Inelastic collision will take place, if a part of incident kinetic energy is utilised in exciting the atom. Here, one atom is to be ionised, i.e. $∆E=13.6 \mathrm{eV}$

Assuming completely inelastic collision.



For an inelastic collision

$v_1=v_2=v$

By momentum conservation,

$mv=2mV$

$\Rightarrow V=\frac{v}{2}$

$\frac{1}{2}m{v}^2=\frac{1}{2}\cdot 2m{V}^2+∆E$

$\begin{array}{cc}=m{\left(\frac{v}{2}\right)}^2+∆E& \left[\because V=v/2\right] \end{array}$

$\frac{1}{4}m{v}^2=∆E$

$v=\sqrt{\frac{4∆E}{m}}$

$v_{\mathrm{m}\mathrm{i}\mathrm{n}}=\sqrt{\frac{4\times 13.6\times 1.6\times {10}^{-19}}{1.67\times {10}^{-27}}}$

$=7.2\times {10}^4\mathrm{ m/s}$