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?
Answer
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.
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