Question

In a silicon semiconductor at room temperature, the intrinsic carrier concentration is \( 1.5 \times 10^{16} \, \text{m}^{-3} \). Calculate the energy band gap of the silicon if the intrinsic carrier concentration is given by: \[ n_i = \sqrt{N_c N_v} e^{-E_g / 2kT} \] Where: - \( N_c = 2.8 \times 10^{25} \, \text{m}^{-3} \) is the effective density of states in the conduction band, - \( N_v = 1.04 \times 10^{25} \, \text{m}^{-3} \) is the effective density of states in the valence band, - \( k = 1.38 \times 10^{-23} \, \text{J/K} \) is the Boltzmann constant, - \( T = 300 \, \text{K} \) is the temperature.

• A. \( 1.1 \, \text{eV} \)
• B. \( 0.9 \, \text{eV} \)
• C. \( 2.0 \, \text{eV} \)
• D. \( 0.7 \, \text{eV} \)

Hint:

The intrinsic carrier concentration \( n_i \) depends exponentially on the energy band gap \( E_g \). This relationship is important when dealing with semiconductors.

Answer 1

The Correct Option is A

We are given the formula for the intrinsic carrier concentration \( n_i \): \[ n_i = \sqrt{N_c N_v} e^{-E_g / 2kT} \] Rearranging the equation to solve for the energy band gap \( E_g \): \[ E_g = -2kT \ln \left( \frac{n_i}{\sqrt{N_c N_v}} \right) \] Substitute the known values: \[ E_g = -2 \times (1.38 \times 10^{-23}) \times 300 \ln \left( \frac{1.5 \times 10^{16}}{\sqrt{2.8 \times 10^{25} \times 1.04 \times 10^{25}}} \right) \] Solving this, we get: \[ E_g \approx 1.1 \, \text{eV} \] Thus, the energy band gap of silicon is \( 1.1 \, \text{eV} \).