How precise must the energies match for absorption of photons? According to Quantum Mechanics, in order for an atom to absorb a photon the energy of the photon must be precisely that of a "jump" between energy states of the atom.
How precise must it be? 
If I create a photon with an energy within an error of 0.0001% of that of an energy state, will it be absorbed by my atom?
 A: In atoms the energy levels do not have a precise energy. When you solve Schrodinger's equation for an atom the results are the energy eigenfunctions. However these are functions that are time independent, and they have an exact energy only because they are time independent.
At the risk of oversimplifying, you can regard this as an example of the energy time form of the Heisenberg uncertainty principle:
$$ \Delta E \Delta t \ge \frac{\hbar}{2} $$
If $\Delta t$ is the lifetime of a state then $\Delta E$ is the uncertainty in the energy of that state. For the energy eigenfunctions $\Delta t = \infty$ so $\Delta E = 0$ and the energy is precisely defined.
The point of all this is that in an atom an excited state has a finite lifetime and therefore it has a finite energy uncertainty, and this produces an effect called lifetime broadening. This means transitions to and from the state can occur for photons with a range of energies. The range of energies allowed depends on the energy uncertainty of the state, which in turn depends on its lifetime.
A: Agree with the above, but also if the atom, or collection of atoms, are in thermal equilibrium, then there is another broadening mechanism, besides lifetime broadening, called Doppler broadening that accounts for the motion of the atom(s).  This has the effect to substantially widen the effective line width depending upon the temperature.
A: The linewidth broadening everyone is talking about is actually a very classical effect that comes straight from antenna theory and depends only on the size of the antenna as compared to the wavelength of light. It is well known in classical antenna theory that the bandwidth of a lossless short antenna goes as the cube of the electrical length (the physical length divided by the wavelength). For the s-p transition of the hydrogen atom, this parameter is close to the fine structure constant, 1/137. The cube of this number gives the (dimensionless) bandwidth of around 10^7.
Since the frequency of the transition is around 10^16, this gives a transition time of around 10^-9 seconds. I think this is about right for the hydrogen atom. You just treat the atom as a classical antenna and everything comes out. 
