Question

The number density of electrons and holes in pure silicon at $27 ^{\circ}C$ are equal and its value is $2.0 \times 10^{16}\, m^{-3}$. On doping with indium the hole density increases to $4.5 \times 10^{22}\, m^{-3}$, the electron density in doped silicon is

• A. $10 \times 10^9\, m^{-3}$
• B. $8 .89 \times 10^9\,m^{-3}$
• C. $11 \times 10^9 \,m^{-3}$
• D. $16.78 \times 10^9\, m^{-3}$

Answer 1

Answer: (B)

Using, $n_e = n_i^2$
Here, $n_i = 2\times 10^{16}\,m^{-3}$,
$\rho = 4.5 \times 10^{22}\,m^{-3}$
$\therefore n = \frac{n_i^2}{\rho}$
$ = \frac{(2\times 10^{16})^2}{4.5 \times 10^{22}}$
$ = 8.89 \times 10^9 \,m^{-3}$