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It's a common understanding that atoms emit or absorb light when the energy of the photons is equal to the the difference in the energy levels in the atom. What I don't understand is how does an electron absorb light inside an atom? Is it that the atom as a whole absorbs light or the individual electrons absorb the light packets? Also I would like to know, what does one mean when you say that an electron absorbs photon. How does an electron absorb a photon??

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  • $\begingroup$ From what I know, energy goes back the the field where it comes from. $\endgroup$ – J. Chomel Sep 22 '16 at 6:33
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    $\begingroup$ Electrons do not absorb light. Bound states of electrons and nuclei absorb light. That's a big difference $\endgroup$ – garyp Sep 22 '16 at 11:53
  • $\begingroup$ I don't think we can really say "how" it happens. We can describe the atom before and after the absorption. In between is not well described. $\endgroup$ – George Herold Sep 23 '16 at 13:20
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An atom is nothing but a bounded state of electrons and a positively charged core called nucleus. The electrons in the atom are in bound state and so their energy levels are quantized. Also, it is possible to have quantized rotational and vibrational energy levels of the molecules. The way in which they differ is in the difference in the energy characterizing the transition from one state to another.

Possible ways in which a photon is absorbed by an atom or a molecule

If the energy level of the incoming photon is such that the electrons can have a transition from a state to some higher permissible state, then the photon energy level will be in the visible or ultraviolet range and we make use of this principle in electronic spectroscopy.

Suppose, a particular electron is in the energy state with energy eigenvalue $E_i$. There exists a higher energy level $E_f$. If the energy levels of the electron bound states are such that it precisely matches with the energy of the photon: $h\nu=E_f-E_i$, then the electron will get excited to the energy state $E_f$.

Now, if the incident photon energy matches the difference in the vibrational energy levels of any pair of states of the molecule, then it can cause transition from that vibrational energy state to the higher energy state. This energy usually lies in the infrared region and the technique is used in infrared spectroscopy.

For example, in the case of diatomic molecules, the vibrational energy levels are quantized and in a good sense they can be approximated to that of a harmonic oscillator: $E_n=\left(n+\frac{1}{2}\right)\bar{h}\omega$. So, if the photon energy is such that $h\nu=E_f-E_i$, the electron transits from the state $E_i$ to $E_f$, where $E_i$ and $E_f$ are given by the above equation of the harmonic oscillator and the states are defined by the quantum number $n=i$ and $n=f$.

Now, if the absorption of a photon can only affect the rotational energy levels of the molecule, then the absorbed photon will be in the microwave region. The spectroscopic technique making use of this principle is the microwave spectroscopy.

For example, the rotational energy levels of a diatomic molecule are given by: $\displaystyle{E_j=\frac{j(j+1){\bar{h}}^2}{2I}}$, where $I$ is the moment of inertia and $j$ is the angular momentum quantum number. In such a case, we can write: $h\nu=E_f-E_i$ and the bound state absorbs the photon and will get excited to the state with energy $E_f$, with $E_f$ and $E_i$ determined by the quantum number $j=f$ and $j=i$.

Now, the energy can be absorbed by the nuclei also. It can be elastic nuclei scattering (analog to very low energy Compton scattering by an electron. In this process, a photon interacts with a nucleon in such a manner that a photon is re-emitted with the same energy), inelastic nuclei scattering (the nucleus is raised to an excited level by absorbing a photon. The excited nucleus subsequently de-excites by emitting a photon of equal or lower energy) and Delbruck scattering (the phenomenon of photon scattering by the Coulomb field of a nucleus, also called nuclear potential scattering, which can be thought of as virtual pair production in the field of the nucleus. i.e., pair production followed by annihilation of the created pair). However, these processes are negligible in photon interactions.

Conclusion:

Absorption of a photon will occur only when the quantum energy of the photon precisely matches the energy gap between the initial and final states of the system. (the atom or a molecule as a whole) i.e., by the absorption of a photon, the system could access to some higher permissible quantum mechanical energy state. If there is no pair of energy states such that the photon energy can elevate the system from the lower to the upper energy state, then the matter will be transparent to that radiation.

So, if any of the above types of energy transition take place, that will affect the quantum state of the system as a whole (transits the system from one state to another). So one could say, as @annav pointed out, it is the atom (or the molecule) that absorbs the radiation and changes the energy levels of its constituent particles, depending on the energy absorbed. Anyway, a change in energy level of the electron, or rotational or vibrational energy levels of the molecules can be seen as changing the quantum state of the molecule. So, it's better to stick with the concept of the molecule as a whole absorbs the energy and changes its state to some higher energy state by changing the quantum state of its constituent particles.

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  • $\begingroup$ I understand that...!! The main point is, what do you mean when you say that an electron absorbs the photon?? How does an electron absorbs photon?? $\endgroup$ – lattitude Sep 22 '16 at 6:36
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    $\begingroup$ I think it confuses the issue to keep talking of an electron absorbing the photon. The whole atom absorbs the energy. Take a hydrogen atom absorbing a photon. You could as well say that the proton absorbed the pboton, but it would be equally fuzzy. In my opinion one should stress that the WHOLE ATOM absorbs the photon and changes energy level for its constituent particles. $\endgroup$ – anna v Sep 22 '16 at 10:35
  • $\begingroup$ @annav: Is there anything that I need to add or remove in my answer. I realize an atom as a bound state of it's constituent particles. But I'am new to the idea that a proton in the nucleus could get interacted with a photon. I knew that it works if the proton is isolated, because a proton is nothing ut a point charge. But inside the nucleus, is that possible? $\endgroup$ – UKH Sep 22 '16 at 10:42
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    $\begingroup$ I would add in the first phrase "Because the mass of the nucleus of an atom, even of a simple hydrogen atom, is more than a thousand times larger than the electron mass, it is usual to describe atoms as the electrons occupying the energy levels, even though the energy levels belong to the total atom and the photon transition happens with the total atom." or some such. I agree that we talk of "electron orbitals" again assuming a stationary nucleus, so it is not wrong, just needs clarification. One should be aware that it is the kinematics of the center of mass that gives a bias to electrons. $\endgroup$ – anna v Sep 22 '16 at 11:40
  • $\begingroup$ @RobJeffries: I know, but I'am afraid, in that sense, I have to describe all the possible ways in which matter interact with radiation, which is too broad to answer $\endgroup$ – UKH Sep 22 '16 at 13:34
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I will leave others to comment on the detailed mechanisms involved - all of which are simply mathematical models that we use to understand the process and which can be accessed at various levels of complexity.

What is certainly true though is that an isolated electron cannot completely absorb a photon (partial absorption is possible and is known as Compton Scattering). There is just no way that both energy and momentum could be conserved in such a process. Therefore the absorption of a photon takes place in the context of the atom as a whole, where the various conservation laws can be satisfied. At its simplest it is perhaps best to think of the atom as an oscillator with discrete energy levels that can be excited or de-excited between these modes by interacting with the electromagnetic field of the light (or alternatively by absorbing or emitting the energy of the photon).

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  • $\begingroup$ Compton scattering is an inelastic scattering, even though the collision is elastic. It can be visualized a an elastic collision between between two particles. But how does this account for absorption of a photon? Scattering and absorption are two different processes. $\endgroup$ – UKH Sep 22 '16 at 13:48
  • $\begingroup$ @Unnikrishnan Compton scattering is the "partial absorption" of a photon by an electron - something that you said cannot occur. It does not address the question asked, which is why I wrote it in brackets. However, you have also said that the electron absorbs a photon, failing to correct the OPs misconception. The electron does not and cannot absorb a photon. $\endgroup$ – Rob Jeffries Sep 22 '16 at 16:09
  • $\begingroup$ But I still don't understand, how scattering$\implies$ partial absorption $\endgroup$ – UKH Sep 22 '16 at 16:22
  • $\begingroup$ I haven't mentioned electron can absorb photon. I stated electron bound states absorb photon $\endgroup$ – UKH Sep 22 '16 at 16:25
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From a physical particle perspective, an electron (free or bound to an atom) cannot and does not absorb a photon. The atom absorbs the photon and takes its kinetic energy. That kinetic energy causes increased vibration of the atom which can emit another photon and/or emit a loosely held/ bound electron and/or that vibration can be transmitted to nearby atoms.

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The electron absorbs energy only when it exactly matches with its energy gap you can apply here conservation of momentum. The wavelength of photon matches nearly with that of electron that's why you can say that electron absorbs energy.

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    $\begingroup$ It's not the case that photon and electron wavelengths must "match" for an energy exchange to occur. $\endgroup$ – rob Apr 23 '17 at 14:08
  • $\begingroup$ I m not saying that it must match but it should fall in its range $\endgroup$ – palash jain Apr 23 '17 at 15:08
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    $\begingroup$ "Fall in its range" Are you saying that the deBroglie wavelength of an electron should be approximately close to that of an incoming photon if the electron is going to absorb the photon? That's not true. The energy corresponding to the photon's wavelength must be at least as much as the electron needs to be excited to the next energy level. In general, you'll find that electrons can absorb photons with wavelengths much longer than their deBroglie wavelength $\endgroup$ – Jim Apr 24 '17 at 13:30
  • $\begingroup$ Sorry for my comment i agree with u Jim $\endgroup$ – palash jain Apr 24 '17 at 14:58

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