Discussion
The chemical potential (\(\mu\)) of a solid coincides with Fermi energy (EF) at absolute zero. At temperature T \(\gt\) 0, which of the following options is correct ?
Answer: C
At T \(\gt\) 0, \(\mu \lt E_{F}\).
For a free-electron metal, the Sommerfeld expansion gives:
\[ \begin{aligned} \mu(T) &\approx E_{F}\Big[1 \\ &\quad - \frac{\pi^{2}}{12}\Big(\frac{kT}{E_{F}}\Big)^{2}\Big] \end{aligned} \]
so the chemical potential falls slightly below \(E_{F}\) as temperature rises, because the density of states increases with energy and the chemical potential must shift down to keep the electron number fixed.
The decrease is tiny at ordinary temperatures (\(kT \ll E_{F}\)), which is why \(\mu\) and \(E_{F}\) are often used interchangeably, but strictly they are not equal, and \(\mu\) is certainly not 0.
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