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Atom/photon energy conservation

  1. Sep 5, 2013 #1
    I wonder of someone could help me understand something about emission and absorption of photons by atoms. I thought I understood the basics of this topic but yesterday a thought experiment occurred to me that seemed to violate conservation of energy.

    Imagine two stationary hydrogen atoms. Atom A, on the left, is in an excited state. Atom B, on the right, is in the ground state. Atom A returns to the ground state by emitting a photon. I recall reading that emission spectra had the same frequencies as absorption spectra, so if atom B absorbs the photon from atom A, then I assume atom B should enter into the same excited state that atom A had been in. From a conservation of energy standpoint, atom B has gained all the energy that atom A has lost, so there is no net energy change to the system. All the energy of atom A’s excited state has been “spent” in raising atom B to the same excited state.

    But I also recall reading that photons have momentum. That implies to me that atom A recoiled to the left when it emitted the photon and that atom B recoiled to the right when it absorbed the photon. Since both previously stationary atoms now are in motion, they have gained some kinetic energy. My question is where did that extra energy come from? It seems to me that the kinetic energy could not have come from atom A’s excited energy, because all of that energy was used to raise atom B to the same excited state that atom A was initially in.

    I know conservation of energy can’t be violated, but where has my reasoning gone astray? Any help would be greatly appreciated.


    Clovis
     
    Last edited: Sep 5, 2013
  2. jcsd
  3. Sep 5, 2013 #2
    Chalk it up to quantum weirdness. A few effects are in play here. There's no such thing as a stationary atom. The atoms have some uncertainty in momentum.
    http://en.wikipedia.org/wiki/Heisenberg_uncertainty_principle
    If the atoms are initially moving toward each other, and they exchange a photon, it could slow them down, apparently decreasing their energy. So you could have an apparent increase or decrease in energy, but only if you know how much momentum they had to start with. But you don't.

    Another aspect of HUP is the natural line width. The energy of the emitted photon is not exactly known. It's somewhere near the energy difference between the energy states in the transition, but the idea of an "energy state" is some kind of approximation, using the time-independent Schroedinger equation when the real problem is time-dependent. Saying that the final state of atom B is in the same state as the initial state of atom A is only approximately true, insofar as energy states exist.

    As a sidenote, due to Doppler shift, atom B will have a lower probability of absorbing the photon if the atoms are moving with faster relative velocity. Partly due to this effect, larger apparent deviations from conservation of energy are less likely to be observed.
     
  4. Sep 5, 2013 #3
    It's better to say that conservation of energy is totally true, but we don't know the initial and final energy exactly, than say that conservation of energy is only approximately true.
     
  5. Nov 19, 2014 #4
    I don't think either reply is correct. Don't we have to consider MASS as energy as well? In your thinking, you have not considered it as part of the energy of the system.
     
  6. Nov 19, 2014 #5

    Nugatory

    User Avatar

    Staff: Mentor

    The total mass of the system remains constant throughout the entire interaction.
     
  7. Nov 19, 2014 #6

    Dale

    Staff: Mentor

    You are correct that the system mass needs to be conserved also, but since the energy and momentum are conserved the mass gets conserved automatically. Note that the system mass is greater than the sum of the masses of the particles.
     
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