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Let's assume an atom consists of the nucleus and electrons as point particles. Take the inertial frame to be that of the fixed laboratory. Its total energy consists of the total kinetic and potential energy of the system of particles.If an electron absorbs a photon of energy $E$, the total energy of the atom increases by $E$. Can it happen that this increase in energy result in an increase of its total potential energy = E + Del(potential energy) and its total kinetic energy decreases by Del(potential energy).

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  • $\begingroup$ Kind of, depending on how big you think "Del(potential energy)" can be. This is the principle behind laser cooling en.wikipedia.org/wiki/Laser_cooling. But note that energy and momentum are conserved. So the atom's momentum only changes by the momentum of the photon. If it was moving toward the photon, it's total speed in our frame decreases, and its kinetic energy goes down. If moving away, the kinetic energy increases. Not writing as an answer because if I wanted to answer I'd be more complete with eqns. $\endgroup$
    – AXensen
    Oct 4, 2023 at 13:58
  • $\begingroup$ But the total KE of a system of particle is:1) 1/2Mv^2, M =atom mass and v = velocity of mass center; + 2) sum of 1/2mv^2, relative KE to mass center. So total ke is not 1/2Mv^2 $\endgroup$
    – itsme
    Oct 4, 2023 at 14:19
  • $\begingroup$ by "system of particles" do you mean the nucleus and the electrons? If so it's much better to think of the atom as a whole as having an "internal quantum state" with a particular energy. By absorbing a photon you can change from one energy state to another. Then to that you just add 1/2 mv^2 kinetic energy for the atom as a whole, where m is the total mass of the atom and v is its velocity. $\endgroup$
    – AXensen
    Oct 4, 2023 at 14:24
  • $\begingroup$ What if I insists on the classical model without talking about "quantum state of an atoms energy". I don't know QM. $\endgroup$
    – itsme
    Oct 4, 2023 at 14:34
  • $\begingroup$ Then replace the "energy state" with just "internal energy" which is kinetic energy relative to the center of mass plus potential energy. So it's more convenient to say that when a photon of energy $E$ is absorbed, the "internal energy" of the atom increases by $E+\delta$, and the kinetic energy of the atom changes by $\delta$. But note that essentially none of this can be properly understood without quantum mechanics - like what is a photon, why are only certain photons absorbed, why are only certain internal energy values possible, why do they then decay and re-emit the photon, etc. $\endgroup$
    – AXensen
    Oct 4, 2023 at 14:38

2 Answers 2

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Yes. For an atom of mass $m$ moving at velocity $v$ in the $x$ direction ($v$ can be negative indicating motion in the negative $x$ direction) and a photon of energy $E$ moving in the $x$ direction (always in the positive $x$ direction). In units where $c=1$, the KE of the atom before absorbing the photon is $$\frac{m}{\sqrt{1-v^2}}-m$$ and the KE of the atom after absorbing the photon is $$E+\frac{m}{\sqrt{1-v^2}}-\sqrt{\frac{m^2+m^2 v + 2 m E\sqrt{1-v^2}}{1+v}}$$

This results in a negative change in KE for any $$v<\frac{-E^2-4Em}{E^2+4Em+8m^2}$$

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  • $\begingroup$ It's a good answer... and a correct answer. But the choice to use the relativistic kinetic energy formula of the atom seems odd to me. And it yields equations that are much harder to qualitatively understand quickly. The photon momentum can also be approximated as small compared to the atom's momentum (unless the atom is already at the laser cooling limit). In which case the condition should reduce to $v<0$. $\endgroup$
    – AXensen
    Oct 4, 2023 at 21:39
  • $\begingroup$ I have a relativistic framework already set up on my computer, so the relativistic calculation was easiest for me $\endgroup$
    – Dale
    Oct 4, 2023 at 23:03
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It it's a free atom and its only absorbing a single photon, then it won't be able to reduce its kinetic energy. It'll have a kick of momentum from the initial absorption and then it'll have another kick of momentum once it's re-emitted. The re-emission occurs at a random direction relative to the absorption so the atom will have a net momentum.

That being said, once can use laser cooling of neutral atoms to reduce its kinetic energy. Imagine an atom being illuminated by a near-resonant laser beam along the 3 orthogonal spatial axes, X, Y, and Z. Along each axis I have 2 counter propagating laser beams. We'll also need to include a spatially changing magnetic field that is zero at the center and increases radially. So as the atom moves away from the center the Zeeman shift will modify the atoms excited state energy.

So as the atom moves along one axis it will come into resonance with that laser beam and will absorb a photon from along that direction, and then it will emit a photon of the same frequency but in a random direction. So, after successive absorptions and re-emissions (these re-emissions average out) the atom will cool and have less kinetic energy.

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  • $\begingroup$ I am more interested in atoms within the body of a solid (say metal). The state after it absorbs a photon and before any re-emission. $\endgroup$
    – itsme
    Oct 4, 2023 at 16:28
  • $\begingroup$ Your first paragraph isn't right. If the laser is far red-detuned, then the atom only absorbs a photon if it moves toward the laser. The re-emission is in a random direction, and the net-effect is a lower energy expectation value. The absorption reduces energy and the emission randomly either reduces or increases energy. In fact only the absolute worst case scenario is an energy break-even. This is the principle behind laser cooling. The second paragraph kind of confusingly adds in elements of the magneto-optical trap. Laser cooling is used in many applications - and many don't involve magnets $\endgroup$
    – AXensen
    Oct 4, 2023 at 16:37
  • $\begingroup$ @itsme you didn't say this in your question. Metals have free carriers and won't absorb photons. $\endgroup$
    – JQK
    Oct 4, 2023 at 16:43
  • $\begingroup$ @AXensen What I've written is correct on a per photon emission event. Your statement is similar to my discussion on laser cooling whereby over many absorption-emission cycles you end up with a reduced kinetic energy $\endgroup$
    – JQK
    Oct 4, 2023 at 16:45
  • $\begingroup$ @LQK, all bodies emits and absorb IR. IR from the sun penetrates a piece of metal and is absorbed by electrons within. $\endgroup$
    – itsme
    Oct 4, 2023 at 16:46

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