Question

Pure Si at $300\text{K}$ has equal electron $\left(n_e\right)$ and hole $\left(n_h\right)$ concentrations of $1.5\times {10}^{16}{\text{m}}^{-3}$. Doping by indium increases $n_h$ to $4.5\times {10}^{22}{\text{m}}^{-3}$. Calculate $n_e$ in the doped Si.

• A. $7.5\times {10}^9{\text{m}}^{-3}$
• B. $2.5\times {10}^9{\text{m}}^{-3}$
• C. $5.0\times {10}^9{\text{m}}^{-3}$
• D. $2\times {10}^9{\text{m}}^{-3}$

Answer 1

The correct option is A $7.5\times {10}^9{\text{m}}^{-3}$

For a doped semiconductor, in thermal equilibrium,

$n_e{n}_h=n_i^2$ (Law of mass action)

$\Rightarrow n_e=\frac{n_i^2}{h_h}=\frac{(1.5\times {10}^{16}{)}^2}{3\times {10}^{22}}=7.5\times {10}^9{\text{m}}^{-3}$

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