Answer to Question #239059 in Physics for Reena Biswas

Question

Given that the Fermi energy of gold is 5.54 eV, the number density of electrons is? ( use me=9.11×10−31 kg; h=6.626×10−34Js; 1eV=1.6×10−19J )

Answer 1

The number density of electrons is

$$n=\frac{m}{\pi \hbar^2} E_F=\\\space\\ =\frac{9.11·10^{-31}}{\pi·[6.626×10^{−34}/(2\pi)]^2}·5.54·1.6·10^{-19}=\\\space\\ =2.31·10^{19}\text{ m}^{-3}.$$