Find the relative number of electrons with energies 0.10 eV, 1.0 eV, and 10 eV above the valence band at room temperature (293 K).

Strategy Equation (9.34) for the Fermi-Dirac factor may be used for this comparison. Note that 1.0 eV $=1.60 \times 10^{-19}$ J. The Fermi energy is at the top of the valence band, so the energy above the valence band is $E-E_{ F }$ .

$F_{ FD }=\frac{1}{\exp \left[\beta\left(E-E_{ F }\right)\right]+1}$ (9.34)

Question

Find the relative number of electrons with energies 0.10 eV, 1.0 eV, and 10 eV above the valence band at room temperature (293 K).

Strategy Equation (9.34) for the Fermi-Dirac factor may be used for this comparison. Note that 1.0 eV [latex]=1.60 imes 10^{-19}[/latex] J. The Fermi energy is at the top of the valence band, so the energy above the valence band is [latex]E-E_{ F }[/latex].

[latex]F_{ FD }=\frac{1}{\exp \left[\beta\left(E-E_{ F } ight) ight]+1}[/latex] (9.34)

Accepted Answer

For [latex]E-E_{ F }=1.0 eV ,\left(E-E_{ F } ight) / k T=\left(1.60 imes 10^{-19} J ight) /\left(1.38 imes 10^{-23} J / K ight)(293 K )=39.61[/latex]. Then

[latex]\begin{aligned}F_{ FD }(0.10 eV ) &=\frac{1}{e^{3.961}+1}=0.019 \F_{ FD }(1.0 eV ) &=\frac{1}{e^{39.61}+1}=6.27 imes 10^{-18} \F_{ FD }(10 eV ) &=\frac{1}{e^{396.1}+1}=1.0 imes 10^{-172}\end{aligned}[/latex]

This example illustrates how strongly the Fermi-Dirac factor depends on the size of the band gap. The number of electrons available for conduction drops off sharply as the band gap increases.

Question 11.1: Find the relative number of electrons with energies 0.10 eV,...

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