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Question:

In an intrinsic semiconductor, the band gap is 1.2 eV. Then, the ratio of the number of charge carriers at 600 K and 300 K is of the order:

104

107

1010

103

Solution:

The ratio is given by the formula:

( \frac{n_1}{n_2} = \frac{e^{-\frac{E_g}{kT_1}}}{e^{-\frac{E_g}{kT_2}}} = e^{-E_g(\frac{1}{kT_1} - \frac{1}{kT_2})} )

Where:

  • (n_1) and (n_2) are the number of charge carriers at temperatures (T_1) and (T_2) respectively.
  • (E_g) is the band gap energy (1.2 eV)
  • (k) is the Boltzmann constant (8.617 x 10^-5 eV/K)
  • (T_1) = 600 K
  • (T_2) = 300 K

Substituting the values:

( \frac{n_1}{n_2} = e^{-1.2eV(\frac{1}{8.617 \times 10^{-5} \times 600} - \frac{1}{8.617 \times 10^{-5} \times 300})} )

( \frac{n_1}{n_2} = e^{-1.2(\frac{1}{0.0517} - \frac{1}{0.02586})} \approx e^{-1.2(19.33 - 38.66)} \approx e^{23.196} \approx 1.176 \times 10^{10} )

Therefore, the ratio of the number of charge carriers at 600 K and 300 K is of the order of 10^10.