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In an intrinsic semiconductor the energy gap $\mathrm{E}_{\mathrm{g}}$ is $1.2 \mathrm{eV}$. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at $600 \mathrm{~K}$ and that at $300 \mathrm{~K}$ ? Assume that the temperature dependence of intrinsic carrier concentration $n_i$ is given by
$n_i=n_0 \exp \left(-\frac{E_g}{2 k_B T}\right)$ where $n_0$ is a constant.
$$
K_B=8.62 \times 10^{-5} \mathrm{eV} \mathrm{k}^{-1}
$$
$n_i=n_0 \exp \left(-\frac{E_g}{2 k_B T}\right)$ where $n_0$ is a constant.
$$
K_B=8.62 \times 10^{-5} \mathrm{eV} \mathrm{k}^{-1}
$$
Solution:
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Verified Answer
Here, $\frac{\Delta \mathrm{E}}{2 \mathrm{~K}}\left[\frac{1}{\mathrm{~T}_1}-\frac{1}{\mathrm{~T}_2}\right]$
$=\frac{1.2}{2 \times 8.62 \times 10^{-5}} \cdot\left[\frac{1}{300}-\frac{1}{600}\right]=11.6$
$\frac{\mathrm{n}_{600}}{\mathrm{n}_{300}}=\mathrm{e}^{\frac{\Delta \mathrm{E}}{2 \mathrm{~K}}\left|\frac{1}{\mathrm{~T}_{\mathrm{l}}}-\frac{1}{\mathrm{~T}_2}\right|}=\mathrm{e}^{11.6}=(2.718)^{11.6}$
Solving using logs, $\frac{\mathrm{n}_{600}}{\mathrm{n}_{300}}=1.089 \times 10^5$
$=1.1 \times 10^5$
$=\frac{1.2}{2 \times 8.62 \times 10^{-5}} \cdot\left[\frac{1}{300}-\frac{1}{600}\right]=11.6$
$\frac{\mathrm{n}_{600}}{\mathrm{n}_{300}}=\mathrm{e}^{\frac{\Delta \mathrm{E}}{2 \mathrm{~K}}\left|\frac{1}{\mathrm{~T}_{\mathrm{l}}}-\frac{1}{\mathrm{~T}_2}\right|}=\mathrm{e}^{11.6}=(2.718)^{11.6}$
Solving using logs, $\frac{\mathrm{n}_{600}}{\mathrm{n}_{300}}=1.089 \times 10^5$
$=1.1 \times 10^5$
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