Question

The number of silicon atoms per $m^3$ is $5\times 10^{28}$. This is doped simultaneously with $5\times 10^{22}$ atoms per $m^3$ of arsenic and $5\times 10^{20}$ per $m^3$ atoms of indium. Given that $n_i=1.5\times 10^{16}{m}^{-3}$. Number of electrons and holes (in per metre cube of sample) are respectively,

• A. $4.95\times 10^{22},4.54\times 10^9$
• B. $4.54\times 10^9,4.54\times 10^9$
• C. $4.54\times 10^9,4.95\times 10^{22}$
• D. $4.95\times 10^{22},4.95\times 10^{22}$

Answer 1

Answer: (A)

$N=5\times 10^{28}$ atoms $/m^3,$
$A=$ acceptor, indium,
$D=$ donor, arsenic
$N_A=0.05\times 10^{22}$ atoms $/m^3$
$\Rightarrow N_D=5\times 10^{22}$ atoms $/m^3$
$n_i=1.5\times 10^{16}{m}^{-2}$
$\Rightarrow N_D-N_A=n_e-n_h$ and $n_e{n}_h=n_i^2$
$\therefore N_D-N_A=n_e-\frac{n_i^2}{n_e}$
$\Rightarrow n_e^2-\left(N_D-N_A\right)n_e-n_i^2=0$
$\Rightarrow n_e=\frac{\left(N_D-N_A\right)+\sqrt{{\left(N_D-N_A\right)}^2+4n_i^2}}{2}$
On substituting values, we get,
$n_e=4.95\times 10^{22}/m^3$
$\therefore n_h=\frac{n_i^2}{n_e}=\frac{{\left(1.5\times 10^{16}\right)}^2}{4.95\times 10^{22}}$
$=4.54\times 10^9{m}^{-3}$
Observe that doping level $\frac{10^{22}}{10^{28}}=10^{-6}ppm$ nearly