Answer to Question #240821 in Physics for Kumaresan

Question

Given that the Fermi energy of gold is 5.54 eV, the number density of electrons is? (use me = 9.11 x 10-31 kg; 2 points

h = 6.626 × 10-34 Js; leV 1.6 × 10-1⁹ J)

Answer 1

The Fermi energy

$$E_F=\frac{\hbar^2}{2m}(3\pi^2n)^{2/3}$$

$$n=\frac{1}{3\pi^2}\sqrt{(2mE_F)^3}/\hbar^3\\=\frac{1}{3\pi^2}\sqrt{(2*9.11*10^{-31}*5.54*1.6*10^{-19})^3}/(1.054*10^{-34})^3\\=59.3*10^{27}\:\rm m^{-3}$$