Hint: \(n_{i}=n_{0} \exp \left[\frac{-E_{g}}{2 K_{B} T}\right]\)
Step 1: Find the carrier concentration at 300 K.
Initial temperature,
The intrinsic carrier concentration at this temperature can be written as:
\(n_{i_{1}}=n_{0} \exp \left[\frac{-E_{g}}{2 K_{B} 300}\right].......(1)\)
Step 2: Find the carrier concentration at 600 K.
Final temperature,
The intrinsic carrier concentration at this temperature can be written as:
\(n_{i_{2}}=n_{0} \exp \left[\frac{-E_{g}}{2 K_{B} 600}\right].......(2)\)
Step 3: Find the ratio between the conductivities at 600 K and at 300 K.
The ratio between the conductivities at 600 K and at 300 K is equal to the ratio between the respective intrinsic carrier
concentration at these temperatures.
\(\begin{aligned} \frac{n_{i _{2}}}{n_{i _{1}}} &=\frac{n_{0} \exp \left[\frac{-E g}{2 k_{B} 600}\right]}{n_{0} \exp \left[\frac{-E_{g}}{2 k_{B} 300}\right]}=\exp \frac{E g}{2 k_{B}}\left[\frac{1}{300}-\frac{1}{600}\right] \\ &=\exp \left[\frac{1.2}{2 \times 8.62 \times 10^{-5}} \times \frac{2-1}{600}\right] \\ &=exp[11.6]=1.09 \times 10^{5} \end{aligned}\)
Therefore, the ratio between the conductivities is
