= Solution
<Quasineutrality> gives $n_e=4\times10^{26}\,\mathrm{m^{-3}}$. Using the supplied $90^\circ$ collision-time expression and $\ln\Lambda=10$ gives
$$
\tau_{ei}=4.63\times10^{-13}\,\mathrm s,\quad
\omega_p=1.128\times10^{15}\,\mathrm{s^{-1}},\quad
\omega=\frac{2\pi c}{0.35\,\mu\mathrm m}=5.382\times10^{15}\,\mathrm{s^{-1}}.
$$
The printed <absorption coefficient> then yields $\kappa=1.62\times10^2\,\mathrm{m^{-1}}$. Since the unabsorbed fraction is $e^{-\kappa L}$,
$$
\boxed{L_{1/2}=\frac{\ln2}{\kappa}=4.28\times10^{-3}\,\mathrm m\simeq4.3\,\mathrm{mm}.}
$$
This uses exactly the coefficient specified by the paper and neglects reflection and changing plasma conditions. With $\nu=1/\tau_{ei}$ instead, the corresponding thickness is $2.14\,\mathrm{mm}$, illustrating the convention dependence discussed in part (c).
Back to article page
