**Question:**

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 600K and that at 300K? Assume that the temperature dependence of intrinsic carrier concentration *n**i*is given by $n_{i}=n_{0} \exp \left[-\frac{E_{g}}{2 k_{\mathrm{B}} T}\right]$ where *n*0 is a constant.

**Solution:**

Energy gap of the given intrinsic semiconductor, *E**g* = 1.2 eV

The temperature dependence of the intrinsic carrier-concentration is written as:

$n_{i}=n_{0} \exp \left[-\frac{E_{g}}{2 k_{\mathrm{B}} T}\right]$

Where,

*k*B = Boltzmann constant = 8.62 × 10−5 eV/K

T = Temperature

*n*0 = Constant

Initial temperature, *T*1 = 300 K

The intrinsic carrier-concentration at this temperature can be written as:

$n_{i 1}=n_{0} \exp \left[-\frac{E_{g}}{2 k_{\mathrm{B}} \times 300}\right]$ ...(1)

Final temperature, *T*2 = 600 K

The intrinsic carrier-concentration at this temperature can be written as:

$n_{i 2}=n_{0} \exp \left[-\frac{E_{g}}{2 k_{\mathrm{B}} \times 600}\right]$ ....(2)

The ratio between the conductivities at 600 K and at 300 K is equal to the ratio between the respective intrinsic carrier-concentrations at these temperatures.

$\frac{n_{i 2}}{n_{i 1}}=\frac{n_{0} \exp \left[-\frac{E_{g}}{2 k_{\mathrm{B}} 600}\right]}{n_{0} \exp \left[-\frac{E_{g}}{2 k_{\mathrm{B}} 300}\right]}$

$=\exp \frac{E_{g}}{2 k_{\mathrm{B}}}\left[\frac{1}{300}-\frac{1}{600}\right]=\exp \left[\frac{1.2}{2 \times 8.62 \times 10^{-5}} \times \frac{2-1}{600}\right]$

$=\exp [11.6]=1.09 \times 10^{5}$

Therefore, the ratio between the conductivities is 1.09 × 105.