--- title: "Energy of particle in infinite well" subject: "Engineering Physics" chapter: 10 key: "semester_1/engineering_physics:particle-infinite-well-energy-eigenvalue" source: https://www.bctengineer.com/q/semester_1/engineering_physics/particle-infinite-well-energy-eigenvalue site: BCT Engineering --- # Energy of particle in infinite well Engineering Physics · Chapter 10: Photon and Matter Waves - Asked in: 2073 Chaitra, 5 marks; 2072 Kartik, 5 marks; 2071 Shrawan, 5 marks; 2070 Chaitra, 5 marks; 2069 Asar, 5 marks; 2068 Shrawan, 5 marks; 2067 Asar, 5 marks (asked 7 times) - Video: [Infinite square well energy eigenstates](https://www.youtube.com/watch?v=gMHkf-107Sw) — MIT OpenCourseWare - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/engineering_physics/particle-infinite-well-energy-eigenvalue - Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-10 ## Question Using the Schrödinger wave equation, derive an expression for the energy (energy eigenvalue) of a particle (electron) confined in a one-dimensional infinite potential well (box) of width 'a'. Show that the energy levels are quantized: E = n²π²ħ²/(2ma²). ## Short answer Solving the Schrödinger equation with V=0 inside and ψ(0)=ψ(a)=0 gives ψ_n = √(2/a) sin(nπx/a) and energy eigenvalues E_n = n²π²ħ²/(2ma²). Energy is quantized, with n = 1,2,3,… ## Answer A particle of mass m moves freely inside a one-dimensional box between two impenetrable walls at x = 0 and x = a. ```text V=inf | | V=inf | V = 0 inside | |____________________| 0 a ``` Potential: V = 0 for 0 < x < a, V = ∞ for x ≤ 0 and x ≥ a. #### Step 1: wave function outside The particle cannot be outside, so ψ = 0 for x ≤ 0 and x ≥ a. #### Step 2: Schrödinger equation inside Time-independent equation with V = 0: −(ħ²/2m) d²ψ/dx² = Eψ → d²ψ/dx² + k²ψ = 0, where **k² = 2mE/ħ²** #### Step 3: solution ψ(x) = A sin kx + B cos kx. #### Step 4: boundary conditions - ψ(0) = 0: B = 0. So ψ = A sin kx. - ψ(a) = 0: A sin ka = 0. A ≠ 0 (else ψ = 0 everywhere), so ka = nπ, k = nπ/a with n = 1, 2, 3, … (n = 0 gives ψ = 0, no particle.) #### Step 5: energy eigenvalues Put k = nπ/a into k² = 2mE/ħ²: **Eₙ = n²π²ħ²/(2ma²) = n²h²/(8ma²)**, n = 1, 2, 3, … Only these discrete values are allowed, so the energy is **quantized**. E ∝ n², and the lowest energy (zero-point energy) is E₁ = π²ħ²/(2ma²), not zero. #### Step 6: normalisation ∫₀ᵃ |ψ|² dx = A² (a/2) = 1, so A = √(2/a). **ψₙ(x) = √(2/a) sin(nπx/a)** ```text E4 ---------- n=4 (16 E1) E3 ---------- n=3 (9 E1) E2 ---------- n=2 (4 E1) E1 ---------- n=1 ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is ψ = 0 outside the box? A. The particle has zero energy there B. The wave function is imaginary there C. The walls absorb the particle D. The walls are infinitely high, so the particle cannot be there Answer: D. Infinite potential means zero probability of finding the particle outside. (Key point: Wave function outside) 2. What does ψ(0) = 0 do to ψ = A sin kx + B cos kx? A. Gives B = 0 B. Gives A = 0 C. Gives k = 0 D. Gives A = B Answer: A. cos 0 = 1, so B must be zero. (Key point: Boundary condition at x = 0) 3. Why is the energy quantized? A. The mass is quantized B. ψ(a) = 0 requires ka = nπ, so k takes only discrete values C. The potential changes with time D. Normalisation fixes E Answer: B. Only k = nπ/a satisfy both boundary conditions. (Key point: Quantization) 4. Which is correct for E_n? A. E_n = n h²/(8ma²), lowest E₀ = 0 B. E_n = n²h²/(8ma), n starting at 0 C. E_n = n²h²/(8ma²), with lowest E₁ not zero D. E_n = h²/(8n²ma²) Answer: C. E ∝ n², n = 0 is excluded as ψ = 0, so there is a zero-point energy. 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