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Q48P

Expert-verifiedFound in: Page 1275

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Show that P(E), the occupancy probability in Eq. 41-6, is symmetrical about the value of the Fermi energy; that is, show that ${\mathit{P}}{\mathbf{(}}{{\mathit{E}}}_{{\mathbf{F}}}{\mathbf{+}}{\mathit{\Delta}}{\mathit{E}}{\mathbf{)}}{\mathbf{+}}{\mathit{P}}{\mathbf{(}}{{\mathit{E}}}_{{\mathbf{F}}}{\mathbf{-}}{\mathit{\Delta}}{\mathit{E}}{\mathbf{)}}{\mathbf{=}}{\mathbf{1}}$.**

It is shown that $P({E}_{F}+\Delta E)+P({E}_{F}-\Delta E)=1$ that is the occupancy probability is symmetrical about the value of the Fermi energy.

**The probability of a state to be occupied by and electron is referred to as the occupancy probability. At 0 K temperature, the states below the Fermi level have occupancy probability equal to and for the states above the Fermi level, its value is . **

Formula:

The occupancy probability of the state with energy *E* is-

$P\left(E\right)=\frac{1}{{e}^{\left(E-{E}_{F}\right)/KT}+1}$ ( i )

Here ${E}_{F}$ is the Fermi energy, $k=8.62\times {10}^{-5}eV/K$ and *T* is the absolute temperature.

Upon expansion in view of equation (i), the LHS value of the given equation can be solved as follows:

$LHS=P\left({E}_{F}+\u2206E\right)+P\left({E}_{F}-\u2206E\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{{e}^{\left({E}_{F}+\u2206E-{E}_{F}\right)/KT}+1}+\frac{1}{{e}^{\left({E}_{F}-\u2206E-{E}_{F}\right)/KT}+1}\phantom{\rule{0ex}{0ex}}=\frac{1}{{e}^{\u2206E/KT}+1}+\frac{1}{{e}^{-\u2206E//KT}+1}\phantom{\rule{0ex}{0ex}}=\frac{{e}^{\u2206E/KT}+1+{e}^{\u2206E/KT}+1}{\left({e}^{\u2206E/KT}+1\right)\left({e}^{-\u2206E/KT}+1\right)}\phantom{\rule{0ex}{0ex}}$

On further solving,

$L.H.S=\frac{{e}^{\u2206E/KT}+1+{e}^{\u2206E/KT}+2}{{e}^{\u2206E/KT}+{e}^{\u2206E/KT}+2}\phantom{\rule{0ex}{0ex}}=1\phantom{\rule{0ex}{0ex}}=R.H.S$

Hence, the given condition is proved and this implies the symmetrical condition for occupancy probability.

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