Answer to Question #240850 in Physics for Alwin

Question #240850

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 )

Expert's answer

n=\frac{2m\pi}{h^2} E_F=\\\space\\ =\frac{2(9.11·10^{-31})·3.14}{(6.626·10^{−34})^2}·5.54·1.6·10^{-19}=\\\space\\ =2.31·10^{19}\text{ m}^{-3}.

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