 Solution

Quasineutrality gives $n_e=4\times 10^{26}{\mathrm{m}}^{-3}$. Using the supplied $90^{\circ }$ collision-time expression and $\ln \Lambda =10$ givesThe printed absorption coefficient then yields $\kappa =1.62\times 10^2{\mathrm{m}}^{-1}$. Since the unabsorbed fraction is $e^{-\kappa L}$,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).