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Just for the sake of an example, let's assume a transition from one energy state to a higher energy state in atomic absorption spectrum

If the energy difference between the ground state and first excited state is E and that between ground state and second excited state is 3 E (just for say...). Now if an incident photon has an energy of 2 E so;

(Assuming the electron is initially in ground state)

  1. Will the electron show transition to the first excited state and emit a photon of energy E

OR

  1. Will the electron not show any transition and not accept the photon at all, since the energy of the photon does not exactly match with the energy difference between any two specific states?

The example of atomic absorption spectrum is taken up just for the sake of explaining the question, the question is rather general and pertains to any kind of spectroscopy experiment in which their occurs transition from one state to another by absorption of EM waves.

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  • $\begingroup$ There is a another aspect to the question. A photon doesn't just have to interact with the least bound electron. A photon can interact with any electron in the atom. So the photon could excite the atom by promoting the second or third least bound electron too, if it has enough energy. $\endgroup$ – MaxW Oct 6 '18 at 17:55
  • $\begingroup$ @MaxW The point I'm trying to make is that if the energy of each and every possible transition, be it from any state to any state doesn't match that of the photon(which has a energy sufficiently greater than the least energetic transition), so is it possible that the electron accepts a little amount of energy from the photon does the transistion and at the same time ejecting a photon of the remainder energy? $\endgroup$ – think__tech Oct 6 '18 at 18:06
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    $\begingroup$ The gist is that all sorts of things are possible, what you need to focus on is the probability of such occurrences. The reason that atomic absorption works is that a photon of exactly the right energy for the transition has a high probability of exciting the atom. If a lower energy photon must be scattered, then the probability is much lower. $\endgroup$ – MaxW Oct 6 '18 at 18:11
  • $\begingroup$ @MaxW Ok so let me conclude it this way(correct me if I'm wrong). Both the possibilities mentioned in the question can take place, but it's just a matter of probability as to which will take place(and mostly the probability of the former is greater). Also if you could tell why is the probability of the former greater. $\endgroup$ – think__tech Oct 6 '18 at 18:12
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If the energy is to be transferred from the photon to your atom one energy gap must match the energy of the photon. Thus absorption may occur from the first to second excited state in your example as the energy in both is 2E. However, there has to be some population in the first excited state for this to happen, zero population zero absorption. In fact it is a bit more subtle in that there has to be a difference in population between the two levels involved. It is possible for the photon to stimulate a transition from the upper level to the lower one, just as from the lower one to the upper one. The reason why this seems odd is that in most cases the population in the upper level is vastly smaller than in lower ones (via Boltzmann distribution) and the population in the upper level can often be set to zero without serious error. This is particularly true in electronic spectroscopy, as in the case of atoms, but far less so for pure rotational transitions in molecules which occur in the microwave region of the spectrum. You will find details in any good spectroscopy book.

Thus neither of your cases occurs but something different; transitions between the first to second excited state. How strong this transition is depends on the population in the two levels and on selection rules involving angular momentum and electron spin changes.

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The transitions between quantum states follow selection rules.

https://en.wikipedia.org/wiki/Selection_rule

During any transition, energy, momentum and quantum angular momentum (spin) must be preserved. Those are the hard rules. Also the incoming EM wave has to create an electric field oscillation that overlaps with the electronic waveform oscillation that would occur during the transition (they couple). This overlap can be significant for a given transition and the EM wave frequency matching its energy difference, which makes for a percievable probability of absorption, or very low, in which case the transiton is forbidden. How strictly forbidden depends on the circumstances (neighbouring atoms, collisions, ...).

The usual transitions absorb the complete energy of an incoming photon. Then the difference in spin quantum number between the two states must be 1, the spin of a single photon. Some variable amount of energy always goes into translational energy, pushing the molecule that is hit. That's why absorption lines always have a width. The momentum of the photon is retained in that push, so the molecule has to fly off in an angle that matches excess energy and the momentum.

Raman IR transitions grab a small amount of energy off of a higher energy EM wave / photon (or add to it). The spin difference between the states must therefore be zero, and the momentum and direction of the leaving photon must also be taken into account, which dictates the energy difference.

Crazier transitions are possible. Two photons could be absorbed at the same time. Very low probability unless you have an extremely bright light source, but this allows totally different transitions, because now the same wavelenght has double the energy, and you have the spin from two photons.

So, answering your actual question, there are several possibilities to absorb a photon with an energy that does not exactly match the given transition.

(And then of course you have the intrinsic bandwidth of the transition, due to brownian motion.)

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