Question

Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an $n-\text{type}$ semiconductor, the density of electrons is ${10}^{19}{\text{m}}^{-3}$ and their mobility is $1.6{\text{m}}^2{\text{V}}^{-1}{\text{s}}^{-1}$ then the resistivity of the semiconductor (since it is an $n-\text{type}$ semiconductor contribution of holes is ignored) is close to:

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

Answer 1

The correct option is C $0.4\Omega \text{m}$

Current density is given by,
$j=\sigma E=ne{v}_d$

$\sigma =ne\frac{v_d}{E}=ne\mu$

$\text{Resistivity,}\rho =\frac{1}{\sigma }=\frac{1}{ne\mu }$

$\rho =\frac{1}{{10}^{19}\times 1.6\times {10}^{-19}\times 1.6}$

$\Rightarrow \rho \approx 0.4\Omega \text{m}$

Hence, option $\left(\text{C}\right)$ is correct.