What an electron in an atom will do when given more energy

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Hello all,

I have a question. Consider an electron in a shell of an atom has energy as 1.0ev and in the next shell it energy should be 2.7eV and the further next level energy is let's say 3.1eV. Means an energy gap of 1.7eV is their between first and second level and an energy gap o 2.1 eV is their between first and third level.My question is what happens when energy of 1.9eV which is more than 1.7eV but less than 2.1eV is given to an electron in first level?

Will it aborsb all energy, go to second level and then release the rest of 0.4eV energy?Or will it do not absorb this amount of energy at all? What exactly will the electron do with the energy supplied?

Regards!
Thanks a bunch in advance
 

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If a 1.9 ev photon strikes the electron then it could move to the second level while giving off a .2 ev photon. But I think this is improbable. More likely the electron would just ignore the incoming photon. All of this is my opinion.
 
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hilbert2
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If you put an atom in an oscillating electromagnetic field where the photon energies do not exactly correspond to a transition energy, the atom will go back and forth between an excited state and the original state because of alternating absorption and stimulated emission.
 
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More likely the electron would just ignore the incoming photon. All of this is my opinion.

Thanks Gene Naden giving your opinion and I also think so :) however just need an answer from someone who can answer this with surety not a mere opinion.
 
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Well actually I am pretty sure of my answer. The absorption of light by an atom is subject to resonance, which means that it is much more likely to occur if the incoming energy matches the transition energy.
 
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hilbert2
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Look at the pages 83-85 in this document:

https://www.physik.hu-berlin.de/de/nano/lehre/copy_of_quantenoptik09/Chapter7

The deviation of the photon frequency from the actual atomic transition energy is called "detuning". My previous comment may have been a bit misleading, as actually an EM wave of the exact transition frequency will also cause alternating excitation and de-excitation. The difference is that a detuned wave will never move the atom to a state where it is excited with 100% probability.

This kind of model is only an approximate semiclassical one, though, but trying to solve the problem with QED and Feynman diagrams would be a lot more complicated than the shell model of an atom that you're assuming in the first place, here.
 
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Look at the pages 83-85 in this document:

https://www.physik.hu-berlin.de/de/nano/lehre/copy_of_quantenoptik09/Chapter7

The deviation of the photon frequency from the actual atomic transition energy is called "detuning". My previous comment may have been a bit misleading, as actually an EM wave of the exact transition frequency will also cause alternating excitation and de-excitation. The difference is that a detuned wave will never move the atom to a state where it is excited with 100% probability.

This kind of model is only an approximate semiclassical one, though, but trying to solve the problem with QED and Feynman diagrams would be a lot more complicated than the shell model of an atom that you're assuming in the first place, here.

Thanks hilbert2 for the answer with rigorous concepts of QED. Form what Gene Naden endorse and what I could conclude from your post is that the answer to my very question in my first post of this thread is that the electron will do nothing with that energy and will not go to next excited state liberating excess energy.

Am I correct.
 
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hilbert2
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The elecromagnetic wave with a "mismatched" frequency can excite the electron to a higher energy state with some probability (which becomes smaller when the mismatch increases), at least in the semiclassical model where the EM wave obeys classical electrodynamics and only the atom is treated quantum mechanically. I don't have the skills to analyze this with QED, but I guess there's some process where an atom can capture only a part of the energy of a photon, resulting with the atom in excited state and a lower-frequency photon.
 
  • #9
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Thanks hilbert2 for the answer with rigorous concepts of QED. Form what Gene Naden endorse and what I could conclude from your post is that the answer to my very question in my first post of this thread is that the electron will do nothing with that energy and will not go to next excited state liberating excess energy.

Am I correct.
The emission/absorption spectra have a finite width and it is not clear cut what happens. These lecture notes may help answer your questions.
Laser Cooling of Atoms
https://web.stanford.edu/~rpam/dropoff/Phys041N/lecture6-lasercooling.pdf
 

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