Question

Find the resistance of an intrinsic $\text{Ge}$ rod $1\text{mm}$ long, $1\text{mm}$ wide and $1\text{mm}$ thick at $300\text{K}$. The intrinsic carrier density is $2.5\times {10}^{19}{\text{m}}^{-3}$. If the mobility of electron and hole are $0.39{\text{m}}^2{\text{V}}^{-1}{\text{s}}^{-1}$ and $0.19{\text{m}}^2{\text{V}}^{-1}{\text{s}}^{-1}$, calculate the resistivity.

• A. $130\Omega$
• B. $230\Omega$
• C. $330\Omega$
• D. $430\Omega$

Answer 1

The correct option is D $430\Omega$

We know that, conductivity,

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

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

$\Rightarrow \sigma =1.6\times {10}^{-19}\times 2.5\times {10}^{19}\times (0.39+0.19)$

$\Rightarrow \sigma =2.32{\Omega }^{-1}{\text{m}}^{-1}$

So, resistivity,

$\rho =\frac{1}{\sigma }=\frac{1}{2.32}=0.43\Omega \text{-m}$

Further, resistance,

$R=\rho \frac{l}{A}=0.43\times \frac{1\times {10}^{-3}}{1\times {10}^{-3}\times 1\times {10}^{-3}}$

$\Rightarrow R\approx 430\Omega$

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