# Confusion on light absorption by electrons

Say an atom's electron can jump between energy levels n = 2 and n = 3 with x joules of energy. If a light beam of frequency f provides x+y joules (where y is a positive number), that is, more than enough to excite the electron to the next energy state but not enough to reach n = 4. The electron should accept this quantized quantity of energy and then later emit it. On emission spectra, however, the extra "y" J of energy are not accounted for, only the minimum value x to transition between n = 2 and n = 3 (the energy difference between orbital levels). Where, then, does the extra y energy go? It cannot simply "not be absorbed" because it exists along with the rest of the energy, x, in quantized packets. If it contributes to raiseing the KE of the electron, it must at some point be emitted, but that would give way to an incorrect value on the emission spectra. What am I missing?

• "The electron should accept this quantized quantity of energy and then later emit it." Why should it?
– Greg
Aug 2 '17 at 2:36

Depending on how large the value $y$ is, the premise of your question can be wrong.

On the other hand, if you jump between two electronic energy levels where $y$ is on the order of vibrational states, it is entirely possible this extra energy will be stored in vibrations of the molecule on the excited state. This energy is then typically radiated away quite quickly until you reach the ground state of the higher electronic energy level, and then the emission takes place where you could measure the energy $x$ again but in an emission spectrum.

It is entirely possible, however, that the emission process will happen from above the $v=0$ state on the excited surface. It is also possible that you will emit down to different vibrational energy levels on the ground state electronic surface. That is, $0(v=0)\leftarrow1(v=0)$, $0(v=1)\leftarrow1(v=0)$, $0(v=2)\leftarrow1(v=0)$, etc. are all possible and will be observed with varying intensities.

So, in general if you pump in some amount of energy to excite an electron, you don't need to see that same amount of energy coming back out from electronic transitions. You would need to look quite closely at infrared radiation and other energy to make it add up.

As another point, it is actually possible for this energy $x+y$ to be absorbed even if $x+y$ is very far away from the energy of a state. This is how scattering Raman spectroscopy works. You have to think about it differently, however, because the state you are excited to is not a stationary state, but rather a virtual state. See the question I asked here.

So, the short answer is that energy very frequently gets added to other degrees of freedom besides the electronic ones. The same thing goes for vibrational transitions where the excess energy gets put into rotational states. There are rules for how these transitions (and how they couple) will take place.

The other answer, which is less interesting, is that it's totally possible for the molecule to simply not absorb this $x+y$ precisely because there is nowhere to put this extra $y$.

While jheindel made a lot of interesting points, for most purposes the best answer is just the last sentence

that it's totally possible for the molecule to simply not absorb this $x+y$ precisely because there is nowhere to put this extra $y$.

In fact that is the reason that absorption spectra with clear lines exist in the first place: if any transition with energy $x$ could just swallow any $x+y$ as well then (regardless of what ultimately happened with the $y$) any gas would block basically the entire visual spectrum. But because only the particular transition lines can efficiently be absorbed, most of the light gets through instead even if the gas is dense – most photons simply won't interact with the molecules, instead they pass right through/past them as it were.

One way to derive why this happens is a semiclassical view: the energy levels that we're always considering are Eigenstates of the Hamiltonian; that means they don't change in time. In particular, there is no movement of the charge density. However, the transition between two such states $E_0,E_1$ can be viewed as a superposition between two eigenstates, and such a superposition is itself not and eigenstate but “vibrates” with frequency $\frac{E_1-E_0}{h}$. It thus acts as an “antenna” tuned just so it can interact with photons of energy $E_1-E_0$.