The conductivity of an intrinsic semiconductor depends on temperature as $$\sigma =\sigma _0 e^{-\Delta \, E/2kT}$$, where $$\sigma _0$$, is a constant. Find the temperature at which the conductivity of an intrinsic germanium semiconductor will be double of its value at T= 300 K. Assume that the gap for germanium is $$0\cdot 650\, eV$$ and remains constant as the temperature is increased.
$$\sigma =\sigma _0 E^{-\Delta E / 2KT}$$
$$\Delta E = 0.650\, eV, T = 300\, K$$
According to question, $$ K = 8.62 \times 10^{-5}\, eV$$
$$ \sigma _0 E^{-\Delta E / 2KT} = 2 \times \sigma _0 E^{\dfrac {-\Delta E}{2 \times K \times 300}}$$
$$\Rightarrow \frac {-0.65}{e^{2 \times 8.6 \times 10^{-5}\times T}}= 6.96561 \times 10^{-5}$$
Taking in on both sides,
We get,
$$ \frac {-0.65}{e^{2 \times 8.6 \times 10^{-5}\times T}}= -11.874525$$
$$\Rightarrow \dfrac {1}{T'}, =\frac {11.574525 \times 2 \times 8.62 \times 10^{-25}}{0.65}$$
$$\Rightarrow T' = 317.51178= 318\, K$$
Let $$\Delta \,E$$ denote the energy gap between the valence band and the conduction band. The population of conduction electrons (and of the holes) is roughly proportional to $$e^{-\Delta \, E /2kT}$$. Find the ratio of the concentration of conduction electrons in diamond to that in silicon at a room temperature 300 K. $$\Delta \,E$$ for silicon is $$1\cdot1\, eV$$ and for diamond is $$6\cdot 0\, eV$$.
In an intrinsic semiconductor the energy gap $$E_g$$ is $$1.2\ eV$$. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at $$600\ K$$ and that at $$300\ K$$? Assume that the temperature dependence of intrinsic carrier concentration $$n_i$$ is given by $$n_i = n_o exp \left(- \dfrac{E_g}{2 k_B T} \right)$$ Where $$n_0$$ is a constant.
Find the angle between the planes whose vector equations are $$\vec { r } .\left( 2\hat { i } +2\hat { j } -3\hat { k } \right) =5$$ and $$\vec { r } .\left( 3\hat { i } -3\hat { j } +5\hat { k } \right) =3$$
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