= Solution
Introduce the electron momentum-relaxation rate $\nu$ through $m_e(\dot{\mathbf v}+\nu\mathbf v)=-e\mathbf E$. For a harmonic field $\mathbf E=\operatorname{Re}(\mathbf E_0e^{-i\omega t})$, the real part of the <electrical conductivity> is $\operatorname{Re}\sigma=n_ee^2\nu/[m_e(\omega^2+\nu^2)]$. At weak damping, the mean heating is $Q=\tfrac12\operatorname{Re}\sigma|E_0|^2$ and the beam <intensity> is $I=\tfrac12\epsilon_0cn|E_0|^2$, where $n=\sqrt{1-\omega_p^2/\omega^2}$. Therefore
$$
\frac{dI}{dx}=-\kappa I,\qquad
\kappa\simeq\frac{\nu}{c}\frac{\omega_p^2/\omega^2}{\sqrt{1-\omega_p^2/\omega^2}}.
$$
With the relaxation convention $\nu=1/(2\tau_{ei})$, this becomes the printed result
$$
\boxed{I(x)=I(0)e^{-\kappa x},\qquad\kappa=\frac{1}{2c\tau_{ei}}\frac{\omega_p^2}{\omega^2}\left(1-\frac{\omega_p^2}{\omega^2}\right)^{-1/2}.}
$$
\b[Convention qualification:] a $90^\circ$ deflection time does not by itself specify a momentum-relaxation rate without a collision model. If instead $\tau_{ei}$ is defined by $\nu=1/\tau_{ei}$, the intensity <absorption coefficient> is twice the printed one. Amplitude and intensity damping also differ by two. The result above states the extra convention needed rather than silently identifying these quantities. It assumes $\omega>\omega_p$, $\nu\ll\omega$ and sufficiently weak damping away from the cutoff; the apparent divergence near cutoff is outside that approximation.
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