# 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?

• Talking about Bohr model: I don't think it's explain it at all; in modern theories the thickness of absorption/emitting lines due to the uncertainty principle (the one about time-energy). And this was discovered far later than the Bohr model. – m0nhawk Mar 29 '15 at 16:01
• Bohr's model was a kludge that barely lasted a decade as the model that scientists actually used. It's still taught, not because it is right but because (a) it's easy and (b) it is a stepping stone to a better theory. With emphasis on the easy part. This question was never tackled in it. – dmckee Mar 29 '15 at 16:02
• I don't think the Bohr model can solve for this, but see this answer for the modern QM physics.stackexchange.com/questions/443054/… – user213887 Dec 7 '18 at 15:59

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 $10^{-4}Å$.

• It's important to emphasize that the natural linewidth is rarely observed directly unless one makes specific efforts to reduce all other types of broadening. In particular, at room temperatures the Doppler broadening - the Doppler shifting of the line according to the thermal distribution of velocities in the gas - will generally swamp most other types of line broadening. – Emilio Pisanty Jun 26 '15 at 11:10
• Surely this answer is quite wrong...you can't calculate the linewidth of an atomic transition simply by applying the Heisenberg relationship to the size of the transition??? – Marty Green Nov 15 '18 at 20:07

The answer is yes, the atom does absorb radiation that does not exactly match the transistion frequency. This is due to the Doppler effect that everyone knows from an ambulance with siren driving by. The frequency you hear is higher if the ambulance moves towards you and lower if it drives away from you.

It's the same with the atom. If the atom moves (and it does unless you cool it down to really low temperatures) the observed frequency of the radiation you shine in is shifted depending on the direction of travel of the light and the direction and velocity the atom moves (on the scalar product of both). The phenomen you described is called Doppler broadening of spectral lines and I would say the effect can be described using Bohr's model, since it is a purely classical effect.

The technique to get rid of these broadend lines is called Doppler free spectroscopy. It makes use of some cool techniques you can easily google.

Edit: There are more effects of broadening (like those m0nhawk mentioned in his answer). But under normal conditions the doppler broadening has the biggest effect of all those and overlays the others.

Edit 2: Wolfram alpha offers a tool to calculate the thermal doppler broadening. It says that the line in the picture above ($486\mathrm{nm}$) at $T=300\mathrm{K}$ is broadend $\Delta\lambda\approx 4\cdot10^{-2}Å$

• So, if I'm correct, what you're saying is that light of frequencies different from that of the transition frequency is doppler shifted to the transition frequency. However, the atom still must 'accept' a range of values, since there is a 0 probability that any frequency will be shifted to the exact transition frequency. – Gerard Mar 29 '15 at 16:31
• Actually the final acceptance is governed by the energy-time uncertainty m0nhawk mentioned in his answer. But normally (in 19th century physics) you do not observe single atoms. When you take $10^{23}$ atoms , their velocities are not the same but distributed (Maxwell&Boltzmann distribution). So you have a good chance that some of your atoms actually match the doppler shifted light frequency. Since they move in different directions with different velocities your observed spectral line is quite broad. – bernd Mar 29 '15 at 16:38
• This is only part of the answer. Even using spectroscopy that eliminates the Doppler shift, there is an innate Lorentzian line spread. – zeldredge Mar 29 '15 at 16:42

The linewidths come out very naturally from Maxwell's Equations by treating the atom as a tiny classical antenna. I do the calculations for the 2p-1s transition in hydrogen on my blogsite here: The Semi-Classical Calculation

The idea is that from the Schroedinger equation, the superposition of the s and p states gives you get a tiny oscillating dipole about 1 Angstrom in length, a frequency of 2.5 x 10^15 Hz, and a wavelength of 1200 Angstroms.

From the ratio of the antenna size to the wavelength, you can easily calculate the radiation resistance: it is on the order of 100 micro-ohms.

The "current" in the antenna can be estimated by taking the charge of the electron times the frequency. This is not quite right but it's close enough for what we're doing here. Oddly enough, for the hydrogen atom it comes out to the seemingly macroscopic value of around 1 milliamp.

The power in the antenna is just I-squared-R, which gives us 100 pico-watts. But what about the energy in the excited state? It's 3/4 Rydberg, which works out to around .000 001 pico-joules. So clearly the "lifetime" of the excited state is around 10-8 seconds.

You can continue with the classical analysis to get your linewidth by taking the ratio of the "lifetime" to the single-cycle time. It's called the Q factor in antenna theory. Or your can use the language of quantum mechanics, as other posters have done here, and express the linewidth broadening in terms of the uncertainty principle. It's exactly the same thing.

But you can't apply the uncertainty principle until you calculate the decay constant, or "lifetime". And that comes straight out of classical antenna theory.

EDIT: Just noticed that the accepted answer claims you calculate the linewidth by simply applying the Heisenberg Uncertainty principle. Surely this is quite incorrect. In the example given, the "lifetime" is taken as given...but it is precisely the lifetime (which is inverse to the linewidth) that we want to calculate. You don't get the linewidth by applying Heisenberg to the lifetime...you get it by dividing the lifetime by the speed of light. And that begs the question...how did you get the lifetime?

As I've explained already, you get the lifetime by applying the classical antenna equations to the vibrating atom. Its a very classical calculation.

• Thanks for the edit, Noah. And thanks for my first lesson in LATEX. Now I know that the code for 10^(-8) is DOLLARSIGN\ 10^{-8}DOLLARSIGN (with actual \$'s instead of DOLLARSIGN). – Marty Green Mar 30 '15 at 2:41
• More advice: on Stack Exchange in general, using CAPITALS for an entire word denotes SHOUTING, and that's the way it is read, as if the readers are getting shouted at. It is unequivocally disliked. :) – 299792458 Mar 30 '15 at 6:35
• Just edited my answer and in doing so I think I screwed up the LATEX corrections made by Noah way back when...sorry about that. – Marty Green Nov 15 '18 at 20:33

## protected by Qmechanic♦Mar 29 '15 at 22:01

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