}}[/itex]


    2. Relevant equations
    E=W+K.E.
    (W represents work function)


    3. The attempt at a solution
    Energy of H-atom in ground state, [itex]E_H=-E[/itex]
    Energy of Li+2 ion in ground state, [itex]E_Li=-2.18*10^{-18}J*3^2=-9E[/itex]

    [tex]K.E.=E-W[/tex]
    [tex]\frac{1}{2}mv^2=E_P-(-9E)[/tex]
    [tex]v=\sqrt{\frac{2(E_P+9E)}{m}}[/tex]

    But the answer in the answer key is option (c). :confused:
     
  2. jcsd
  3. Jul 31, 2011 #2
    The sign of the energy is a little confusing in this problem. Remember, energy is conserved, so the total energy before the photon is absorbed is that same as the total energy after. "Total" means you add all the relevant energies. But remember also that the energy of the atom before absorption of the photon is negative.

    I would recommend that, instead of the photoelectric equation, you use the conservation of energy equation.
     
  4. Jul 31, 2011 #3
    Sorry, i didn't get you.
    Which is the conservation of energy equation?
     
  5. Jul 31, 2011 #4
    Total energy before = total energy after
     
  6. Aug 1, 2011 #5
    I know this but what formulae should i use here?
     
  7. Aug 1, 2011 #6
    I have found the mistake in my solution. :smile:
    -9E is the energy of Li2+ ion.
    Therefore i need +9E energy to remove an electron.
    So the work function becomes +9E rather than -9E and i get the answer as (c) option.

    But would you please tell me your way to solve it? :smile:
     
  8. Aug 2, 2011 #7

    I like Serena

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    Homework Helper

    Hi Pranav-Arora! :smile:

    The energy before (initial energy Ei) is the energy of the photon plus the energy of the Lithium ion, which is Ei = Ep + (-9E).

    The energy after (final energy Ef) is the kinetic energy of the electron plus the new energy of the Lithium ion, which is Ef = (1/2)mv2 + 0.

    Energy conservation says Ei = Ef.
    So Ep + (-9E) = (1/2)mv2 + 0.

    From this you can deduce v.
     
  9. Aug 2, 2011 #8
    Got it. Thanks for the explanation. :smile:
     
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