What you're describing does happen to some extent. It's called "Doppler broadening": absorption and emission lines in hot materials are wider (in wavelength) than absorption and emission lines in cool materials, because atoms in the hot material occupy a broader range of velocities.
An atom at rest can't absorb any old photon and convert the extra energy into energy of motion, because such collisions must conserve both energy and momentum. Let's start with the decay of an excited atom by photon emission. An $n=1$ excited hydrogen atom at rest has energy
$$E_\text{excited} = m_\text{ground}c^2 + \Delta E \approx \rm 1\,GeV + 10\,eV$$
and decays to a ground-state atom nearly at rest and a photon, with energies
$$
E_\text{ground} = mc^2 + \frac{p^2}{2m} \approx \rm 1\,GeV
\qquad \it E_\gamma = h\nu \approx \rm 10\,eV
$$
Conservation of energy tells us that
$$
\Delta E = \frac{p^2}{2m} + h\nu
$$
By conservation of momentum we must have the same magnitude of momentum $p$ for the atom and for the photon. The photon's momentum is $p_\gamma = h\nu/c$, so we get
$$
\Delta E = \frac{1}{2m} \left(\frac{h\nu}{c}\right)^2 + h\nu = h\nu \left( \frac{h\nu}{2mc^2} + 1 \right) \approx h\nu
$$
So you can see that the atom does get a kick from emitting the photon, but it's at the parts-per-billion level of the photon's energy.
You're asking about absorption. In the center-of-mass reference frame, every absorption must look exactly like the process we've just gone through, only with the initial and final states switched. If you want your ground-state hydrogen atom to absorb a 12 eV photon, it must be "running away" from the photon so the the Doppler-shifted energy in the center-of-mass frame corresponds to 10 eV.
This is different from elastic collisions between billiard balls, because there is no photon in the final state, and different from inelastic collisions like the "ballistic pendulum" because the atom can only absorb energy from the electromagnetic field, and thus momentum from the electromagnetic field, in specially-sized lumps.
We can be even more specific. Let's start with an incident photon of unspecified energy and a ground-state atom at rest,
$$
E_\text{ground} = mc^2 \qquad E_\gamma = h\nu
$$
and try to end up with a "kicked" excited atom,
$$
E_\text{excited} = mc^2 + \Delta E + \frac{p^2}{2m}.
$$
From conservation of energy we have
\begin{align}
E_\text{excited} &= E_\text{ground} + E_\gamma \\
\Delta E + \frac{p^2}{2m} &= h\nu. \tag{A}
\end{align}
By the same logic as above we have $|p| = h\nu$ (though now the directions are parallel, rather than antiparallel, since we are not in the center-of-momentum frame), which gives
\begin{align}
\Delta E &= h\nu \left( 1 - \frac{h\nu}{2mc^2} \right) \\
0 &= \Delta E - h\nu + \frac{1}{2mc^2} (h\nu)^2 \\
h\nu &= mc^2 \left(
1 \pm \sqrt{ 1 - \frac{2\Delta E}{mc^2} }
\right)
\end{align}
This quadratic equation has exactly two solutions: $h\nu \approx 2mc^2$ (a relativistic photon interaction that we're not interested in) and the one we expected, $h\nu \approx \Delta E$. If we keep two terms in the binomial expansion for the square root we get
\begin{align}
\frac{h\nu}{mc^2} &\approx 1 - 1 + \frac12 \frac{2\Delta E}{mc^2} + \frac18 \left(\frac{2\Delta E}{mc^2} \right)^2
\\
h\nu & \approx \Delta E \left( 1 + \frac12 \frac{\Delta E}{mc^2} \right), \tag{B}
\end{align}
which is more than the "transition energy" by an extra amount $\epsilon = (\Delta E)^2/2mc^2$.
That's the only extra energy that you're allowed to impart to the atom at rest, and it's required by momentum conservation and small enough to usually be negligible.
When a photon hits a boundary condition , three things can happen: a) it can scatter elastically, which means it retains its frequency but changes angle, b)it can scatter inelastically, which means it changes frequency, or c) it can be absorbed raising the energy level of an electron ( in a lattice, in a molecule, in an atom) and a different photon is emitted and phases are lost.
Q1: If a photon with 10.1 eV energy (insufficient to excite electron) would hit the atom of the hydrogen what would happen? Will the photon be absorbed by the atom and immediately emitted and the emitted photon (or photons?) will have the same 10.1 eV energy? Or the photon will pass through the atom or what would happen?
The hydrogen atom hit with a photon of energy lower than an energy level transition falls under a) or b) The photon will scatter elastically in the center of mass with the total atom and go on its way at adifferent angle, or inelastically giving kinetic energy to the whole atom and changing frequency.
Q2: Same question as the above one in this case our photon has a slightly more energy lets say it has 10.3 eV. Again what would happen? Will the atom absorb the photon and excite the electron but since the energy of the photon exceeds the required energy to excite the electron will the atom emit a photon with 0.1 eV energy or what will happen in this case?
If the extra energy of the photon is not within the energy width of the hydrogen energy level, again it will go on its way scattering elastically or inelastically in the center of mass "photon atom" . If the energy of the photon is higher than the ionization energy of the atom, the work function, the electron may be kicked off and the ion proton remain. The photoelectric effect.
One has to realized that at the quantum mechanical level it is probabilities that are important. The probability for a photon of the correct energy difference to raise the electron of an atom is very high, with the wrong energy difference. very very small.
For bulk matter interaction see this answer of mine here.
Best Answer
No, it is sufficient for the photon energy to exceed the band gap. Any excess energy is transformed into kinetic energy for the electron in the new band. You get exactly the same effect when ionizing an atom - the excess energy simply powers the electron into a faster continuum state.
You should also take into account that photon energies are never exactly defined except for monochromatic beams with infinite temporal duration. This is exactly because of the energy-time uncertainty relation: the only way to have a perfectly defined photon frequency, and hence energy, is to observe it for an infinitely long time. Thus, the photon energy is always spread out over a finite bandwidth.
A similar effect holds for atomic bound-bound transitions, which will always have a finite natural linewidth. This is caused by spontaneous emission, which means that if you leave the atom in an excited state for long enough then it will eventually return to the ground state. This then limits the amount of time in which you can coherently probe the frequency of the transition, and in turn limits the precision to which you can measure this frequency.