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);

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

$E = E_0 - k$;

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

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

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

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

Case A: For $\varepsilon \geq 0.88$ the screening effect is negligible, $\psi(\varepsilon)=19.19-4ln\varepsilon$ and hence $C_B = 1$.