Question

The electron and hole mobilities in a silicon sample are \(0.135 \, \text{m}^2/\text{V·s}\) and \(0.048 \, \text{m}^2/\text{V·s}\), respectively. If the intrinsic carrier concentration is \(1.5 \times 10^{16} \, \text{atoms/m}^3\), the conductivity at 300 K is:

• A. \(\sigma = 4.39 \times 10^{-1} \, \text{S·m}^{-1}\)
• B. \(\sigma = 6.39 \times 10^{-1} \, \text{S·m}^{-1}\)
• C. \(\sigma = 7.39 \times 10^{-1} \, \text{S·m}^{-1}\)
• D. \(\sigma = 9.39 \times 10^{-1} \, \text{S·m}^{-1}\)

Hint:

For intrinsic semiconductors, conductivity depends on both electron and hole mobilities. Add μe and μh to calculate σ.

Answer 1

The Correct Option is A

The conductivity is calculated as:
\[\sigma = qn_i(\mu_e + \mu_h)\]
Substituting $q = 1.6 \times 10^{-19} \text{ C}$, $n_i = 1.5 \times 10^{16} \text{ m}^{-3}$, $\mu_e = 0.135 \text{ m}^2\text{V}^{-1}\text{s}^{-1}$, and $\mu_h = 0.048 \text{ m}^2\text{V}^{-1}\text{s}^{-1}$:
\[\sigma = 1.6 \times 10^{-19} \times 1.5 \times 10^{16} \times (0.135 + 0.048) \approx 4.39 \times 10^{-4} \text{ }(\Omega \cdot \text{m})^{-1}\]