Why does the Fermi energy \(\left(E_{\mathrm{F}}\right)\) increase with the number of electrons in a metal?

The Fermi energy is the highest energy level that an electron can occupy in a metal at absolute zero temperature (0 K). According to the Pauli Exclusion Principle, every electron in the metal should occupy a unique quantum state. In a metal, each energy level can contain two electrons, one with spin up and the other with spin down. So, as the number of electrons in a metal increases, the energy levels that these electrons can occupy also need to increase to ensure every electron has a unique quantum state.

The density of states is a measure of how many different quantum states are available for each energy level. In a metal, the density of states for electrons increases with energy, meaning there are more states available for electrons at higher energy levels. This means that as the number of electrons increases, they fill up the energy levels from the bottom and move towards higher energy states.

The Fermi energy can be calculated using the relationship between electron concentration (n) and Fermi wave vector (k_F). The electron concentration (n) is the number of electrons per unit volume. We also know that the number of states available within the volume is given by 4πk^3/3, where \(k = k_{\mathrm{F}}\). We can then relate electron concentration (n) to the Fermi wave vector (k_F) with the following equation: \[n = \frac{2(4\pi k_{\mathrm{F}}^3)}{3(2\pi)^3}\] Solving for k_F, we get: \[k_{\mathrm{F}} = \left(\frac{3n\pi^2}{8}\right)^{\frac{1}{3}}\] Now, we can find the Fermi energy (E_F) using the relationship between the Fermi wave vector and Fermi energy: \[E_{\mathrm{F}} = \frac{\hbar^2 k_{\mathrm{F}}^2}{2m}\] Substituting the expression for k_F calculated above, we get the Fermi energy in terms of electron concentration: \[E_{\mathrm{F}}(n) = \frac{\hbar^2}{2m}\left(\frac{3n\pi^2}{8}\right)^{\frac{2}{3}}\]

From the Fermi energy equation derived above, we can see that the Fermi energy (E_F) is directly proportional to the electron concentration (n) raised to the power of 2/3. As the number of electrons in a metal increases, the Fermi energy also increases to accommodate the increasing electron population filling higher energy states in the conduction band. So, the Fermi energy increases with the number of electrons in a metal.