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 solve this problem, we need to find the number density of holes in the doped silicon semiconductor. When pentavalent atoms are introduced into silicon, they donate extra electrons to the conduction band, increasing the electron concentration \( n \).

First, calculate the number density of donor atoms introduced. Given the pure silicon crystal has \( 5 \times 10^{28} \) atoms/m³ and is doped with 1 pentavalent atom per \( 10^5 \) silicon atoms:

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

This is the concentration of electrons contributed by the doping. Assuming each donor atom donates one electron, the electron concentration \( n \) in doped silicon will be approximately equal to \( N_d \).

The intrinsic carrier concentration \( n_i \) is given as \( 1.5 \times 10^{16} \text{ m}^{-3} \). In equilibrium, the product of electron and hole concentrations is equal to the square of the intrinsic carrier concentration:

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

Where:
\( n \) is the concentration of electrons, approximately equal to \( N_d \).
\( p \) is the concentration of holes.

Substituting the values:

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

Calculate \( n_i^2 \):

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

Now substitute back to find \( p \):

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

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

Answer 2

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

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