A Microscopic View of Ohm's Law — Question 9

Question 9

A mean free path uses microscopic speed rather than drift speed

Use $e=1.602176634\times10^{-19}$ C, exact. Use $\ell=v_r\tau$ with random speed $v_r=1.5\times10^6$ m/s and $\tau=2.0\times10^{-14}$ s. A film has thickness 20 nm. A weak-field criterion is $eE\tau/m\le .01v_r$, with $m=9.11\times10^{-31}$ kg.

1. Find the mean free path and assess whether bulk scattering alone is likely adequate for the film.

2. Find the maximum E allowed by the stated drift criterion.

Question 9 - Solution

A mean free path uses microscopic speed rather than drift speed

(a) Method and calculation

$$\ell=(1.5\times10^6)(2.0\times10^{-14})=30.0\,\mathrm{nm}.$$$$\ell/h=30/20=1.50.$$Surface scattering may matter because the thickness is not large compared with the path.

(b) Interpretation and further calculation

$$eE\tau/m\le.01v_r\Rightarrow E\le\frac{.01mv_r}{e\tau}.$$$$E_{\max}=4.26451\times10^6\,\mathrm{V/m}.$$

Physical check

This field bound checks only the stipulated drift criterion; it is not a material breakdown rating.