If we have an atom with a valency of 1, say hydrogen, and strip the electron from the nucleus, what speed would a free electron have to be travelling near the nucleus to prevent a fall into the Coulomb potential well?
~2.2*10
^{3} m/s Km/s, if I computed well.
The ionization energy of hydrogen is about 1.3*10
^{6} joule/mol:
http://en.wikipedia.org/wiki/Hydrogenit means that is the minimum kinetic energy an electron shoud have not to bind with the proton (if, paradoxically, it would bind having a greater energy than that, the atom would ionize immediately, releasing the electron again).
From E
_{kinetic} = 1/2 m v
^{2}, m = electron mass, you find the answer.
I have used the non-relativistic equation for kinetic energy because I have verified, a posteriori, that the computed speed for the electron is negligible with respect to light speed. If instead the result were comparable with light speed, I would have discovered that my approximation were false, and I would have re-done my computation using the (a bit more complex) relativistic equation. Physicists' trick
--
lightarrow