I'm currently studying atomic physics and have encountered a topic that I find challenging. My question is regarding the collision between a photon with an energy of, let's say, 12.20 eV and an atom. During this collision, the electron in the ground state can be excited to reach either the second or third excited state. However, there is a condition: the photon must have an energy equivalent to 12.09 eV or 10.2 eV in order to achieve this. Now, my question is: How can the photon accomplish this? If it cannot, then why? In electron-atom collisions, electrons can transition by providing the necessary energy and retaining their leftover kinetic energy. Can a photon do the same? Please explain the reasons for your answer.


3 Answers 3


An atom can take in a photon of one energy and give off a photon of another energy. This is called inelastic scattering or Raman scattering. It is a two-photon process and therefore under typical conditions it is much less likely than any available single-photon process.

  • $\begingroup$ I thought that Raman scattering involved excitations of the vibrational spectra of a molecule, and so it didn't really apply to atoms. $\endgroup$ Jun 8 at 21:14
  • $\begingroup$ No; you can have Raman scattering with any system and it is much observed and used in atoms. For example we use a controlled version to run an ion trap quantum computer where single (ionised) atoms are the basic element. $\endgroup$ Jun 9 at 10:56
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    $\begingroup$ You learn something new every day! $\endgroup$ Jun 9 at 11:45
  • $\begingroup$ If it absorbs only one photon then why it is called a two-photon process? $\endgroup$ Jun 10 at 12:57
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    $\begingroup$ In atomic physics, an absorption when atom goes from one state to another and two photons are absorbed is indeed a two-photon process. You can also have two-photon emission. If, in a single scattering event, two photons come in and one goes out then you have a third-order process called a 3-photon process. The reflection example does not have a standard terminology I think. It depends on how you count the modes of the field. In one way of counting only a single mode is involved throughout; in another way one mode loses a photon and another gains a photon. $\endgroup$ Jun 11 at 11:16

No, a photon can not do the same because a photon is fundamentally different from an electron. In electron-atom collisions, the electron hits an electron in the atom, but the electron does not absorb the other electron, because of conservation of lepton number and so the electron just transfers enough energy into the electron in the atom to send the electron into the next energy level with the other electron with less energy.

In the case of photon-atom collisions, the electron is required to fully absorb all of the photon's energy whenever it collides, and a free electron is free to take on however much energy it needs but since the electron is in the atom, there are rules regarding how much energy the electron can take because there are no half-energy levels. So if the photon is too energetic, the electron is not allowed to absorb it.

The difference lies in that the photon is absorbed if it has the right energy and is not allowed to be absorbed if it has the wrong energy but in the case of electron atom collisions, the other electron can not be absorbed, and scattering occurs between the two electrons electromagnetically where the other electron transfers energy into the atom with the other electron free to go.


In standard scenario when photon is absorbed by an atom with electron say in a ground state, and electron passes to the next energy level, then photon must cover energy differences in energy levels $$E_1 - E_0 = h\nu \tag 1$$

That's why atom energy level system is quantized. If all photons could do it (passing additional energy to electron kinetic energy, heat in the lattice, etc), there would be no point in quantum mechanics.

Albeit, some exceptions exist to this rule :

  1. Non-linear processes. When light intensity is very big, there can happen a multi-photon absorption, when electron passes to the next level using intermediate virtual energy level. So that some photon pushes electron to the intermediate virtual level and another - completes the job. Coupling to the lattice, i.e. absorbing $\text{photon+phonon}$ in one go of a non-linear process,- is also an option.

  2. Photoionization. When photon energy is big enough to free electron from atom at all,- then whatever photon which passes ionization threshold can do the ionization job,- redundant energy passing to free electron kinetic energy, as photoelectric effect states: $$ \mathcal{E}_{ionization} + K_e = h\nu \tag 2$$

  3. Quantum tunnelling. Next energy level can be interpreted for electron as a potential barrier, which from time to time can be by-passed by electron tunneling in a new state without energy input at all.

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    $\begingroup$ Photons don't need to have "exact" energies to be absorbed. Lines in nature are broadened both due to basic principles (uncertainty, this is referred to as "natural line shape") and experimental considerations (Doppler broadening, Pressure broadening). Wikipedia has a quick overview en.wikipedia.org/wiki/Spectral_line_shape A quick googling finds that in typical experimental conditions the hydrogen $\beta$ line has a width of slightly below $1\,\textrm{nm}$. $\endgroup$
    – tobi_s
    Jun 9 at 4:21
  • $\begingroup$ I never said that spectral lines can't be broadened due to some reasons, like you have mentioned, of course they do. However, basic reasons why spectral lines are centered around $\lambda_0 \pm \Delta \lambda$ are these that I have explained. Line broadening does not conflict with electron energy levels quantization. $\endgroup$ Jun 9 at 6:25
  • $\begingroup$ I don't know what you mean when you are saying "exact" then. Also, "broadened due to some reason" doesn't really describe the natural line shape. $\endgroup$
    – tobi_s
    Jun 10 at 14:59
  • $\begingroup$ @tobi_s Natural broadening is not the only cause of spectral line width, hence no need to exaggerate just one phenomena as you do. There are plenty of broadening reasons. $\endgroup$ Jun 11 at 9:04
  • $\begingroup$ Frankly, I have no idea what you are trying to tell me as you are now pointing to the same things that I pointed to in my original reply. Have a nice day! $\endgroup$
    – tobi_s
    Jun 12 at 3:12

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