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 $1.6m^{2}Vs$ then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to

Answer 1

Step 1: Given values

Density of electrons, $n={10}^{19}{m}^{-3}$

Mobility, ${\mu }_e=1.6m^{2}Vs$

We know, Charge of electron, $e=1.6\times {10}^{-19}C$

Step 2: Formula used and substitution

Charge density,

$$\begin{gathered} J=\sigma E \\ J=ne{v}_d \\ \because \sigma =\frac{ne{v}_d}{E} \\ \Rightarrow \sigma =ne{\mu }_e\left(\because {\mu }_e=\frac{v_d}{E}\right) \\ \Rightarrow \frac{1}{\sigma }=\rho =\frac{1}{ne{\mu }_e} \\ \Rightarrow \rho =\frac{1}{{10}^{19}\times 1.6\times {10}^{19}\times 1.6} \\ \Rightarrow \rho =0.39\Omega m \end{gathered}$$

The resistivity of the semiconductor is close to $0.39\Omega m$