Question

$A \;pure \;silicon \;crystal \;with \; 5 \times 10^{28} \, \text{atoms/m}^3 \text{ has } n_i = 1.5 \times10^{16}\text{m}^{-3}.\text{ It is doped with a concentration of 1 in } 10^5 \text{ pentavalent atoms.}\text{The number density of holes (per m}^3\text{) in the doped semiconductor will be:}$

• A. $\quad 4.5 \times 10^3 \\\\$
• B. $\quad 4.5 \times 10^8 \\\\$
• C. $\quad \left( \frac{10}{3} \right) \times 10^{12} \\\\$
• D. $\quad \left( \frac{10}{3} \right) \times 10^7$

Answer 1

Answer: (B)

$\quad 4.5 \times 10^8 \\\\$

Answer 2

$\text{ Given: - The intrinsic carrier concentration } n_i = 1.5 \times 10^{16} \, \text{m}^{-3}, \\\\$
$\text{- Doping concentration of pentavalent atoms } N_d = \frac{5 \times 10^{28}}{10^5} = 5 \times 10^{23} \, \text{m}^{-3}.\\\\$
$\text{The number of holes } p \text{ in the doped semiconductor is given by:}$
$p = \frac{n_i^2}{N_d} = \frac{(1.5 \times 10^{16})^2}{5 \times 10^{23}} = 4.5 \times 10^8 \, \text{m}^{-3}$

$\text{Thus, the number density of holes is } 4.5 \times 10^8 \, \text{m}^{-3}.$