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Is this explanation correct:

When a photon with the appropriate energy hits an atom, the electron will make a transition from the ground state to a excited state. This will make the potential energy of the atom higher. Also, momentum is conserved, and the velocity of the atom will change in the direction from the incoming photon. When the electron returns back to it’s ground state, a photon is emitted in a random direction. Again, momentum is conserved and the velocity of the atom changes in the opposite direction from the emitted photon.

I am a bit confused because if that's the case, the atom will gain kinetic energy. Doesn’t that violate the conservation of energy? Since the energy is already conserved by absorbing and emitting a photon with the same wavelength.

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  • $\begingroup$ The energy of "photon + atom" before the absorption and "photon + atom" after emission is the same. Energy is a property of a system - the atom's kinetic energy you observe depends on your relative motion to the atom, but the energy of the "photon + atom" system is something all the observers agree on. Talking about the kinetic energy of the atom alone only makes sense if you setup some preferred frame of reference. And just to be clear, the energy of the emitted photon is not the same as the absorbed photon - for one, the same reason as above, two, it's not always just one photon :) $\endgroup$ – Luaan Nov 18 at 8:34
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When a photon with the appropriate energy hits an atom, the electron will make a transition from the ground state to a excited state. This will make the potential energy of the atom higher.

This is mostly correct, but it's not the potential energy of the atom which is higher, it's its internal energy

Also, momentum is conserved,

this is correct, but

and the velocity of the atom will change in the opposite direction from the incoming photon.

this is the wrong way round. The photon momentum ($h/\lambda$ in the direction of propagation) is absorbed, and this will increase the velocity of the atom in the direction of propagation of the photon

When the electron returns back to it’s ground state, a photon is emitted in a random direction. Again, momentum is conserved and the velocity of the atom changes in the opposite direction from the emitted photon.

This is correct, and indeed it is the basis for Doppler cooling.

I am a bit confused because if that's the case, the atom will gain kinetic energy and therefore will gain thermal energy. Doesn’t that violate the conservation of energy?

Gaining kinetic energy does not mean that it will gain thermal energy $-$ kinetic energy is only thermal energy when it is in randomized directions. You can have objects which are very cold but moving very fast, ranging from ice cubes fired from a potato cannon all the way to atoms in particle accelerators with a high velocity but a thin velocity spread.

That said, the photon kick to the atom's center of mass does not mean that energy is somehow non-conserved. Instead, if the final state of motion (after absorption of the photon's momentum) has a higher kinetic energy than the state of motion before the photon absorption, this energy deficit is provided by the photon: in other words, the transition frequency gets blue-detuned, and the photon's energy needs to provide both for the change in internal energy and the change to the center of mass's kinetic energy.

Since the energy is already conserved by absorbing and emitting a photon with the same wavelength.

There is no requirement that the wavelengths of the absorbed and emitted photons be precisely equal, and in general, if absorption-then-emission cycle changes the center-of-mass motion, the two wavelengths won't match. Again, this is what powers Doppler cooling.


I've written up the details of how this looks like within the framework of quantum mechanics at How does one account for the momentum of an absorbed photon?, though that thread might lie outside of what your background allows you to tackle in full for now.

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    $\begingroup$ I think the "in the opposite direction from the incoming photon" may be intended as "it will move away from the incoming photon; i.e., in the direction it was traveling". Unclear, though. $\endgroup$ – chrylis -on strike- Nov 18 at 6:32
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I am a bit confused because if that's the case, the atom will gain kinetic energy.

In the way you've set it up (atom is initially at rest in our reference frame), then that is correct.

Doesn’t that violate the conservation of energy?

That depends on the energy inputs and outputs. We haven't described them completely yet.

Since the energy is already conserved by absorbing and emitting a photon with the same wavelength.

That's not correct here. Because of the recoil, the emitted photon will be less energetic. If you instead chose the frame where the center of momentum were at rest (same as the excited atom at rest), then the photons would have the same energy. But in that frame, the atom has identical kinetic energy before and after the interaction.

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When the electron returns back to it’s ground state, a photon is emitted in a random direction

Not exactly. If it's a spontaneous emission - then YES. If it's a stimulated emission - then NO, photon is emitted in the same direction of incident photon which has hit the atom. LASER or MASER works according to this principle. Btw, stimulated emission was predicted by Einstein and first laser was constructed on May 16, 1960 by Theodore H. Maiman. Also in case of stimulated emission not only direction of emitted photon coincides with incident photon, but phase, frequency and polarization as well.

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