An intrinsic semiconductor has equal electron and hole concentrations of \( 6\times10^{8}\,\mathrm{m^{-3}} \). On doping it with certain impurity, electron concentration increases to \( 9\times10^{12}\,\mathrm{m^{-3}} \). The new hole concentration is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 4\times10^{4}\,\mathrm{m^{-3}} \)
In thermal equilibrium the product of the electron and hole concentrations of a semiconductor is constant (the mass action law):
\( n_en_h = n_i^{2} \)
\( n_i = 6\times10^{8}\,\mathrm{m^{-3}} \Rightarrow n_i^{2} = 36\times10^{16}\,\mathrm{m^{-6}} \)
\( \displaystyle n_h = \frac{n_i^{2}}{n_e} = \frac{36\times10^{16}}{9\times10^{12}} = 4\times10^{4}\,\mathrm{m^{-3}} \)
Doping raises the electron concentration, so the hole concentration falls below \( n_i \).
Note: A similar question was also asked in CUET (UG) 2026 on 30-05-2026 (Slot 2).