A Microscopic View of Ohm's Law — Question 10

Question 10

Conductivity and mobility measurements separate density from scattering

Two states A and B of the same one-carrier material have the same effective mass. Measurements give $\sigma_B=3\sigma_A$ and $\mu_B=2\mu_A$. Use $\sigma=ne\mu$ and $\mu=e\tau/m$.

1. Find the density and relaxation-time ratios.

2. At the same imposed current density, find the field and drift-speed ratios, and distinguish their causes.

Question 10 - Solution

Conductivity and mobility measurements separate density from scattering

(a) Method and calculation

$$\frac{n_B}{n_A}=\frac{\sigma_B/\sigma_A}{\mu_B/\mu_A}=3/2.$$$$\tau_B/\tau_A=\mu_B/\mu_A=2.$$

(b) Interpretation and further calculation

$$E_B/E_A=\sigma_A/\sigma_B=1/3.$$$$|v_B|/|v_A|=n_A/n_B=2/3.$$Density controls drift at fixed J; both density and scattering affect the field needed.

Physical check

Conductivity alone cannot identify whether carrier density or mobility changed.