Question

The intrinsic carrier concentration of a semiconductor is $1.2\times10^{16}$ m$^{-3}$. On doping with an impurity, the electron concentration becomes $20\times10^5$ times initial concentration. The concentration of holes in the doped semiconductor is:

• A. $6\times10^{12}$ m$^{-3}$
• B. $12\times10^{16}$ m$^{-3}$
• C. $12\times10^{9}$ m$^{-3}$
• D. $6\times10^{9}$ m$^{-3}$

Hint:

In doped semiconductors: $n \cdot p = n_i^2$.
Use intrinsic carrier concentration formula.
Electron concentration increases → holes decrease.
Always square root only for intrinsic before doping.

Answer 1

The Correct Option is A

• Intrinsic carrier concentration $n_i = \sqrt{n_0 p_0}$.
• Doped: $n = 20 \times 10^5 \cdot n_i$, $p = n_i^2 / n$.
• $p = (1.2\times10^{16})^2 / (20 \times 10^5 \cdot 1.2 \times 10^{16}) = 6 \times 10^{12}$ m$^{-3}$.
• Hence hole concentration = $6 \times 10^{12$ m$^{-3}$}.