Question

In a semiconductor mobility of electron, i.e. drift velocity per unit applied electric field is $1.6m^2/Vs$. Density of electrons is ${10}^{19}/m^3$. (Neglect holes concentration). Resistivity of semiconductor is:

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

Answer 1

The correct option is A

$0.4\Omega m$

Step 1: Given

Density of electrons, $n_e={10}^{19}$

Drift velocity, ${\mu }_e=1.6m^2/Vs$

Step 2: Determine the conductivity

This is an n-type semiconductor because electrons are charge carriers here.

Conductivity of n-type semiconductor is given by,

$$\begin{gathered} \sigma =n_e{\mu }_{e}e \\ ={10}^{19}\times 1.6\times 1.6\times {10}^{-19} \\ =2.56{\left(\Omega m\right)}^{-1} \end{gathered}$$

Step 3: Determine the resistivity

We know that resistivity is given by

$$\begin{gathered} \rho =\frac{1}{\sigma } \\ =\frac{1}{2.56} \\ =0.4\Omega m \end{gathered}$$

Hence, option A is the correct option.