The number densities of electrons and holes in pure silicon at 273 K are equal and their values are \( 1.5\times10^{16}\,\mathrm{m^{-3}} \). On doping, the hole density increases to \( 4.5\times10^{27}\,\mathrm{m^{-3}} \). The electron density in the doped silicon will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 5\times10^{4}\,\mathrm{m^{-3}} \)
In a semiconductor the product of the electron and hole densities stays constant at a given temperature, \( n_e n_h = n_i^{2} \), where \( n_i \) is the density of either carrier in the pure (intrinsic) semiconductor. Doping changes \( n_e \) and \( n_h \) but not their product.
\( \displaystyle n_e = \frac{n_i^{2}}{n_h} = \frac{(1.5\times10^{16})^{2}}{4.5\times10^{27}} = \frac{2.25\times10^{32}}{4.5\times10^{27}} \)
\( n_e = 5\times10^{4}\,\mathrm{m^{-3}} \)