--- title: "Energy levels in well and copper wire" subject: "Electrical Engineering Material" chapter: 1 key: "semester_3/electrical_engineering_material:potential-well-copper-wire" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/potential-well-copper-wire site: BCT Engineering --- # Energy levels in well and copper wire Electrical Engineering Material · Chapter 1: Theory of Metals - Asked in: 2071 Shrawan, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/potential-well-copper-wire - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1 ## Question Derive the relation of energy level inside a potential well of width L. Show mathematically that energy level in a copper wire of length L is quantized similar to energy level inside a potential well. ## Short answer For an infinite well of width L, solving the Schrödinger equation yields discrete energies E_n = n²h²/(8mL²). A conduction electron in a copper wire of length L experiences nearly infinite potential barriers at the ends, so the same boundary condition gives identical quantized energy levels E_n = n²h²/(8mL²). ## Answer #### Energy levels in a potential well of width L Consider a particle of mass $m$ in a one-dimensional infinite well: $$ V(x) = 0 \ \text{ for } 0 < x < L, \qquad V(x) = \infty \ \text{ elsewhere} $$ ```text V = inf | | V = inf | V = 0 | | | +------------------+ 0 L x ``` Outside, $\psi = 0$. Inside, the Schrödinger equation is $$ \frac{d^2\psi}{dx^2} + \frac{2mE}{\hbar^2}\psi = 0 \quad\Rightarrow\quad \frac{d^2\psi}{dx^2} + k^2\psi = 0, \quad k = \frac{\sqrt{2mE}}{\hbar} $$ General solution: $\psi(x) = A\sin kx + B\cos kx$. Boundary conditions ($\psi$ continuous, so zero at the walls): - $\psi(0) = 0 \Rightarrow B = 0$ - $\psi(L) = A\sin kL = 0 \Rightarrow kL = n\pi$, $n = 1, 2, 3, \ldots$ $$ \begin{aligned} k_n &= \frac{n\pi}{L} \\ E_n &= \frac{\hbar^2k_n^2}{2m} = \frac{n^2\pi^2\hbar^2}{2mL^2} = \frac{n^2h^2}{8mL^2} \end{aligned} $$ Normalizing, $\psi_n(x) = \sqrt{2/L}\,\sin(n\pi x/L)$. Only discrete energies $E_1, 4E_1, 9E_1, \ldots$ are allowed. #### Copper wire of length L In the free electron model, a conduction electron in a copper wire: - moves freely inside the wire, where the potential of the ion cores is nearly uniform (take $V = 0$); - cannot escape at the ends, because leaving the metal needs energy equal to the work function (about 4.6 eV for Cu), which is very large compared with the energy changes involved. The ends act as nearly infinite walls. ```text copper wire, length L surface ====================== surface barrier e- -> <- e- -> barrier (high) free electrons, V = 0 (high) 0 L ``` So the wire is exactly a one-dimensional potential well of width $L$, and the electron's wave function must vanish (be zero) at $x = 0$ and $x = L$. Applying the same Schrödinger equation and boundary conditions gives the same result: $$ E_n = \frac{n^2h^2}{8mL^2}, \qquad n = 1, 2, 3, \ldots $$ Hence **the energy of an electron in a copper wire is quantized**, just like a particle in a potential well. #### Why it seems continuous For a wire of $L = 1$ cm: $$ E_1 = \frac{(6.626\times10^{-34})^2}{8\times9.1\times10^{-31}\times(10^{-2})^2} = 6.03\times10^{-34}\ \text{J} = 3.8\times10^{-15}\ \text{eV} $$ Near the Fermi level of copper (about 7 eV), $n \approx 4.3\times10^{7}$ and the spacing between neighbouring levels is $\Delta E \approx (2n + 1)E_1 \approx 3\times10^{-7}$ eV, far smaller than $kT = 0.026$ eV. The levels are quantized but so closely spaced that they form a practically continuous band. Quantization becomes visible only when $L$ is a few nanometres. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What conditions on psi are applied at the walls of the infinite well? A. psi = 0 at x = 0 and x = L B. psi maximum at the walls C. d psi/dx = 0 at the walls D. psi = 1 at x = L Answer: A. psi is zero outside and continuous, so it vanishes at both walls: B = 0 and sin kL = 0. (Key point: Well boundary conditions) 2. What gives the discrete values of k? A. kL = n pi, n = 1, 2, 3... B. kL = n pi/2, n = 0, 1, 2... C. kL = 2n D. k = L/n pi Answer: A. sin kL = 0 requires kL = n pi, so k_n = n pi/L. (Key point: Quantisation condition) 3. What is the energy of the nth level? A. n h^2 / 8mL^2 B. n^2 h^2 / 8mL^2 C. h^2 / 8 n^2 m L^2 D. n^2 h^2 / 2mL Answer: B. E_n = n^2 pi^2 hbar^2 / 2mL^2 = n^2 h^2 / 8mL^2, so levels go as E1, 4E1, 9E1. (Key point: Energy levels) 4. Why can a copper wire be treated as an infinite well? A. Electrons are immobile in copper B. Ion-core potential is nearly uniform inside, and the work function (about 4.6 eV) makes the ends act as nearly infinite walls C. The wire has infinite resistance D. The wire is a perfect insulator at the ends Answer: B. Electrons move freely (V about 0) but cannot leave because of the large surface barrier. 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