1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    xts

    User Avatar

    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

    xts

    User Avatar

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook