Physics - Semiconductors Question with Solution | TestHub

The conductivity of a semiconductor is given by the formula $\sigma =e(n_e{\mu }_e+n_h{\mu }_h)$, where $e$ is the elementary charge, $n_e$ and $n_h$ are electron and hole concentrations, and ${\mu }_e$ and ${\mu }_h$ are electron and hole mobilities, respectively. Given values are: $e=1.6\times 10^{-19}\text{ C}$, $n_e=10^{19}{\text{ m}}^{-3}$, ${\mu }_e=1.25{\text{ m}}^2{\text{V}}^{-1}{\text{s}}^{-1}$, $n_h=10^{15}{\text{ m}}^{-3}$, ${\mu }_h=0.05{\text{ m}}^2{\text{V}}^{-1}{\text{s}}^{-1}$.$\sigma =1.6\times 10^{-19}((10^{19}\times 1.25)+(10^{15}\times 0.05))$ $\sigma =1.6\times 10^{-19}(1.25\times 10^{19}+0.000005\times 10^{19})$

Since $1.25\times 10^{19}$ is much larger than $0.000005\times 10^{19}$, the hole contribution is negligible for an n-type semiconductor.$\sigma \approx 1.6\times 10^{-19}\times 1.25\times 10^{19}=1.6\times 1.25=2.0\text{ (}\Omega \text{m}{)}^{-1}$. The conductivity is 2.