Solution
= Solution
Substitution in part (a) gives
$$
\boxed{k_F=\sqrt{2\pi(4.77\times10^{19})}\simeq1.73\times10^{10}\ {\rm m^{-1}},}
$$
and
$$
\boxed{E_F=\frac{\pi\hbar^2(4.77\times10^{19})}{m_e}
\simeq1.83\times10^{-18}\ {\rm J}\simeq11.4\ {\rm eV}.}
$$
The conversion uses $1\ {\rm eV}=1.60218\times10^{-19}\ {\rm J}$. These are the predictions of the stipulated two-dimensional <free electron model>, with the vacuum <electron mass> and one carrier per inferred square cell.