14.13 In an intrinsic semiconductor the energy gap Eg is 1.2eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What are the ratio between conductivity at 600K and that at 300K? Assume that the temperature dependence of intrinsic carrier concentration ni is given by:

\(n_{i}=n_{0} \exp \left[\frac{-E_{g}}{2 K_{B} T}\right]\) where n0 is a constant.

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, Ti=300 K
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, T2=600 K
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 1.09 ×105.