The collision will be complete inelastic collision in which 13.6eV of energy is absorbed by one. 
$Mv+0=(m+m)V^1$
$V^1=V/2$
Loss of K.E =  $\frac{1}{2}{\text{mV}}^2\text{-}\left(\frac{1}{2}\text{m}{\left(\frac{\text{V}}{2}\right)}^2\text{×2}\right)$
Loss of K.E = 13.6 eV

So,   $\frac{1}{4}m{V}^2=13.6eV$

$V=\sqrt{\frac{4\times 13.6\times 1.6\times {10}^{-19}}{1.67\times {10}^{-27}}}=7.2\times {10}^4 m/s$