Suppose a moving hydrogen atom makes a head-on collision with stationary H atom. Both atoms are in ground state before collision, and after collision they move together. The minimum value of kinetic energy of the moving hydrogen atom is $(10.2\times \lambda )eV,$ such that one of the atoms reaches the 1st excitation state. Find  $\lambda$.

Applying momentum conservation

$$\begin{gathered} mv+0=(m+m)v^{\text{'}} \\ \Rightarrow v^{\text{'}}=\frac{v}{2} \end{gathered}$$

Now energy balance gives 

$\frac{1}{2}m{v}^2=\frac{1}{2}\times 2m\times v^{\text{'}2}+E$

Where E is the amount of energy liberated in the collision and absorbed by one of the atoms.

$\frac{1}{2}m{v}^2=\frac{1}{4}m{v}^2+E\Rightarrow \frac{1}{4}m{v}^2=E$

$\frac{1}{2}m{v}^2=2E$

$E=10.2eV$  is required to excite one of the atoms to Ist excitation state.
$\therefore$ Minimum KE of the moving hydrogen atom  $=2\times 10.2=20.4eV$