MHT CET · Physics · Semiconductors
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:
- A \(5 \times 10^9\)
- B \(2.25 \times 10^{10}\)
- C \(3 \times 10^{12}\)
- D \(9 \times 10^6\)
Answer & Solution
Correct Answer
(A) \(5 \times 10^9\)
Step-by-step Solution
Detailed explanation
In an extrinsic semiconductor:
\(\begin{aligned} & n_e n_h=\left(n_i\right)^2 \\ & n_e \times 4.5 \times 10^{22}=\left(1.5 \times 10^{16}\right)^2 \\ & n_e=\frac{2.25 \times 10^{32}}{4.4 \times 10^{22}} \\ & \therefore n_e=5 \times 10^9\end{aligned}\)
\(\begin{aligned} & n_e n_h=\left(n_i\right)^2 \\ & n_e \times 4.5 \times 10^{22}=\left(1.5 \times 10^{16}\right)^2 \\ & n_e=\frac{2.25 \times 10^{32}}{4.4 \times 10^{22}} \\ & \therefore n_e=5 \times 10^9\end{aligned}\)
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