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When an atom has an electron in an excited energy level and it transitions to a lower level it emits a photon. What direction is it likely to emit the photon in? Are all directions equally likely, even toward the nucleus?

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    $\begingroup$ Have a look, might be along the same lines: physics.stackexchange.com/q/22001 actually, how would you explain a mirror, or Snell's law, in the above case? $\endgroup$ – user81619 Sep 6 '15 at 17:24
  • $\begingroup$ Take a look at the dipole emission pattern. This is a good approximation for luminescent molecules (and probably for atoms too). You end up with a $sin^2(\theta)$ profile if memory serves. $\endgroup$ – boyfarrell Sep 10 '15 at 18:30
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In the general case, "no" as there is a angular momentum transfer involved (meaning there are preferred directions relative the original and/or final angular momentum of the atom).

That said, for most matter at room temperature the atoms have random orientation so you can treat the answer as "yes" for experimental purposes.


Now, I see that you are wondering about the nucleus in the question. There are two thing to keep in mind here.

  • First you may be thinking that the photon is emitted by the electron and that the electron is on a particular side of the atom at the time of emission. (That is, you may have some version of the almost completely incorrect Rutherford/Bohr/de Broglie atom in your head.)

    It is the system of nucleus and electron(s) that undergoes a change and emits a photon, and even from the beginning asking "from what part of the atom is the photon released?" has no precise answer.

  • Second the nucleus is also a quantum system and can't absorb arbitrary amounts of energy in internal changes (it wouldn't be able to absorb it in arbitrary translational motion either because that would violate the conservation of momentum---it drags the electron with it, after all). The nuclear energy splittings are generally too large to absorb an atomic photon.

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    $\begingroup$ "'no' as there is a angular momentum transfer involved". I think that this is wrong. Yes, there is angular momentum transfer involved, but that would correspond to the polarization of the photon emitted, but not the direction of it's propagation. $\endgroup$ – aquirdturtle Sep 6 '15 at 17:40
  • $\begingroup$ @aquirdturtle Photons don't have degrees of polarization. Each one is circularly polarized either parallel or anti-parallel to their direction of motion. This is a feature of their being massless (which also precludes them from exhibiting the third state that would be expected of a spin=1 particle). You can make experimental use of this limit by building polarized targets. $\endgroup$ – dmckee Sep 6 '15 at 17:45
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    $\begingroup$ "Each one is circularly polarized either parallel or anti-parallel to their direction of motion." It's better to think about it relative to the atom. Relative to the axis of the atom you have $\sigma+$, $\sigma-$, or $\pi$ polarization. These different configurations have different directions of angular momentum relative to the atom, allowing the photon emitted to carry away angular momentum from the particle via it's polarization orientation, not it's direction of propagation. $\endgroup$ – aquirdturtle Sep 6 '15 at 17:50
  • $\begingroup$ @aquirdturtle True , the wavefunction of the photon has a component of polarization but the measurement of a single photon can only show + or - spin . The polarization is a mathematical construct that adds up in an emergent em beam, not measurable on a single photon. Going away from the atom it takes spin + or -1 away in its direction of motion and leaves the atom minus that spin. $\endgroup$ – anna v Sep 6 '15 at 18:02
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    $\begingroup$ @aquirdturtle this might help , feynmanlectures.caltech.edu/III_18.html $\endgroup$ – anna v Sep 7 '15 at 3:31
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It depends on the mechanism by which the photon is emitted.

  • Stimulated Emission: Yes, the emitted photon will inherit the characteristics of the photon that stimulated it, including its propagation direction. That's how lasers get coherent light.
  • Spontaneous Emission: No, this should be random orientation. Another way to think about spontaneous emission is as stimulated emission where the stimulating photons are virtual photons, who have random orientation themselves.
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