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As simple as in the title.. I would like to know also some mathematics about it!

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It cannot. This is because energy and momentum are not both conserved if a free charged particle (say, an electron) emits a photon. It needs interaction with at least a second charged particle in order to do so (as in Bremsstrahlung). The mathematic involved is that of the energy of a photon $E=\hbar \omega$, energy of a particle $E^2 = m^2 c^4 + p^2 c^2$, momentum of a photon $p = \hbar \omega /c$ and simple trigonometry and basic algebra, very much as in the classical version of Compton scattering.

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    $\begingroup$ Note that if the particle has internal structure, this argument can fail. For example, atomic nuclei emit gamma rays. This is because they have more than one internal state with different energies. $\endgroup$
    – user4552
    Oct 20, 2013 at 19:58
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    $\begingroup$ That is certainly true. Nice addition to the discussion. $\endgroup$ Oct 20, 2013 at 22:14
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    $\begingroup$ @BenCrowell In the context of the shell model, one can argue that the gamma is emitted by a nucleon falling to an orbital of lower energy, i.e. by a bound particle. $\endgroup$
    – user154997
    Jul 17, 2017 at 2:45
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    $\begingroup$ I know this was a long time ago, but I am just wondering why we have to consider the conservation of energy and momentum for the particle alone, and not the particle + photon after the hypothetical emission happens. $\endgroup$
    – newtothis
    Jun 26, 2021 at 8:40
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    $\begingroup$ @newtothis As the answer points out, it is precisely considering the energy and momentum (already given) of the photon + particle system that makes it impossible. So they not only are being considered, they must be considered. $\endgroup$ Jul 1, 2021 at 6:40
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Let's assume it could then $\frac{1}{2} mv^2=\frac{hc}{\lambda}$ also $P=mv=\frac{h}{\lambda}$ by replacing $mv$ in first equation with$\frac{h}{\lambda}$ we get $\frac{1}{2} \frac{h}{\lambda}v=\frac{hc}{\lambda}$ now $v=2c$ which is of course impossible.

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No. Photon does not change electron state ever.

Atom or molecule can have excited or no-excited state - depending how far electron is from nucleus.

According to nowadays view on the Space, it is not the emptiness. Space is heavily filled-in with waves (dark energy) and debris (like quarks, nutrino and other unknown yet particles - dark mass).

Nucleus makes a disturbance in the space and because of it creates spherical or more complex "wave boxes" (named shells or orbitals) around itself; for more than one boson in nucleus it can be cube-like-shape or some other wave combinations

Electron orbital in hydrogen is the gap between two spherical wave bumps of the space around nucleus; wave nature of electron makes orbital conductive and electrically atom does not radiate -- nucleus proton and electron shell are static to each other. Speed of electron is really does not matter and it makes random jerky moves inside the orbital. Keep in mind, orbital can have very different shapes. In molecule it is a tube between two nuclei.

Photon makes a disturbance in the atom adding electrical energy as a short pulse.

This energy pushes electron and proton apart; it is not always enough energy to push electron absolutely out of atom; it may simply push it from the lower orbital to the higher.

So the answer is: only atom, molecule or two particles with different electrical properties can emit/absorb photon. Photon is electromagnetic pulse and IS NOT a particle.

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    $\begingroup$ New age-ish gibberish. $\endgroup$
    – user154997
    Jul 17, 2017 at 2:31

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