Question

At a given temperature, the number of intrinsic charge carriers in a semiconductor is \( 2.0 \times 10^{10} \) cm\(^{-3}\). It is doped with pentavalent impurity atoms. As a result, the number of holes in it becomes \( 8 \times 10^3 \) cm\(^{-3}\). The number of electrons in the semiconductor is:

• A. \( 2 \times 10^{24} \) m\(^{-3} \)
• B. \( 4 \times 10^{23} \) m\(^{-3} \)
• C. \( 1 \times 10^{22} \) m\(^{-3} \)
• D. \( 5 \times 10^{22} \) m\(^{-3} \)

Hint:

In a doped semiconductor, the product of electron and hole concentrations remains equal to \( n_i^2 \).

Answer 1

The Correct Option is D

Carrier Concentration Calculation
- The relationship between electron and hole concentrations in a semiconductor is defined by: \[n_e \cdot n_h = n_i^2\]Where: \( n_i = 2.0 \times 10^{10} \) cm\(^{-3} \) (intrinsic carrier concentration), \( n_h = 8 \times 10^3 \) cm\(^{-3} \) (hole concentration post-doping), \( n_e \) = electron concentration post-doping.Calculation of \( n_e \):\[n_e = \frac{n_i^2}{n_h} = \frac{(2.0 \times 10^{10})^2}{8 \times 10^3}\]\[n_e = \frac{4.0 \times 10^{20}}{8 \times 10^3} = 5 \times 10^{16} \text{ cm}^{-3}\]Conversion to m\(^{-3}\):\[n_e = 5 \times 10^{22} \text{ m}^{-3}\]