Particle in a One-Dimensional Box: Energy Levels and Wavefunctions
The particle in a one-dimensional box is the simplest problem in quantum mechanics that can be solved exactly. It also appears again and again in physical chemistry papers for IIT-JAM, GATE, CSIR-NET and CUET-PG, because it shows quantisation of energy, zero-point energy and nodes in a few lines. It is also used as a rough model for π electrons in conjugated molecules. Here are the formulas, the reasons behind them, and worked numbers.
The model
A particle of mass m moves freely between two walls at x = 0 and x = L. Inside the box the potential energy is zero. At the walls and outside it is infinite, so the particle cannot be found there. The wavefunction must therefore be zero at x = 0 and x = L. Only waves that fit an exact number of half-wavelengths into L satisfy this, and that condition quantises the energy.
En = n²h² / (8mL²)
Here h = 6.626 × 10−34 J s is the Planck constant, m is the particle's mass in kg, and L is the box length in metres. For an electron, m = 9.109 × 10−31 kg. One electron volt is 1.602 × 10−19 J.
What the formula tells you
- E ∝ n². Levels are not evenly spaced; they grow as 1 : 4 : 9 : 16.
- E ∝ 1/L². Making the box half as long raises every level by a factor of 4. Smaller boxes mean larger energies.
- E ∝ 1/m. A heavier particle has smaller spacing between levels. That is why quantisation is invisible for everyday objects.
- Zero-point energy. The lowest level is n = 1, not n = 0. The particle can never have zero energy. With n = 0 the wavefunction would vanish everywhere, which would mean there is no particle.
- Nodes. ψn has (n − 1) nodes inside the box (points where ψ = 0, not counting the walls).
- Spacing. En+1 − En = (2n + 1) h² / (8mL²). It increases with n.
Step 1: Compute the constant h² / (8mL²). h² = (6.626 × 10−34)² = 4.390 × 10−67. 8mL² = 8 × 9.109 × 10−31 × (1.0 × 10−9)² = 7.287 × 10−48.
Step 2: E1 = 4.390 × 10−67 / 7.287 × 10−48 = 6.02 × 10−20 J = 0.376 eV.
Step 3: E2 = 4 × 0.376 = 1.504 eV. E3 = 9 × 0.376 = 3.384 eV.
Answer: E1 = 6.02 × 10−20 J (0.376 eV), E2 = 1.50 eV, E3 = 3.38 eV.
Step 1: ΔE = E2 − E1 = 3 × E1 = 3 × 6.02 × 10−20 = 1.807 × 10−19 J.
Step 2: λ = hc / ΔE = (6.626 × 10−34 × 2.998 × 108) / 1.807 × 10−19 = 1.099 × 10−6 m.
Answer: λ ≈ 1099 nm, which is in the infrared region.
Step 1: E ∝ 1/L². L is halved, so E1 is 4 times larger than in the 1.0 nm box.
Step 2: E1 = 4 × 0.376 = 1.504 eV.
Answer: 1.50 eV. Notice that this equals E2 of the 1.0 nm box, as expected.
Step 1: Each level holds two electrons, so four electrons fill n = 1 and n = 2. The highest occupied level is n = 2, and the lowest empty level is n = 3.
Step 2: ΔE = E3 − E2 = (9 − 4) × E1 = 5 E1.
Step 3: E1 at 0.60 nm = 0.376 / (0.60)² = 0.376 / 0.36 = 1.044 eV. So ΔE = 5 × 1.044 = 5.22 eV.
Step 4: λ = hc / ΔE = 1239.8 eV nm / 5.22 eV = 237 nm.
Answer: a predicted absorption near 237 nm in the ultraviolet. The model is crude. It shows the trend that a longer chain gives a smaller gap and a longer wavelength, but it should not be used for exact values.
Step 1, n = 1, a = L/4: P = 0.25 − sin(π/2) / (2π) = 0.25 − 1/6.283 = 0.25 − 0.1592 = 0.0908.
Step 2, n = 2, a = L/4: P = 0.25 − sin(π) / (4π) = 0.25 − 0 = 0.2500.
Answer: 9.1 % for n = 1 and 25 % for n = 2. The n = 1 particle is mostly near the middle; the classical expectation would be 25 % for any level.
Common mistakes
- Using nm for L directly. Convert to metres (1 nm = 10−9 m) before squaring.
- Using the proton or atomic mass in g. Use kg for m with h in J s.
- Setting n = 0. It gives no particle. The lowest level is n = 1.
- Forgetting to square n. E ∝ n², not n.
- Counting the walls as nodes. The number of nodes inside is n − 1.
- Using the electron mass for a heavier particle. Replace m with the correct mass.
- Finding the HOMO wrongly. With N electrons and two per level, the HOMO is n = N/2 (for an even N), and the LUMO is n = N/2 + 1.
Exam relevance
| Question type | Key step |
|---|---|
| Energy of level n | En = n²h²/8mL² with SI units |
| Transition wavelength | ΔE from the n² difference, then λ = hc/ΔE |
| Scaling with L or m | E ∝ 1/(mL²) |
| Normalisation or expectation values | Use ψ = √(2/L) sin(nπx/L) |
| Conjugated systems | Fill electrons two per level, then use HOMO to LUMO |
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