 Solution

Use collisional inverse bremsstrahlung in a weakly coupled, unmagnetized plasma. The specified density is $n_e=10^{26}{\mathrm{m}}^{-3}$, and $k_{B}T=1000\mathrm{eV}$. The electron Debye length is $2.35\times 10^{-8}\mathrm{m}$, the plasma frequency is $5.64\times 10^{14}{\mathrm{s}}^{-1}$, and the laser angular frequency is $\omega =2\pi c/\lambda =1.79\times 10^{15}{\mathrm{s}}^{-1}$. Thus $X={\omega }_p^2/{\omega }^2\simeq 0.099<1$: the plasma is underdense and a propagating wave exists.

Estimate $b_{\min }$ as the larger of the classical Coulomb-deflection length $Z{e}^2/(4\pi {ϵ}_0{k}_{B}T)$ and the electron quantum length $\hslash /\sqrt{m_e{k}_{B}T}$. Both are about $9\times 10^{-12}\mathrm{m}$ here, giving $\ln {\Lambda }_C\simeq 8$. With the usual thermal collision convention,The $Z$ factor follows from $n_i{Z}^2=n_{e}Z$.^[ref]

The weakly collisional Drude model gives ${ϵ}_r=1-{\omega }_p^2/[\omega (\omega +i{\nu }_{ei})]$ for fields proportional to $e^{-i\omega t}$. Expanding at ${\nu }_{ei}\ll \omega$ yields a real refractive index $n_r=\sqrt{1-X}\simeq 0.949$ and intensity attenuation coefficientConsequently the intensity e-folding absorption length isi.e. of order a centimetre. Taking a Coulomb logarithm between about 5 and 10 gives roughly $5$–$10\mathrm{mm}$. The electric-field amplitude decay length is twice this intensity length. This estimate assumes a uniform slab thick enough to define an absorption length and neglects gradients, nonlinear quiver heating and other absorption mechanisms.