Question

Calculate the conductivity of an n-type semiconductor from the following data: Density of conduction electrons = \( 8 \times 10^{13} \, \text{cm}^{-3} \) Density of holes = \( 5 \times 10^{12} \, \text{cm}^{-3} \) Mobility of electrons = \( 2.3 \times 10^4 \, \text{cm}^2/\text{V-s} \) Mobility of holes = \( 100 \, \text{cm}^2/\text{V-s} \)

Hint:

The conductivity of a semiconductor depends on the density and mobility of both electrons and holes.

Answer 1

\textit{Conductivity of N-Type Semiconductor:} The electrical conductivity (\( \sigma \)) of a semiconductor is given by: \[ \sigma = q \cdot (n_e \mu_e + n_h \mu_h), \] where:
- \( q \) is the charge of an electron,
- \( n_e \) is the density of conduction electrons,
- \( n_h \) is the density of holes,
- \( \mu_e \) is the mobility of electrons,
- \( \mu_h \) is the mobility of holes. Using the given values, the conductivity can be calculated as: \[ \sigma = (1.6 \times 10^{-19}) \cdot \left( (8 \times 10^{13}) \cdot (2.3 \times 10^4) + (5 \times 10^{12}) \cdot (100) \right). \]