Question

A pure semiconductor crystal has \(8\times10^{28}\ \text{atoms m}^{-3}\). It is doped by \(2\ \text{ppm}\) concentration of pentavalent atoms. The number of holes formed in the semiconductor crystal is
\[ (\text{Intrinsic carrier concentration, } n_i=1\times10^{16}\ \text{m}^{-3}) \]

• A. \(4.5\times10^9\ \text{m}^{-3}\)
• B. \(6.25\times10^8\ \text{m}^{-3}\)
• C. \(2.5\times10^9\ \text{m}^{-3}\)
• D. \(1.25\times10^8\ \text{m}^{-3}\)

Hint:

For doped semiconductors, use the mass action law: \[ np=n_i^2. \] For an \(n\)-type semiconductor, \[ n\approx N_D, \] so the hole concentration is \[ p=\frac{n_i^2}{N_D}. \]

Answer 1

The Correct Option is B

Step 1: Find donor (electron) concentration.
Doping: 2 ppm of \( 8\times10^{28} \) atoms/m³.
\( n_e = 2\times10^{-6}\times8\times10^{28} = 1.6\times10^{23}\text{ m}^{-3} \).

Step 2: Find hole concentration using mass action law.
\( n_e n_h = n_i^2 \Rightarrow n_h = \frac{n_i^2}{n_e} = \frac{(10^{16})^2}{1.6\times10^{23}} = \frac{10^{32}}{1.6\times10^{23}} = 6.25\times10^8\text{ m}^{-3} \)

\[ \boxed{n_h = 6.25\times10^{8}\text{ m}^{-3}} \]