# How exact does the wavelength of light have to be to be absorbed by an atom? [duplicate]

I have been reading sources such as this on absorption and emission which make statements such as:

The interesting thing is that each atom will only absorb photons with exactly the right energy.

I am struggling to understand this because how exact does the energy/wavelength have to be? Would an atom absorb light with a wavelength of 455 nm but not 455.5 nm? or 455.05 nm etc?

• That depends on the width of the transition... Commented Feb 27, 2016 at 21:50
• Commented Feb 27, 2016 at 21:51
• As an absolute statement that is certainly incorrect. The natural line widths of atoms can be extremely narrow, but the absorption and emission lines are not exactly at one frequency. That's impossible, since otherwise the absorption/emission process would take an infinite amount of time. Commented Feb 27, 2016 at 21:51
• Possible duplicate of How precise must the energies match for absorption of photons? Commented Feb 28, 2016 at 11:56

I will try and put the comments into a little more context. When an atom absorbs a photon, or energy package, the electron makes a transition from a lower (energy-) state (usually "ground" state) to an excited state. An excited state is not stable, that means the atom will not stay in that (energy-) state, simply for reasons of energy minimization. Therefore, the energy will be emitted. This is not an instantaneous process, i.e. it takes some time before the energy is emitted. This time is called the life-time of the excited (energy-) state. This can be combined with Heisenbergs uncertainty principle:

$\Delta E\cdot\Delta \tau >= \hbar/2$

$\Delta E = \frac{\hbar}{2\Delta\tau}$

($\tau$ = lifte-time, $\Delta E$ energy "width")

This tells us that the energy of the excited state has a certain "width" which is called natural linewidth. The energy of an absorbed photon can be written as $E=hf$. Therefore the energy width of our excited state can be converted to a frequency uncertainty $\Delta E = h\Delta f, \Delta f = \frac{\hbar}{h\Delta\tau}$. This does not only imply that the excited state has a certain width. It also tells us that our transition frequency may fluctuate by $\Delta f$ and we can still do the transition.

Thus, we also see what was mentioned in the comments: If our excited state was stable (infinite life-time) then you our frequency spread $\Delta f = 0$ and you have to exactly match the frequency.

As Jon Custer said in a comment, the answer is "it depends".

For a perfectly cold atom that is suspended away from all other atoms / interactions, and that has been in a certain state "forever", the line width of the transition would indeed be extremely narrow.

But for a real atom, in a gas, the line width is finite. The phenomenon of spectral broadening can be used to determine things like temperature and pressure of gases in distant stars, and is the reason why the light from certain "high pressure sodium" lamps looks whiter, where the low-pressure version is "quite yellow".

According to the article linked, the natural linewidth depends on the lifetime of the transition - if that is on the order of $10^{-8}~\rm{s}$, the line width is on the order of $10^{-7}~\rm{eV}$. In a high pressure gas, frequent collisions shorten the lifetime of the transition, and cause significant line broadening; similarly, Doppler shifting (due to the velocity of the atom relative to the observer) will cause temperature-dependent line broadening, with the greatest effect for the lightest atoms (e.g. hydrogen) which have the highest velocity at a given temperature. The Doppler line broadening is given by

$$\frac{\Delta \lambda}{\lambda}=2\sqrt{2\ln 2\frac{kT}{m_0c^2}}$$

A more detailed analysis can be found at this link