Pure silicon at \( 300\,\mathrm{K} \) has equal electron \( (n_{e}) \) and hole \( (n_{h}) \) concentrations of \( 1.6\times10^{16}\,\mathrm{m^{-3}} \). It is doped by indium, which increases hole concentration to \( 6.4\times10^{22}\,\mathrm{m^{-3}} \). The number of electrons in the doped silicon will be:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 4.0\times10^{9}\,\mathrm{m^{-3}} \)
In a semiconductor in thermal equilibrium the mass action law \( n_en_h = n_i^{2} \) holds. Indium is trivalent, so it raises the hole concentration, and the electron concentration must fall to keep the product equal to \( n_i^{2} \).
\( \displaystyle n_e = \frac{n_i^{2}}{n_h} = \frac{(1.6\times10^{16})^{2}}{6.4\times10^{22}} = \frac{2.56\times10^{32}}{6.4\times10^{22}} = 4.0\times10^{9}\,\mathrm{m^{-3}} \)
The “number of electrons” asked for is the number per cubic metre, as the options show.