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:

For intrinsic semiconductors, use the ratio of conductivity at two temperatures to eliminate the prefactor and directly solve for the band gap.

Answer 1

Step 1: Recall the conductivity relation.
For an intrinsic semiconductor: \[ \sigma = \sigma_0 \exp\left(-\frac{E_g}{2k_B T}\right) \] Step 2: Write ratio of conductivities.
\[ \frac{\sigma_2}{\sigma_1} = \exp\left(\frac{E_g}{2k_B} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\right) \] Given: $\sigma_1 = 100$ at $T_1 = 300$ K, $\sigma_2 = 300$ at $T_2 = 500$ K. \[ \frac{\sigma_2}{\sigma_1} = \frac{300}{100} = 3 \] Step 3: Substitute values.
\[ \ln(3) = \frac{E_g}{2 \times 8.6 \times 10^{-5}} \left(\frac{1}{300} - \frac{1}{500}\right) \] Compute inside: \[ \frac{1}{300} - \frac{1}{500} = \frac{500 - 300}{150000} = \frac{200}{150000} = 0.001333 \] Denominator factor: \[ 2 \times 8.6 \times 10^{-5} = 1.72 \times 10^{-4} \] So: \[ \ln(3) = 1.099 = \frac{E_g \times 0.001333}{1.72 \times 10^{-4}} \] \[ E_g = \frac{1.099 \times 1.72 \times 10^{-4}}{0.001333} \approx 0.72 \, eV \] Final Answer: \[ \boxed{0.72 \, eV} \]