A pure semiconductor crystal has $8 \times 10^{28} \frac{\text { atoms }}{\mathrm{m}^3}$. It is doped by $2…
A pure semiconductor crystal has $8 \times 10^{28} \frac{\text { atoms }}{\mathrm{m}^3}$. It is doped by $2 \mathrm{ppm}$ concentration of pentavalent atoms. The number of holes formed in the semiconductor crystal is (Intrinsic carrier concentration, $\left.n_i=1 \times 10^{16} \mathrm{~m}^{-3}\right)$.
$4.3 \times 10^9 \mathrm{~m}^{-3}$
$6.25 \times 10^8 \mathrm{~m}^{-3}$
$2.5 \times 10^9 \mathrm{~m}^{-3}$
$125 \times 10^8 \mathrm{~m}^{-3}$
Solution
Number of atoms per $\mathrm{m}^3$ in semiconductor crystal $=8 \times 10^{28}$ atoms $/ \mathrm{m}^3$
Number of doped atoms $=2 \mathrm{ppm}=2$ atoms out of $10^6$ atoms.
$\therefore$ Numebr of doped atoms in $8 \times 10^{28}$ atoms $/ \mathrm{m}^3$
$
=2 \times \frac{8 \times 10^{28}}{1 \times 10^6}=16 \times 10^{22} \text { atoms } / \mathrm{m}^3
$
As, each pentavalent impurity atom produces 1 -electron in conduction band, number of conduction electrons $=n_e=4 \times 10^{22}$ electrons $/ \mathrm{m}^3$
As $n_e \cdot n_n=n_i^2,=$, we have
$
n_n=\frac{n_i^2}{n_e}=\frac{\left(1 \times 10^{16}\right)^2}{16 \times 10^{22}}=6.25 \times 10^8 \text { holes } / \mathrm{m}^3
$