In an intrinsic semiconductor band gap is $1.2 \mathrm{eV}$ then ratio of number of charge carriers at $600…

In an intrinsic semiconductor band gap is $1.2 \mathrm{eV}$ then ratio of number of charge carriers at $600 \mathrm{~K}$ and $300 \mathrm{~K}$ is
  1. $10^4$
  2. $10^7$
  3. $10^5$
  4. $10^3$

Solution

The energy gap \(E_g=1.2 \mathrm{eV}\) Now the relation in charge carrier and temperature is \(n_i=n_o \exp \left[-E_g / 2 K_B T\right]\) where \(K_B\) is Boltzmann constant and its value is \(8.62 \times 10^{-5} \mathrm{eV} / \mathrm{K}\) So at the given temperature of 300 K the relation is \(n_{i 1}=n_o \exp \left[-E_g / 2 K_B \times 300\right]\) Similarly at the given temperature of 600 K the relation is \(n_{i 2}=n_o \exp \left[-E_g / 2 K_B \times 600\right]\) Now the ratio of the given two is \(\begin{aligned} & \frac{n_{i 2}}{n_{i 1}}=\frac{n_o \exp \left[-E_g / 2 K_B \times 600\right]}{n_o \exp \left[-E_g / 2 K_B \times 300\right]} \\ & =\exp \frac{E_g}{2 K_B}\left[\frac{1}{300}-\frac{1}{600}\right] \\ & =\exp [11.6] \\ & =1.09 \times 10^5 \\ & \approx 10^5 \end{aligned}\)

Asked in: NEET 2010 (Mains)

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