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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.)
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Solution
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
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