First slide

Types of semiconductors

Question

Assume that the number of hole – electron pair in an intrinsic semiconductor is proportional to $e^{\frac{△ E}{2 k T}}$ . $H e r e, △ E =$ energy . gap and $k = 8.62 \times 10^{- 5} e V / k e l v i n$ . The energy gap for silicon is 1.1 eV. The ratio of electron hole pairs at 300 K and 400 K is $e^{- x}$ . Find x

Difficult

Solution

$T_{1} = 300 K$
$T_{2} = 400 K$
$∴ r a t i o = \frac{e^{- △ E / 2 k T_{1}}}{e^{- △ E / 2 k T_{2}}} = e^{- \frac{△ E}{2 k T_{1}} + \frac{△ E}{2 k T_{2}}} = e^{- \frac{△ E}{2 k} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]} = e^{- \frac{△ E}{2 k} [\frac{1}{300} - \frac{1}{400}]}$
$= e^{- \frac{1.1 e v}{2 \times 8.62 \times 10^{- 5} e v} [\frac{4 - 3}{1200}]} = e^{\frac{- 1.1}{17.24} [\frac{1}{1200 \times 10^{- 5}}]} = e^{- \frac{110}{1724 \times 12 \times 10^{- 3}}} = e^{- \frac{110000}{1724 \times 12}} = = e^{- 5.32}$

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