Calculation of radiation yield

The number of photons per MeV per incident electron per $g/cm^2$ of radiator (Z,A) is given by [*]:

$\frac{d^2n}{d\kappa dt} = \frac{3.495 \times 10^{-4}}{A\kappa}[Z^2\Phi_n(Z,E_0,k)+Z\Phi_e(Z,E_0,k)](MeV^{-1}g^{-1}cm^2)$,

where $\kappa$ - photon kinetic energy in MeV;

$E_0$ - incident electron total energy (in units of the electron rest mass);

$k$ - incident photon energy (in units of the electron rest mass);

Calculation of $\Phi_e(Z,E_0,k)$

$\Phi_e(Z,E_0,k) = C_B\{2[1-\frac{2E}{3E_0}+(\frac{E}{E})^2][L-\sqrt{\eta}]+\sqrt{\eta}[1-\frac{L^2}{2\rho}-\frac{1}{\rho^2}(\frac{1}{2}L-[\frac{\rho(\rho+2)(E_0+1)}{E_0-1}]^{\frac{1}{2}})^2]\}$;

$E = E_0 - k$;

$\rho = E_0 -k(1+E_0-\sqrt{E^2 - 1})$;

$\eta = \rho/(\rho+2)$;

$L = 2 ln(\frac{(E_0-1)^{\frac{1}{2}}+[\eta(E_0+1)^{\frac{1}{2}}]}{(E_0-1)^{\frac{1}{2}}-[\eta(E_0+1)^{\frac{1}{2}}]})$;

$C_B = \frac{\frac{1}{4}\psi(\varepsilon)-1-lnZ^{\frac{2}{3}}}{3.798-ln\varepsilon-lnZ^{\frac{2}{3}}}$;

$\varepsilon = 100k/E_0EZ^{2/3}$;

Case A: For $\varepsilon \geq 0.88$ the screening effect is negligible, $\psi(\varepsilon)=19.19-4ln\varepsilon$ (free electron form) and in this case $C_B = 1$.

Case B: For $\varepsilon \lt 0.88$ we have $\psi(\varepsilon) = 19.70 + 4.117(0.88-\varepsilon)-3.806(0.88-\varepsilon)^2 + 31.84(0.88-\varepsilon)^3-58.63(0.88-\varepsilon)^4+40.77(0.88-\varepsilon)^5$

Calculation of $\Phi_n(Z,E_0,k)$

1.a $\gamma(=100k/E_0EZ^{1/3}) \leq 15$ , $k\lt k_x$:

$\Phi_n(Z,E_0,k) = 4([1+ (\frac{E}{E_0})^2][\frac{1}{4}\phi_1(\gamma)-\frac{1}{3}lnZ - f(Z)]-\frac{2E}{3E_0}[\frac{1}{4}\phi_2(\gamma)-\frac{1}{3}lnZ-f(Z)])$

$\phi_1(\gamma),\phi_2(\gamma)$ - screening functions;

$\phi_1(\gamma) = 19.24 - 4ln(\gamma + \frac{2}{\gamma+3})-0.12\gamma e^{-\frac{1}{3}\gamma}$

$\phi_2(\gamma)=\phi_1(\gamma)-0.027-(0.8-\gamma)^2$, for $\gamma\leq0.80$;

$\phi_2(\gamma)=\phi_1(\gamma)$, for $\gamma \gt 0.80$;

1.b $\gamma \gt 15$, $k\lt k_x$:

$\Phi_n(Z,E_0,k) = \frac{p}{p_0}(\frac{4}{3}-2EE_0(\frac{p^2+p^2_0}{p^2p^2_0})+\frac{\omega_0E}{p^3_0}+\frac{\omega E_0}{p^3}-\frac{\omega\omega_0}{pp_0}+l[\frac{k}{2pp_0}(\omega_0(\frac{EE_0+p^2_0}{p^3_0})-\omega(\frac{EE_0+p^2}{p^3})+\frac{2kEE_0}{p^2p^2_0})+\frac{8EE_0}{3pp_0}+\frac{k^2(E^2E^2_0+p^2p^2_0)}{p^3p^3_0}])$

$p_0 = \sqrt{E^2_0-1}$

$p = \sqrt{E^2-1}$

$\omega_0 = ln(\frac{E_0+p_0}{E_0-p_0})$

$l=2ln(\frac{EE_0+pp_0-1}{k})$

2. $k_x \leq k \leq T_0$

$\Phi_n(Z,E_0,k) = \frac{\Phi_{n(1.a or 1.b)}(k_x)-\Phi_{tip}(T_0)}{k_x - T_0}(k-k_x)+\Phi_{n(1.a or 1.b)}(k_x)$

$\Phi_{tip} = 4\pi ae^{-\pi a}F(\zeta)P(\beta)[1-0.838aR(\beta)+0.650a^2]$

$F(\zeta)=[2\frac{\Gamma(\zeta)}{\Gamma(2\zeta+1)}(2a)^{\zeta-1}]^2$, $\zeta = \sqrt{1-(\frac{Z}{137})^2}$

$P(\beta)\rightarrow 1$, $R(\beta)\rightarrow 1$ when $\beta = \frac{p_0}{E_0}\rightarrow 1$

$P(\beta) = \frac{\beta^3\delta^3}{T^4_0}exp[-2a(\frac{1}{\beta}-1)cos^{-1}a]\times M(\beta)$

$R(\beta)=\frac{K(\beta)}{M(\beta)}$

$M(\beta) = \frac{4}{3}+\frac{(\delta-2)(\delta-1)}{\beta^2 \delta}[1+\frac{1}{2\beta^2 \delta}ln(\frac{1-\beta}{1+\beta})]$

$K(\beta)=\frac{1}{\beta^3}[\delta - \frac{17}{2}+\frac{63}{4\delta}-\frac{25}{4\delta^2}-\frac{2}{\delta^3}-\frac{15}{8\beta \delta^3}(\delta-2)(\delta-1)ln(\frac{1-\beta}{1+\beta})]$

$\delta = (1-\beta^2)^{-\frac{1}{2}}$

---

Reference: [*] J.L. Matthews, R.O. Owens, Accurate Formulae For the Calculation of High Energy Electron Bremsstrahlung Spectra, NIM III (1973) I57-I68.