## Electron after ionizion

If I have, lets say, light of 70nm wavelength. I should be able to calculate the energy like this:
$E_p = h\nu = \frac{hc}{\lambda}$

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

Wish gives the speed:

$v = \sqrt (\frac{2}{m_e} \frac{hc}{\lambda} )$
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 You forgot about the potential energy of binding the electron in the atom.
 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. $\frac{13.6*num protons}{1^2} = 13.6 (eV)$ ; number of protons = 1 $v = \sqrt{\frac{2}{m_e} \left( \frac{hc}{\lambda}-13.6\right)}$

## Electron after ionizion

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.