Theoretical analysis of electron beam scattering

Note: This is the write-up associated with https://codeberg.org/elaraproject/elara-labs/issues/35. Moreover we should link to this page on the safety review page (Project Elara Safety Review and Precautions).

Introduction

Electron beams are essential to many technologies, including electron microscopy, free electron lasers, synchrotrons, among others. When electron beams penetrate a medium, the electrons within the beam interact with the particles of the medium in complicated ways that are very different from macroscopic particle collisions. These interaction processes are collectively known as scattering and can be intentional (for instance, using electron beams to bombard a test sample for microscale imaging, or produce neutrons from a lead target), or unintentional (for instance, electron beams in imperfect vacuum chambers which still have some residual air). In this page, we go over the dynamics of electron beam scattering.

Scattering terminology

We will first introduce some terms that are used extensively through this page.

Scattering, broadly speaking, refers to the interaction between two particles that results in an exchange of energy, momentum, or both. Macroscopically, scattering can be described using the classical notion of a collision (elastic or inelastic); microscopically, it is a fundamentally quantum process and cannot be understood in terms of colliding particles. Notably, scattering is typically random, with the probability of two particles scattering through some sort of interaction dictated by statistical laws.

A cross-section is a measure of how likely an interaction is to occur. It has units of area, which can seem counterintuitive at first, but an analogy is a shooting target in archery; the size of the target will determine how likely it is to score a hit. Since scattering is a probabilistic process, it can be understood through the same perspective: the cross-section describes the probability of a scattering event in terms of the equivalent probability of hitting an (imaginary) target.

Bremsstrahlung (“breaking radiation”) refers to the the emission of photons due to a charged particle slowing down in an extern electron field. Generally, bremsstrahlung occurs as a result of a charged particle approaching an atomic nucleus; the electrostatic field of the nucleus has a repulsive effect on the charged particle to rapidly decelerate, leading to the emission of electromagnetic radiation (photons).

Ionization most generally refers to the removal of an electron from an atom by a high-energy particle. In this page, we mostly focus on photons ejecting electrons out of atoms, although there are other ways for ionization to occur. The K-edge refers to ionization of the innermost electron shell (called the K shell in X-ray spectroscopy), which leads to a sharp spike in the radiation spectrum. Ionizing radiation generally refers to any particle with energies high enough to cause ionization, which is in general around 10 eV, corresponding to photons in the UV range (with wavelength $\lambda \le 124\mathrm{nm}$). However, lower-energy photons can nonetheless cause ionization depending on the type of atom/molecule, so this definition is less clear-cut than it may appear to be.

Hard X-rays and soft X-rays both refer to photons in the X-ray spectrum, although they refer to photons of different energies. Soft X-rays generally have photon energies of 5 keV at maximum, corresponding to a minimum wavelength of $\text{0.1 nm}$; hard X-rays have photon energies above 5 keV and have shorter wavelengths than soft X-rays. This difference is manifest in scattering, since hard X-rays can penetrate materials much further than soft X-rays. For the same reason, hard X-rays are much more dangerous to organisms as compared to soft X-rays.

Types of scattering media

We will mostly focus on scattering processes in gaseous media (e.g. air), although we will also devote some discussion on scattering process in bulk media (such as solids). In particular, our focus will be on a low-pressure gas (such as air in a partially-evacuated vacuum tube).

Effect of density

The amount and type of scattering also depends on the density of the particles (atoms, molecules, etc.) within the medium. For an ideal gas, the density $\rho$ is directly proportional to the gaseous pressure $P$. As the density of a medium is decreased (for instance, by lowering the air pressure if the medium is air), the number of scattering events decreases dramatically. In the limit of zero density (for instance, if the electron beam is within an ultra-high vacuum chamber with almost all air removed) then no radiation of any type will be produced and the electron beam continues to travel forever, assuming no external electromagnetic fields are present. If there are external magnetic fields, synchrotron radiation will be produced, and if there are external electric fields, bremsstrahlung will be produced.

Scattering processes

To understand scattering, we must first differentiate between several regimes, which result in very different scattering behavior:

Regime	Beam energies
Low-energy regime	<100 keV
Mid-energy regime	100 keV - 1 MeV
High-energy regime	>1 MeV

In the low-energy regime, initial photon production will come from K-shell ionization (ejection of a valence electron by a photon) and bremsstrahlung, both of which primarily emitting photons in the soft X-ray and UV range, although given the small bremsstrahlung cross-sections at low energies (in the keV range) relatively few of these photons are produced.

As electrons scatter and lose energy due to collisions with gas molecules, fluorescence (light caused by atomic/molecular transitions) becomes possible, with the primary emission of UV light from the highest-energy atomic transitions, followed by blue/violet light (next-highest energy), aided by secondary photons (and secondary electrons) emitted from scattering events. The color emerges from the strong spectral lines of nitrogen and oxygen in the blue/violet region of the visible spectrum, such as the 408 nm and 441 nm lines of oxygen as well as the 400 nm and 496 nm lines of nitrogen. This online visualizer can be used to “see” the colors corresponding to these spectral lines. Some infrared photons will also be produced as a result of vibrational and rotational transitions in gas molecules once the electrons slow down further, although later than UV and visible light as IR transitions generally take place at much lower energies compared to visible and UV transitions. Lastly, infrared photons will also be indirectly produced from the electron beam — emitted from the gas (as opposed to the electron beam itself), due to collisional energy transfers between electrons and gas molecules causing them to gain kinetic energy.

Hence, other than the bremsstrahlung spectrum (which is formally continuous up to a cutoff energy equal to the beam energy, although very weak) the general spectrum is characterized by discrete spectra, in the UV, visible, and (to limited extent) infrared range. Besides (some) high-energy photons, the other source of ionizing radiation are the primary electrons (the electrons in the beam itself) and secondary electrons (delta rays), although they lose energy quickly to the gaseous medium and generally cannot pass through vacuum chamber walls.

In the mid-energy regime, the picture (quite literally) changes. Above 100 keV, a gaseous medium cannot slow down the electrons fast enough to enable UV and optical transitions, so the beam becomes mostly invisible (the figure of 100 keV is from this article. Meanwhile, Compton scattering produces photons of longer wavelengths in addition to secondary electrons. Bremsstrahlung and ionization become far more significant sources of radiation, with both soft and hard X-ray photons being produced.

In the high-energy regime, nuclear effects become significant as the produced bremsstrahlung, with wavelengths in the gamma ray range, will have sufficient energy to cause photonuclear reactions, such as photodisintegration, producing neutron flux. Moreover, pair production becomes possible, resulting in the production of positrons. These conditions are similar to those found in extreme lightning, which have been known to produce terrestrial gamma-ray bursts.

Calculating electron travel range

The electron travel range, also called the stopping distance, measures the average distance an electron in an electron beam can travel before being fully stopped in the medium. It is directly related to the energy electrons lose to the medium, which decreases their kinetic energy and thus causes the electrons to slow down.

Approximate calculation

The mean free path can be used to provide an order-of-magnitude estimate for the average range an electron can travel in the surrounding medium, whether that be a gas or a solid/bulk medium. In the case of an ideal gas, the mean free path $\ell$ is given by:

$$\ell =\frac{k_{B}T}{\sqrt{2}\pi d^{2}p}$$

Where $k_B$ is the Boltzmann constant, $T$ is the temperature, $d$ is the kinetic diameter of a particle, and $p$ is the pressure. The kinetic diameter of various particles is listed in the table below (based on this Wikipedia article and NOAA), according to their respective concentrations in air:

Molecule	Kinetic diameter	Mole fraction in air
Nitrogen	364 pm	78.084%
Oxygen	346 pm	20.946%
Water (vapor)	265 pm	0% (ideal), <4% (typical)
Argon	340 pm	0.934%
Carbon dioxide	330 pm	0.042%
Neon	275 pm	0.0018%
Ozone	460 pm	0.0005%
Methane	380 pm	0.0002%
Krypton	360 pm	0.0001%
Hydrogen	289 pm	<0.0001%

Note: The following is only true for perfectly dry air. Typically, water vapor occupies a significantly higher fraction of air, which can be up to 4% (according to NOAA), although in desert climates air can become nearly perfectly dry.

Crucially, for low energy electron beams the mean free path dictates the mean distance that electrons in an electron beam can travel straight without scattering. While a single collision with an atom/molecule in the surrounding does not necessarily stop an electrons, it will cause energy loss and possibly change the direction of motion of the electron, which macroscopically leads to the electron beam losing collimation as the electrons scatter in all directions.

Detailed calculation

A more accurate method of calculating the mean electron range than the mean free path is the stopping power formula, which gives the mean range in the following form:

$$\Delta x={\int }_0^{E_{max}}\frac{dE}{|S(E)|}$$

Where $S(E)$ is called the stopping power of the medium and depends on both the medium’s properties (such as its density) and the energy of the electrons.

Effects on surrounding medium

Other than causing electron beams to diverge (spread) and lose energy, scattering also causes various effects on the surrounding medium, including thermal, radiative, and structural effects. Indeed, electron beams have been used for applications as varied as generating neutrons for materials testing to changing the color of gemstones by exploiting their effects on different media. We will analyze some of these effects here.

Electron beam heating

Besides emitting radiation, the scattering of the electron beam also generates heat. This comes primarily from low-energy collisions between the electrons and gas molecules, which increases the kinetic energy of the gas molecules and thus the overall temperature. Thus, the electron beam can heat its surrounding medium to very high temperatures, depending on the composition of the medium.

First, we’ll introduce our notational conventions. Let $\Delta T$ be the increase in temperature of the medium surrounding the electron beam, relative to the outside environment, due to electron beam heating. Let us denote the radius of the electron beam by $r_b$, while the boundary walls containing the medium (for instance, the glass walls of a vacuum tube) be at radius $R_0$.

Letting the boundary walls at $r=R_0$ be kept at constant temperature $T_0$ (e.g. room temperature), while letting the gas particles immediately at the edge of the beam ($r=r_b$) have temperature $T=T_0+\Delta T$, one may solve Laplace’s equation ${\nabla }^{2}T=0$ for steady-state heat condition, with the boundary conditions $T(r_b,0)=T_0+\Delta T$ and $T(R_0,0)=T_0$. Assuming cylindrical symmetry and assuming the medium is homogeneous along $z$ (the electron beam axis), Laplace’s equation takes the form:

$$\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial T}{\partial r}\right)+\frac{1}{r^2}\frac{{\partial }^{2}T}{\partial {\theta }^2}=0$$

We solve via separation of variables. Let $T(r,\theta )=f(r)g(\theta )$. After separation of variables it is not hard to verify that $g(\theta )=e^{in\theta }$ for arbitrary integer $n$. Meanwhile, $f(r)$ satisfies the following ODE:

$$\frac{1}{r}\frac{d}{dr}\left(r\frac{df}{dr}\right)=0$$

Upon expanding, this results in the following ODE:

$$r^2\frac{d^{2}f}{d{r}^2}+r\frac{df}{dr}=0$$

This ODE can be solved exactly, giving us $f(r)=C_1+C_2\ln (r)$ where $C_1,C_2$ are undetermined coefficients. The general solution is:

$$T(r)=(C_1+C_2\ln r)e^{in\theta },n=1,2,3,\dots$$

Upon substituting our boundary conditions we have:

$$\begin{array}{rl}T(r_b,\theta )& =C_1+C_2\ln r_b\\ & =T_0+\Delta T\\ T(R_0,\theta )& =C_1+C_2\ln R_0\\ & =T_0\end{array}$$

These two equations can be reduced to one equation:

$$C_1+C_2\ln r_b-\Delta T=C_1+C_2\ln R_0$$

Upon simplifying we have:

$$C_2\ln r_b-\Delta T=C_2\ln R_0$$

Hence:

$$C_2(\ln r_b-\ln R_0)=C_2\ln \left(\frac{r_b}{R_0}\right)=\Delta T$$

Therefore we obtain:

$$C_2=\frac{\Delta T}{\ln (r_b/R_0)}$$

From which $C_1+C_2\ln R_0=T_0$, our second boundary condition, reduces to:

$$C_1+\Delta T\frac{\ln (R_0)}{\ln (r_b/R_0)}=T_0$$

Thus, we have:

$$C_1=T_0-\Delta T\frac{\ln (R_0)}{\ln (r_b/R_0)}$$

And our solution to the boundary-value problem becomes:

$$\begin{array}{rl}T(r,\theta )& =\left(T_0-\Delta T\frac{\ln (R_0)}{\ln (r_b/R_0)}+\Delta T\frac{\ln (r)}{\ln (r_b/R_0)}\right)e^{in\varphi }\\ & =\left(T_0-\Delta T\frac{\ln (R_0/r)}{\ln (r_b/R_0)}\right)e^{in\varphi },r_b\le r\le R_0\end{array}$$

Note that this result is very general and does not assume a priori a specific form for $\Delta T$, other than symmetry and homogeneity. This temperature distribution is axisymmetric and falls off sharply initially, then slowly decreases, meaning that if you were to image an electron beam passing through a medium, you would be able to “see” a high-temperature region immediately surrounding the beam, and a (comparatively) cooler region closer to the boundary walls.

Electron beam heating in solid media

In the case of solid media, we may use Castaing’s formula to calculate the maximum temperature increase $\Delta T$ caused by beam heating, given (in modern notation) by¹:

$$\Delta T=\frac{3W_0}{4\pi \kappa r_b}$$

Where $r_b$ is (in this case) equal to the electron beam radius, $\kappa$ is the thermal conductivity of the medium, and $W_0$ is the power carried by the beam (which, in the case of an ideal electron beam, is given by $W_0=IV$ where $I$ is the beam current and $V$ is the accelerating voltage). For high-density polyethylene (a very common type of plastic), the thermal conductivity is $\kappa \approx 0.5\mathrm{W}\cdot {\mathrm{m}}^{-1}\cdot {\mathrm{K}}^{-1}$. Hence, for a 2 keV electron beam with a current of $I=\text{1 mA}$ and a beam radius of $r_b\approx \text{1 mm}$ one finds that $\Delta T\approx \text{954 K}$, a very high temperature indeed, and enough to melt plastic! Note that for metals, which have high thermal conductivities, $\Delta T$ is substantially smaller, since $\Delta T\sim {\kappa }^{-1}$.

In the case of thin solid media, one must use the alternative formula²:

$$\frac{I⟨E⟩t}{\lambda }=\frac{4\pi \kappa t\Delta T}{0.58+2\ln (2R_0/d)}+\frac{\pi d^2}{2}\epsilon \sigma (T^4-T_0^4)$$

Where $⟨E⟩$ is the average energy loss per collision, $\lambda$ is the mean free path, $t$ is the thickness of the medium, $R_0$ is the same as defined earlier, $d=2r_b$ is the beam diameter, $\epsilon$ is the emissivity of the medium, and $\sigma$ is the Stefan-Boltzmann constant. Note that $\Delta T=T-T_0$, and hence $T=\Delta T+T_0$, so the above expression may be rewritten independently of $T$ as:

$$\frac{I⟨E⟩t}{\lambda }=\frac{4\pi \kappa t\Delta T}{0.58+2\ln (2R_0/d)}+\frac{\pi d^2}{2}\epsilon \sigma ([\Delta T+T_0{]}^4-T_0^4)$$

Note that the second term (which comes from the Stefan-Boltzmann law) can generally be neglected, giving us the following expression for the temperature:

$$\Delta T=\frac{I⟨E⟩}{4\pi \kappa \lambda }\left(0.58+2\ln \left(\frac{2R_0}{d}\right)\right)$$

Electron beam heating in gaseous media

Gaseous media (in our case, we are concerned primarily with low-pressure gases) behave differently when heated by an electron beam, since they have a far lower particle density than solids. For a general fluid (including gases), the heat transfer coefficient $\chi$ relates the change in temperature $\Delta T$ with the thermal power per unit area $P/A$, as follows³:

$$|q|\equiv \frac{P}{A}={\chi }_a\nabla T\approx \frac{{\chi }_{\alpha }\Delta T}{R_0}$$

Where ${\chi }_{\alpha }$ is the heat transfer coefficient, and for an ideal gas, it is given by³:

$${\chi }_{\alpha }=\frac{1}{3}\rho c_V\lambda \overline{v}$$

Where $\overline{v}=\sqrt{\frac{8RT}{\pi M}}$ is the mean velocity of the gas molecules, $M$ is the molar mass of the gas, $R\approx 8.314\mathrm{J}\cdot \mathrm{mo}{\mathrm{l}}^{-1}\cdot {\mathrm{K}}^{-1}$ is the ideal gas constant, $\lambda$ is the mean free path, $\rho$ is the density of the gas, and $c_V$ is the specific heat capacity (at constant volume). However, in the case of real gases, a modification of the above formula is required (via replacing $\lambda$ with $R_0$)³, giving us:

$${\chi }_{\alpha }=\frac{1}{3}\rho c_V{R}_0\overline{v}$$

Hence, we have:

$$\Delta T=\frac{P{R}_0}{{\chi }_{\alpha }A}=\frac{3W_0}{\pi r_b^2\rho c_V}\sqrt{\frac{\pi M}{8RT}}$$

Where we assume that $A=\pi r_b^2$ (as we consider the temperature in the immediate vicinity of the electron beam) and $P\approx W_0=IV$ is the power of the electron beam (with $W_0$, the beam power, being the same as defined previously). Notice that this is an implicit equation since $T=T_0+\Delta T$, hence there is a dependence on $\Delta T$ on both sides of the equation.

An alternative form of the above equation is sometimes more useful. The density of an (ideal) gas is related to its pressure by $\rho =MP/(RT)$, where $P$ is the pressure, $M$ is the molar mass of the gas, and $T$ is the temperature. Therefore, we may express $\Delta T$ in terms of the pressure as:

$$\Delta T=\frac{3W_0}{\pi r_b^2{c}_{V}P}\sqrt{\frac{\pi RT}{8M}}=\frac{3W_0}{\pi r_b^2{c}_{V}P}\sqrt{\frac{\pi R(T_0+\Delta T)}{8M}}$$

Solving this implicit equation for $\Delta T$ gives us:

$$\Delta T=\frac{\alpha \left(\alpha \beta +\sqrt{\beta \left(4T_0+{\alpha }^2\beta \right)}\right)}{2},\alpha =\frac{3W_0}{\pi r_b^2{c}_{V}P},\beta =\frac{\pi R}{8M}$$

The result we find is that $\Delta T$ is approximately inversely proportional to the pressure. For an electron beam with parameters $E=\text{2 keV}$, $I=\text{1 mA}$, $r_b\approx \text{1 mm}$ with an ambient of $T_0=\text{300 K}$, we find that $\Delta T\approx \text{3.7 K}$ for air at a pressure of 1 torr, dramatically increasing to $\Delta T\approx \text{654 K}$ for air at a pressure of $10^{-2}$ torr.

Thermal radiation from beam heating

As the electron beam heats up the surrounding medium, the medium emits thermal radiation (as per the blackbody spectrum). The peak emission wavelength is given by Wien’s law:

$${\lambda }_{\mathrm{peak}}=\frac{b}{T}=\frac{b}{T_0+\Delta T}$$

Where $b\approx 2898\mu \mathrm{m}\cdot \mathrm{K}$ is Wien’s displacement constant. For most low-energy electron beams, this produces a peak wavelength in the infrared range, and is thus (mostly) invisible. However, since the blackbody spectrum is continuous, smaller amounts of visible light are emitted alongside the much larger proportion of infrared light, with the specific ratio between the two dependent on the temperature. For high $\Delta T$ one may be able to see a faint red or orange-red glow from the incandescence of the medium, alongside the blue-violet glow of the electron beam.

Emission of radiation

Electron beams are almost always ionizing radiation themselves, having more than enough energy to ionize atoms. However, they can also create other forms of secondary radiation, primarily through bremsstrahlung, which emits photons in the extreme UV and X-ray range. See Theoretical analysis of bremsstrahlung from electron beam for more details.

Numerical methods

While this page has primarily focused on analytical methods, electron beam scattering is most accurately modelled using numerical algorithms. Because scattering of subatomic particles (like electrons) is fundamentally probabilistic in nature, these algorithms typically use Monte-Carlo algorithms: they sample particles with randomized energies (among other properties) from a chosen probability distribution, and then propagate them through a medium. Such methods can yield extremely accurate values of cross-sections, particle ranges, and other particle transport information.

Meanwhile, for problems involving heat transport, steady-state heat conduction can be determined numerically via solving Laplace’s equation on a grid (e.g. method of relaxation). This is important for irregular geometries, in which case a simple analytical solution often cannot be found. Moreover, if we are also interested in the time evolution of the heat distribution rather than simply the steady-state, we can find it by solving the heat equation (a parabolic PDE) numerically.

Footnotes

1. See eq. 20 of Electron Probe Microanalysis by Castaing (1960). Note that the original formula also contained a factor of $1/J$, where $J=4.186\frac{\mathrm{J}}{\mathrm{cal}}$ is the mechanical equivalent of heat, essentially a conversion factor between joules and calories when working with mixed units (redundant when using SI units, where $J=1$) and used the symbol $C$ rather than $\kappa$ (as is modern practice) for thermal conductivity. We also replaced $r_0$ with $r_b$ and ${\theta }_m$ with $\Delta T$ for consistency with the notation used in the rest of the page. ↩

2. See Radiation damage in the TEM and SEM by Egerton et al. (2004) ↩

3. See Effect of working gas on the electron-beam heating of a ceramic target in the fore-vacuum pressure range by Zolotukhin et. al. (2020) ↩ ↩² ↩³