I understand that the electron needs a specific quantized amount of energy in order to be excited to another state. For example, hydrogen requires $10.2\ \mathrm{eV}$ for its electron to jump from $n=1$ to $n=2$.

Scenario 1:
What happens if the photon that it has collided with has an energy above $10.2\ \mathrm{eV}$, let's say $10.3\ \mathrm{eV}$? Would the electron still jump from $n=1$ to $n=2$, but the remaining $0.1\ \mathrm{eV}$ be kept within the photon? If so, would Compton's effect occur where the photon is scattered in another direction with a different frequency?

Scenario 2:
What happens if the light is emitting photons with energy of $13\ \mathrm{eV}$? Would it be possible for the electrons to be absorbing different amounts of energy? i.e some electrons absorb energy to be excited to $n=3$ or some to $n=2$? I would assume that this is the case since the emission spectra plays on this idea by having different types of “light” created with the electrons emitting different frequencies of light.

I understand that similar questions have been posted on this site, but I do not understand the wordings of some of them.

  • $\begingroup$ I have not studied this very much but I do recall an article somewhere a few years ago. The idea was that protons emit energy waves from the nucleus radiating outward spherically. When the electrons absorb energy and rise to higher level they become unstable but are more stable in the valleys or trough of the radiating energy wave. I have always assumed the energy levels of the valence electrons rise and fall together and settle into these troughs. I also assumed the small difference of energy would either radiate away or settle back into the electrons again. $\endgroup$ Commented Sep 7, 2019 at 14:56
  • $\begingroup$ @BillAlsept Could you please provide a further explanation on this? I also thought that the small difference in energy would radiate away due to Compton's effect. $\endgroup$
    – S. Lee
    Commented Sep 8, 2019 at 2:48
  • $\begingroup$ Look up "Compton defect". $\endgroup$
    – John Doty
    Commented Jul 1, 2022 at 19:22

2 Answers 2


If the photon energy is off significantly, like in the example, it won’t be absorbed: those atoms are transparent to light of that wavelength.

This is why gases show a spectrum of absorption lines, with only specific wavelengths absorbed.

  • $\begingroup$ I understand that the electron won't absorb the photon's energy if it is significantly off, as it only accepts quantised amount of energy. However, what I'm questioning is what happens if the energy given by the photon is higher than the quantised energy needed? Are you suggesting that the electron won't accept it at all? Then I don't understand how come an element can emit different wavelengths, as that can only happen if the amount of energy accepted by the electrons varies. i.e, in discharge tube of Hydrogen, some electrons accept 12.09eV's, whilst some accepts 12.75eV's. $\endgroup$
    – S. Lee
    Commented Sep 8, 2019 at 2:46
  • $\begingroup$ 12.09 can be absorbed or emitted by one transition, others valued by others. But if the energy doesn’t match a transition, it doesn’t happen. $\endgroup$ Commented Sep 8, 2019 at 2:54
  • $\begingroup$ Yes, but I'm just confused because in a question it stated; "In an experiment similar to that of Franck and Hertz, electrons of energy 12 eV are fired into a gas. Electrons penetrating the gas are collected and their energies measured at 12 eV, 1.4 eV and x eV. If the spectrum of the light emitted from the gas is also analysed and found to contain photon energies of 11.4 eV, 10.6 eV and y eV, deduce the values of x and y." In this example, it's connoting to the idea that an electrons absorb a quantised amount of energy from a larger portion of energy, leaving behind unused energies. $\endgroup$
    – S. Lee
    Commented Sep 8, 2019 at 5:03
  • $\begingroup$ Or would this scenario be different as it's absorbing energy from an electron and not a photon per se? However, wouldn't the same theory of absorbing quantised amount of energy apply? $\endgroup$
    – S. Lee
    Commented Sep 8, 2019 at 5:06
  • 1
    $\begingroup$ Many things in that statement are wrong, unfortunately. The line spectrum is done with light. Photons give energies to atoms that match any one of many transitions. The energies that atoms can absorb are the same regardless of whether that energy comes come electrons or photons. Unfortunately, i don’t think StackExchange is the right place to get the kind of interactive tutorial about this physics that you’re looking for; this is not what comments are meant for. $\endgroup$ Commented Sep 8, 2019 at 6:39

Compton scattering allows photons to lose partial amounts of energy to electrons. I presume that if the energy acquired by the electron is insufficient to ionize it, then the electron will transition to a different orbital. When the electron returns at some point to a lower energy orbital, it will emit a longer wavelength photon than the original photon. That is why the car interior heats up. The visible light photons are losing energy to the material inside the car. Since the material is radiating infrared, which is trapped by the glass inside the car, the interior material cannot lose energy through radiation, or conduction until it gets sufficiently hot. Also, since most infrared radiation comes from the vibration of chemical bonds, I suspect that the electrons of chemical bonds are also capable of absorbing a small fraction of the energy of the visible light photons and thus lengthening the wavelengths of the original photons into the infrared range. I suspect that the new longer wavelength photons then created would become even more effectively absorbed by the molecular bonds of the material inside the car to increase the molecular bond vibrations velocities, which elevates the kinetic energy of the material to make the material increase in temperature. I am not a physicist, but am just trying to get the general idea of the process. Please correct me if I am mistaken.


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