In a pure silicon, number of electrons and holes per unit volume are $1.6 imes 10^{16} \mathrm{~m}^{-3}$. If silicon is doped with Boron in a way that on doping hole density increases to $4 imes 10^{22} \mathrm{~m}^{-3}$. Then electron density in doped semiconductor will be

Question

In a pure silicon, number of electrons and holes per unit volume are $1.6 	imes 10^{16} \mathrm{~m}^{-3}$. If silicon is doped with Boron in a way that on doping hole density increases to $4 	imes 10^{22} \mathrm{~m}^{-3}$. Then electron density in doped semiconductor will be

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{n}_{\mathrm{i}}=1.6 	imes 10^{16} \mathrm{~m}^{-3}, \mathrm{n}_{\mathrm{h}}=4 	imes 10^{22} \mathrm{~m}^{-3} \ & \mathrm{n}_{\mathrm{e}} \mathrm{n}_{\mathrm{h}}=\mathrm{n}_{\mathrm{i}}^2 \ & 	herefore \mathrm{n}_{\mathrm{e}}=\frac{\mathrm{n}_{\mathrm{i}}^2}{\mathrm{n}_{\mathrm{h}}}=\frac{\left(1.6 	imes 10^{16}ight)^2}{4 	imes 10^{22}}=\frac{1.6 	imes 1.6 	imes 10^{32}}{4 	imes 10^{22}} \ & =6.4 	imes 10^9 \mathrm{~m}^{-3} \end{aligned}$

In a pure silicon, number of electrons and holes per unit volume are $1.6 \times 10^{16} \mathrm{~m}^{-3}$.…

Solution