Let $Δ E$ 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 $e^{- Δ E / 2 k T} .$ Find the ratio of the concentration of conduction electrons in diamond to that in silicon at room temperature 300 K. $Δ E$ for silicon is 1.1 eV and for diamond is 6.0 eV. How many conduction electrons are likely to be in one cubic metre of diamond ?

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Solution

Given $n = e^{- Δ E / 2 K T}$

$Δ E$ (Diamond) = $6 e V$

$Δ E (S i) = 1.1 e V .$

Now, $n_{1} = e^{\frac{- Δ E_{1}}{2 K T}}$

$= e^{\frac{- 6}{2 \times 300 \times 8.62 \times 10^{- 5}}}$

$n_{2} = e^{\frac{- Δ E_{2}}{2 K T}}$

$= e^{\frac{- 11}{2 \times 300 \times 8.62 \times 10^{- 5}}}$

$\frac{n_{1}}{n_{2}} = \frac{4.14722 \times 10^{- 51}}{5.7979 \times 10^{- 10}}$

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

Due to more $Δ E,$ the conduction electrons per cubic metre in diamond is almost zero.

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Given n=e−ΔE/2KT ΔE (Diamond) = 6 eV ΔE (Si)=1.1 eV. Now, n1=e−ΔE12KT =e−62×300×8.62×10−5 n2=e−ΔE22KT =e−112×300×8.62×10−5 n1n2=4.14722×10−515.7979×10−10 =7.15×10−42 Due to more ΔE, the conduction electrons per cubic metre in diamond is almost zero.