# Electron after ionizion

1. Aug 12, 2011

### liquidFuzz

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:

$v = \sqrt (\frac{2}{m_e} \frac{hc}{\lambda} )$

2. Aug 12, 2011

### xts

You forgot about the potential energy of binding the electron in the atom.

3. Aug 12, 2011

### liquidFuzz

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

$v = \sqrt{\frac{2}{m_e} \left( \frac{hc}{\lambda}-13.6\right)}$

4. Aug 12, 2011

### xts

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.