In an intrinsic semiconductor,the energy gap $E_{g}$ is $1.2 \; eV$. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at $600 \; K$ and that at $300 \; K$? Assume that the temperature dependence of intrinsic carrier concentration $n_{i}$ is given by $n_{i} = n_{0} \exp \left(-\frac{E_{g}}{2 k_{B} T}\right)$,where $n_{0}$ is a constant.

(D) The conductivity $\sigma$ of an intrinsic semiconductor is given by $\sigma = e n_{i} (\mu_{e} + \mu_{h})$. Since $\mu_{e} \gg \mu_{h}$ and both are independent of temperature,$\sigma \propto n_{i}$.
The intrinsic carrier concentration is $n_{i} = n_{0} \exp \left(-\frac{E_{g}}{2 k_{B} T}\right)$.
At $T_{1} = 300 \; K$,$n_{i1} = n_{0} \exp \left(-\frac{E_{g}}{2 k_{B} \times 300}\right)$.
At $T_{2} = 600 \; K$,$n_{i2} = n_{0} \exp \left(-\frac{E_{g}}{2 k_{B} \times 600}\right)$.
The ratio of conductivities is $\frac{\sigma_{2}}{\sigma_{1}} = \frac{n_{i2}}{n_{i1}} = \exp \left[ \frac{E_{g}}{2 k_{B}} \left( \frac{1}{300} - \frac{1}{600} \right) \right]$.
Substituting $E_{g} = 1.2 \; eV$ and $k_{B} = 8.62 \times 10^{-5} \; eV/K$:
Ratio $= \exp \left[ \frac{1.2}{2 \times 8.62 \times 10^{-5}} \times \frac{1}{600} \right] = \exp \left[ \frac{1.2}{10.344 \times 10^{-2}} \right] = \exp [11.601] \approx 1.09 \times 10^{5}$.