In an intrinsic semiconductor the energy gap

Solution:

Energy gap of the given intrinsic semiconductor, Eg = 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,

kB = Boltzmann constant = 8.62 × 10−5 eV/K

T = Temperature

n0 = Constant

Initial temperature, T1 = 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, T2 = 600 K

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

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