When light is shined through hydrogen gas, three colors of light appear. The issue I have with this is that hydrogen has one electron, meaning somehow the electron has to be emitting all three of these colors simultaneously. This, however, would be impossible since a single electron can only make one orbital jump at a time, which in turn means one electron can only emit one color at a time, not three. If anyone can offer an explanation, I welcome it.
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$\begingroup$ The electron can inhabit multiple energy states, and transition between them. $\endgroup$– JMLCarterCommented Jul 12, 2018 at 19:59
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$\begingroup$ I understand that, but what I don't understand is how one electron can inhabit 2 orbitals at the same time. $\endgroup$– Anthony DucharmeCommented Jul 12, 2018 at 20:04
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5$\begingroup$ There are many hydrogen atoms in hydrogen gas. Each of these contains an electron, and each electron can transition independently of the others. Not every electron in the gas has to be going through a given transition in order for it to be visible. $\endgroup$– probably_someoneCommented Jul 12, 2018 at 20:04
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$\begingroup$ @probably_someone you are right. each electron can accomodate different energy level according to what amount of energy it has absorbed. $\endgroup$– Gurbir SinghCommented Jul 13, 2018 at 15:11
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2 Answers
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I'm assuming you're talking about the four (the fourth is often faint) Balmer lines in the visible part of the spectrum, at about 656, 486, 434, and 410 nm, often seen emitted from a hydrogen discharge tube. There are other Balmer lines, but at shorter wavelengths, so they're not readily visible (or visible at all) to the human eye.
The tube contains a large number of hydrogen atoms, all with electrons being excited to higher energy levels. These electrons don't all get excited to the same state; they're excited to different energy levels - in some cases, they may be freed all together - and then deexcite in different ways. Some of these produce the four Balmer lines you see. Therefore, no single electron undergoes multiple transitions at the same time; you just have a large number of electrons undergoing different transitions and excitations simultaneously.
The same phenomenon is also at work for tubes of other gases, like helium and neon; I don't know (quantitatively or qualitatively) to what extent having additional electrons changes the emission and strengths of different lines. Hydrogen is commonly used for demonstrations, though (from what I've experienced), and it's the case you asked about.
The probability of a given transition depends on something called Fermi's Golden Rule, which allows you to calculate the probability of a given transition per unit time, $\Gamma$. The mean lifetime of a state is then $\tau=1/\Gamma$. You might have seen this embodied in something called the first Einstein coefficient, which governs the rate of spontaneous emission and shows up in many formulae for calculating line strengths. The greater the transition probability, the shorter the lifetime, and the stronger the line. In general, $\Gamma$ is different for the various transitions from a given energy level.
If we designate the probability of all transitions from a state $i$ per unit time as $\Gamma$, we have $$\Gamma P_i(t)=-\frac{dP_i}{dt}$$ where $P_i(t)$ is the probability that the electron will remain in this state until some time $t$. This leads to natural exponential decay, and the solution $$P_i(t)=P_0e^{-\Gamma t}=P_0e^{-t/\tau}$$ with $\tau$ again being the lifetime of the state.1 In this case, we're assuming that $\Gamma$ is a summation over a whole bunch of transition probabilities. For a more in-depth treatment, read from these notes onwards.
Other quantum mechanical effects cause phenomena like forbidden lines, which have low Einstein coefficients and low transition probabilities (and thus long lifetimes). They're not important for these lines, but do show up in the $\text{O II}$ and $\text{O III}$ forbidden lines in astronomical spectroscopy. The 21-cm hydrogen line - another forbidden transition - is a key indicator of neutral hydrogen, and is therefore great for mapping things like HI regions, which are largely composed of neutral hydrogen. Again, though, these aren't important for the Balmer lines you're interested in.
1 You can easily see that as $t\to\infty$, $P_i(t)\to0$ - that is, the probability of staying in a given state for longer and longer times goes down to $0$.
answered Jul 12, 2018 at 20:07
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$\begingroup$ @pela Good idea! I've filled stuff in as best I could; my knowledge here is really just from studying astronomical spectra, and I couldn't give rigorous derivations. But I'm guessing the OP isn't looking for a whole lot more detail than that. :-) $\endgroup$ Commented Jul 12, 2018 at 21:16
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$\begingroup$ Astronomical spectra are the best spectra :) $\endgroup$– pelaCommented Jul 12, 2018 at 22:29
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$\begingroup$ Is there a rule of thumb for predicting these intensities? I know you've loosely said that they are proportional to their transition probabilities, but isn't there a lot of complexity to where the states transition to? Do they always decay to the ground state? $\endgroup$ Commented Jul 13, 2018 at 16:48
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$\begingroup$ Suppose there is only one hydrogen atom in the discharge tube, what will be the spectrum? $\endgroup$– Cang YeCommented Jul 13, 2019 at 18:18
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You make the Assumption that all the visible light emitted by Hydrogen is emitted at the same time and that a singular atom would even have an emission spectrum
Transitions are instantaneous and photons travel at the speed of light. There is plenty of time in the time it takes to record data for multiple absorptions and emissions of different photons. Not that this matters becasue we cannot isolate and observe the emission spectrum of a singular hydrogen atom; more important is this second point.
Even though there is only one electron, we analyze emission spectrums as a function of a gas which is composed of MOLS of Hydrogen atoms and therefore MOLS of electrons. 10^23 electrons is more than enough to emit three wavelengths of photons.
