Pure Silicon crystal at $300 \mathrm{~K}$ has equal electron and hole concentration of $1.5 \times 10^{16}…

Pure Silicon crystal at $300 \mathrm{~K}$ has equal electron and hole concentration of $1.5 \times 10^{16} \mathrm{~m}^{-3}$. Doping by indium increases $n_h=4.5 \times 10^{22} \mathrm{~m}^{-3}$. The $n_e$ in the doped silicon is:
  1. $5 \times 10^9$
  2. $2.25 \times 10^{10}$
  3. $3 \times 10^{12}$
  4. $9 \times 10^6$

Solution

In an extrinsic semiconductor: $\begin{aligned} & n_e n_h=\left(n_i\right)^2 \\ & n_e \times 4.5 \times 10^{22}=\left(1.5 \times 10^{16}\right)^2 \\ & n_e=\frac{2.25 \times 10^{32}}{4.4 \times 10^{22}} \\ & \therefore n_e=5 \times 10^9\end{aligned}$ .

Asked in: MHT CET 2022 (07 Aug Shift 2)

Practice more Semiconductors questions on Aicharya