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

Given:

• Silicon atom density (N_Si) = 5 × 10²⁸ m^-3
• Intrinsic carrier concentration (n_i) = 1.5 × 10¹⁶ m^-3
• Doping concentration = 1 pentavalent atom per 10⁵ Si atoms

Step 1: Calculate donor concentration (n_d):

n_d = N_Si/10⁵ = (5 × 10²⁸)/10⁵ = 5 × 10²³ m^-3

Step 2: For n-type semiconductor, electron concentration (n) ≈ n_d:

n ≈ 5 × 10²³ m^-3

Step 3: Calculate hole concentration (p) using mass-action law:

n_i² = n × p

p = n_i²/n = (1.5 × 10¹⁶)²/(5 × 10²³)

p = (2.25 × 10³²)/(5 × 10²³) = 0.45 × 10⁹ = 4.5 × 10⁸ m^-3

The number density of holes in the doped semiconductor is:

(2) 4.5 × 10⁸ m^-3

Answer 2

\(\text{ Given: - The intrinsic carrier concentration } n_i = 1.5 \times 10^{16} \, \text{m}^{-3}, \\\)
\(\text{- Doping concentration of pentavalent atoms } N_d = \frac{5 \times 10^{28}}{10^5} = 5 \times 10^{23} \, \text{m}^{-3}.\\\)
\(\text{The number of holes } p \text{ in the doped semiconductor is given by:}\)
\(p = \frac{n_i^2}{N_d} = \frac{(1.5 \times 10^{16})^2}{5 \times 10^{23}} = 4.5 \times 10^8 \, \text{m}^{-3}\)

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