Question

The ratio of charge carrier densities in a pure silicon crystal at \(27^\circ \text{C}\) and \(57^\circ \text{C}\) is 1.25. The energy gap of the semiconductor is:

• A. \(E_g = 1.11 \, \text{eV}\)
• B. \(E_g = 2.15 \, \text{eV}\)
• C. \(E_g = 6.1 \, \text{eV}\)
• D. \(E_g = 7.1 \, \text{eV}\)

Hint:

Use the Boltzmann factor exp (−Eg/2k) for temperature-dependent carrier density ratios.

Answer 1

The Correct Option is A

The ratio of charge carrier densities is related to the energy gap by:
\[\frac{n_2}{n_1} = \exp\left(-\frac{E_g}{2k}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\right)\]
Given $\frac{n_2}{n_1} = 1.25$, $T_1 = 27 + 273 = 300 \text{ K}$, $T_2 = 57 + 273 = 330 \text{ K}$, and $k = 8.617 \times 10^{-5} \text{ eV/K}$:
\[1.25 = \exp\left(-\frac{E_g}{2 \times 8.617 \times 10^{-5}}\left(\frac{1}{330} - \frac{1}{300}\right)\right)\]
Taking the natural logarithm and solving:
\[E_g \approx 1.11 \text{ eV}\]