This is going to take a bit, so be prepared. Also, some of what I say will not be exactly correct, in order to get the larger point across, but bear with me.
First, the Rutherford model really did not speak to the question of emission lines. It simply noted that, on the basis of Rutherford's scattering experiments, all of the protons in an atom had to be concentrated in a dense clump. Before this, the fact that protons repel each other suggested that an atom was made of separate protons (which repelled each other) and electrons (which also repelled each other), while the attraction between the protons and electrons kept the whole assembly from flying apart. This was called the Plum Pudding Model.
In this model if you try to fire an $\alpha$ particle through something like gold foil, the $\alpha$ particle ought to hit a large number of small obstacles, and this would produce a specific scattering effect. When Rutherford actually did this, he discovered that this isn't so - the experiment behaved as if there were a small number of very large obstacles. Rutherford concluded that, although it made no immediate sense, the protons were somehow clumped together.
Shortly thereafter, Niels Bohr pointed out that this would make sense if the atom's electrons, being attracted to the clump (which shortly became known as the nucleus) of protons, could be thought of as orbiting the nucleus, just as planets orbit a star. This became the Rutherford-Bohr model. At it happens, it made all kinds of sense out of the known behavior of atoms, and was quickly adopted. It is still taught in grade schools and high schools, even though it has been overshadowed by the more accurate and powerful ideas of quantum mechanics. You can think of it as being like the difference between Newtonian and relativistic physics. For some cases, Newtonian physics works just fine. For some analyses the Rutherford-Bohr model works just fine, too.
Thinking of an atom as a little solar system, you can think about the energy of the planets/electrons. Let's take Mercury, for instance. If you put a really big rocket on Mercury, you could speed it up and cause it to move to a larger orbit. The increase in speed means an increase in energy, so you can say that larger orbits have more energy than smaller orbits.
This leads to the idea that, when an excited electron emits light, it loses energy and falls to a lower (smaller) orbit. With me so far? Now a problem comes up. If you look at the emission spectrum in anna v's answer, you may notice something: there are only a few sharp lines visible. This means that, in an excited gas which is emitting light, billions of electrons are all emitting exactly the same amount of energy when they "de-excite". Think about what this means in the Rutherford-Bohr model.
First: Each hydrogen has one proton in its nucleus, and all protons are identical, so for every hydrogen "solar system" the central sun is exactly the same - check.
Second: Each hydrogen has one electron as a planet, and all electrons are identical, so for every hydrogen "solar system" the planet is exactly the same - check.
Third: Since each "de-excited" planet gives off one of a small number of specific energies, the orbits found in each hydrogen "solar system" must be the same - che - wait, what?
Why in the world would this be true? In a macroscopic solar system, you can get any orbit you want by tweaking the speed of the moving body. So why would electron orbits be restricted to a small number of possibilities?
Well, that's what quantum theory is all about. The energy that a bound electron can have is restricted to integral multiples of some base energy - it is quantized.
And that is why the Rutherford-Bohr model does not explain emission lines. In the case of the hydrogen atom, an electron can switch between different energy levels as it emits light, and the relationships between the possible energy states of the electron produce a well-defined relationship between the different emission spectra. The Balmer and Lyman series are examples of different classes of orbit-switching.
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.
Best Answer
According to Bohr model, the absorption and emission lines should be infinitely narrow, because there is only one discrete value for the energy.
There are few mechanism on broadening the line width - natural line width, Lorentz pressure broadening, Doppler broadening, Stark and Zeeman broadening etc.
Only the first one isn't described in Bohr theory - it's clearly a quantum effect, this is a direct consequence of the time-energy uncertainty principle:
$$\Delta E\Delta t \ge \frac{\hbar}{2}$$
where the $\Delta E$ is the energy difference, and $\Delta t$ is the decay time of this state.
Most excited states have lifetimes of $10^{-8}-10^{-10}\mathrm{s}$, so the uncertainty in the energy sligthly broadens the spectral line for an order about $10^{-4}Å$.