Given –

$\rho=2\times10^{-4} \Omega cm$

$\mu _e=6m^2/V-sec$

$\mu_h=0.2m^2/V-sec$

To find – $n_i$

Solution –

For intrinsic semiconductor,

$n_e=n_i-n_h$

we know that,

$\sigma =\dfrac {1}{\rho}$

$\sigma_{in}=\dfrac {1}{2\times 10^{-4}}$

$\sigma_{in}=50 ohm/cm$

We know that, for intrinsic semiconductor,

$\sigma_{in}=n_{i} \ e(\mu_{e}+\mu_{h})$

$n_{i}=\dfrac {\sigma_{in}}{e\cdot(\mu_{e}+\mu_{n})}$

$=\dfrac {50}{1.6\times 10^{-19}[6+0.2]}$

  $=\dfrac {50}{1.6\times 10^{-19}[6.2]}$

$n_{i}=5.04 \times 10^{19}$

∴ Intrinsic carrier density is $5.04 \times 10^{19}$.