Question

A pure Si crystal having \( 5 \times 10^{28} \) atoms m\(^{-3}\) is doped with 1 ppm concentration of antimony. If the concentration of holes in the doped crystal is found to be \( 4.5 \times 10^9 \) m\(^{-3}\), the concentration (in m\(^{-3}\)) of intrinsic charge carriers in the Si crystal is about:

• A. \( 1.2 \times 10^{15} \)
• B. \( 1.5 \times 10^6 \)
• C. \( 3.0 \times 10^{15} \)
• D. \( 2.0 \times 10^6 \)

Hint:

The intrinsic carrier concentration \( n_i \) in semiconductors is important for understanding charge transport. In doped semiconductors, the carrier concentration is influenced by both the intrinsic carriers and the dopants.

Answer 1

The Correct Option is B

We are given the following information: - The concentration of silicon atoms in pure Si is \( 5 \times 10^{28} \) m\(^{-3}\). - The doping concentration of antimony is 1 ppm, which means \( 1 \, \text{ppm} = \frac{1}{10^6} \). - The concentration of antimony in the doped Si is therefore: \[ \text{Concentration of Sb} = \frac{1}{10^6} \times 5 \times 10^{28} = 5 \times 10^{22} \, \text{m}^{-3} \] - The concentration of holes in the doped crystal is \( 4.5 \times 10^9 \) m\(^{-3}\). Now, the intrinsic carrier concentration \( n_i \) in a semiconductor is given by the product of electron and hole concentrations: \[ n_i = \sqrt{n_e \times n_h} \] where: - \( n_e \) is the concentration of electrons, - \( n_h \) is the concentration of holes. In the doped Si crystal, the number of electrons is approximately equal to the concentration of the dopant atoms (antimony), i.e., \( n_e \approx 5 \times 10^{22} \, \text{m}^{-3} \). The concentration of holes \( n_h \) is \( 4.5 \times 10^9 \, \text{m}^{-3} \). Thus, the intrinsic concentration of charge carriers \( n_i \) is: \[ n_i = \sqrt{5 \times 10^{22} \times 4.5 \times 10^9} \] \[ n_i = \sqrt{2.25 \times 10^{32}} = 1.5 \times 10^{16} \, \text{m}^{-3} \] Therefore, the concentration of intrinsic charge carriers is approximately \( 1.5 \times 10^{16} \, \text{m}^{-3} \).