I’m watching the Great Courses Chemistry course and am confused about a few things about atomic emission spectra and can’t readily find an answer via Google so I’m hoping to get a little help here. Thanks in advance!
Why isn’t there a spectral line for every permutation of potential energy orbital drops? So one for every n to ground state, and then one for, say, n4 to n3, n4 to n2, n3 to n2, etc.?
If every element has a characteristic spectrum, why do I see various emission spectrum charts online for sodium that have a different numbers of lines? I see they vary from between 1 and 7 different lines ( though the majority show 2 yellow).
Not directly related to to emission spectrum, but why not - When an electron, say on S1, gets hit with a photon giving it energy to jump to n=3, what happens if the n=3 subshells are already filled?
Can a hydrogen electron only exist at 1s, 2s, 3s, etc? At what n will it get ejected from the atom?
Answering in order:
[ol]
[li]There are, with the caveat that some transitions are forbidden by selection rules which essentially say that the process is possible energetically but has zero or small probability of happening. [/li][li]You can find the sodium spectrum at NIST. The bright lines are at around 5900 Angstroms and the rest of the lines (the ones marked “Na I”) are much fainter (lower “Intensity”) and/or not in the range of visible light (about 3800 - 7400 Angstroms).[/li][li]Caveats about indistinguishability aside, the Pauli exclusion principle tells me that I can’t excite an electron from the n=1 shell to the n=3 shell if the n=3 shell is full… but if the photon has enough energy to do that, it also has enough energy to kick an electron from the n=3 shell off the atom entirely. So what happens is that the atom is ionized.[/li][li]No, it can also be in 2p, or 4f, or whatever. It gets ejected from the atom as n tends to infinity.[/li][/ol]
The selection rules and forbidden transitions are due to angular momentum. The electrons in an atom are quantum mechanical, and hence each electron state has a specific well-defined angular momentum associated with it. And the photon it absorbs or emits is also quantum mechanical, and so only has a limited number of possible angular momentum states. If the change in angular momentum between two states doesn’t correspond to an angular momentum allowed to a photon, then an electron can’t transition directly between those two states.
And while a hydrogen atom (or any atom, really) in principle has an infinite number of possible states, in practice, you’ll only see a few. There’s very little difference in energy between a photon capable of bumping an electron up to n=3 or n=4, and one capable of stripping the electron out entirely (at which point the atom’s energy levels become irrelevant).
When you have two atoms, the electron levels split into two. When you have many atoms, the electron levels split into many. A band. When you have many electrons, each of which can move to any level in the band, and you have a low-energy photon with a wavelength large enough to be much larger than the typical distance between electrons, it becomes hard to distinguish between “affecting an electron” and “affecting the electrons”.
When you have two atoms, the electron levels split into two. When you have many atoms, the electron levels split into many. A band. When you have many electrons, each of which can move to any level in the band, and you have a low-energy photon with a wavelength large enough to be much larger than the typical distance between electrons, it becomes hard to distinguish between “affecting an electron” and “affecting the electrons”.
But doesn’t the photon’s quantum energy have to affect only one electron to still be considered a quantum?
Thank you! A few quick follow-ups if you’ll indulge:
So when you see high atomic number elements with just a few spectral lines, it’s because the others are fainter, or outside the visible light spectrum (as you mentioned with sodium)?
If a line is fainter, then are the brighter lines the result of electrons from the same shell constructively interfering thus creating a higher intensity? That would mean the fainter lines are likely from the valence shell where few compatriot electrons might exist?
Why would there be no theoretical limit to n? Wouldn’t the shell with the energy level right below the ionization energy be the theoretical limit?
1.) It could be that other lines are in the ultraviolet or the infrared, as you note. It could also be that under the circumstances of the experiment there are other mechanisms besides photon emission by which the energy can be lost. this happens frequently with species that are isolated inside crystals, rather than in gas phase – the energy goes into phonons that heat the crystal, rather than into light emission. (Although you can get energy loss by non-photon means in gas phases, too – by collisions, for instance)
2.) I’ve never heard of a case of electron interference affecting line intensity, and don’t think that can even happen. In any system, the strength of a line depends upon the population of the excited state and the branching ratio between modes of de-excitation. But see what happens in Rydberg atoms at high values of n:
3.) As the energy increases n goes up, and the levels get closer together. For a simple model, the energy of the nth level is inversely proportional to n squared, so as the energy increases the electron gets kicked into higher, although more closely spaced levels. ( Ionization energy - Wikipedia )
as a practical matter, real atoms start to depart from the model. Besides, as the levels get higher they get "floppier’ and the energy binding them to the nucleus drop off to very tiny values, and you have to take extraordinary effeorts to keep thermal energy and optical energy from ionizing your atom. In practice, scientists have beenb able to send hydrogen-like “Rydberg atoms” to outrageously high values of n
I was actually speaking of photon interference. Since in the double-slit experiment the intensity of each line in the interference pattern is dictated by the degree of constructive/destructive interference from different photons, I was making the assumption that the intensity of the spectral lines from elements was a result of electrons from the same shell (or similar energy difference?) emitting photons of the same wavelength and constructively interfering to create lines of relatively higher intensity. Along the same lines, I was asking if the fainter lines in the emission spectrum were from photons created by electrons in the valence shell where there are possibly fewer electrons available to drop and create photons of the same wavelength to constructively interfere thereby creating a fainter line.
