Calculation of fluorescence of electron beam

Unlike bremsstrahlung (see Theoretical analysis of bremsstrahlung from electron beam), the faint color of electron beams comes from fluorescence¹. This happens when the electrons in the beam collide with residual air molecules, exciting them to a higher energy level before a decay to a lower energy level, leading to the emission of a photon. Fluorescence only occurs when the electron has sufficient energy to excite an atom (or molecule), but not so much energy that the electron ends up ionizing it. Thus, it is only possible after the electrons have slowed down somewhat such that they do not end up ionizing all the atoms they collide with.

Air primarily consists of nitrogen and oxygen. Both gases have a variety of spectral lines in the UV and visible spectrum. The wavelength(s) ${\lambda }_{\mathrm{em}}$ of the emitted radiation due to electron beam fluorescence are equivalent to the spectral lines of the constituents of air, since in order for the beam to be visible, we must leave a bit of residual air in the chamber – sufficient to slow the electrons down enough that electronic excitations can happen more often than pure elastic scattering or ionization. Effectively speaking, this means that the upper limit for electron scattering is 12.5 eV for molecular oxygen and 15.6 eV for molecular nitrogen², corresponding to photon wavelengths of 99.2 nm and 79.5 nm respectively (which are in the vacuum UV range). Of course, we cannot see wavelengths shorter than ~400-410 nm, hence the visible beam comes from longer-wavelength (and thus lower-energy) photons.

We will now give a basic mathematical analysis. Assuming that the electron beam is in thermodynamic equilibrium, we can model the electrons via a Fermi-Dirac distribution with a continuous distribution of energies, such that the average number of electrons $n(\epsilon )$ with energy between $\epsilon$ and $\epsilon +d\epsilon$ is given by:

$$n(\epsilon )=\frac{1}{e^{(\epsilon -\mu )/(k_{B}T)}+1}$$

Where $\mu$ is the chemical potential, $k_B$ is Boltzmann’s constant, and $T$ is the temperature. The temperature $T$ can be estimated from the (average) kinetic energy $⟨E_K⟩$ of the beam via $⟨E_K⟩\approx \frac{3}{2}{k}_{B}T$ giving us $T=\frac{2}{3k_B}⟨E_K⟩$ (the average kinetic energy is what we typically just call the “beam energy”). Upon substitution we obtain:

$$n(\epsilon )=\frac{1}{e^{3(\epsilon -\mu )/(2⟨E_K⟩)}+1}$$

This tells us that the number of electrons with higher energies slowly decreases with increasing energy, meaning that most electronic excitations occur at “low” energies. However, since $⟨E_K⟩\gg E_I$ (where $E_I$ are the ionization energies of the molecular species in air), the electrons must first lose enough energy before it is even possible for them to emit photons due to electronic excitation (instead of causing ionization).

It can be reasonably deduced that once electrons have slowed down enough such that electron excitations become possible, we will first detect UV light (having the highest energy) and then violet-blue light (the shortest-wavelength light in the visible spectrum) as the electrons gradually slow down³. Given we cannot see UV, this explains why violet/deep blue is the dominant color seen in electron guns (such as cathode-ray tubes) that have enough residual air inside to emit a (faint) glow.

Footnotes

1. A less technical explanation can be found at https://euro-fusion.org/faq/where-does-the-plasma-colour-come-from/ ↩

2. See table from https://www.researchgate.net/figure/Ionization-energy-of-diatomic-molecules-of-the-single-and-double-ionization_tbl2_340942819. ↩

3. This is a simplified explanation; reality is more complex because atomic decays are not always a single decay, where an atom transitions from one distinct energy level to another one. Instead, fluorescence often comes from multiple transitions resulting in the emission of longer-wavelength light than what one would expect from a single decay (this is why phosphors can absorb UV light and emit visible light e.g. in a fluorescent lamp). Such secondary decays are typically illustrated using a Jablonski diagram. For more information consult these slides. ↩