Answer to Question #239057 in Physics for Reena Biswas

Question

If 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

Find the relaxation time:

$$\tau=\frac{m^*\gamma}{q^2n}.$$

Fermi velocity:

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

The Fermi mean free path for copper is

$$\Lambda=\nu_F\tau.\\\space\\ \Lambda=\sqrt{\frac{2E_F}{m^*}}·\frac{m^*\gamma}{q^2n}=\frac{\gamma\sqrt{2E_Fm^*}}{q^2n},\\\space\\ \Lambda=3.22·10^{21}\text{ m}.$$