# Excited state line width and thermodynamics

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1. Feb 21, 2015

### Gruxg

I would appreciate very much if you could help me with a puzzling question that seems a paradox to me. Every excited state of an atom (every line of the characteristic spectrum) has a “width”: the higher the width, the shorter the lifetime of the state, and when the atom de-excitates we get radiation with a range of frequencies, not with a single well-defined one.

Well, let's suppose we excite some type of atoms using monochromatic light with a very well defined frequency (photon energy) that is near the low-energy edge of the line width. I'm not sure of this but I suppose when the atoms de-excitate by spontaneous emission they will produce a line with the characteristic width, regardless the precise frequency used for the excitation. Therefore, sometimes the photon emitted in the de-excitation has more frequency, and hence more energy, than the photon used for the excitation? Where does this energy come from?

I do not want to go off on a tangent discussing about the energy-time uncertainty or technical details about the structure of energy levels in an atom. Just: how can we reconcile this with thermodynamics?

Last edited: Feb 21, 2015
2. Feb 21, 2015

### Bystander

3. Feb 21, 2015

### Gruxg

Thanks, could you explain it a bit? By temperature do you mean that in the excitation and de-excitation process we would be reducing the kinetic energy of the atoms?

4. Feb 21, 2015

### Bystander

Anything in the excited state is able to "borrow" energy from its local environment, other excited moieties.

5. Mar 1, 2015

### DrDu

I don't see how your question is related to thermodynamics.
Second, this problem is a scattering process. The setup you have in mind will be exactly energy conserving.
The point is that the excitation and emission can take place at different times and the emitted frequencies will destructively interfere if they don't preserve energy.