Explanation:
D Given,
Kinetic energy of electron $=5.5 \mathrm{eV}$
We know that,
For $\mathrm{n}=1, \mathrm{E}_{1}=-13.6 \mathrm{eV}$
And, $\mathrm{n}=2, \mathrm{E}_{2}=-3.4 \mathrm{eV}$
Again, $\Delta \mathrm{E}=\mathrm{E}_{2}-\mathrm{E}_{1}$
$\Delta \mathrm{E}=-3.4-(-13.6)$
$\Delta \mathrm{E}=10.2 \mathrm{eV}$
When an electron collides with hydrogen atom in the ground state with energy of $5.5 \mathrm{eV}$ which is less than $10.2 \mathrm{eV}$ then it will not be able to excite hydrogen atom into first excited state. Therefore, electron will not give any energy to the hydrogen atom.
Hence, total kinetic energy of electron remains conserved and its collision with hydrogen atom is elastic.