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.

VKC Notes · Civil Services Chemistry · Paper I · Unit 1 Atomic structure

1.3 Particle in a One-Dimensional Box

★★★ Must-do. Asked in every Paper I from 2021 to 2025: 6 parts, 75 marks, plus one indirect use (F-centres, 2022).

Syllabus line: “particle in one-dimensional box”, read together with “Heisenberg's uncertainty principle” and “interpretation of wave function” from the same unit.

Short notes for revision

Download PDF PLAN

PDF edition, print-ready. It opens with the VKC plan, together with every locked answer, note and test on the site.

CSE 2025 Q2(a) = asked there; tap to open the solved answerSN = line kept in the short notes

1.3.1 The model

A particle of mass m moves along a line between two rigid walls at x = 0 and x = L. Inside, V = 0. Outside, V = ∞, so the particle can never be there.

It is the simplest system where the Schrödinger equation produces quantisation. Everything you need for the harder systems (boundary conditions, normalisation, nodes, zero-point energy) appears here in its cleanest form.
It is also a working model of real chemistry: π electrons in conjugated chains, electrons trapped in F-centres, quantum dots, and the translational energy of gas molecules (1.3.8).

Fig. 1.3.1 The infinite square well.

Draw it in 60 s: two vertical walls and a floor → hatch the outside of the walls → write V = ∞ outside, V = 0 inside → mark 0 and L on the x-axis.

Boundary conditions

ψ must be continuous. Outside the box ψ = 0, so at the walls ψ(0) = 0 and ψ(L) = 0.SN These two conditions are the whole reason the energy comes out quantised.

1.3.2 Solving the Schrödinger equation ★★★ derivation

Equation inside the box. With V = 0 the energy is all kinetic:

(1.3.1)
General solution. Any function whose second derivative is −k² times itself:

(1.3.2)
Left wall. ψ(0) = A·0 + B·1 = 0, so B = 0 and ψ = A sin kx.
Right wall. ψ(L) = A sin kL = 0. A cannot be zero (no particle), so sin kL = 0:

(1.3.3)

n = 0 is rejected: it gives ψ = 0 everywhere. Negative n only flips the sign of ψ, which is the same state.SN
Energy. Put k = nπ/L into k² = 2mE/ħ² and use ħ = h/2π:

(1.3.4)
Normalise. The particle is certainly somewhere between 0 and L:

(1.3.5)

Result

Remember both lines as one result.SN

Marks gainer: two-line cross-check

A standing wave fits the box only when L holds a whole number of half waves: L = nλ/2, so λ = 2L/n. Then p = h/λ = nh/2L and E = p²/2m = n²h²/8mL². Same answer. Add it after the derivation as a check. Do not use it as the derivation when the question says “solve the Schrödinger equation”.

Asked: CSE 2023 Q2(b) 10m why the box is quantised and a free particle is not. The set-up is also the first half of CSE 2022 Q2(b) 20m and CSE 2025 Q2(a) 15m.

1.3.3 Reading ψ and ψ²

Each higher state adds one half wave. ψ_n has n − 1 nodes (walls not counted), at x = kL/n, k = 1 … n − 1.SN
Shorter wavelength means sharper curvature, and curvature is kinetic energy. That is why E rises with n.
ψ² is the probability per unit length (Born). Its maxima sit at x = (2k − 1)L/2n: at L/2 for n = 1, at L/4 and 3L/4 for n = 2.
Symmetry about the centre: odd n (1, 3, …) are symmetric about L/2; even n (2, 4, …) are antisymmetric. This one fact gives ⟨x⟩ = L/2 and orthogonality almost for free (1.3.6, 1.3.7).SN

Fig. 1.3.2 ψ_n (left) and ψ_n² (right) for n = 1, 2, 3, each drawn on its own energy level. Open red circles are the nodes.

Draw it in 60 s: two boxes side by side → three baselines at heights 1, 4, 9 → left: half, one, one-and-a-half sine loops → right: the same loops squared, all above the line → dot the nodes → label n = 1, 2, 3.

Subtle point · node at the centre

For n = 2 the probability of finding the particle at L/2 is zero, yet it is found on both sides. There is no contradiction: the particle is a standing wave, not a ball travelling through L/2. Examiners like this as a short “explain” part.

Subtle point · box from −L/2 to +L/2

If the origin is put at the centre, ψ becomes √(2/L) cos(nπx/L) for odd n and √(2/L) sin(nπx/L) for even n. The energies do not change. Moving the origin changes how ψ looks, never E.

1.3.4 The energy ladder ★★★

E ∝ n²: E₁, 4E₁, 9E₁, 16E₁ …SN
Gap between neighbours:
(1.3.6)
The gaps widen as n rises, but the gap relative to the energy, (2n + 1)/n², falls to zero. At high n the ladder looks continuous: the correspondence principle.SN
E ∝ 1/m and E ∝ 1/L². Light particle + small box = big gaps = visible quantum effects.SN
- Electron in a 1 nm box: E₁ = 6.0 × 10⁻²⁰ J ≈ 0.38 eV. Chemically significant.
- O₂ molecule in a 1 cm box: E₁ ≈ 1 × 10⁻³⁸ J, against kT ≈ 4 × 10⁻²¹ J at 298 K. The levels are so close that translational energy is treated as continuous in kinetic theory.

Fig. 1.3.3 The energy ladder. Levels go as n², gaps as 2n + 1, and the lowest level is not zero.

Draw it in 60 s: a box → four lines at heights 1, 4, 9, 16 → label E₁, 4E₁, 9E₁, 16E₁ → double arrows for the gaps 3E₁, 5E₁, 7E₁ → write “E = 0 not allowed” at the floor.

Zero-point energy

The lowest energy is E₁ = h²/8mL² ≠ 0.SN Two reasons, write both:

Uncertainty. The particle is confined, so Δx ≤ L. Then Δp cannot be zero, so the kinetic energy cannot be zero.
Curvature. ψ must be zero at both walls but not zero inside, so it has to bend. Bending means kinetic energy.

Free particle versus particle in a box

The Schrödinger equation is the same in both cases. The difference is the boundary.

Free particle: no walls, so every k is allowed. E = ħ²k²/2m takes any value from zero up: a continuum. ψ = e^±ikx, a particle with definite momentum ±ħk.
Box: the walls admit only k = nπ/L, so only the points E₁, 4E₁, 9E₁ … of the same parabola survive.

Confinement causes quantisation. Let L → ∞ in eq. (1.3.6) and the gaps vanish: the box becomes a free particle.SN

Fig. 1.3.4 E against k. Free particle: the whole dashed parabola. Box: only the red points.

Draw it in 60 s: axes E and k → dashed parabola from the origin → mark π/L, 2π/L, 3π/L → dots on the parabola above them → read off E₁, 4E₁, 9E₁.

Solved PYQ CSE 2023 Q1(a)(i)

Find the wavelength of light emitted when a 1.0 × 10⁻²⁷ g particle in a one-dimensional box of length 3 Å goes from n = 2 to n = 1.

Convert first: m = 1.0 × 10⁻³⁰ kg, L = 3.0 × 10⁻¹⁰ m.
ΔE = (2² − 1²) h²/8mL² = 3h²/8mL² = 1.83 × 10⁻¹⁸ J
λ = hc/ΔE = 8mL²c/3h = 1.09 × 10⁻⁷ m = 109 nm, far ultraviolet.

Closing line that earns the mark: the particle is about as heavy as an electron and the box is atom-sized, so the line falls near the Lyman region of hydrogen.

1.3.5 Probability in a region ★★★ most asked · 35 marks

By the Born interpretation, ψ²dx is the probability of finding the particle between x and x + dx. Over a stretch x₁ to x₂:

(1.3.7)
Use sin²θ = ½(1 − cos 2θ) and integrate:

SN(1.3.8)

Read eq. (1.3.8) as classical share − a wave term. The wave term shrinks as 1/n.

The two forms UPSC has used

(1.3.9)

(1.3.10)

Fig. 1.3.5 Shaded area under ψ² from 0 to L/4. The green dashed line is the uniform classical density.

Draw it in 60 s: three small boxes side by side → draw ψ² with one, two, three humps → a dashed vertical at L/4 → shade the strip from 0 to L/4 → write P under each.

Marks gainer: two checks to write under the answer

Half box: P(0 → L/2) = ½ for every n, by symmetry. Put x₂ = L/2 in (1.3.8) and the sine term is zero.
Correspondence: as n → ∞, P → (x₂ − x₁)/L, the classical value (0.25 for a quarter). The sequence 0.091 → 0.250 → 0.303 swinging about 0.25 shows it.SN

Another standard number: n = 1, middle third (L/3 to 2L/3): P = 1/3 + √3/2π ≈ 0.61. The ground state crowds the centre.
Probability at a point is zero. A probability always needs a range dx. If a question asks “probability at x”, answer with ψ² dx or with a small range.

Asked: CSE 2022 Q2(b) 20m first 1/a of the width. CSE 2025 Q2(a) 15m 0 ≤ x ≤ a/4 for n = 1, 2, 3. Full strategy in the blog post Particle in a box: probability in part of the box.

1.3.6 ⟨x⟩, ⟨p⟩ and the uncertainty principle ★★

Expectation value = average of many measurements: ⟨A⟩ = ∫ψ* Â ψ dx for normalised ψ.

Position

(1.3.11)

Short route: ψ² is symmetric about L/2, so its balance point is L/2. Write the integral, then this sentence.SN
⟨x⟩ is a mean, not a location. For n = 2 the mean is L/2, yet ψ² is zero there.

(1.3.12)

n = 1: Δx = 0.181 L. As n → ∞, Δx → L/√12 = 0.289 L, the spread of a classical particle bouncing at constant speed.

Momentum

⟨p⟩ = 0: moving left and right are equally likely.
⟨p²⟩ = 2mE = n²h²/4L², so Δp = nh/2L.
ψ_n is not an eigenfunction of p̂, because sin kx = (e^ikx − e^−ikx)/2i. A measurement gives +nh/2L or −nh/2L, each half the time. ψ_n is an eigenfunction of Ĥ and of p̂².SN

Heisenberg check

(1.3.13)

Ground state: Δx · Δp = 0.181L × h/2L = 0.090h, which is 1.14 × ħ/2. The principle is obeyed, and the ground state sits close to the minimum. The product grows with n.SN

Statement and origin, for the “explain” part

Position and momentum cannot both be known exactly at the same time. The origin is the wave nature of matter: a wave with one exact wavelength (exact p = h/λ) spreads over all space; squeezing it into a small Δx needs many wavelengths added together, so p spreads. Link it to the zero-point energy of the box in one line.

Asked: CSE 2021 Q1(a) 10m ⟨x⟩ in the ground state. CSE 2023 Q1(a)(ii) and CSE 2024 Q2(a) explain the uncertainty principle.

1.3.7 Orthogonality ★★

Two states are orthogonal when ∫ψ_mψ_n dx = 0 (m ≠ n). Show it for ψ₁ and ψ₂:

(1.3.14)
Use sin A sin B = ½[cos(A − B) − cos(A + B)]:

(1.3.15)

With normalisation the full statement is ∫ψ_mψ_n dx = δ_mn: the box states are orthonormal.SN

Fig. 1.3.6 ψ₁ψ₂ is positive in the left half and negative in the right half, with equal areas.

Draw it in 60 s: box with a dashed centre line → dashed ψ₁ (one hump) and dotted ψ₂ (up then down) → solid product, positive left, negative right → shade both lobes and mark + and −.

Marks gainer: the one-line symmetry proof

ψ₁ is even about L/2 and ψ₂ is odd, so their product is odd and its integral over the box is zero. Write it after the integral, with Fig. 1.3.6. General rule: eigenfunctions of the same Hermitian operator with different eigenvalues are orthogonal.

Asked: CSE 2024 Q2(a) 10m

1.3.8 Applications ★★ use as the conclusion

Conjugated polyenes (free-electron model)

Treat the π electrons as particles in a box the length of the conjugated chain. With N π electrons, two per level, the HOMO is n = N/2 and the LUMO is n = N/2 + 1:

(1.3.16)

Fig. 1.3.7 Butadiene as a box: four π electrons fill n = 1 and 2; light lifts one to n = 3.

Draw it in 60 s: zig-zag of four carbons → a box under it with L marked → a ladder of four levels spaced as n² → two pairs of arrows in the bottom two → red arrow HOMO → LUMO.

Worked example (butadiene)

N = 4. Take L as the three bonds plus half a C–C bond beyond each end: L ≈ 135 + 154 + 135 + 2 × 77 = 578 pm.
ΔE = 5h²/8m_eL² = 9.0 × 10⁻¹⁹ J
λ = hc/ΔE ≈ 220 nm. Observed λ_max ≈ 217 nm.

Longer chain → larger L → smaller ΔE → red shift. That is why β-carotene, with 11 conjugated C=C, absorbs in the visible and looks orange.SN
Be honest about the limit: for long chains the model gets the trend but not the number. Atkins works β-carotene and gets about 1240 nm against 497 nm observed, because real bonds alternate in length.

Other places the box appears

F-centres: an electron trapped in an anion vacancy behaves as a particle in a box the size of the site. The colour depends on the host crystal, not on how the electron got there. CSE 2022 Q1(c)
Quantum dots: in a semiconductor nanocrystal the size sets the gap. Smaller dot → larger gap → bluer light.
Gas molecules: levels in a laboratory flask are so close (1.3.4) that translation is treated as continuous.

1.3.9 Beyond one dimension ★ add-on

The CSE syllabus names only the one-dimensional box. Keep this short; it is useful when a question asks about degeneracy.

(1.3.17)

(n_x, n_y, n_z)	E in h²/8mL²	Degeneracy
(1,1,1)	3	1
(2,1,1), (1,2,1), (1,1,2)	6	3
(2,2,1) and permutations	9	3
(3,1,1) and permutations	11	3
(2,2,2)	12	1

Degeneracy comes from symmetry. Stretch one side of the cube and the three-fold levels split.

1.3.10 How UPSC asks this

Question type	Asked	What earns the marks	Time
Probability in a region	2022 Q2(b) 20m 2025 Q2(a) 15m	Eq. (1.3.8), values to 3 decimals, Fig. 1.3.5, the two checks	14 / 11 min
Box vs free particle, why quantised	2023 Q2(b) 10m	Same equation, boundary conditions, steps 3–5, Fig. 1.3.4	7 min
Wavelength of a transition	2023 Q1(a)(i)	Units in SI first, ΔE, region of the spectrum	≈ 4 min
Expectation value	2021 Q1(a) 10m	Integral steps, symmetry sentence, mean ≠ location	7 min
Orthogonality + uncertainty	2024 Q2(a) 10m	Product-to-sum integral, Fig. 1.3.6, Heisenberg statement	7 min

Time at 0.72 min per mark (180 min for 250 marks).

1.3.11 Traps

✗ Mass left in grams: 10⁻²⁷ g is 10⁻³⁰ kg. One wrong power of ten ruins the whole numerical.

✗ Writing E_n = n²h²/8mL² without “n = 1, 2, 3 …”. n = 0 is not allowed.

✗ Using P = (x₂ − x₁)/L for n = 1, 2, 3. That is only the large-n limit.

1.3.12 Where to read

Any one book is enough.

Book 1: Atkins, Physical Chemistry, 9th ed., §8.1 “A particle in a box” (p. 289), §8.2 (p. 293) and Impact I8.1 on quantum dots (p. 295). In the 11th ed. the same material is Topic 7D, sections 7D.2–7D.3, with the β-carotene example as Example 7D.1. Best for the physical picture and applications.
Book 2: R.K. Prasad, Quantum Chemistry, Ch. 4 “Translational Motion of a Particle”: §4.1 free particle (p. 81), §4.2.1 one-dimensional box (p. 83), §4.2.2 utility of the model (p. 95), §4.2.3–4.2.5 three-dimensional box and degeneracy (pp. 98–101). Best for step-by-step derivations.
Book 3: K.L. Kapoor, A Textbook of Physical Chemistry, Vol. 4, Ch. 1, §1.4 “Quantization of Translational Energy” (p. 37), after §1.3 on the Schrödinger wave theory (p. 30).

Online: LibreTexts: Particle in a 1-dimensional box · MIT OpenCourseWare 5.61 Physical Chemistry (lecture notes).

Solved PYQs, short notes and books: vkcscience.comTelegram: t.me/upsc_chemistry_vkc