Question

The conductivity of silver is 6.5 x 10⁷per Ohm per m and number of conduction electrons per m³ is 6 x 10²⁸. Find the mobility of conduction electrons and the drift velocity in an electric field of 1 V/m. Given m = 9.1 x 10^–31 kg and e = 1.602 x 10^–19 C.

Answer 1

First of all, find the relaxation time:

$$\tau=\frac{m_e\sigma}{ne^2}.$$

Now write the expression for drift velocity:

$$v=\frac{eE}{m_e}·\tau=\frac{E\sigma}{ne}=6.76\text{ mm/s}.$$

Find the mobility of conduction electrons:

$$\mu=\frac{v}{E}=6.76·10^{-3}\space\frac{\text{m}^2}{\text{V}·\text s}$$