Q. Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is $10^{19} m^{-3}$ and their mobility is $1.6 m^2/(V.s)$ then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to:

Solution:

$j = \sigma E = nev_{d} $
$ \sigma =ne \frac{v_{d}}{E} $
$=ne \mu$
$ \frac{1}{\sigma} = \rho = \frac{1}{n_{e}e\mu_{e}} = \frac{1}{10^{19} \times1.6 \times10^{-19} \times1.6} $
$ = 0.4 \Omega m$

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