Discussion
Consider ideal Fermi gas in three dimensions. The density of energy states ε is proportional to :
Answer: A
The density of states is proportional to \(\varepsilon^{1/2}\).
For free particles in three dimensions, the number of states with momentum below p is proportional to the volume of a sphere in momentum space, \(p^{3}\). With \(\varepsilon = p^{2}/2m\), that number goes as \(\varepsilon^{3/2}\), and its derivative gives the density of states:
\[ \begin{aligned} g(\varepsilon) &= \frac{V}{2\pi^{2}}\left(\frac{2m}{\hbar^{2}}\right)^{3/2} \\ &\quad\times\varepsilon^{1/2} \end{aligned} \]
(times 2 for spin). In two dimensions the density of states is constant (\(\varepsilon^{0}\)), and in one dimension it goes as \(\varepsilon^{-1/2}\), so those options describe lower-dimensional gases. \(\varepsilon^{2}\) belongs to no free-particle gas; photons, with \(\varepsilon \propto p\), give \(\varepsilon^{2}\) dependence, not a non-relativistic Fermi gas.
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