Free-electron theory of metals: The Fermi level (energy) of a metal is 5.5 eV. What is the number of conduction electrons per unit volume for this metal? (m_el = 9.11 × 10^-31 kg, 1 eV = 1.60 × 10^-19 J, h = 6.626 × 10^-34 J ∙ s) A) 5.8 × 10²⁸/ m³ B) 6.2 × 10²⁸/ m³ C) 4.1 × 10²⁸/ m³ D) 8.4 × 10²⁸/ m³ E) 3.3 × 10²⁸/ m³

Question

Quizplus · Accepted Answer

According to the free-electron theory of metals, the number of conduction electrons per unit volume is given by:
$$n = \frac{ho}{m} = \frac{N}{V} = \frac{\frac{N_A ho}{A}}{\frac{M}{ho}} = \frac{N_A ho^2}{MA}$$
where $N_A$ is Avogadro's number, $ho$ is the mass density of the metal, $A$ is the atomic mass of the metal, and $M$ is the molar mass of the metal. In this problem, we are given the Fermi level of the metal, which is a measure of the energy of the highest occupied electronic state at absolute zero temperature. The Fermi level is related to the density of states per unit energy interval $g(E)$ as:
$$n = \int_{-\infty}^{E_F}g(E)f(E)dE$$
where $f(E)$ is the Fermi-Dirac distribution function, which is equal to 1 for energies below $E_F$ and 0 for energies above $E_F$. For the free-electron model, $g(E)$ is a constant, and we can evaluate the integral to get:
$$n = \frac{1}{3}\frac{8\pi V}{h^3}\sqrt{2m_e}(E_F)^{\frac{3}{2}}$$
where $m_e$ is the mass of an electron, and $V$ is the volume of the metal. Substituting the given values and using the conversion factors, we get:
$$n = \frac{1}{3}\frac{8\pi V}{h^3}\sqrt{2m_e}(5.5	ext{ eV})(1.60	imes 10^{-19}	ext{ J/eV})^{\frac{3}{2}} = 5.8	imes 10^{28}	ext{ m}^{-3}$$
Therefore, the answer is (A).

Free-Electron Theory of Metals: the Fermi Level (Energy) of a Metal