Two H atoms in the ground state collide in elastically. The maximum amount by which their combined kinetic energy is reduced, is

Total energy of two H-atom in ground state $=2(-13.6)=-27.2\mathrm{eV}$.
The maximum amount by which their combined kinetic energy is reduced when any one H-atom goes into first excited state after the inelastic collision, that is, the total energy of two H-atom after inelastic collision :
$\begin{array}{l}\mathrm{E}=\frac{13.6}{{\mathrm{n}}^2}+13.6\\ \left.=\frac{13.6}{2^2}+13.61\text{ [For excited state }(\mathrm{n}=2)\right]\\ =3.4+13.6=17.0\mathrm{eV}\end{array}$
So that the loss in kinetic energy due to inelastic collision will be,$=27.2-17.0=10.2\mathrm{eV}$