Short Note

Indium antimonide has a band gap of 0.23 eV between the valence and the conduction band. Find the temperature at which kT equals the band gap.

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#### Solution

Given:

Band gap between the conduction band and the valence band, E = 0.23 eV

Boltzmann's constant, k = 1.38 × 10^{−23} J/K

We need to find the temperature at which thermal energy kT becomes equal to the band gap of indium antimonide.

∴ kT = E

\[\Rightarrow 1 . 38 \times {10}^{- 23} \times T = 0 . 23 \times 1 . 6 \times {10}^{- 19} \]

\[ \Rightarrow T = \frac{0 . 23 \times 1 . 6 \times {10}^{- 19}}{1 . 38 \times {10}^{- 23}}\]

\[ \Rightarrow T = \frac{0 . 23 \times 1 . 6 \times {10}^4}{1 . 38}\]

\[ \Rightarrow T = 0 . 2666 \times {10}^4 \approx 2670 \] K

Concept: Energy Bands in Conductors, Semiconductors and Insulators

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