Q 12-14-024NEETJEE MainMedium
Pure silicon has an intrinsic carrier concentration of $1.5 \times 10^{16}\ \text{m}^{-3}$. It is doped with $4.5 \times 10^{22}$ donor atoms per m$^3$. The hole concentration in the doped crystal is
Answer: (D) $5 \times 10^{9}\ \text{m}^{-3}$
Each donor gives one electron, and the donors far outnumber the intrinsic carriers, so $n_e \approx 4.5 \times 10^{22}\ \text{m}^{-3}$.
Using $n_en_h = n_i^2$:
$$n_h = \frac{(1.5 \times 10^{16})^2}{4.5 \times 10^{22}} = \frac{2.25 \times 10^{32}}{4.5 \times 10^{22}} = 5 \times 10^{9}\ \text{m}^{-3}$$
Solution by Sreeraj P, M.Sc Physics