I understand that atoms emit photons of wavelength $\lambda$ when electrons transition to lower orbitals according to the equation $E = \frac{hc}{\lambda}$. Based on my understanding, those orbitals have fixed energies, and simple elements like oxygen should have a small number of orbitals. The number of transitions possible, and thus the number of wavelengths, should be quite small. How then is the large numbers of lines on the oxygen line spectrum explained?


2 Answers 2


There are an infinite number of orbitals. Even for simple one-electron Hydrogen, there are an infinite number of energy levels, $E=-E_0/n^2$ for $n=1,2,3,...$ all the way to infinity.

So there are many possible transitions between energy levels, each corresponding to a spectral frequency. Some transitions have a much higher probability than others, so you see the most likely transitions, not all of them.

Addition: Some transitions are actually forbidden, based on “selection rules”. For example, in the dipole approximation for Hydrogen, the $\ell$ quantum number must increase or decrease by 1 in a transition.

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    $\begingroup$ Thanks a lot! Makes more sense now! Would you have a pointer to a resource explaining why some transitions have higher probabilities than other? $\endgroup$ Nov 21, 2018 at 23:14
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    $\begingroup$ You could take a look at this: farside.ph.utexas.edu/teaching/qmech/Quantum/node110.html. Basically, when the atom transitions between two states, it acts as an oscillating electric dipole. The intensity of the spectral line depends on the matrix element $\langle f|\vec{d}|i\rangle$ of the dipole moment operator $\vec{d}$ between the initial state $i$ and the final state $f$. $\endgroup$
    – G. Smith
    Nov 21, 2018 at 23:27
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    $\begingroup$ The term "forbidden transition" is a bit of a misnomer, since these transitions actually happen. For instance, the 21-cm hyperfine transition of hydrogen is "forbidden", and yet its observation is the basis of a good fraction of radio astronomy. $\endgroup$
    – Kyle Oman
    Nov 22, 2018 at 8:38
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    $\begingroup$ @Kyle It's not too much of a misnomer - it just needs to be prefaced correctly. A better term is dipole-forbidden, which means that the transition cannot happen in the dipole approximation. It might still be possible in quadrupole or higher-order approximations, but those are much weaker and you'll likely need a harder, dedicated experiment (or astronomical amounts of gas) to detect it. Wikipedia has more information. $\endgroup$ Nov 23, 2018 at 9:47
  • $\begingroup$ (not that you don't know that. But OP could use the information, as might G.Smith.) $\endgroup$ Nov 23, 2018 at 9:48

The first effect that causes that just follows from oxygen nuclei having a higher charge. Observe the hydrogen spectrum below (by Wikipedia user OrangeDog, CC BY-SA 3.0):

The spectral lines of Hydrogen.

You'll notice that there are a lot of spectral lines out there in the infrared, that exist but that you can't actually see with your eyes. Oxygen has a nucleus with eight times higher charge, so in the Bohr model all the energies are sixty-four times higher (but not in reality because the Bohr model is bad for Oxygen; more on that later), so all those infrared spectral lines are shifted into the visible region of the spectrum, where you can see them.

Another effect is that the lovely hydrogenic spectrum you're familiar with, $E_n = R hc \frac{Z^2}{n^2}$, is, like all good things in physics, an approximation. Specifically, it's the approximation that the potential an electron is in is $U(r) \propto \frac{Ze}{r}$, .

This works wonderfully when you have one electron, because in that case, the potential really is exactly that; the hydrogenic spectrum is, thankfully, the spectrum we see in hydrogen. It works almost as well when you have one outer electron, because in that case all the inner electrons are in a spherically symmetric configuration. But if you have more than one electron, things start getting thorny.

This is because the potential an electron is really in is $U_i(r) = -\frac{Ze^2}{4 \pi \epsilon_0 r_i} + \sum_{ij} \frac{e^2}{8 \pi\epsilon_0 r_{ij}} $; you have the nice coulomb potential from the nucleus, and you have that additional, very displeasing term from the interactions with other electrons. You can't usually just treat that term as small and Taylor expand or what have you, either, because the distance between electrons is about the same as the distance between an electron and a nucleus, and the electron and the nucleus have similar charges.

Approximately, what happens is something called "screening"; electrons end up feeling a small overall charge from the center of the atom, equal to the charge of the nucleus minus the proportion of it that's cancelled out by the other electrons, and you proceed as normal using the adjusted nuclear charge $ Z^* e$. But this charge isn't a constant; electrons far away from the nucleus are well-screened, and might only experience a charge of $e$ (the nuclear charge $Ze$, minus the charge of the remaining electrons of $\left(Z-1\right)e$), and electrons close to the nucleus aren't screened much, so they feel nearly the full charge $Ze$. In between, $Z^*$ is a complex nonlinear function that can't be determined analytically.

(There's still a small part of the potential that doesn't depend only on the distance from the nucleus, because screening is just an approximation, but this really is small so you can just Taylor expand it.)

Since electron spend different amounts of time near the nucleus depending on their angular momentum, this means that the energy of an electron is, in generally, a function of $n$ and $\ell$. Since $\ell$ ranges from $0$ to $n-1$, this means that more complicated atoms have lots more spectral lines.


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