GATE Quantum Chemistry — Particle in a Box Essentials
The particle in a box is the one exactly solvable quantum model that GATE Chemistry returns to again and again, because it can be examined in so many different ways: energy gaps, node counting, degeneracy in three dimensions, expectation values, probability integrals and the free-electron model of conjugated dyes. All of it comes from a single energy expression and a single wavefunction, so this is very high return for the time invested. This guide sets out both, then works through the standard question types with the arithmetic actually done.
The model and its two results
A particle of mass m is confined to a one-dimensional region 0 ≤ x ≤ L with V = 0 inside and V = ∞ outside. The infinite walls force the wavefunction to vanish at both ends, and those two boundary conditions are what quantise the energy.
ψn(x) = √(2/L) · sin(nπx / L)
Read four facts straight off these: energy scales as n², it scales as 1/L² (a smaller box means larger gaps), it scales as 1/m (a proton in the same box has far closer levels than an electron), and the lowest allowed energy is not zero. That non-zero E₁ = h²/(8mL²) is the zero-point energy, and it is a direct consequence of the uncertainty principle — a particle pinned to a finite length cannot have exactly zero momentum.
The wavefunction ψn has n − 1 nodes strictly inside the box. The walls at x = 0 and x = L are not counted as nodes. So ψ₁ has none, ψ₂ has one (at L/2), ψ₃ has two (at L/3 and 2L/3). Node counting is a favourite one-mark question.
Worked example 1 — electron in a 1 nm box
Take m = 9.109 × 10⁻³¹ kg, L = 1.00 × 10⁻⁹ m, h = 6.626 × 10⁻³⁴ J s.
h² = 4.390 × 10⁻⁶⁷ J² s²
8mL² = 8 × 9.109 × 10⁻³¹ × (1.00 × 10⁻⁹)² = 7.287 × 10⁻⁴⁸ kg m²
E₁ = 4.390 × 10⁻⁶⁷ / 7.287 × 10⁻⁴⁸ = 6.02 × 10⁻²⁰ J
In electronvolts: 6.02 × 10⁻²⁰ / 1.602 × 10⁻¹⁹ = 0.376 eV.
The n = 1 → n = 2 gap is (2² − 1²)E₁ = 3E₁ = 1.81 × 10⁻¹⁹ J. The photon that supplies it has λ = hc/ΔE = (6.626 × 10⁻³⁴ × 2.998 × 10⁸) / 1.81 × 10⁻¹⁹ = 1.10 × 10⁻⁶ m, i.e. ≈ 1100 nm, in the near infrared.
Worked example 2 — probability in a region
Probability is the integral of |ψ|², never of ψ. For the ground state the indefinite integral is standard and worth memorising:
Find the probability of locating a ground-state (n = 1) particle in the middle third of the box, L/3 to 2L/3.
At x = 2L/3: 2/3 − sin(4π/3)/(2π) = 0.6667 − (−0.8660)/6.2832 = 0.6667 + 0.1378 = 0.8045
At x = L/3: 1/3 − sin(2π/3)/(2π) = 0.3333 − 0.8660/6.2832 = 0.3333 − 0.1378 = 0.1955
P = 0.8045 − 0.1955 = 0.609, i.e. about 60.9%.
Sanity check: classically the particle would be equally likely anywhere, giving exactly 1/3. The quantum answer is much larger because ψ₁² peaks at the centre — a good sign the working is right.
Expectation values you should be able to quote
By symmetry ⟨x⟩ = L/2 for every state, and ⟨p⟩ = 0 for every state (equal amplitudes of momentum +p and −p). The one that needs real integration is
From these, Δx = √(⟨x²⟩ − ⟨x⟩²) and Δp = √(⟨p²⟩ − ⟨p⟩²) = nh/(2L). For n = 1 the product ΔxΔp works out to about 0.568 ħ, comfortably above the ħ/2 floor, as it must be.
Three dimensions and degeneracy
Degeneracy appears whenever different triplets give the same sum of squares. This is almost always asked as "what is the degeneracy of the k-th level of a cubic box".
| nx²+ny²+nz² | Quantum numbers | Degeneracy |
|---|---|---|
| 3 | (1,1,1) | 1 |
| 6 | (2,1,1) and permutations | 3 |
| 9 | (2,2,1) and permutations | 3 |
| 11 | (3,1,1) and permutations | 3 |
| 12 | (2,2,2) | 1 |
| 14 | (3,2,1) and permutations | 6 |
Note that the level at 9 is accidentally degenerate in a different sense from the level at 6 — but for a cube the counting rule is simply "how many ordered triplets of positive integers give this sum of squares". If the box is rectangular with unequal sides, the permutation degeneracy is destroyed.
Worked example 3 — the free-electron model of butadiene
Treat the π electrons of a linear conjugated polyene as particles in a one-dimensional box spanning the conjugated chain. With N π electrons, two per level, the HOMO is n = N/2 and the LUMO is n = N/2 + 1, so
1,3-butadiene has 4 π electrons; take the effective box length as L = 578 pm.
8mL² = 8 × 9.109 × 10⁻³¹ × (5.78 × 10⁻¹⁰)² = 8 × 9.109 × 10⁻³¹ × 3.341 × 10⁻¹⁹
= 2.435 × 10⁻⁴⁸
h²/(8mL²) = 4.390 × 10⁻⁶⁷ / 2.435 × 10⁻⁴⁸ = 1.803 × 10⁻¹⁹ J
HOMO n = 2, LUMO n = 3, so ΔE = (3² − 2²) × 1.803 × 10⁻¹⁹ = 5 × 1.803 × 10⁻¹⁹ = 9.02 × 10⁻¹⁹ J
λ = hc/ΔE = 1.986 × 10⁻²⁵ / 9.02 × 10⁻¹⁹ = 2.20 × 10⁻⁷ m = 220 nm, against an observed λmax near 217 nm. A crude model landing that close is why it survives in the syllabus.
The qualitative consequence matters more than the number: as the chain lengthens, L grows, ΔE falls, and λmax shifts to longer wavelength. That is exactly why extended conjugation produces coloured compounds.
Common mistakes in the exam hall
- Using ħ where the formula wants h. En = n²h²/(8mL²) uses h. The equivalent form with ħ is n²π²ħ²/(2mL²). Mixing the two costs a factor of about 39.5.
- Counting the walls as nodes. ψn has n − 1 interior nodes; the boundary zeros are required by the boundary condition, not nodes of the oscillation.
- Forgetting to square the wavefunction before integrating a probability. ∫ψ dx has no physical meaning here.
- Reporting a gap as En rather than En − Em. The 1 → 2 gap is 3E₁, not 4E₁.
- Assuming the 3D degeneracy pattern survives a rectangular box. Different side lengths remove the permutation degeneracy entirely.
- Unit slips — nanometres left unconverted into metres is the single most common arithmetic loss in this topic.
Quick reference
| Quantity | Result | Comment |
|---|---|---|
| Energy, 1D | n²h²/(8mL²) | n = 1, 2, 3 …; no n = 0 |
| Normalised ψ | √(2/L) sin(nπx/L) | Normalisation constant √(2/L) |
| Interior nodes | n − 1 | Walls excluded |
| ⟨x⟩ | L/2 | All n, by symmetry |
| ⟨p⟩ | 0 | All n |
| Δp | nh/(2L) | Since ⟨p²⟩ = 2mEn |
| 3D cube | (h²/8mL²)(nx²+ny²+nz²) | Degeneracy from permutations |
| Polyene ΔE | (N+1)h²/(8mL²) | N = number of π electrons |
Check the arithmetic, not the concept. Every number on this page is an exponent-heavy multiplication or division of the kind that is easy to fumble under time pressure. Run your h², 8mL² and hc/ΔE steps through the Scientific Calculator and confirm the powers of ten before you commit an answer.
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