Question

An intrinsic semiconductor has conductivity of 100 Ω\(^{-1}\) m\(^{-1}\) at 300 K and 300 Ω\(^{-1}\) m\(^{-1}\) at 500 K. The band gap of the semiconductor is _________ eV (rounded off to two decimal places). Given: Boltzmann constant \( k_B = 8.6 \times 10^{-5} \, {eV K}^{-1} \)

Hint:

To calculate the band gap of an intrinsic semiconductor, use the conductivity equation that involves temperature and the exponential dependence of conductivity on the band gap. The ratio of conductivity at two different temperatures can help isolate and solve for the band gap.

Answer 1

The electrical conductivity \( \sigma \) of an intrinsic semiconductor is given by the relation:

\[ \sigma = A \cdot e^{-\frac{E_g}{2k_B T}} \] where:
- \( E_g \) is the band gap energy,
- \( k_B \) is the Boltzmann constant,
- \( T \) is the temperature,
- \( A \) is a constant that depends on the material (and cancels out in the ratio).

We are given the conductivity at two different temperatures, 300 K and 500 K. To find the band gap \( E_g \), we use the ratio of conductivities at two temperatures:

\[ \frac{\sigma_2}{\sigma_1} = \frac{e^{-\frac{E_g}{2k_B T_2}}}{e^{-\frac{E_g}{2k_B T_1}}} \] Substituting the given values, we have:

\[ \frac{300}{100} = \frac{e^{-\frac{E_g}{2 \times 8.6 \times 10^{-5} \times 500}}}{e^{-\frac{E_g}{2 \times 8.6 \times 10^{-5} \times 300}}} \] Simplifying the expression:

\[ 3 = e^{\frac{E_g}{2 \times 8.6 \times 10^{-5}} \left(\frac{1}{300} - \frac{1}{500}\right)} \] After solving for \( E_g \), we find:

\[ E_g \approx 0.14 \, {eV} \] Thus, the band gap is between 0.13 eV and 0.15 eV.

Answer: 0.13 to 0.15 eV.