Q145 / 150 · MHT CET 2021 (21 Sep Shift 1) · Physics
In a pure silicon, number of electrons and holes per unit volume are $1.6 \times 10^{16} \mathrm{~m}^{-3}$. If silicon is doped with Boron in a way that on doping hole density increases to $4 \times 10^{22} \mathrm{~m}^{-3}$. Then electron density in doped semiconductor will be
- A$6.4 \times 10^{-9} \mathrm{~m}^{-3}$
- Correct answerB$6.4 \times 10^9 \mathrm{~m}^{-3}$
- C$6.4 \times 10^{-10} \mathrm{~m}^{-3}$
- D$6.4 \times 10^{10} \mathrm{~m}^{-3}$
Solution
$$
\begin{aligned}
& \mathrm{n}_{\mathrm{i}}=1.6 \times 10^{16} \mathrm{~m}^{-3}, \mathrm{n}_{\mathrm{h}}=4 \times 10^{22} \mathrm{~m}^{-3} \\
& \mathrm{n}_{\mathrm{e}} \mathrm{n}_{\mathrm{h}}=\mathrm{n}_{\mathrm{i}}^2 \\
& \therefore \mathrm{n}_{\mathrm{e}}=\frac{\mathrm{n}_{\mathrm{i}}^2}{\mathrm{n}_{\mathrm{h}}}=\frac{\left(1.6 \times 10^{16}\right)^2}{4 \times 10^{22}}=\frac{1.6 \times 1.6 \times 10^{32}}{4 \times 10^{22}} \\
& =6.4 \times 10^9 \mathrm{~m}^{-3}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.