 Solution

For one spin-degenerate parabolic two-dimensional band, $k_F=\sqrt{2\pi n}=1.3729\times 10^8$ m${}^{-1}$. The free-electron model with $m=m_e$ givesThe given $10^3\Omega$ is a sheet resistivity, not a bulk resistivity in ohm-metres. The two-dimensional Drude sheet conductance is $n{e}^2\tau /m_e$, soThus $v_F=\hslash k_F/m_e=1.589\times 10^4$ m/s andThe sample temperature is not given. This is specifically the Fermi-surface mean free path, appropriate in the degenerate regime $T\ll 8.3$ K for the bare-mass model. At higher temperatures the characteristic thermal speed should replace $v_F$ for a thermal transport distance. If a real two-dimensional band has mass $m^{\ast }$, the stated results scale as $E_F,T_F\propto 1/m^{\ast }$, $\tau \propto m^{\ast }$ and $v_F\propto 1/m^{\ast }$; the inferred ${\ell }_F=\hslash k_F/(n{e}^2{\rho }_{\square })$ is mass independent under these same Drude assumptions. No semiconductor effective mass or valley degeneracy is silently introduced.