Question:
In a semiconductor, the number density of intrinsic charge carriers at $27^{\circ} \mathrm{C}$ is $1.5 \times 10^{16} / \mathrm{m}^{3}$. If the semiconductor is doped with impurity atom, the hole density increases to $4.5 \times 10^{22} / \mathrm{m}^{3}$. The electron density in the doped semiconductor is __________ $\times 10^{9} / \mathrm{m}^{3}$.
Solution:
$\mathrm{n}_{\mathrm{e}} \mathrm{n}_{\mathrm{h}}=\mathrm{n}_{\mathrm{i}}^{2}$
$\mathrm{n}_{\mathrm{c}}=\frac{\mathrm{n}_{\mathrm{i}}^{2}}{\mathrm{n}_{\mathrm{h}}}=\frac{\left(1.5 \times 10^{16}\right)^{2}}{4.5 \times 10^{22}}$
$=\frac{1.5 \times 1.5 \times 10^{32}}{4.5 \times 10^{22}}$
$5 \times 10^{9} / \mathrm{m}^{3}$