Question

The conductivity of an intrinsic semiconductor depends on temperature as σ = σ₀e^−ΔE/2kT, where σ₀ is a constant. Find the temperature at which the conductivity of an intrinsic germanium semiconductor will be double of its value at T = 300 K. Assume that the gap for germanium is 0.650 eV and remains constant as the temperature is increased.

(Use Planck constant h = 4.14 × 10^-15 eV-s, Boltzmann constant k = 8·62 × 10^-5 eV/K.)

Answer 1

Let the conductivity at temperature T₁ be \[\sigma_1\]  and the conductivity at temperature T be \[\sigma_2\] .

Given: \[T_1    =   300  K\]

Band gap, E = 0.650 eV
Now,
According to the question,

\[\sigma =  \sigma_0 e -^\frac{\Delta E}{2KT}\]

\[\sigma_2    =   2 \sigma_1\]

\[\Rightarrow  \sigma_0  e^\frac{- \Delta E}{2kT}    =   2 \times  \sigma_0  e^\frac{- \Delta E}{2 \times k \times T_1} \] 

\[ \Rightarrow  \sigma_0  e^\frac{- \Delta E}{2kT}  =     2 \times  \sigma_0  e^\frac{- \Delta E}{2 \times k \times 300} \] 

\[ \Rightarrow  e^\frac{- 0 . 650}{2 \times 8 . 62 \times {10}^{- 5} \times T}    =   2 \times  e^\frac{- 0 . 650}{2 \times 8 . 62 \times {10}^{- 5} \times 300} \] 

\[ \Rightarrow  e^\frac{- 0 . 650}{2 \times 8 . 62 \times {10}^{- 5} \times T}    =   6 . 96561 \times  {10}^{- 6} \] 

On taking natural natural log on both sides, we get

\[\frac{- 0 . 650}{2 \times 8 . 62 \times {10}^{- 5} \times T}   =    - 11 . 874525\] 

\[ \Rightarrow \frac{1}{T}   =   \frac{11 . 874525 \times 2 \times 8 . 62 \times {10}^{- 5}}{0 . 65}\] 

\[ \Rightarrow T   =   317 . 51178   \approx   318\] K