--- title: "Population density and temperature-independent Fermi energy" subject: "Electrical Engineering Material" chapter: 2 key: "semester_3/electrical_engineering_material:fermi-energy-temperature-independent" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/fermi-energy-temperature-independent site: BCT Engineering --- # Population density and temperature-independent Fermi energy Electrical Engineering Material · Chapter 2: Free Electron Theory of Conduction in Metal - Asked in: 2073 Chaitra, 2+6 marks - Video: [Density of states and Derivation of Fermi Energy at 0 K](https://www.youtube.com/watch?v=M7Z6B9MJOb4) — Ravi C.B - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/fermi-energy-temperature-independent - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-2 ## Question Define population density. Prove that fermi energy in a metal is independent of temperature and depends only in its electron concentration. ## Short answer Population density n_E(E) = g(E)f(E) is the number of electrons per unit volume per unit energy. At 0 K, integration gives E_{F0} = (h²/8m_e)(3n/π)^{2/3}, depending only on n. For T>0, the Sommerfeld expansion yields E_F(T) = E_{F0}[1 − (π²/12)(kT/E_{F0})²]; the correction is ~10⁻⁵, so E_F is essentially temperature independent. ## Answer #### Population density The **population density** $n_E(E)$ is the number of electrons per unit volume per unit energy at energy $E$. It is the product of the number of available states and the probability that they are occupied: $$ n_E(E) = g(E)\, f(E), \qquad g(E) = \frac{8\pi\sqrt{2}\, m_e^{3/2}}{h^3} E^{1/2}, \quad f(E) = \frac{1}{1 + e^{(E-E_F)/kT}} $$ The area under the $n_E$ versus $E$ curve gives the total electron concentration $n$. ```text nE(E) ^ T=0: sharp cut at EF | ..---. | .-' |\ <- T>0: small tail | / | \ | / | '.. |/___________|______> E 0 EF ``` #### Fermi energy depends only on n At $T = 0$ K, $f(E) = 1$ for $E < E_{F0}$ and $0$ above it. Hence $$ n = \int_0^{E_{F0}} g(E)\, dE = \frac{8\pi\sqrt{2}\, m_e^{3/2}}{h^3} \cdot \frac{2}{3} E_{F0}^{3/2} = \frac{8\pi}{3}\left(\frac{2m_e E_{F0}}{h^2}\right)^{3/2} $$ Solving, $$ E_{F0} = \frac{h^2}{8m_e}\left(\frac{3n}{\pi}\right)^{2/3} $$ Thus $E_{F0}$ contains only constants and $n$, the free electron concentration. A metal with more free electrons per volume has a higher Fermi energy (Cu, $n = 8.5 \times 10^{28}$ m⁻³: 7.0 eV; Na, $2.5 \times 10^{28}$: 3.1 eV). #### Temperature has almost no effect At $T > 0$, the electron number must stay the same: $$ n = \int_0^{\infty} g(E) f(E)\, dE $$ Evaluating this integral (Sommerfeld expansion) gives $$ E_F(T) = E_{F0}\left[1 - \frac{\pi^2}{12}\left(\frac{kT}{E_{F0}}\right)^2\right] $$ For copper at 300 K: $kT = 0.0259$ eV, $E_{F0} = 7.0$ eV, so $kT/E_{F0} = 3.7 \times 10^{-3}$ and the correction is $\frac{\pi^2}{12}(3.7 \times 10^{-3})^2 \approx 1.1 \times 10^{-5}$, a change of only about 0.001%. #### Physical reason Only electrons within a few $kT$ of $E_F$ can be thermally excited, because states deeper down have all their neighbouring states already filled (Pauli principle). Since $kT \ll E_F$, only a tiny fraction of electrons are disturbed, and the electrons moved above $E_F$ match the holes left below $E_F$ (the Fermi function is symmetric about $E_F$). Also $n$ in a metal does not change with $T$. Hence the Fermi energy of a metal is practically independent of temperature and depends only on its electron concentration. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Population density n_E(E) is A. g(E) + f(E) B. g(E) f(E): number of states times occupation probability C. f(E)/g(E) D. g(E) only Answer: B. It gives electrons per unit volume per unit energy. (Key point: Population density) 2. At 0 K, E_F0 is obtained by A. Integrating g(E) from 0 to E_F0 and setting it to n B. Setting f(E) = 0.5 at E = 0 C. Setting E_F0 = kT D. Using the work function Answer: A. n = (8π/3)(2mE_F0/h^2)^(3/2), solved for E_F0. (Key point: E_F0 depends on n) 3. E_F0 contains A. n and temperature B. work function and n C. lattice constant and T D. only constants and n Answer: D. E_F0 = (h^2/8m)(3n/π)^(2/3). (Key point: Dependence) 4. Why is E_F nearly temperature independent in metals? A. All electrons are thermally excited B. n increases exactly with T C. kT << E_F, only electrons near E_F are excited, and the Fermi function is symmetric about E_F D. Pauli principle does not apply Answer: C. Excited electrons above E_F match vacated states below, so E_F barely shifts (~0.001%). 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