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Let's say we have an electron around atom. Let's say the electron drops into a lower electron shell. Is 100% of the energy difference converted to a photon? Does the atom recoil at all? Is the any of the energy lost to any other means?

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There is always energy loss due to recoil. Considering a single atom that is in free space, the total momentum is conserved, and the total reaction energy is constrained by the transition energy $E_0$ (ignoring the natural line width due to the energy-time uncertainty), so we get in the centre of mass frame: $$0 = \hbar k + mv $$ $$E_0 = \frac 1 2 mv^2 + \hbar \omega = \frac 1 2 mv^2 + \hbar c k$$ Leading to the equation $$E_0 = \frac {\hbar^2 k^2}{2m} + \hbar c k.$$ So $$k = - \frac{mc}{\hbar} \pm \sqrt{ \frac{m^2c^2}{\hbar^2} + \frac{2mE_0}{\hbar^2} } = \frac{mc}{\hbar} \left( \sqrt{1 + \frac{2E_0}{mc^2}} - 1 \right) \approx \frac{E_0}{\hbar c} - \frac 1 4 \frac{E_0^2}{\hbar mc^3}. $$ This corresponds to an energy correction for the photon of $$\frac{\Delta E}{E_0} = -\frac{E_0}{4 mc^2}.$$ So the relative correction is of the order of the transition energy compared to the rest energy of the recoiling mass.

If the atom interacts with other atoms (and is not in free space) the process becomes complex and energy may be transferred to the interacting objects (e.g. in a crystal lattice) the possible recoil energies are determined by the phonon spectrum, here, at low temperatures, the Mößbauer effect becomes important wherein the recoil momentum is transferred to the entire crystal, leading to a virtually recoil free emission (since the reaction mass is macroscopically large).

Further, there are other ways an electron can loose its energy – e.g. in Auger processes another electron is ejected from the atom, or the energy can be transferred ("coherently") to another atom (exciting an electron there). Further, combination processes, such as emitting a photon and ejecting an electron are allowed if all quantum numbers are conserved.

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  • $\begingroup$ Would it not take at least as much time for the recoil to be cancelled out by the macro mass, making the emission still recoil on the micro level? $\endgroup$ – Cees Timmerman May 9 '19 at 11:51
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It depends on the reference system used.

In the center of mass of the excited atom, momentum concervation should assure that the photon and the atom have equal and opposite momenta, so that will diminish the energy taken by the photon. The mass of the nucleus though is so much larger than the ev transitions of the electons that in effect the center of mass is the same as the rest frame of the nucleus.

The energy levels have a width, thus there is a probability of a photon with a smaller or higher frequency than the average of the energy level should appear.

Have a look here for a discussion of broadening of lines due to the motion of the atoms, and also the doppler shift in spectra due to the motion of the source or the detector.

It all depends on the frame of reference.

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  • $\begingroup$ You mention in the center of mass frame the atom and the photon will have opposite momenta... That will be true in the ground frame too... Wouldn't it... Essentially the conservation done by Sebastian Riese is in ground frame only.. Isn't it... And secondly you mention the probability of a photon with a higher frequency than the average of the energy level to appear... Where will this extra energy come from because the recoil will always decrease the energy never increase... $\endgroup$ – Shashaank Jan 1 at 6:56
  • $\begingroup$ I do not know what the ground frame is. As I say because of the large mass of the nucleus, the laboratory frame is within measurement errors the cms frame too. The widths of the lines are there , in the probailities describing the orbitals of the electrons around the atoms. An electron in a higher energy level orbital, can relax to a lower state and the energy plots of a large number of similar photons will show the width of the energy level it came from.. There is an uncertainty of the energy levels. www-star.st-and.ac.uk/~kw25/teaching/nebulae/… $\endgroup$ – anna v Jan 1 at 7:56
  • $\begingroup$ yep I meant laboratory by ground... Isn't the analysis done above by Sebastian Reise and also what you say not valid in lab frame.. In lab frame by conservation of momentum the momentum of the photon and the recoiled atom should be equal and opposite... Is this wrong.. Because you are doing the analysis in cms frame.. Regarding the 2nd part I understand the origin of smaller photon energies ( because of the recoil) what physical process ( I understand the probabilistic nature) gives rise to higher energy of photons.. What is the physical process... I don't understand why higher energy $\endgroup$ – Shashaank Jan 1 at 9:01
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    $\begingroup$ I am trying to say that because of the ** large** mass of the nucleus the laboratory frame and the cms frame of the atom is the same, i.e. a stationary nucleus in the lab is within errors the same as in the cms. Think of a ball hitting a wall.The mass of the wall is so large, that the center of mass frame ball-wall is the same as the laboratory frame . The electron in its orbital plays with a delta E) with the nucleus.. over all energy is conserved, but the particular probability if getting the electron to fall to a lower state has a width in energy, the nucleus orbital $\endgroup$ – anna v Jan 1 at 10:07
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    $\begingroup$ yes to the latter, except you must remember that there exists the uncettainty principle at each measurement: if you measure the momentum of the nucleus it will be within the unceertainty dpdx, so depending on the measuring instruments its kinetic energy after the decay will also be within an uncertanty. It is quantum mechanics we are discussing. Also the end state energy level has an uncertainty. $\endgroup$ – anna v Jan 1 at 14:29
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If there are only two levels involved and the atom transitions entirely from the excited state down to the ground state then yes, all of the energy that was stored in the electron excited state (this energy was in the form of the electron's kinetic energy and potential energy due to being further away from the nucleus on average) is converted into a photon, i.e. into electromagnetic field excitation.

When this happens yes, the atom does recoil due to the momentum carried away by the electromagnetic field. This is a key part of the principle behind Doppler cooling of atomic clouds down to microkelvin temperatures.

Note that in many cases (spontaneous emission for example) the photon is not emitted in a single direction but is rather emitted radially in many directions. This is a little weird to think about and can lead to confusion. Here is how you should think about it. When the atom decays to the ground state it creates an electromagnetic field pattern similar to that of a dipole antenna (i.e. outgoing radially). However, this whole electromagnetic field pattern only contains 1 photon. The natural question then is 1) what is the momentum of the photon and thus which direction does the photon recoil? The answer is the dipole pattern should be thought of as the photon being in a quantum superposition of having been emitted as a plane wave in many directions at once (note that a dipole pattern can be decomposed into plane waves). Thus, the atom also becomes in a superposition of having recoiled in many directions at once. If you put photon detectors all around the atom then you will see that only one of the detectors clicks. If the detector to the right of the atom clicks then you will see that the atom recoiled to the left. If the detector to the left of the atom clicks then you will see that the atom recoiled to the right.

When considering an atom the energy can either be electronic or photonic. That is, all of the energy is either in the EM field or with the electron. The system can be in a variety of superpositions of having the energy in different places but there aren't other options. If instead of considering an atom you consider a molecule then there are now vibrational rotational configurations of the nuclei involved which can store energy. In this way energy can be lost to relative motion of the nuclei rather than held purely in the electron configuration or the electromagnetic field.

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    $\begingroup$ I am a bit confused my your answer. In the first paragraph you mentioned that all of the energy is converted to a photon. In the second paragraph you mentioned that the atom does recoil too. So it doesn't like the atom DOES recoil and this takes away some energy from the outgoing photon? $\endgroup$ – Marc DiNino May 7 '19 at 16:26
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    $\begingroup$ Ah, yes that is a very good point that I overlooked and it is also a critical point regarding the doppler cooling that I referenced. I will amend my answer to say that ALMOST ALL of the energy is carried away by the photon. But it is true that some of the "internal" energy of the electron is also converted into "external" energy in the form of the whole atom now recoiling with some velocity. Later I will update my answer to be more clear on this point and I will provide estimates for the energy scales involved. $\endgroup$ – jgerber May 7 '19 at 17:30
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    $\begingroup$ To anticipate a possible next question: yes, this recoil energy does show up as a slight shift in the frequency of the photon. But I'll also point out, and refer to the answer from anna v, that at this point Doppler shifts become important. $\endgroup$ – jgerber May 7 '19 at 17:32

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