Question

The electron density of intrinsic semiconductor at room temperature is ${10}^{16}$ $m^{-3}$. When doped with a trivalent impurity, the electron density is decreased to ${10}^{14}$ $m^{-3}$ at the same temperature. The majority carrier density is:

• A. ${10}^{16}{m}^{-3}$
• B. ${10}^{18}{m}^{-3}$
• C. ${10}^{21}{m}^{-3}$
• D. ${10}^{20}{m}^{-3}$
• E. ${10}^{19}{m}^3$

Answer 1

The correct option is B ${10}^{18}{m}^{-3}$

$n_e\times n_n=n_1^2$
$n_i={10}^{18}{m}^{-3}$
$n_e={10}^{14}m-3$ given
$\Rightarrow n_n=\frac{n_i^2}{n_e}=\frac{({10}^{16}{)}^2}{{10}^{14}}=\frac{{10}^{32}}{{10}^{14}}$
$={10}^{18}{m}^{-3}$.
Given semiconductor is doped with a trivalent impurity, therefore majority charge carriers are holes.