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I know once a photon hits an electron it moves from the ground state to an excited state, then come back down to ground by releasing that energy as a photon. But the the electron "and the atom" also increase in kinetic energy or does the energy get released too fast before it has a chance to increase the electron "and atom" kinetic energy?

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  • $\begingroup$ Do you mean to ask if a photon that hits an atom causes it to gain velocity? $\endgroup$
    – eranreches
    Commented Nov 13, 2017 at 18:54
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    $\begingroup$ Yeah. And by how much? $\endgroup$
    – DeusIIXII
    Commented Nov 13, 2017 at 19:05

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Consider an atom with two levels initially at rest. The energy of the ground state is $E_{g}=0$ and the energy of the excited state is $E_{e}=\hbar \omega_{0}$. A photon of wavevector $\vec{k}$ and frequency $\omega$ is aimed at the atom. To see what happens, we need to write equations for the conservation of energy and momentum. If we assume the atom absorbs the photon then from the conservation of momentum

$$\hbar\vec{k}=m\vec{v}$$

where $m$ is the mass of the atom and $\vec{v}$ is its velocity. Moreover, from conservation of energy

$$\hbar\omega=\hbar\omega_{0}+\frac{1}{2}mv^2$$

Using the dispersion relation of the photon $\omega=ck$ these two equations can be solved. But even without solving the equations you can see some interesting results. First, you clearly observe that the absorption of photon causes the atom to gain velocity. This means that your statement is correct. But more interesting, because the atoms gains velocity and velocity is related to energy, the photon must have a frequency larger then the frequency of transition for absorption to happen. This is because some of the energy goes to the internal degrees of freedom of the atom (exited state), and some goes to external degrees of freedom (velocity of the center of mass).

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  • $\begingroup$ I didn't even know the energy break up like that and goes to different things. That's cool. Thanks for the detailed answer $\endgroup$
    – DeusIIXII
    Commented Nov 14, 2017 at 7:10
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Electrons and photons and nuclei are described with quantum mechanical equations. A photon can interact with a free electron, scattering elastically or inelastically called Comtpon scattering, where part of the energy of the photon turns into kinetic energy of the electron.

An electron bound to a nucleus forms an atom , and usually occupies the ground energy level. If a photon with the appropriate energy hits the atom and it has an energy covering the energy levels of the atom, the system goes to a higher excited energy level. The electron is located at a higher energy level, the momentum of the photon is transferred to the atom. In the semiclassical Bohr model one can say that the electron has higher energy, though the rigorous quantum mechanical solution is about probable values of the energy if measured.

When the atom relaxes to the lower level emitting a photon , momentum has to be conserved and the whole atom has to take part in the exercise. The incoming and outgoing photons will not have the same direction since the decay is probabilistic.

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  • $\begingroup$ Brilliant answer and it matches up with what I've been reading so it makes sense. Thanks for your help $\endgroup$
    – DeusIIXII
    Commented Nov 13, 2017 at 19:16
  • $\begingroup$ May I clarify the part about momentum transfer? You say that the "momentum of the photon is transferred to the atom", which, of course, must be the case. However, I am curious as to the exact distribution of additional momentum when the photon is absorbed. My teacher says that most of the additional momentum goes to the nucleus, would that be true? $\endgroup$ Commented Oct 2, 2019 at 9:48
  • $\begingroup$ I think that momentum conservation would require that the whole atom takes this extra momentum I think your teacher is thinking about momentum , mp,, and the mass of the atom is mostly the mass of the nucleus but I think that one cannot separate in this case the electron clouds from the nucleus. , as the atom is one whole quantum state. $\endgroup$
    – anna v
    Commented Oct 2, 2019 at 13:42
  • $\begingroup$ I did find the statement here, ocw.mit.edu/courses/nuclear-engineering/…. so it may be a common one, though not really accurate imo $\endgroup$
    – anna v
    Commented Oct 2, 2019 at 13:48
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Momentum is conserved, so the momentum of the atom must be different after the photon is absorbed. Since the momentum is changed, the kinetic energy of the atom is changed as well. Even after a photon is emitted and the electron returns to the ground state, the atom does not usually have the same kinetic energy as it started with, since the photon is probably emitted in a different direction.

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  • $\begingroup$ That's what i thought as well. Sorry if it seem like a noob question. Learning physics on my own and kind of want to make sure i understand things before moving on. Better to learn slow now then relearn later $\endgroup$
    – DeusIIXII
    Commented Nov 13, 2017 at 19:10

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