Question

the effective mass of electron in copper is m∗=1.01 me(me = mass of an electron) and the electric conductivity of copper is 5.76 ×107 Ω−1 m−1 with EF=7 eV then calculate the Fermi mean free path for copper.

Answer 1

The number density of electrons:

$$n=\frac{N_A}{\Mu}\rho=8.43·10^{28}\text{ m}^{-3}.$$

Find Fermi velocity:

$$v_F=\sqrt{\frac{2E_F}{m^*}}.$$

Find relaxation time:

$$\tau=\frac{m^*\sigma}{e^2n}.$$

Thus, the Fermi mean free path $\lambda$ and conductivity are connected within the following equation:

$$\lambda=\tau v_F=\frac{m^*\sigma}{e^2n}\sqrt{\frac{2E_F}{m^*}}=\\\space\\ =\frac{\sigma\sqrt{2m^*E_F}}{ne^2}=\\\space\\=\frac{(5.76·10^7)\sqrt{2(1.01·9.109·10^{-31})(7·1.602·10^{-19})}}{(8.43·10^{28})(1.602·10^{-19})^2}=\\\space\\ =3.82·10^{-8}\text{ ms}^{-1}.$$