Consider an atom in the excited state radiating a photon and goes to the lower energy state. But photons have a certain angular momentum, the momentum itself is not defined. In this case, will there be a recoil of the atom due to photon emitted?
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1$\begingroup$ Interesting reading on these lines: Mossbauer effect. Sometimes it is a larger system that recoils. $\endgroup$– dmckee --- ex-moderator kittenCommented Aug 25, 2019 at 0:39
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3 Answers
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The momentum of a photon is not only defined, but is defined very well in the famous Einstein's quation:
$$ E = \sqrt {(mc^2)^2 + (pc)^2 } $$
that leads for massless photons to
$$p=\frac Ec =\frac h{\lambda}$$
Therefore, atoms recoil when emitting photons.
The opposite phenomena, an atom recoiling during photon absorption, is used by laser cooling near $0 \mathrm{K}$.
The laser frequency is set just below a chosen atomic absorption line. Due the Doppler effect, absorption occurs only for those atoms with a particular velocity component toward the laser.
Absorption of a photon and it's momentum decreases this velocity component, what means decreasing the atom kinetic energy. That leads in large scale to decreasing of temperature.
Effectively, the thermal energy is spent to be added to otherwise insufficient energy of photons.If the gained energy is released by emission of other photon, it has in average the nominal absorption line energy with negative net energy outcome.
As @dmckee has noted, recoilless scenario can be achieved in solid matrices, if the momentum is distributed within the whole solid matrix.
Mossbauer effect (mentioned in Laser cooling page)
The Mössbauer effect, or recoilless nuclear resonance fluorescence, is a physical phenomenon discovered by Rudolf Mössbauer in 1958. It involves the resonant and recoil-free emission and absorption of gamma radiation by atomic nuclei bound in a solid. Its main application is in Mössbauer spectroscopy.
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Poutnik did a very good job at answering your question with regards to linear momentum, but your original question mentions angular momentum.
We know electrons can occupy discrete energy levels in the atom, and they can also occupy discrete angular momentum (and spin momentum) states. These are labelled by the $l$ quantum number (and $m$ for spin). You may have seen the term $2p$ or $2P_{1/2}$ before, and here the letter labels the angular momentum of the electron (P is for $l$ = 1). In the most basic situation, an electron in an energy level n could be in one of n different angular momentum states. So a transition between energy levels can also be a transition between states of different angular momentum, and conservation of angular momentum is satisfied because the photon carries that angular momentum. This is (partly) the origin of polarised light.
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Momentum is conserved, so when a photon is emitted its momentum must be exactly counterbalanced by a change in the momentum of the emitter. The recoil velocity of the emitter will depend on its mass. If the photon is emitted by an isolated atom, then its momentum must be offset by a change in the momentum of that atom alone, in which case the recoil will be much larger than if the emitting atom is part of a crystalline solid, where the recoil velocity may be entirely negligible.
answered Oct 16, 2019 at 21:55