Physics - Semiconductors Question with Solution | TestHub

Boron is a trivalent impurity, so doping silicon with boron creates a p-type semiconductor.

The hole concentration is approximately equal to the acceptor concentration:

$n_h\approx N_A=5\times 10^{17}{\text{cm}}^{-3}$

According to the mass action law:

$n_e{n}_h=n_i^2$

Substituting the given values:

$n_e=\frac{n_i^2}{n_h}$

$n_e=\frac{(1.5\times 10^{10}{)}^2}{5\times 10^{17}}$

$n_e=\frac{2.25\times 10^{20}}{5\times 10^{17}}$

$n_e=0.45\times 10^3{\text{cm}}^{-3}$

$n_e=4.5\times 10^2{\text{cm}}^{-3}$

Answer: $4.5\times 10^2{\text{cm}}^{-3}$