A doped silicon crystal has a hole concentration of \(4.5\times10^{22}\ \mathrm{m^{-3}}\) in thermal equilibrium. The intrinsic carrier concentration of silicon is \(1.5\times10^{16}\ \mathrm{m^{-3}}\). The ratio of the hole concentration to the electron concentration, \(n_h/n_e\), is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(9\times10^{12}\)
In thermal equilibrium, the product of the electron and hole concentrations is equal to \( n_i^{2} \). This holds for a doped crystal too. So we first find the electron concentration and then take the ratio.
\( n_h=4.5\times10^{22}\ \mathrm{m^{-3}} \)
\( n_i=1.5\times10^{16}\ \mathrm{m^{-3}} \)
\( \begin{aligned} n_e &= \frac{n_i^{2}}{n_h} \\ &= \frac{(1.5\times10^{16})^{2}}{4.5\times10^{22}} \\ &= 5\times10^{9}\ \mathrm{m^{-3}} \end{aligned} \)
\( \begin{aligned} \frac{n_h}{n_e} &= \frac{4.5\times10^{22}}{5\times10^{9}} \\ &= 9\times10^{12} \end{aligned} \)
From the above, the crystal has \( 9\times10^{12} \) holes for every free electron. Hence, holes are the majority carriers here.
NCERT (2026-27):
Extrinsic Semiconductor – “… The electron and hole concentration in a semiconductor in thermal equilibrium is given by”
\( n_en_h=n_i^{2} \)\( \qquad (14.5) \)