Answer to Question #240575 in Physics for Kumaresan

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 x107-

Answer 1

The number density n of electrons for copper of mass density $\rho$ and molar mass M is

$$n=\frac{N_A}{\Mu}\rho=\frac{6.022·10^{23}}{64·10^{-3}}·8960=8.43·10^{28}\text{ m}^{-3}.$$

From this data (effective mass of electron m* and conductivity $\sigma$), we can find the relaxation time:

$$\tau=\frac{\sigma m^*}{e^2n}=\frac{(5.76·10^{7})(1.01·9.109·10^{-31})}{(1.602·10^{-19})^2(8.43·10^{28})}=\\\space\\ =8.50·10^{-14}\text{ s}.$$