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Here's a quote from Wikipedia:

As an example, the ground state configuration of the sodium atom is 1s22s22p63s, as deduced from the Aufbau principle (see below). The first excited state is obtained by promoting a 3s electron to the 3p orbital, to obtain the 1s22s22p63p configuration, abbreviated as the 3p level. Atoms can move from one configuration to another by absorbing or emitting energy. In a sodium-vapor lamp for example, sodium atoms are excited to the 3p level by an electrical discharge, and return to the ground state by emitting yellow light of wavelength 589 nm.

Usually the excitation of valence electrons (such as 3s for sodium) involves energies corresponding to photons of visible or ultraviolet light. The excitation of core electrons is possible, but requires much higher energies generally corresponding to x-ray photons. This would be the case for example to excite a 2p electron to the 3s level and form the excited 1s22s22p53s2 configuration.

The questions this prompted for me are this:

1) Does each atom have a finite number of possible state transitions, or could any electron in any atom theoretically be excited to some absurdly high orbital, if the photon it absorbed was energetic enough?

2) Does a given transition (e.g. the 3p to 3s sodium transition) always emit a photon of the exact same energy for a given atom (e.g. the 589 nm example above)? Does the same transition in a different atom emit the same photon (i.e. does a 3p to 3s transition always emit a photon of 589 nm)?

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2 Answers 2

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  1. There's an infinite number of orbitals, and thus an infinite number of possible state transitions. However, the energy of the orbitals asymptotes to a finite value - for example, a hydrogen atom has energy levels given by the formula $E_n = -\frac{13.6\text{ eV}}{n^2}$ (ignoring some very tiny quantum corrections). If you let $n$ go up to infinity, the energy approaches zero. So a photon with enough energy, namely 13.6 eV, will knock the electron past all those infinite energy levels and out of the atom entirely.

    Furthermore, if you look at some of the numerical values that come out of that formula: -13.6 eV, -3.4 eV, -1.51 eV, -0.85 eV, -0.54 eV, etc., you'll notice that they get closer and closer together as $n$ gets higher. At some point, the difference between one energy level and the next becomes so small that you can't measure it.

  2. All instances of a particular atom are exactly the same, and so a given transition will always have the same energy for a particular kind of atom (including the isotope). This is how we are able to identify elements by their spectral lines. However, different atoms have different energy levels, because of the different nuclear charge and mass, and also because of multiple electron interactions, so the energy will vary from one atom to another, even for the same transition.

To address the question in your title, there is a list of the spectral lines of most elements in the CRC Handbook of Chemistry and Physics. I'm not sure if it lists the corresponding state transitions, though. Unfortunately the CRC website itself is restricted, so you'd need to pay to access it, but the table itself is probably available elsewhere online. It's quite extensive, listing hundreds of spectral lines for each of the elements.

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    $\begingroup$ I'd just point out that this is for a single "free" hydrogen atom. A hydrogen molecule, or a hydrogen atom bound in some other molecule would be different. $\endgroup$
    – MaxW
    Commented May 4, 2018 at 17:37
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A typical highly excited atom has an outer electron that sees a charge +1 remainder far away, essentially at a point, and such a configuration is indistinguishable from a Hydrogen atom, in terms of the transitions of the outer electron. Instead of orbiting the proton, it orbits the core of remaining electrons.

The hydrogen atom in a zero temperature environment has infinitely many states, indexed by n,l,m, and the spin of the electron, and the energy goes as $R/n^2$ where $R=13.6eV$. Thermal energies at room temperature are 1/30 eV, so if the atom is excited to the n=20 state or above, it will ionize with thermal energy, and this is the limit of n. Beyond this limit, the Hydrogen atom at room temperature will not stay together.

For an effective hydrogen atom in these configurations, where one electron is orbiting the rest, the limit on n is about the same as at room temperature, except the asymptotic approximate-hydrogen "n" is not the same "n" as the n appropriate to the nucleus you are considering. At a temperature of 4K, you can go to n=100 before you ionize the atom.

There are restrictions to how a single photon can excite an atom, which are called the selection rules. These follow from the fact that the X operator in quantum mechanics has a lot of matrix elements which are exactly zero. Absorbing a photon can only link states with orbital angular momentum l different by at most 1 unit, and "m" different by at most one unit.

In addition, there are also qualitative selection rules, because the matrix elements for the X operator decay away from the diagonal. They decay quickly for circular orbits (l the same size as n), which are classically differentiable, and they decay slowly away from the diagonal for highly elliptical orbits (l much much smaller than n), depending on the orbit. So to excite an atom from a circular orbit (large l relative to n) to a much higher n with the l in the range of the selection rule is practically difficult. But l=0 and l=1 states at high n are highly elliptical orbits, which are classically singular (l=0) or nearly singular (l=1), and these have powerlaw decaying X matrix elements to different states at large n. These things are studied in the experimental field of Rydberg spectroscopy.

The precise answers:

  1. You can theoretically excite an atom to very high levels, but there are restrictions on how you can do it using dipole absorption of a single photon, you have to keep knocking it up in energy gradually, through several intermediate states. The high energy levels correspond well to classical motions, and include integrable motions where one electron is far from everything, and solar-system states, where several eletrons are orbiting at very different distances, and chaotic many body states, where a few high-n electrons are interacting strongly. At zero temperature, there are always infinitely many levels
  2. No, the energy emitted in different atoms is not the same, not even close. One difference is that the nuclear charge is bigger, so that the 3p state is a 3p state of a different nucleus, but there is also the fact that electrons interact, so adding more electrons shifts all the level energies profoundly. The energies of the outermost electrons in atoms which fill up to n=3 (those on the third line of the periodic table) is more like the energies of the outermost electrons in atoms which fill up to n=2, this is Mendeleev's correspondences.

It isn't even true that the same atom with the same state will emit exactly the same wavelength each time it is knocked into an excited state. The range of frequency the atom emits is called the natural line-width, and it is the reciprocal of the lifetime of the state.

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