First slide

Types of semiconductors

Question

Let $△ E$ denotes the energy gap between the valence band and the conduction band. The population of conduction electrons (and of the holes) is roughly proportional to $e^{- △ E / 2 k T}$ . Find the ratio of the concentration of conduction electrons in diamond to that in silicon at room temperature 300 K . $△ E$ for silicon is 1.1 eV and for diamond is 6.0 eV.

Very difficult

A

$7.10 \times 10^{- 42}$
B

$7.12 \times 10^{- 42}$
C

$7.15 \times 10^{- 42}$
D

$7.17 \times 10^{- 42}$

Solution

Given $η = e^{- △ E / 2 k T}$
$△ E$ (diamond) = 6 eV
$△ E$ (Si) = 1.1 eV
Now, $η_{1} (D i a m o n d) = e^{\frac{- △ E_{1}}{2 k T}}$
$= e^{\frac{- 6 e V}{2 \times 1.38 \times 10^{- 23} \times 300}} = e^{\frac{- 6 \times 1.6 \times 10^{- 19}}{2 \times 1.38 \times 10^{- 23} \times 300},}$
$η_{1} = e^{- \frac{6}{2 \times 300 \times 8.62 \times 10^{- 5}}} = e^{- 116} = 4.14722 \times 10^{- 51}$
$S i m i l a r l y, η_{2} (S i l i c o n) = e^{\frac{- △ E_{2}}{2 k T}} = e^{\frac{- 1.1}{2 \times 1.38 \times 10^{- 23} \times 300}} = e^{\frac{- 1.1 \times 1.6 \times 10^{- 19}}{2 \times 1.38 \times 10^{- 23} \times 300}} = e^{\frac{- 1.1}{2 \times 8.62 \times 10^{- 5} \times 300}} = e^{- 21.2683} = 5.7979 \times 10^{- 10}$
$∴ r a t i o = \frac{η_{1}}{η_{2}} = \frac{4.14722 \times 10^{- 51}}{5.7979 \times 10^{- 10}} = 7.15 \times 10^{- 42}$

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