The conductivity of a pure semiconductor is approximately proportional to T^ 3/2 e^ -Δ E/2kT , where Δ E represents the band gap. For germanium, the band gap is 0.74 eV at 4 K and 0.67 eV at 300 K . Determine the factor

Correct answer: C. Approximately 10^ 463 x 8.625 × 10^ -5 eV/K : (Δ E_1)/(2kT_1) = 0.74 2 × 8.625 × 10^ -5 × 4 1072.5 (Δ E_2)/(2kT_2) = 0.67 2 × 8.625 × 10^ -5 × 300 13.0 The difference in the exponent is 1072.5 - 13.0 = 1059.5 . The exponential term is e^ 1059.5 . Converting to base 10 using e 10^ 0.4343 : e^ 1059.5 = 10^ 1059.5 × 0.4343 10^ 460.1 The pre-exponential factor is: ((300)/(4))^ 3/2 = ( 75)^ 3/2 650 10^ 2.8 Multiplying the two factors gives the total ratio: (σ_2)/(σ_1) 10^ 2.8 × 10^ 460.1 = 10^ 462.9 10^ 463

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: Semiconductors and Semiconductor Devices

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Semiconductors and Semiconductor Devices

The conductivity of a pure semiconductor is approximately...

The conductivity of a pure semiconductor is approximately proportional to

T^{3/2} e^{-\Delta E/2kT}

, where

\Delta E

represents the band gap. For germanium, the band gap is

0.74 \text{ eV}

4 \text{ K}

and

0.67 \text{ eV}

300 \text{ K}

. Determine the factor by which the conductivity of pure germanium increases when the temperature is elevated from

4 \text{ K}

300 \text{ K}

Options

AApproximately $10^{75}$
BApproximately $10^{231}$
CApproximately $10^{463}$
DApproximately $10^{1063}$

Correct answer

C. Approximately $10^{463}$

Step-by-step solution

x 8.625 \times 10^{-5} \text{ eV/K}

:<br><br>\dfrac{\Delta E_1}{2kT_1} = \dfrac{0.74}{2 \times 8.625 \times 10^{-5} \times 4} \approx

1072.5

<br><br>\dfrac{\Delta E_2}{2kT_2} = \dfrac{0.67}{2 \times 8.625 \times 10^{-5} \times 300} \approx

13.0

<br><br>The difference in the exponent is

1072.5 - 13.0 = 1059.5

.<br><br>The exponential term is

e^{1059.5}

. Converting to base

using

e \approx 10^{0.4343}

:<br><br>

e^{1059.5} = 10^{1059.5 \times 0.4343} \approx 10^{460.1}

<br><br>The pre-exponential factor is:<br><br>\left(\dfrac{300}{4}\right)^{3/2} = (

75)^{3/2} \approx 650 \approx 10^{2.8}

<br><br>Multiplying the two factors gives the total ratio:<br><br>\dfrac{\sigma_2}{\sigma_1} \approx

10^{2.8} \times 10^{460.1} = 10^{462.9} \approx 10^{463}

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