quantum mechanics - Why do lines in atomic spectra have thickness? (Bohr's Model)

Consider the atomic spectrum (absorption) of hydrogen.

The Bohr's model postulates that there are only certain fixed orbits allowed in the atom. An atom will only be excited to a higher orbit, if it is supplied with light that precisely matches the difference in energies between the two orbits.

But how precise does 'precisely' mean. Of course, if we need energy $E$ to excite the electron to a higher energy level, and I supply a photon with just $E/2$ I would expect nothing to happen (since the electron cannot occupy an orbit between the allowed ones). But what if I supplied a photon with energy $0.99E$, or $1.0001E$ or some such number. What will happen then?

I think that the electron should still undergo excitation precisely because the lines we observe in the line spectrum have some thickness. Which means that for a given transition, the atom absorbs frequencies in a certain range.

Is my reasoning correct? If not, why? How does Bohr's model explain this? How about modern theory?

If I'm right, what is the range of values that an atom can 'accept' for a given transition?

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 $10^{-4}Å$.