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So I've recently learned about photoelectric effect. It made sense at first, but when I try to combine it with what I thought I knew about emission and reflection of photon, things become rather messy.

My questions :

1)When photons interact with electrons, they can either be absorbed or reflected or pass right through. Is that correct?

2)When light photon excites an electron in the inner shell, it is absorbed. A photon is subsequently given off during de-excitation. Does the photon given off have the same energy as the photon absorbed earlier?

3)If the answer to the above question is "yes"(which is what I have been taught in school, though it's a struggle to reason it), then how is the process different from reflection? Say it absorbs light photon and then re-emits the exact same photon.

4)to add to question 2 and 3,how come in some cases, photons with shorter wavelengths are apparently radiated as photons with longer wavelength? (Earth absorbing uv but radiates it as infrared).

There are likely to be major mistakes in my understandings, which would explain why I can't seem to puzzle all these phenomena together... Any correction would be greatly appreciated!

Thank you.

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  • $\begingroup$ Don’t forget non-radiative relaxation... $\endgroup$
    – Jon Custer
    Apr 14, 2020 at 1:34
  • $\begingroup$ Is non-radiative relaxation one of the reasons behind why we experience heat? $\endgroup$
    – Clara
    Apr 14, 2020 at 3:26
  • $\begingroup$ If the energy always stayed as either an excited electronic state or a photon, there would be no 'heat' to experience. $\endgroup$
    – Jon Custer
    Apr 14, 2020 at 18:21

1 Answer 1

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Photons when absorbed put the electron in a higher energy level (read Bohr orbits to understand this better). But high energy orbits are unstable as every system tends to minimize its energy. So the electron loses energy and returns to the ground state in less than 10^-8 s. Reflection is when a photon with the same energy (and correspondingly, the wavelength) as that of the absorbed one is emitted. This doesn't always happen.

It does not necessarily have to be the same energy as the photon absorbed. Let's say the electron was in n=1 before getting kicked into a higher orbit. Then it gets upped to n=3. The electron may transition from n=3 to 1 directly, emitting the same energy photon as was absorbed (more precisely, with energy equal to the energy difference between n=3 and n=1 levels). But it could also take a different path and go from n =3->2->1 emitting two photons as it finally reaches ground state, one in each successive transition.

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  • $\begingroup$ But re-emission of photon can be in any direction, right? While reflection has a fixed direction. So are we able to say that "Reflection is when a photon with the same energy (and correspondingly, the wavelength) as that of the absorbed one is emitted."? PS: thanks for the reply $\endgroup$
    – Clara
    Apr 14, 2020 at 3:30
  • $\begingroup$ @Clara check out the answers to physics.stackexchange.com/questions/83105/… for a detailed explanation of how reflection works at the molecular scale. $\endgroup$
    – cartilage
    Apr 14, 2020 at 3:51
  • $\begingroup$ So we are able to say that reflection = absorption + emission? $\endgroup$
    – Clara
    Apr 15, 2020 at 22:54
  • $\begingroup$ @Clara Sorry for the very late reply, I had forgotten about this. In a gist yes, but try not to memorize it as reflection = absorption + emission Becuase you might end up confusing some terms such as "reflectivity" and "emissivity" which are thermal physics terms (used for things like a blackbody) which have different meaning. $\endgroup$
    – cartilage
    Jun 9, 2020 at 7:28

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