The band gap between the valence and conduction bands for indium antimonide is 0.23 eV . Determine the temperature at which kT becomes equal to this band gap.

Correct answer: B. 2670 K Given band gap E_g = 0.23 eV Boltzmann constant k = 1.38 × 10^ -23 J/K Equating the thermal energy to the band gap: kT = E_g T = (E_g)/(k) Substituting the values in standard units: T = 0.23 × 1.6 × 10^ -19 1.38 × 10^ -23 T = 0.368 × 10^ -19 1.38 × 10^ -23 T = (3680)/(1.38) 2667 K The closest value among the given options is 2670 K .

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 band gap between the valence and conduction bands for indium...

The band gap between the valence and conduction bands for indium antimonide is

0.23 \text{ eV}

. Determine the temperature at which

kT

becomes equal to this band gap.

Options

A $267 \text{ K}$
B $2670 \text{ K}$
C $1335 \text{ K}$
D $5340 \text{ K}$

Correct answer

B. $2670 \text{ K}$

Step-by-step solution

Given band gap

E_g = 0.23 \text{ eV}

Boltzmann constant

k = 1.38 \times 10^{-23} \text{ J/K}

Equating the thermal energy to the band gap:

kT = E_g

T = \dfrac{E_g}{k}

Substituting the values in standard units:

T = \dfrac{0.23 \times 1.6 \times 10^{-19}}{1.38 \times 10^{-23}}

T = \dfrac{0.368 \times 10^{-19}}{1.38 \times 10^{-23}}

T = \dfrac{3680}{1.38} \approx 2667 \text{ K}

The closest value among the given options is

2670 \text{ K}

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