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# The Conductivity of an Intrinsic Semiconductor Depends on Temperature as σ = σ0e−δE/2kt, Where σ0 is a Constant. - Physics

Short Note

The conductivity of an intrinsic semiconductor depends on temperature as σ = σ0eΔE/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

Concept: Energy Bands in Conductors, Semiconductors and Insulators
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 23 Semiconductors and Semiconductor Devices
Q 14 | Page 419
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