Question

Suppose a pure Si crystal has $5\times {10}^{28}$ atoms $m^{-3}$. it is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that $n_1=1.5\times {10}^{16}{m}^{-3}$.

Answer 1

1 atom of Si doped out of ${10}^6$ atoms (1 ppm of As)

$\Rightarrow$ In $5\times {10}^{28}$ atoms net doped =$\frac{5\times {10}^{28}}{{10}^6}=5\times {10}^{22}$ atoms

1 As atom creates $1e^{⊝}$ excen

so, no. of excen electron $=5\times {10}^{22}=n_e$

also, $n_e{n}_n=n_1^2$

$\Rightarrow n_n=\frac{n_1^2}{n_e}=\frac{(1.5\times {10}^{16}{)}^2}{5\times {10}^{22}}=4.5\times {10}^9$

$\therefore$ No. of holes formed $=4.5\times {10}^9.$

as $n_e>n_n\to$ It is a N - type semiconductor formed.