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Imagine an electron around an atom absorbs a photon and becomes excited, it has now jumped to a higher orbital. At this point in time, where does the momentum of the photon goes?


marked as duplicate by John Rennie, Community Dec 27 '18 at 6:16

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    $\begingroup$ Can we reopen this question because I'm having difficulty following the Q&A written by Emilio? $\endgroup$ – user6760 Dec 27 '18 at 11:19
  • $\begingroup$ What level of rigor were you looking for? $\endgroup$ – bRost03 Dec 27 '18 at 15:42

Emilio's answer (How does one account for the momentum of an absorbed photon?) is great but quite technical. The simplified version is some of the momentum is accounted for by the different linear momentum associated with different orbitals. From the Bohr model we have $$p_n=\dfrac{\hbar}{a_0 n}$$ where $n$ is the principal quantum number which changes when an atom is excited. The rest of the momentum causes the atom itself to recoil. This is the mechanism whereby a photon absorption can excite a phonon (it kicks the atom which sets the lattice vibrating).

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    $\begingroup$ That doesn't actually account for the linear momentum, though. You should probably take a look at the linked Q & A by Emilio. $\endgroup$ – PM 2Ring Dec 27 '18 at 7:17
  • $\begingroup$ It actually does, it is just a huge simplification. Perhaps my answer is too obvious. The electron absorbs the photon and thus absorbs its energy and momentum. The details of how this process works and how the nucleus is involved (free electrons cannot absorb photons because both energy and momentum cannot be conserved) is handled in gory detail by Emilio but, as the asker seems to lament, that answer requires an advanced undergraduate/graduate level of quantum mechanics. $\endgroup$ – bRost03 Dec 27 '18 at 15:41
  • $\begingroup$ The recoil affects the linear momentum of the whole atom, not just the electron(s) relative to the nucleus. It might be easier to see this by looking at the reverse process. If an atom emits a photon in the X direction, the recoil forces the atom to move in the -X direction. The atom's velocity is very tiny though, around $10^{-9}$ the speed associated with its thermal motion at room temperature. $\endgroup$ – PM 2Ring Dec 27 '18 at 17:05
  • $\begingroup$ I'm well aware of all of this, I've personally worked through the math in my graduate quantum 2 course. Given the phrasing of the question I assumed the asker wanted an introductory level answer. Would you be in agreement if I said the momentum is transferred to the entire atom instead of saying it is transferred to the electron? Of course the electron is bound to the atom so "kicking" the electron is tantamount to "kicking" the entire system. $\endgroup$ – bRost03 Dec 27 '18 at 17:14
  • $\begingroup$ Sure. Saying that the momentum is transferred to the electron implies that the motion of the CoM of the atom is unchanged by the photon's momentum. And that's what I object to. (BTW, I didn't downvote). $\endgroup$ – PM 2Ring Dec 27 '18 at 17:22

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