A Microscopic View of Ohm's Law — Question 2

Question 2

Measured resistivity determines relaxation time only after carrier density is known

Use $e=1.602176634\times10^{-19}$ C, exact. A one-carrier metal has $n=5.00\times10^{28}\,\mathrm{m^{-3}}$, $m=9.11\times10^{-31}$ kg and resistivity $\rho=2.00\times10^{-8}\,\Omega\,\mathrm m$. Use $\sigma=ne^2\tau/m$ and mobility $\mu=e\tau/m$.

1. Infer relaxation time and mobility.

2. Find drift speed in a field of .100 V/m and compare it with the supplied random speed $v_r=1.00\times10^6$ m/s.

Question 2 - Solution

Measured resistivity determines relaxation time only after carrier density is known

(a) Method and calculation

$$\tau=\frac{m}{ne^2\rho}=3.54893\times10^{-14}\,\mathrm s.$$$$\mu=\frac1{ne\rho}=6.24151\times10^{-3}\,\mathrm{m^2/(V\,s)}.$$

(b) Interpretation and further calculation

$$|v_d|=\mu E=6.24151\times10^{-4}\,\mathrm{m/s}.$$$$|v_d|/v_r=6.24151\times10^{-10}.$$

Physical check

Large microscopic speed is compatible with a tiny mean drift; random directions nearly cancel.