Question

Indium antimonide has a band gap of $0.23\text{eV}$ between the valence and the conduction band. Find the temperature at which the thermal energy $kT$ equals the indium antimonide’s band gap.
$(k=\text{Boltzmann constant},T=\text{temperature})$
$(\text{Take}k=1.38\times {10}^{-23}{\text{JK}}^{-1})$

• A. $1250\text{K}$
• B. $1800\text{K}$
• C. $2670\text{K}$
• D. $3125\text{K}$

Answer 1

The correct option is C $2670\text{K}$

It is to be found that the temperature at which the thermal energy $kT$ equals the indium antimonide’s band gap.

$\therefore kT=E$

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

$\Rightarrow T=\frac{0.368\times {10}^{-19}}{1.38\times {10}^{-23}}$

$\Rightarrow T=0.2666\times {10}^4\approx 2670\text{K}$

Hence, option $\left(C\right)$ is correct.