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## Q. 11.1

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)

## Verified Solution

For $E-E_{ F }=1.0 eV ,\left(E-E_{ F }\right) / k T=\left(1.60 \times 10^{-19} J \right) /\left(1.38 \times 10^{-23} J / K \right)(293 K )=39.61$. Then

\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 \times 10^{-18} \\F_{ FD }(10 eV ) &=\frac{1}{e^{396.1}+1}=1.0 \times 10^{-172}\end{aligned}

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.