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Electron after ionizion

  1. Aug 12, 2011 #1
    If I have, lets say, light of 70nm wavelength. I should be able to calculate the energy like this:
    [itex]E_p = h\nu = \frac{hc}{\lambda}[/itex]

    Now, if the photon ionize a hydrogen atom. Can can I calculate the electron speed using. [itex]E_e = \frac{mv^2}{2}[/itex] And say that the energy is preserved in the process?
    [itex]E_e = E_p[/itex]
    [itex]\frac{m_ev^2}{2} = \frac{hc}{\lambda}[/itex]

    Wish gives the speed:

    [itex]v = \sqrt (\frac{2}{m_e} \frac{hc}{\lambda} )[/itex]
  2. jcsd
  3. Aug 12, 2011 #2


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    You forgot about the potential energy of binding the electron in the atom.
  4. Aug 12, 2011 #3
    Ahh... You're right. The electron has the be, sort of, pulled away. I take it I can discard all but the first energy level, or? Giving me something like this.

    [itex]\frac{13.6*num protons}{1^2} = 13.6 (eV)[/itex] ; number of protons = 1

    [itex]v = \sqrt{\frac{2}{m_e} \left( \frac{hc}{\lambda}-13.6\right)}[/itex]
  5. Aug 12, 2011 #4


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    For Hydrogen you have only one electron to kick out, which require 13.6eV. For bigger atoms your photon may kick out any of electrons (if ots energy is sufficient) - not only 1s one, so you'll have choice of energies to subtract.
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