Question

A pure silicon crystal has \(5 \times 10^{28} \, \text{atoms/m}^3\). It is doped with \(1 \, \text{ppm}\) concentration of pentavalent As. The number of holes is:

• A. \(n_p = 0.5 \times 10^{19} \, \text{m}^{-3}\)
• B. \(n_p = 2.5 \times 10^{19} \, \text{m}^{-3}\)
• C. \(n_p = 1.5 \times 10^{19} \, \text{m}^{-3}\)
• D. \(n_p = 4.5 \times 10^{19} \, \text{m}^{-3}\)

Hint:

For doped semiconductors, remember nh = n^2i/ne , where ne is the majority carrier density.

Answer 1

The Correct Option is D

The number of dopant atoms is calculated as:
\[N_D = 5 \times 10^{28} \times 10^{-6} = 5 \times 10^{22} \text{ m}^{-3}\]
For silicon, the intrinsic carrier concentration $n_i = 1.5 \times 10^{16} \text{ m}^{-3}$, and the relation $n_h = \frac{n_i^2}{n_e}$ applies:
\[n_h = \frac{(1.5 \times 10^{16})^2}{5 \times 10^{22}} \approx 4.5 \times 10^{19} \text{ m}^{-3}\]