Question

The intrinsic carrier density is $1.5\times {10}^{16}{\text{m}}^{-3}$. If the mobility of electron and hole are $0.13{\text{m}}^2{\text{V}}^{-1}{\text{s}}^{-1}$ and $0.05{\text{m}}^2{\text{V}}^{-1}{\text{s}}^{-1}$, calculate the conductivity.

• A. $2.32\times {10}^{-4}{\Omega }^{-1}{\text{m}}^{-1}$
• B. $4.32\times {10}^{-4}{\Omega }^{-1}{\text{m}}^{-1}$
• C. $5.32\times {10}^{-4}{\Omega }^{-1}{\text{m}}^{-1}$
• D. $6.32\times {10}^{-4}{\Omega }^{-1}{\text{m}}^{-1}$

Answer 1

The correct option is B $4.32\times {10}^{-4}{\Omega }^{-1}{\text{m}}^{-1}$

We know that, conductivity,

$\sigma =e(n_e{\mu }_e+n_h{\mu }_h)$

$\Rightarrow \sigma =e{n}_i({\mu }_e+{\mu }_h)$

$\Rightarrow \sigma =1.6\times {10}^{-19}\times 1.5\times {10}^{16}\times (0.13+0.05)$

$\Rightarrow \sigma =4.32\times {10}^{-4}{\Omega }^{-1}{\text{m}}^{-1}$

Hence, option $\left(B\right)$ is the correct answer.