Answer to Question #240580 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;

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

Answer 1

$$E_f=\frac{h^2}{8m}\left(\frac{3}{\pi}\right)^\frac{2}{3}n^\frac{2}{3}\\ 5.54(1.6\cdot10^{-19})=\frac{(6.63\cdot10^{-34})^2}{8(9.11\cdot10^{-31})}\left(\frac{3}{\pi}\right)^\frac{2}{3}n^\frac{2}{3}\\n=8.9\cdot10^{28}\frac{1}{m^3}$$