A doped semiconductor sample is in thermal equilibrium at room temperature. Its electron concentration is \(9\times10^{21}\ \mathrm{m^{-3}}\) and its hole concentration is \(1\times10^{11}\ \mathrm{m^{-3}}\). The intrinsic carrier concentration of the semiconductor is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(3\times10^{16}\ \mathrm{m^{-3}}\)
In thermal equilibrium, the product of the electron and hole concentrations is equal to the square of the intrinsic carrier concentration. This holds for a doped sample too.
\( n_e=9\times10^{21}\ \mathrm{m^{-3}} \)
\( n_h=1\times10^{11}\ \mathrm{m^{-3}} \)
\( \begin{aligned} n_i^{2} &= n_en_h \\ &= 9\times10^{21}\times1\times10^{11} \\ &= 9\times10^{32}\ \mathrm{m^{-6}} \end{aligned} \)
\( \begin{aligned} n_i &= \sqrt{9\times10^{32}} \\ &= 3\times10^{16}\ \mathrm{m^{-3}} \end{aligned} \)
NCERT (2026-27):
Extrinsic Semiconductor – “… The electron and hole concentration in a semiconductor in thermal equilibrium is given by”
\( n_en_h=n_i^{2} \)\( \qquad (14.5) \)