The hole and the free electron concentrations in a pure silicon at room temperature are given by $1.4 \times…

The hole and the free electron concentrations in a pure silicon at room temperature are given by $1.4 \times 10^{16} \mathrm{~m}^{-3}$ each under equilibrium. When it is doped with indium and the hole concentration is $\mathrm{n}_{\mathrm{h}}=4 \times 10^{22} \mathrm{~m}^{-3}$, the electron concentration is
  1. $0.49 \times 10^{10} \mathrm{~m}^{-3}$
  2. $0.14 \times 10^{10} \mathrm{~m}^{-3}$
  3. $0.36 \times 10^{10} \mathrm{~m}^{-3}$
  4. $0.72 \times 10^{10} \mathrm{~m}^{-3}$

Solution

Here, $\mathrm{n}_{\mathrm{i}}=1.4 \times 10^{16} \mathrm{~m}^{-3}$ $\begin{aligned} & \mathrm{n}_{\mathrm{h}}=4 \times 10^{22} \mathrm{~m}^{-3} \\ & \mathrm{n}_{\mathrm{i}}^2=\mathrm{n}_{\mathrm{h}} \mathrm{n}_{\mathrm{e}}\end{aligned}$ Electron concentration, $=\frac{\left(1.4 \times 10^{16}\right)^2}{4 \times 10^{22}}=0.49 \times 10^{10} \mathrm{~m}^{-3}$

Asked in: AP EAMCET 2023 (17 May Shift 2)

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