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I am beginning to learn some very basic electronics. I was learning how and why lightbulbs light up. It turns out it happens because they have a very thin filament which makes the passage very narrow for electrons so they lose a lot of energy colliding with the molecules making up the filament instead of using it to drive the passage of the current. This causes three things.

  • Voltage drop
  • Decrease in intensity
  • Transfer of the electrons' energy to the atoms making up the filament so they vibrate causing it to heat up, and the energy is also used to excite electrons to a higher energy level momentarily where they release the energy in the form of photons where the lower the wavelength, the brighter the light color and the greater the energy. So the filament also glows and lights up.

I am interested in the latter and I have a couple of questions.

  1. Why don't the electrons remain in the higher energy levels?
  2. Why do they choose to release the energy as photons?
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4 Answers 4

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  1. Why don't the electrons remain in the higher energy levels?

In principle, for an atom in complete isolation, the electron can remain in the higher energy level indefinitely. The higher energy level is a "stationary state," of the (completely isolated) system.

However, when there is any coupling (however small) to the environment the atomic energy levels are no longer true stationary states.

Furthermore, since real electrons do couple to photons, it is quite hard to keep individual atoms in an excited state for very long. (Since it is quite hard to achieve total isolation from electromagnetic fields.)

Therefore, one expects that any real atomic electron will eventually transition from a higher energy level to a lower energy level by emitting a photon. (There are other caveats, though, for example, the electronic transition can not violate the Pauli exclusion principle, the electronic transition can't violate conservation of angular momentum, etc.)

  1. Why do they choose to release the energy as photons?

It is a basic fact of nature that electrons are charged and thus are coupled to the electromagnetic field. The quanta of the electromagnetic field are called "photons." Thus, electrons can release energy that can be carried away by photons. Electrons also couple to other charged particle as well as the lattice (in a solid) as well as to other more exotic particles (muons, quarks, taus, etc). So, electrons do not have to release the energy as photons, but it is a very common way for the energy to be transported away, since usually the other couplings are smaller or otherwise repressed.

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    $\begingroup$ Well, it is not true that an electron can stay indefinitely in an energy level which is not the fundamental (lowest possible level). Indeed, even an atom in complete isolation is coupled to the possibility of photons. When there are photons around, the decay is faster (because of the "stimulated emission" en.wikipedia.org/wiki/Stimulated_emission) but the so-called "spontaneous emission" takes place even in complete vacuum. en.wikipedia.org/wiki/Spontaneous_emission $\endgroup$
    – Alfred
    Commented Oct 11, 2023 at 19:24
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    $\begingroup$ @Alfred I think the statement about isolation was assuming the Schrödinger's Hamiltonian where there exists only a static electric field by which the nucleus and electron interact. While it's impossible to get rid of the EM vacuum even in principle in QED (since this would remove the electrostatic interaction too), this simplified model does have its uses. $\endgroup$
    – Ruslan
    Commented Oct 11, 2023 at 20:00
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    $\begingroup$ Even if totally isolated from other atoms, an atom in an excited state WILL decay through spontaneous emission. The Schrödinger equation is a first approximation. QED is not the final word, but it is a much better approximation and it does predict decay by photon emission because the atom is coupled to the EM spectrum even if there is not yet any photon in it. Do check the wiki page en.wikipedia.org/wiki/Spontaneous_emission $\endgroup$
    – Alfred
    Commented Oct 12, 2023 at 7:59
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    $\begingroup$ @hft I guess the problem is the word "isolated". Some might think of it as "a limit of imaginable things in this universe" and being isolated from the EM field is not possible even as a limit, so they don't realize you mean that by "isolated". $\endgroup$
    – JiK
    Commented Oct 12, 2023 at 19:50
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    $\begingroup$ A normal person reading your answer would reasonably assume that "complete isolation" means "this excited atom is the only thing in an entire universe (at the moment)" as @JiK says, but that physics still works normally in that universe. So nothing prevents photon emission, unless you declare that impossible (so physics doesn't work normally in this hypothetical almost-empty universe). Your answer doesn't say that, so it's at best unclear and misleading to a casual audience with a loose qualitative understanding of the relevant physics, at worst wrong. $\endgroup$ Commented Oct 13, 2023 at 6:51
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  1. The short answer is that the state where the electrons are on a higher level isn't stable. Therefore the atoms in the higher energy states decay (i.e. the electrons in the atoms return to the ground state), kind of like when a radioactive nucleus decays. But if we wanted to model this behavior accurately, we would need to model the situation with quantum mechanics.
  2. They don't always release the energy as photons, some atoms release the energy in other ways (so called non-radiative transitions). One way of releasing their energy would be by creating lattice vibrations in the solid which in our case essentially means that the filament heats up. So statistically speaking, some of the excited atoms release their energy as photons, some release them in other ways, like creating lattice vibrations.
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Would you be surprised that a rock you put a meter above ground falls down to the ground when you let go? And yet the same rock can stay up when you support it with other rocks all the way.

Now, of course, electrons are slightly different from rocks. They can stay in the excited state for somewhat longer (or shorter) - though it also needs to be noted that they do not "accelerate" down, so on average, you'll find they do actually behave much the same way if you correct for the energies and forces involved, of course. If you look closely enough, you'll find the rock to fall through distinct energy levels too, though of course a lot more than any electron in an atom has available :)

As the rock falls, the Earth-rock system loses energy, just like the nucleus-electron system when an electron that "falls" to a lower energy level. Incidentally, both tend to ultimately release that energy through electro-magnetism, and in both cases, it isn't always radiation. Photons are involved in either case, if you count virtual photons as photons (and mind, there's some good reasons not to - they are very different).

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It is true that excited states are unstable, and in general the system drops down to the ground state almost immediately, as mentioned in other answers. But there are cases where the system remains in a higher energy state far longer than one would expect.

Metastable states of an atom are excited states that have a lifetime that's longer than usual excited states.

For example, the lifetime of the "F-state" in an ytterbium ion is approximately six years. Because of this long lifetime, the F-state of $\text {Yb}^+$ has a very narrow resonance. In fact, with some smart engineering scientists have been able to develop an atomic clock that has an uncertainty of $\approx 3\times 10^{-23}$.

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