Question

The 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 $1.6m^2/(V.s)$ then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to:

• A. $2\Omega m$
• B. $0.4\Omega m$
• C. $4\Omega m$
• D. $0.2\Omega m$

Answer 1

Answer: (B)

$0.4\Omega m$

$j=\sigma E=ne{v}_d$
$\sigma =ne\frac{v_d}{E}$
$=ne\mu$
$\frac{1}{\sigma }=\rho =\frac{1}{n_{e}e{\mu }_e}$
$=\frac{1}{{10}^{19}\times 1.6\times {10}^{-19}\times 1.6}$
$=0.4\Omega m$