Theoretical analysis of bremsstrahlung from electron beam

When electrons undergo scattering due to an atomic nucleus¹, two sources of energy loss are present: Coulomb losses, which are perfectly elastic and dominate at low energies, and Bremsstrahlung, which is (typically) inelastic and dominate at high energies. The latter case is the primary emitter of high-energy photons from electron-nuclei collisions, hence we are interested in the latter.

Bremsstrahlung emits photons of all wavelengths above a minimum wavelength of ${\lambda }_{\min }=\frac{hc}{E_K}$, where $E_K$ is the kinetic energy of the incident electron (equal to the beam energy if the electrons come from an electron beam). The radiation spectrum of Bremsstrahlung is therefore continuous, and may be expressed using Kramers’ formula²:

$$I(\lambda )d\lambda =\frac{K}{{\lambda }^2}\left(\frac{\lambda }{{\lambda }_{\min }}-1\right)d\lambda ,\lambda >{\lambda }_{\min }$$

Where $I(\lambda )$ is the intensity (photon count) emitted between wavelengths $\lambda$ and $d\lambda$, ${\lambda }_{\min }=\frac{hc}{E_K}$, and $K$ is a constant (which we will discuss more in-depth later). One may also obtain the normalized radiation spectrum given by:

$$\frac{I(\lambda )}{K}=\frac{1}{{\lambda }^2}\left(\frac{\lambda }{{\lambda }_{\min }}-1\right),\lambda >{\lambda }_{\min }$$

Both the unnormalized and normalized spectrum have a peak at $\lambda =2{\lambda }_{\min }$; hence, while the bremsstrahlung photons can have any energy $0<E<E_K$, Kramers’ formula predicts that most of the photons will have an energy of:

$$E_{brem.}\approx \frac{hc}{2{\lambda }_{\min }}=\frac{1}{2}{E}_K$$

Where $E_{brem.}$ denotes the typical photon energy of bremsstrahlung. While Kramers’ result was derived prior to modern quantum theory and is not fully consistent with empirical data, this prediction is quite accurate. Indeed, a general heuristic is that bremsstrahlung photons are most likely to have approximately one-third to one-half of the energy of the incident electron³⁴.

However, there will always be outlier photons with both very high (relative to $E_K$) and very low energies. The average energy will therefore be skewed by these extremes. To compute it, we integrate $I(\lambda )d\lambda$ over all wavelengths ${\lambda }_{\min }<\lambda <{\lambda }_{\max }$, where we set a cutoff wavelength to avoid the integral diverging as ${\lambda }_{\max }\to \infty$, leading to an unphysical result. We set ${\lambda }_{\max }$ to be the wavelength of a photon with the minimum energy above the binding energy of the innermost electron shell (K-shell)³; this is known as the “K-edge”⁵ and is given (exactly for hydrogen-like atoms, approximately otherwise) by:

$$E_{\mathrm{bind}}=Rhc{Z}^2=(\text{13.6 eV})\cdot Z^2,{\lambda }_{\max }=\frac{hc}{E_{\mathrm{bind}}}$$

Where $R$ is the Rydberg constant, and $Z$ is the atomic mass. Thus the average of the radiation spectrum is given by:

$$\begin{array}{rl}⟨I⟩& =\frac{1}{{\lambda }_{\max }-{\lambda }_{\min }}{\int }_{{\lambda }_{\min }}^{{\lambda }_{\max }}\frac{K}{{\lambda }^2}\left(\frac{\lambda }{{\lambda }_{\min }}-1\right)d\lambda \\ & =\frac{K}{{\lambda }_{\min }({\lambda }_{\max }-{\lambda }_{\min })}\left(\ln \left(\frac{{\lambda }_{\max }}{{\lambda }_{\min }}\right)+\frac{{\lambda }_{\min }}{{\lambda }_{\max }}+1\right)\end{array}$$

The constant $K$ can be empirically-determined, but Kramers also gave an analytical expression in his original paper, albeit in a different form. Using modern notation, Kramers’ original (1923) formula⁶ is given by:

$$\tilde{I}(\nu )d\nu =\frac{8\pi \dot{N}}{3\sqrt{3}l}\frac{Z{e}^{2}h}{m{c}^3}({\nu }_{\max }-\nu )d\nu \approx \dot{N}Z\cdot (5\times 10^{-50})({\nu }_{\max }-\nu )d\nu$$

Where $d\epsilon =I(\nu )d\nu$ is the energy emitted (in the form of Bremsstrahlung) between frequencies $\nu$ and $\nu +d\nu$ across all solid angles (hence no angular dependence), $\tilde{I}(\nu )$ is the radiated energy at frequency $\nu$, $\dot{N}$ is the number of incident electrons per unit time, $l\sim 10^6$ is a dimensionless number, and ${\nu }_{\max }$ is the cutoff (maximum) frequency of the Bremsstrahlung photons. However, this result is in CGS units; to convert to modern SI units we perform the replacement $e^2\to \frac{e^2}{4\pi {\epsilon }_0}$ in the formula. This gives us:

$$\tilde{I}(\nu )d\nu =\frac{2}{3\sqrt{3}l}\frac{\dot{N}Z{e}^{2}h}{{\epsilon }_{0}m{c}^3}({\nu }_{\max }-\nu )d\nu$$

Written in terms of wavelength, one first notes that since $\nu =c/\lambda$ we have $|d\nu |=\frac{c}{{\lambda }^2}|d\lambda |$, although this divides out from both sides of the equation and hence does not appear in the final result. One also makes the substitution $h\nu \to \frac{hc}{\lambda }$, giving us (after simplifications):

$$\tilde{I}(\lambda )d\lambda =\frac{2}{3\sqrt{3}l}\frac{\dot{N}Z{e}^{2}h}{{\epsilon }_{0}m{c}^2}\left(\frac{1}{{\lambda }_{\min }}-\frac{1}{\lambda }\right)d\lambda$$

Where ${\lambda }_{\min }\equiv \frac{c}{{\nu }_{\max }}$. To convert this to photon count $I(\lambda )$, we divide by the photon energy $E_{\mathrm{ph}}=\frac{hc}{\lambda }$ of a single photon of wavelength $\lambda$:

$$I(\lambda )=\frac{\tilde{I}(\lambda )}{E_{\mathrm{ph}}}=\tilde{I}(\lambda )\frac{\lambda }{hc}$$

Hence, upon substitution, we have:

$$\begin{array}{rl}I(\lambda )d\lambda & =\frac{2}{3\sqrt{3}l}\frac{\dot{N}Z{e}^2}{{\epsilon }_{0}m{c}^3}\lambda \left(\frac{1}{{\lambda }_{\min }}-\frac{1}{\lambda }\right)d\lambda \\ & =\underset{K/{\lambda }^2}{\underbrace{\frac{2}{3\sqrt{3}l}\frac{\dot{N}Z{e}^2}{{\epsilon }_{0}m{c}^3}}}\left(\frac{\lambda }{{\lambda }_{\min }}-1\right)d\lambda \end{array}$$

Hence we obtain a value of:

$$\frac{K}{{\lambda }^2}=\frac{2}{3\sqrt{3}l}\frac{\dot{N}Z{e}^2}{{\epsilon }_{0}m{c}^3}\approx \dot{N}Z\cdot (4.546\times 10^{-29}{\mathrm{Hz}}^{-1})$$

Which, for a given homogeneous material with a known electron flux (number of electrons per second), is a constant. In practice, however, the normalized version of $I(\lambda )$ is frequently used instead, and as mentioned before, the predicted photon energy of $E_{\mathrm{ph}}=E_K/2$ is independent of which formula is used, hence $K$ need not necessarily be known.

Note that while we have mostly focused on bremsstrahlung, there is also a small contribution spectrum from characteristic radiation⁵. It comes as a result of high-energy photons ejecting an electron, leaving a hole; as the hole is filled through an atomic transition, a photon is emitted. Characteristic radiation, unlike bremsstrahlung, occurs at discrete wavelengths, hence (after spectral broadening) it appears as sharp peaks in the full radiation spectrum. Note that characteristic radiation only occurs for electrons at or beyond the K-edge; that is, those with initial energies greater than the binding energy of the innermost shell, i.e. $E_{\mathrm{ini}}>(\text{13.6 eV})\cdot Z^2$ where $Z$ is the atomic number of the element. Thus, it is not significant at low energies, except for very light nuclei. We will thus consider it negligible.

Shielding requirements

Bremsstrahlung often produces photons in the X-ray range, which are biologically-harmful. Thus, shielding is typically required whenever bremsstrahlung is present; this is a particular concern because photons are typically much harder to block than massive particles (that is, particles with mass) such as alpha and beta particles.

To compute the shielding required for protection from bremsstrahlung, we must calculate the mean range $⟨\Delta x⟩$ that a bremsstrahlung photon produces, which is based on the cross-section for the scattering event. Assuming electron energies that are not too high we may use the photoelectric cross section⁷, which can be derived from quantum mechanics via time-dependent perturbation theory. The differential cross-section is given by:

$$\frac{d\sigma }{d\Omega }={\alpha }^4{r}_e^2{\left(\frac{Z}{E_{\mathrm{ph}}}\right)}^5\frac{{\beta }^3}{\gamma }\frac{{\sin }^2\theta }{(1-\beta \cos \theta {)}^4}\left[1+\frac{1}{2}\gamma (\gamma -1)(\gamma -2)(1-\beta \cos \theta )\right]$$

Where $r_e$ is the classical electron radius. Taking the nonrelativistic limit ($\gamma \approx 1$) we have:

$$\frac{d\sigma }{d\Omega }={\alpha }^4{r}_e^2{\beta }^3{\left(\frac{Z}{E_{\mathrm{ph}}}\right)}^5\frac{{\sin }^2\theta }{(1-\beta \cos \theta {)}^4}$$

Integrating over all solid angles, we obtain the total cross-section as a function of frequency $\omega$:

$$\sigma (\omega )=\oint d\Omega \left(\frac{d\sigma }{d\Omega }\right)=2\pi {\alpha }^4{r}_e^2{\beta }^3{\left(\frac{Z}{E_{\mathrm{ph}}}\right)}^5{\int }_0^{\pi }d\theta \sin \theta \frac{{\sin }^2\theta }{(1-\beta \cos \theta {)}^4}=\frac{8\pi }{3}{\left(\frac{Z}{\hslash \omega }\right)}^5\frac{{\alpha }^4{r}_e^2{\beta }^3}{({\beta }^2-1{)}^2}$$

Where as before, the photon energy is $E_{\mathrm{ph}}=\hslash \omega$. The absorption (that is, attenuation) coefficient is $\mu =n\sigma$, where $n$ is the atomic density of the target medium. The photon flux thus satisfies $\Phi (x)={\Phi }_0{e}^{-\mu x}$, and since the energy of the beam is linearly-proportional to the flux, it thus satisfies $E(x)=E_0{e}^{-\mu x}$. Translating this in terms of the energy, we obtain the following expression for the energy loss per unit length:

$$-\frac{dE}{dx}=-\mu E_0{e}^{-2\mu x}=\mu E$$

For $N$ photons the total energy is $E=N\hslash \omega$. Thus we obtain:

$$-\frac{dE}{dx}=N\hslash \omega \sigma (\omega )n$$

The characteristic length scale is given by:

$$⟨\Delta x⟩={\int }_0^{E_{\max }}dE\frac{1}{-dE/dx}={\int }_0^{{\omega }_{\max }}\frac{d\omega }{N\omega \sigma (\omega )n}=\frac{3}{40\pi }\frac{({\beta }^2-1{)}^2}{N{\alpha }^{4}n{r}_e^2{\beta }^3}{\left(\frac{\hslash {\omega }_{\max }}{Z}\right)}^5$$

Where we integrate over all possible photon frequencies from $\omega =0$ to $\omega =\frac{E_{\mathrm{ini}}-E_{\mathrm{fin}}}{\hslash }$, and utilize the substitution $dE=\hslash d\omega$. One finds that $⟨\Delta x⟩\sim Z^{-5}$, hence heavier nuclei are more effective as shielding material. Moreover, $⟨\Delta x⟩\sim {\omega }_{\max }^5\sim E_{\mathrm{ph}}^5$, meaning that lower-energy photons are stopped much more quickly by shielding materials than higher-energy photons (as we would expect).

Derivation from first principles

The most general expression for the (classical) radiation spectrum due to electron scattering is given by⁸:

$$I(\omega ,\mathbf{n})=\frac{1}{4{\pi }^{2}c}\frac{e^2}{4\pi {\epsilon }_0}{\left|\frac{\mathbf{n}\times \left(\mathbf{n}\times {\vec{\mathbf{\beta }}}_{\mathrm{fin}}\right)}{1-{\vec{\mathbf{\beta }}}_{\mathrm{fin}}\cdot \mathbf{n}}-\frac{\mathbf{n}\times \left(\mathbf{n}\times {\vec{\mathbf{\beta }}}_{\mathrm{ini}}\right)}{1-{\vec{\mathbf{\beta }}}_{\mathrm{ini}}\cdot \mathbf{n}}\right|}^2$$

Where $I=\frac{d^{2}E}{d\Omega d\omega }$ is the spectral power density (energy per unit solid angle per unit frequency), ${\vec{\mathbf{\beta }}}_{\mathrm{fin}}$ is the final (normalized) electron velocity, ${\vec{\mathbf{\beta }}}_{\mathrm{ini}}$ is the initial (normalized) electron velocity, and $\mathbf{n}$ is the normal vector in the direction of the observer. In the non-relativistic limit this reduces to:

$$I(\omega ,\mathbf{n})=\frac{1}{4{\pi }^{2}c}\frac{e^2}{4\pi {\epsilon }_0}\frac{g^2}{m^2{c}^2}{\sin }^2\theta ,\theta ={\cos }^{-1}\left(\frac{\mathbf{g}\cdot \mathbf{n}}{g}\right)$$

Where $\mathbf{g}\equiv mc\left({\vec{\mathbf{\beta }}}_{\text{fin }}-{\vec{\mathbf{\beta }}}_{\text{ini }}\right)$ is the transferred momentum, $g=|\mathbf{g}|$, ${\beta }_{\mathrm{ini}}=\left|{\vec{\mathbf{\beta }}}_{\text{ini }}\right|$, ${\beta }_{\mathrm{fin}}=\left|{\vec{\mathbf{\beta }}}_{\text{fin}}\right|$, and $\omega$ is the frequency of the emitted photon, which is defined by $E_{\mathrm{ini}}-E_{\mathrm{fin}}=\hslash \omega$, where $E_{\mathrm{fin}}$ is the final energy of the electron after the collision. In the case of elastic collisions, ${\beta }_{\mathrm{fin}}={\beta }_{\mathrm{ini}}$, that is, the speed of the electron remains constant. This does not necessarily mean $\mathbf{g}=0$ since $\mathbf{g}$ also depends on the angle between the initial and final velocity vectors; ${\vec{\mathbf{\beta }}}_{\text{fin }},{\vec{\mathbf{\beta }}}_{\text{ini }}$ are vector quantities after all. Meanwhile, as mentioned before, the photon energy can take any value from $\hslash \omega =0$ to $\hslash \omega =E_{\mathrm{fin}}$. We will now derive the more general result We, however, are interested in the general case, where ${\beta }_{\mathrm{fin}}$ can be anything from zero (the perfectly-inelastic limit) to ${\beta }_{\mathrm{ini}}$ (the perfectly-elastic limit).

We choose a coordinate system in which ${\beta }_{\mathrm{ini}}$ is purely on-axis and ${\vec{\mathbf{\beta }}}_{\mathrm{ini}},{\vec{\mathbf{\beta }}}_{\mathrm{fin}}$ are in the same plane, that is, ${\vec{\mathbf{\beta }}}_{\mathrm{ini}}=({\beta }_{\mathrm{ini}},0{)}^T$, and ${\vec{\mathbf{\beta }}}_{\text{fin}}=({\beta }_{\text{fin}}\cos {\theta }^{\prime },{\beta }_{\text{fin}}\sin {\theta }^{\prime }{)}^T$ where ${\theta }^{\prime }$ is the angle between ${\vec{\mathbf{\beta }}}_{\text{fin}}$ and ${\vec{\mathbf{\beta }}}_{\mathrm{ini}}$, which we will refer to as the scattering angle. Note that ${\theta }^{\prime }$ is generally random and cannot be predicted from first principles⁹. However, we may integrate over all possible values of ${\theta }^{\prime }$ to find an average of $I(\omega ,\mathbf{n})$. To begin, we must explicitly write out our vector quantities: note that $\mathbf{n}=(\cos \theta ,\sin \theta {)}^T$ where $\theta$ is the angle of the observer ($\theta$ is entirely independent of the scattering angle ${\theta }^{\prime }$). Invoking the Pythagorean identity we have:

$${\sin }^2\theta =1-{\cos }^2\theta =1-{\left(\frac{\mathbf{g}\cdot \mathbf{n}}{g}\right)}^2=1-\frac{(\mathbf{g}\cdot \mathbf{n}{)}^2}{g^2}$$

Thus, we obtain:

$$\begin{array}{rl}I(\omega ,\mathbf{n})& =\frac{1}{4{\pi }^{2}c}\frac{e^2}{4\pi {\epsilon }_0}\frac{g^2}{m^2{c}^2}\left(1-\frac{(\mathbf{g}\cdot \mathbf{n}{)}^2}{g^2}\right)\\ & =\frac{e^2}{16{\pi }^3{\epsilon }_{0}c}\frac{1}{m^2{c}^2}\left(g^2-(\mathbf{g}\cdot \mathbf{n}{)}^2\right)\end{array}$$

Where $g^2$ and $\mathbf{g}\cdot \mathbf{n}$ are respectively given by:

$$\begin{array}{rl}\mathbf{g}\cdot \mathbf{n}& =mc\left(({\beta }_{\mathrm{fin}}\cos {\theta }^{\prime }-{\beta }_{\mathrm{ini}})\cos \theta +{\beta }_{\mathrm{fin}}\sin {\theta }^{\prime }\sin \theta \right)\\ g^2& =|\mathbf{g}{|}^2=m^2{c}^2{\left({\vec{\mathbf{\beta }}}_{\text{fin }}-{\vec{\mathbf{\beta }}}_{\text{ini }}\right)}^2\\ & =m^2{c}^2\left(({\beta }_{\mathrm{fin}}\cos {\theta }^{\prime }-{\beta }_{\mathrm{ini}}{)}^2+{\beta }_{\mathrm{fin}}^2{\sin }^2{\theta }^{\prime }\right)\end{array}$$

Hence, $I(\omega ,\mathbf{n})$ can now be written as a function of frequency $\omega$, viewing angle $\theta$, and scattering angle ${\theta }^{\prime }$:

$$\begin{array}{rl}I(\omega ,\theta ,{\theta }^{\prime })& =\frac{e^2}{16{\pi }^3{\epsilon }_{0}c}[({\beta }_{\mathrm{fin}}\cos {\theta }^{\prime }-{\beta }_{\mathrm{ini}}{)}^2+{\beta }_{\mathrm{fin}}^2{\sin }^2{\theta }^{\prime }\\ & -{\left(({\beta }_{\mathrm{fin}}\cos {\theta }^{\prime }-{\beta }_{\mathrm{ini}})\cos \theta +{\beta }_{\mathrm{fin}}\sin {\theta }^{\prime }\sin \theta \right)}^2]\\ & =\frac{e^2}{16{\pi }^3{\epsilon }_{0}c}({\beta }_{\mathrm{fin}}^2{\sin }^2\left({\theta }^{\prime }\right)+{\left({\beta }_{\mathrm{fin}}\cos \left({\theta }^{\prime }\right)-{\beta }_{\mathrm{ini}}\right)}^2-{\left({\beta }_{\mathrm{fin}}\cos \left(\theta -{\theta }^{\prime }\right)-{\beta }_{\mathrm{ini}}\cos \left(\theta \right)\right)}^2)\end{array}$$

To make the dependence on frequency $\omega$ clearer one may rewrite ${\beta }_{\mathrm{fin}}$ in terms of $\omega$ as follows:

$$E_{\mathrm{fin}}=E_{\mathrm{ini}}-\hslash \omega =\frac{m{c}^2}{\sqrt{1-{\beta }_{\mathrm{fin}}^2}}⟹{\beta }_{\mathrm{fin}}=\sqrt{1-\frac{m^2{c}^4}{(E_{\mathrm{ini}}-\hslash \omega {)}^2}}$$

In the non-relativistic limit this reduces to:

$$E_{\mathrm{fin}}\approx \frac{1}{2}m{c}^2{\beta }_{\mathrm{fin}}^2=E_{\mathrm{ini}}-\hslash \omega ⟹{\beta }_{\mathrm{fin}}\approx \sqrt{\frac{2(E_{\mathrm{ini}}-\hslash \omega )}{m{c}^2}}$$

Hence, we have:

$$\begin{array}{rl}I(\omega ,\theta ,{\theta }^{\prime })& =\frac{e^2}{16{\pi }^3{\epsilon }_{0}c}({\beta }_{\mathrm{fin}}^2{\sin }^2\left({\theta }^{\prime }\right)+{\left({\beta }_{\mathrm{fin}}\cos \left({\theta }^{\prime }\right)-{\beta }_{\mathrm{ini}}\right)}^2-{\left({\beta }_{\mathrm{fin}}\cos \left(\theta -{\theta }^{\prime }\right)-{\beta }_{\mathrm{ini}}\cos \left(\theta \right)\right)}^2),\\ {\beta }_{\mathrm{fin}}(\omega )& =\sqrt{\frac{2(E_{\mathrm{ini}}-\hslash \omega )}{m{c}^2}},{\beta }_{\mathrm{ini}},{\beta }_{\mathrm{fin}}\ll 1\end{array}$$

Integrating over all possible scattering angles $d{\Omega }^{\prime }=2\pi \sin {\theta }^{\prime }d{\theta }^{\prime }$ with $|{\theta }^{\prime }|\in [0,\pi ]$ (since the scattered electron can recoil backwards i.e. ${\theta }^{\prime }>\pi /2$) as well as over all possible final velocities ${\beta }_{\text{fin}}\in [0,{\beta }_{\mathrm{ini}}]$ and dividing by the respective ranges for the solid angle and velocity, one obtains the averaged spectral density $⟨I⟩$ given by:

$$\begin{array}{rl}⟨I(\theta )⟩& =\frac{1}{4\pi }\oint d{\Omega }^{\prime }\frac{1}{{\beta }_{\mathrm{ini}}}\int d{\beta }_{\mathrm{fin}}I(\omega ,\theta ,{\theta }^{\prime })\\ & =\frac{1}{2{\beta }_{\mathrm{ini}}}{\int }_{{\theta }^{\prime }=0}^{{\theta }^{\prime }=\pi }\sin {\theta }^{\prime }d{\theta }^{\prime }{\int }_{{\beta }_{\mathrm{fin}}=0}^{{\beta }_{\mathrm{fin}}={\beta }_{\mathrm{ini}}}I(\omega ,\theta ,{\theta }^{\prime })d{\beta }_{\mathrm{fin}}\end{array}$$

Evaluating this type of integral is ideal using Monte-Carlo methods, but it can also be evaluated analytically; the solution is:

$$⟨I(\theta )⟩=\frac{e^2}{16{\pi }^3{\epsilon }_{0}c}\frac{{\beta }_{\mathrm{ini}}^2}{36}\left(\frac{9\pi \sin \left(2\theta \right)}{2}-16\cos \left(2\theta \right)+24\right)$$

We now solve for the value of ${\beta }_{\mathrm{fin}}$ at which $I(\omega ,\theta ,{\theta }^{\prime })$ is equal to $⟨I⟩$. This is mathematically incredibly-complicated in the general case, so we will restrict our analysis to the case $\theta =0$ (that is, the observation angle is along the axis) in which case we have:

$${\beta }_{\mathrm{fin}}={\beta }_{\mathrm{ini}}\frac{2\sqrt{2}}{3}\frac{\sin {\theta }^{\prime }}{1-\cos (2{\theta }^{\prime })},a\equiv \frac{e^2}{16{\pi }^3{\epsilon }_{0}c}\approx 1.21667\times 10^{-19}\mathrm{eV}\cdot {\mathrm{s}}^{-1}$$

The average energy of an emitted photon from Bremsstrahlung, which we denote as $⟨E_{\mathrm{ph}}⟩$, is thus given by:

$$\begin{array}{rl}⟨E_{\mathrm{ph}}⟩& =E_{\mathrm{ini}}-\frac{1}{2}m{\beta }_{\mathrm{fin}}^2{c}^2,{\beta }_{\mathrm{ini}}\ll 1\\ & =E_{\mathrm{ini}}-\frac{4}{9}m{c}^2{\beta }_{\mathrm{ini}}^2{\left(\frac{\sin {\theta }^{\prime }}{1-\cos (2{\theta }^{\prime })}\right)}^2,{\beta }_{\mathrm{ini}}\ll 1\end{array}$$

Where ${\beta }_{\mathrm{ini}}$ is defined as:

$${\beta }_{\mathrm{ini}}=\sqrt{1-{\left(1+\frac{E_B}{m{c}^2}\right)}^{-2}}$$

Note: Our analysis utilized a purely classical treatment and ignored quantum mechanical effects. To account for quantum-mechanical corrections, the Bethe-Heitler formula should be used instead (see Zhu (2020) for more details).

Footnotes

1. We assume energies are not so high that we enter into the Compton scattering regime (where all particles are ultra-relativistic, i.e. $\beta \approx 1$). Moreover, we assume a “heavy” nucleus in which the mass of the nucleus is much greater than the electron mass and thus the motion of the nucleus is essentially negligible in the scattering process. ↩

2. See Laguitton & Parrish (1977) ↩

3. See https://doi.org/10.53347/rID-14248 ↩ ↩²

4. See https://oncologymedicalphysics.com/bremsstrahlung-braking-radiation/ ↩

5. See Radiopaedia, Characteristic Radiation ↩ ↩²

6. See On the theory of X-ray absorption and of the continuous X-ray spectrum (1923); this is equation 10 of Kramers’ paper. We have made the replacements $i\to \dot{N}$, $N\to Z$ as they are the standard modern convention. Moreover, we denote the radiated energy as $\tilde{I}(\nu )$ as this is not the same as the photon count $I(\lambda )$. ↩

7. See https://indico.esa.int/event/48/contributions/2538/attachments/2027/2375/photoelectric.pdf ↩

8. See Classical Electrodynamics (Likharev) Ch. 10.4 ↩

9. See Classical Electrodynamics (Likharev) Ch. 10.4, where Likharev writes that “In most applications of the bremsstrahlung theory (as in most scattering problems)…the scattering angle ${\theta }^{\prime }$…[must] be considered random.” ↩