I was wondering, if the Energy of a Photon which is absorbed by an Electron, hast to be exactly the Energy of the bound gap.

So if i have two energy levels in an atom $E_2$ and $E_1$, does my Electron have to have exactly the Energy $$h\nu = E_2 - E_1$$ or is it sufficient if the photon has a bigger Energy than that ?

I was wondering because if one assumes the spectrum to be continuous the chance of finding a photon with just the right energy of lets say $h\nu=10.2\text{eV}$ should be rather small.

  • $\begingroup$ What happens if the energy is not sufficient? What happens with the photon in that case? Also what do you mean with was wondering because if one assumes the spectrum to be continuous the chance of finding a photon with just the right energy of lets say $h\nu=10.2\text{eV}$ should be rather small. $\endgroup$
    – iharob
    Jul 3, 2015 at 12:59
  • $\begingroup$ If I have one hydrogen atom, how long would I have to wait until it absorbs a photon and goes into that excited state ? The chance of finding the 'proper' photon seemed very small to me. One would basically have to take the spectral density of the light $u(\nu)$ so the chance of exciting one Electron would be something like $$dW_{1\rightarrow2} = u(\nu) B_{1\rightarrow2} dt$$ $\endgroup$
    – bobo
    Jul 3, 2015 at 13:13

1 Answer 1


No, it is sufficient for the photon energy to exceed the band gap. Any excess energy is transformed into kinetic energy for the electron in the new band. You get exactly the same effect when ionizing an atom - the excess energy simply powers the electron into a faster continuum state.

You should also take into account that photon energies are never exactly defined except for monochromatic beams with infinite temporal duration. This is exactly because of the energy-time uncertainty relation: the only way to have a perfectly defined photon frequency, and hence energy, is to observe it for an infinitely long time. Thus, the photon energy is always spread out over a finite bandwidth.

A similar effect holds for atomic bound-bound transitions, which will always have a finite natural linewidth. This is caused by spontaneous emission, which means that if you leave the atom in an excited state for long enough then it will eventually return to the ground state. This then limits the amount of time in which you can coherently probe the frequency of the transition, and in turn limits the precision to which you can measure this frequency.

  • $\begingroup$ Thank you for your comment ! This indeed clarifies a lot. So one could think of the absorption as a kind of measurement in quantum mechanical terms ? While the state of the Photon might be a superposition of many Eigenstates of the Hamiltonian, the absorption prepares the Photon into the Eigenstate with the appropriate energy ? $\endgroup$
    – bobo
    Jul 3, 2015 at 13:06
  • 2
    $\begingroup$ Please clarify that in the case of raising or lowering an electron from a specific energy level of an atom, ( the way we know what stars are made of) there is a quantum mechanical width to that energy level, so "exact" means "within that width the probability is high" for an electron to be kicked up to the higher energy level by the atom absorbing the photon. $\endgroup$
    – anna v
    Jul 3, 2015 at 13:10
  • $\begingroup$ Probability that the Energy of the photon is exactly equal to band gap is zero. $\endgroup$
    – Paul
    Jul 3, 2015 at 13:14
  • 1
    $\begingroup$ Well, not always: yes, it's possible to eject an electron with an overenergetic photon, but only the "exact" wavelength will send an electron from one bound state to another. Consider, for example, any laser system or the beautiful case of Mossbauer spectroscopy. $\endgroup$ Jul 3, 2015 at 17:04
  • 1
    $\begingroup$ @CarlWitthoft Mossbauer spectroscopy, along with all wave phenomena, is still limited by the bandwidth theorem. The linewidth is bounded by the inverse of the observation time; put another way, you can excite a Mossbauer transition at $\omega$ with light of central frequency $\omega+\Delta\omega$ with arbitrarily large $\Delta \omega$, as long as the pulse length is short enough. $\endgroup$ Jul 3, 2015 at 17:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.