In an intrinsic semiconductor at room temperature, the number of free electrons is $n_e=10^{16}{\text{m}}^{-3}$. If the mobility of electrons is ${\mu }_e=0.13{\text{m}}^2/\text{V}\cdot \text{s}$ and the mobility of holes is ${\mu }_h=0.05{\text{m}}^2/\text{V}\cdot \text{s}$, what is the conductivity $\sigma$ of the semiconductor? (Take $e=1.6\times 10^{-19}\text{C}$)

Solution

1. Given: $$\begin{array}{rl}{n}_e& =n_h\\ & =n_i\\ & =10^{16}{\text{m}}^{-3}\end{array}$$, ${\mu }_e=0.13{\text{m}}^2/\text{V}\cdot \text{s}$, ${\mu }_h=0.05{\text{m}}^2/\text{V}\cdot \text{s}$, and $e=1.6\times 10^{-19}\text{C}$.
2. The formula for the conductivity of an intrinsic semiconductor is $\sigma =e{n}_i({\mu }_e+{\mu }_h)$.
3. Substituting the given values: $\sigma =(1.6\times 10^{-19})\times 10^{16}\times (0.13+0.05)$.
4. Calculating the result: $$\begin{array}{rl}\sigma & =1.6\times 10^{-3}\times 0.18\\ & =0.288\text{S/m}\end{array}$$.
5. Hence the answer is (A).