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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper I covers physical and inorganic chemistry.

All short notes
VKC Short Notes · Civil Services Chemistry · Paper I · Unit 1

1.3 Particle in a Box: Short Notes

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Result from 1.3.1–1.3.2

  • V = 0 inside, ∞ outside → ψ(0) = ψ(L) = 0. These boundary conditions cause quantisation.
  • n = 0 not allowed (ψ = 0 everywhere). Negative n = same state.
  • Check: L = nλ/2 → p = nh/2L → E = p²/2m. Same result.

ψ and ψ² 1.3.3

  • n − 1 nodes; one extra half wave per level.
  • Odd n symmetric, even n antisymmetric about L/2.
  • ψ² = probability per unit length (Born).
ψn0Lψn20Ln = 1E1n = 24E1n = 39E1node (n − 1 of them)levels drawn at E ∝ n2
Fig. 1.3.2

Energy 1.3.4

  • E ∝ n²: E₁, 4E₁, 9E₁ …
  • ΔEn→n+1 = (2n + 1)h²/8mL²; ΔE/E → 0 at high n (correspondence).
  • E ∝ 1/m, 1/L²: light particle, small box → big gaps.
  • ZPE = h²/8mL² ≠ 0: (i) Δx ≤ L so Δp ≠ 0; (ii) ψ must curve to vanish at both walls.
  • Free vs box: same equation. No walls → any k → continuous E. Walls → k = nπ/L only. Confinement causes quantisation.
n = 1E1n = 24E1n = 39E1n = 416E13E15E17E1E = 0 not allowed (n ≠ 0)gaps (2n + 1)E1widen with nE1 = h2/8mL2 ≠ 0zero-point energybigger L or m:levels crowd
Fig. 1.3.3

Probability in a region ★★★ 35 marks 1.3.5

(1.3.8)
  • 0 to L/4: P = ¼ − sin(nπ/2)/2nπ → 0.091, 0.250, 0.303 for n = 1, 2, 3.
  • Checks to write: P(0 → L/2) = ½ for all n; P → (x₂ − x₁)/L as n → ∞.
  • n = 1, middle third: P ≈ 0.61.
  • Probability at a single point is zero; always use a range.
n = 1L/40LP = 0.091n = 2L/40LP = 0.250n = 3L/40LP = 0.303classical (uniform) density → P = 0.25
Fig. 1.3.5

⟨x⟩, ⟨p⟩, Δx Δp 1.3.6

  • ⟨x⟩ = L/2 for all n (symmetry). A mean, not a location.
  • ⟨x²⟩ = L²(1/3 − 1/2n²π²); Δx(n = 1) = 0.181L → L/√12 classically.
  • ⟨p⟩ = 0; Δp = nh/2L.
  • n = 1: Δx Δp = 0.090h = 1.14 × ħ/2. Obeys Δx Δp ≥ h/4π.
  • ψn is not an eigenfunction of p̂: gives ±nh/2L, half the time each.

Orthogonality 1.3.7

  • ∫ψmψn dx = δmn (orthonormal).
  • Proof: sin A sin B = ½[cos(A − B) − cos(A + B)] → sin π, sin 3π = 0.
  • One line: ψ₁ even × ψ₂ odd = odd function → integral 0.

Applications 1.3.8

  • Polyene: ΔE = (N + 1)h²/8mL². Butadiene (L ≈ 578 pm) → ≈ 220 nm (obs. 217).
  • Longer chain → red shift (β-carotene coloured). Trend right, numbers rough for long chains.
  • Also: F-centres, quantum dots, gas translation.

Traps 1.3.11

✗ 10−27 g = 10−30 kg. Convert before you calculate.
✗ P = (x₂ − x₁)/L is the large-n limit only.

PYQs: 2021 Q1(a) 2022 Q2(b) 2023 Q1(a) 2023 Q2(b) 2024 Q2(a) 2025 Q2(a)

Read: Atkins 9th ed. §8.1 (11th ed. Topic 7D) · R.K. Prasad Ch. 4 §4.2 · K.L. Kapoor Vol. 4 §1.4. Any one.

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