Question

A pure silicon crystal with 5 × 10²⁸ atoms m⁻³ has n_i = 1.5 × 10¹⁶ m⁻³. It is doped with a concentration of 1 in 10⁵ pentavalent atoms, the number density of holes (per m³) in the doped semiconductor will be:

• A. \(\quad 4.5 \times 10^3 \\\)
• B. \(\quad 4.5 \times 10^8 \\\)
• C. \(\quad \left( \frac{10}{3} \right) \times 10^{12} \\\)
• D. \(\quad \left( \frac{10}{3} \right) \times 10^7\)

Answer 1

The Correct Option is B

To determine the hole number density in the doped silicon semiconductor, we first quantify the donor atom concentration. Pure silicon contains \( 5 \times 10^{28} \) atoms/m³. Doping with one pentavalent atom per \( 10^5 \) silicon atoms yields the donor atom density:

\[ N_d = \frac{5 \times 10^{28}}{10^5} = 5 \times 10^{23} \text{ m}^{-3} \]

This \( N_d \) represents the concentration of electrons added to the conduction band. Assuming each donor atom contributes one electron, the electron concentration \( n \) in the doped silicon is approximated as \( n \approx N_d \).

Given the intrinsic carrier concentration \( n_i = 1.5 \times 10^{16} \text{ m}^{-3} \), the product of electron and hole concentrations at equilibrium is:

\[ n \cdot p = n_i^2 \]

Where \( n \) is the electron concentration and \( p \) is the hole concentration. We can now calculate \( p \):

\[ p = \frac{n_i^2}{n} = \frac{(1.5 \times 10^{16})^2}{5 \times 10^{23}} \]

First, calculate \( n_i^2 \):

\[ n_i^2 = (1.5 \times 10^{16})^2 = 2.25 \times 10^{32} \text{ m}^{-6} \]

Substituting this value back into the equation for \( p \):

\[ p = \frac{2.25 \times 10^{32}}{5 \times 10^{23}} = 0.45 \times 10^9 = 4.5 \times 10^8 \text{ m}^{-3} \]

Therefore, the number density of holes in the doped semiconductor is \( 4.5 \times 10^8 \text{ m}^{-3} \).