# Hydrogen excitation by collision

Which is the minimum kinetic energy that a Hydrogen atom need to have to excite by collision a second Hydrogen atom initially at rest. Before collision both atoms are in the ground state. The ionization energy of Hydrogen atom is $E_{ionization} =13.6 eV$

The equation for the __conservation of energies__ before and after collision is

$m v^2/2 = m v_1^2/2 +mv2^2/2 + \epsilon^*$

where $\epsilon^*=3hcR/4$ is the first excitation energy of hydrogen atom from ground state

$m=1\rightarrow n=2$ and $R=10.974*10^6 (1/m)$ is the Rydberg constant

But since for all collisions also __the total momentum need to be conserved__ we have

$v_2 =v -v_1$ since both atoms have the same mass $m$

From both momentum and energy conservation equations above we get

$2v_1 =v+\sqrt{v^2-(4\epsilon^*)/m}$

The condition of possible excitation is that the speed of the hydrogen atom after collision is real so that

$v^2\geq 4\epsilon^*/m$

$m v^2/2 \geq 2\epsilon^*$

which is equivalent to

$E_{min} =3*E_{ionization}/2 =20.4 eV$