PYQ explainedUnit 1 · Atomic structure13 October 2026

Particle in a box: probability in part of the box

By Vimal Swarnkar (VKC) · UPSC Chemistry · Civil Services, Combined Geo-Scientist and IFoS

It looks like a long question, but it is really one integral, one identity and three numbers. In this post I will show you how often UPSC asks it, which book it comes from, how I plan the answer, and how the answer should look in your booklet.

The question as set in UPSC Civil Services (Main) 2025, Chemistry Paper I, Q2(a), 15 marks

Find the probability of existence of a particle in a one-dimensional box of length ‘a’ in the region 0 ≤ x ≤ a/4 for the states n = 1, 2 and 3.

Short answer

With ψₙ = √(2/a) sin(nπx/a), the probability in 0 ≤ x ≤ a/4 is Pₙ = 1/4 − (1/2nπ) sin(nπ/2). So P₁ = 0.091, P₂ = 0.250 and P₃ = 0.303. As n grows, Pₙ approaches 1/4, the classical value (correspondence principle).

Full topic notes: Particle in a one-dimensional box · two-page short notes

A word from me

I have taught this topic to many batches, and I see the same thing every year. Students read “∫ψ² dx” in the question and move straight to the next one. I understand that fear. But let me be honest with you: this is one of the safest scoring questions in the whole paper. The steps never change. Once you have written it two or three times, you will finish it well within the time.

So my advice is simple. Do not collect material on this topic. Pick one book, solve its worked example, then close everything and write the answer below once by hand. Wherever you get stuck, that is the only step you need to revise. And if something still does not make sense, write to me on Telegram or at [email protected]. I would rather you ask than skip 15 marks.

How often UPSC asks the particle in a box

From 2021 to 2025, the particle in a box appeared in 13 of 15 papers across Civil Services, Combined Geo-Scientist and IFoS. It came 3 times directly, as this exact probability question: Civil Services 2022 (20 marks), Civil Services 2025 (15 marks) and Geo-Scientist 2021 (10 marks). The other 10 times it came indirectly, as energy levels, degeneracy, the wavelength of emitted light or a property of the wavefunction.

The topic is not new either. IAS 1996 asked for the length of a box from the ground-state energy of an electron, for 20 marks.

Syllabus and books

In Civil Services and IFoS it is part of Paper I, Unit 1, Atomic structure. In the Geo-Scientist exam it is in the quantum chemistry part of Mains Paper II. The Civil Services syllabus reads:

“Heisenberg's uncertainty principle, Schrodinger wave equation (time independent); interpretation of wave function, particle in one-dimensional box, quantum numbers, hydrogen atom wave functions; shapes of s, p and d orbitals.”

Two phrases in it decide this question: interpretation of wave function and particle in one-dimensional box.

You do not need all of these books. Any one of them is enough for this question. I have listed three so that you can use whichever you already have.

Book 1: Atkins, Physical Chemistry (9th ed.), Chapter 8, Section 8.1 “A particle in a box”. Example 8.1 derives the general formula used here, and Self-test 8.4 is the Geo-Scientist 2021 question word for word. If you have this book, this is the one I would use.
Book 2: Puri, Sharma and Pathania, Principles of Physical Chemistry, the chapter on quantum mechanics. The standard Indian text, with the same derivation.
Book 3: McQuarrie, Quantum Chemistry, Chapter 3, “The Schrödinger equation and a particle in a box”. Clear if you want more practice problems.

Keyword map: how to build the answer

Read the tree from left to right. Every leaf is a point to write, in that order. The highlighted words are the ones the examiner looks for.

Probability of the particle in 0 ≤ x ≤ a/4, n = 1, 2, 3

Introduction

Given information

Box of length a
Region 0 to a/4
States n = 1, 2, 3

Assumptions

V = 0 inside, V = ∞ outside
Point particle of mass m
Stationary state: time-independent equation
ψ = 0 at x = 0 and x = a

Key law

Schrödinger equation → ψₙ, Eₙ
Normalisation → A = √(2/a)
Born interpretation: ψ² dx = probability

Body

Calculation

P = ∫ ψ² dx from 0 to a/4
sin²θ = ½(1 − cos 2θ)
General result Pₙ = 1/4 − sin(nπ/2)/2nπ

Values

Table of n, sin(nπ/2), Pₙ
0.091, 0.250, 0.303

Check

Whole box gives P = 1

Sketch

ψ² for n = 1, 2, 3
Shade the region 0 to a/4

Conclusion

Reading the result

n = 1: piled at the centre
n = 2: node at a/2
n = 3: first lobe ends at a/3

Principle

Correspondence principle: Pₙ → 1/4

Application

π electrons in a polyene
Longer chain, longer λ, colour of carotene

Final answer

P₁, P₂, P₃, double underlined

Diagrams that prove it

Figure 1. ψ² for n = 1, 2 and 3 in a one-dimensional box. The shaded area from 0 to a/4 is the probability asked for. For n = 1 most of the area lies in the middle, so the first quarter holds only 9 %. For n = 3 the first hump ends at a/3 and lies mostly inside the region, so the share rises to 30 %.

Figure 2. Pₙ for the first quarter of the box against n. Even values of n give exactly 1/4. Odd values swing above and below and close in on the classical value 1/4. This is the correspondence principle, drawn.

What gains marks

Numbered assumptions and a labelled sketch of the box, before any equation.
The general result Pₙ before any number. One formula answers all three states and shows method.
Equation numbers, so each step says what was put into what.
A check: the whole box gives P = 1.
A table of n, sin(nπ/2) and Pₙ. It is quick to mark.
A shaded ψ² sketch, and one line on why each value is small or large.
The words correspondence principle, and one application.

Match the length to the marks

I always tell my students to let the marks decide the length. If this comes for 5 marks, I would write ψₙ, the integral, the general result and the three values, about half a page. For 10 marks, add the numbered assumptions, the box sketch and the ψ² sketch. For 15 or 20 marks, write all three pages below, with the correspondence principle and the application.

Model answer, as written in the booklet

This is the full 15-mark answer, in the order of the tree above. Page 1 sets up the problem, page 2 does the calculation and page 3 explains it.

Page 1 of 3 · IntroductionGiven information, assumptions, key law

Q.)Find the probability of existence of a particle in a one-dimensional box of length ‘a’ in the region 0 ≤ x ≤ a/4 for the states n = 1, 2 and 3. [CSE 2025], 15M

Ans.Given: length of box = a, region 0 ≤ x ≤ a/4, states n = 1, 2, 3.

For a particle in a one-dimensional box, assume that

i) the box has length a, walls at x = 0 and x = a.
ii) potential energy (V) of the particle inside the box = 0, and V = ∞ outside.
iii) particle of mass m, size ignored.
iv) particle is in a stationary state, so the time-independent equation applies.

The Schrödinger equation for the particle inside the box is

- \frac{ℏ^{2}}{2 m} \frac{d^{2} ψ}{d x^{2}} = E ψ (V = 0)

or,

\frac{d^{2} ψ}{d x^{2}} + \frac{2 m E}{ℏ^{2}} ψ = 0

Boundary conditions: ψ(0) = 0 and ψ(a) = 0, as the particle cannot be outside the box.

We get the following solutions (physically acceptable):

ψ_{n} = A \sin \frac{n π x}{a}, n = 1, 2, 3, \dots

and the eigenvalues E as $E_{n} = \frac{n^{2} h^{2}}{8 m a^{2}}$

where m = mass of the particle, and all other symbols have their usual meaning.

Here n starts from 1, not 0, because n = 0 gives ψ = 0 everywhere: no physical significance.

Page 2 of 3 · BodyNormalisation, key law, calculation, check

Normalisation: the particle is certainly somewhere in the box, so

\int_{0}^{a} A^{2} \sin^{2} \frac{n π x}{a} d x = A^{2} \frac{a}{2} = 1 \Rightarrow A = \sqrt{\frac{2}{a}}

So, ψ_{n} = \sqrt{\frac{2}{a}} \sin \frac{n π x}{a}

— — (i)

Key law: by the Born interpretation, ψ² dx is the probability of finding the particle between x and x + dx. So the probability in 0 ≤ x ≤ a/4 is

P_{n} = \int_{0}^{a / 4} ψ_{n}^{2} d x

— — (ii)

So, putting the value of (i) in (ii), we get

P_{n} = \frac{2}{a} \int_{0}^{a / 4} \sin^{2} \frac{n π x}{a} d x

Using sin²θ = ½(1 − cos 2θ),

P_{n} = \frac{1}{a} {[x - \frac{a}{2 n π} \sin \frac{2 n π x}{a}]}_{0}^{a / 4}

P_{n} = \frac{1}{4} - \frac{1}{2 n π} \sin \frac{n π}{2}

— — (iii) general result

Putting n = 1, 2, 3 in (iii):

n	sin(nπ/2)	Pₙ
1	+1	1/4 − 1/2π = 0.091
2	0	1/4 = 0.250
3	−1	1/4 + 1/6π = 0.303

Check: taking the limits 0 to a instead, P = 1 − (1/2nπ) sin 2nπ = 1 ✓ (the particle is surely in the box).

Page 3 of 3 · ConclusionSketch, reading the result, principle, application

Sketch of ψₙ² (shaded area = probability in 0 to a/4):

Reading the result:

i) n = 1: ψ² is piled up at the centre of the box, so the end quarter holds only 9 %.
ii) n = 2: the node at a/2 splits the box into two equal halves, so the first quarter holds exactly 1/4.
iii) n = 3: the first lobe ends at a/3 and lies mostly inside the region, so P rises above 1/4, to 0.303.

Principle: as n increases, the term (1/2nπ) sin(nπ/2) falls as 1/n, so Pₙ → 1/4. This is the classical value for a particle equally likely to be anywhere in the box: Bohr’s correspondence principle.

Application: the π electrons of a conjugated polyene (butadiene, β-carotene) are treated as particles in a box of the chain length. The area under ψ² over part of the chain gives the π-electron density there, and the gap between levels gives λ_max: the longer the chain, the longer the wavelength absorbed, which is why β-carotene is coloured.

So, P₁ = 0.091, P₂ = 0.250, P₃ = 0.303

The same question in other exams

Combined Geo-Scientist 2021, Chemistry Paper II, Q7 (10 marks): What is the probability of locating a particle in one-dimensional box between L/4 and 3L/4, where L is the length of the box? Assume the particle to be in the lowest energy state.

Use the same integral and change only the limits. For n = 1, P = 1/2 + 1/π = 0.818. The middle half of the box holds 82 % of the particle. The two end quarters hold 9.1 % each, and 9.1 + 81.8 + 9.1 = 100.

Civil Services 2022 (Paper I, Q2(b), 20 marks) asked the general case, the first 1/a of the width: Pₙ = 1/a − (1/2nπ) sin(2nπ/a). Putting a = 4 gives the 2025 result. Learn the general form once and all three questions become the same question.

Questions students ask

What is the probability of finding a particle in the first quarter (0 to a/4) of a one-dimensional box?

With ψₙ = √(2/a) sin(nπx/a), the probability is Pₙ = ∫ψₙ² dx from 0 to a/4 = 1/4 − (1/2nπ) sin(nπ/2). This gives P₁ = 0.091, P₂ = 0.250 and P₃ = 0.303. This was asked in UPSC Civil Services 2025, Chemistry Paper I, Q2(a), for 15 marks.

Why is the probability for n = 1 less than 1/4 in the first quarter of the box?

In the ground state ψ² has a single hump that is largest at the centre of the box and zero at the walls. The end quarter holds only 9.1 % of the probability, well below the classical 25 %.

What is the probability of finding a particle between L/4 and 3L/4 in the ground state?

For n = 1 the probability is 1/2 + 1/π = 0.818, so the middle half of the box holds about 82 % of the particle. This was asked in UPSC Combined Geo-Scientist 2021, Chemistry Paper II, Q7, and appears as Self-test 8.4 in Atkins' Physical Chemistry (9th edition).

What is the general formula for the probability in the first 1/a of the width of the box?

Pₙ = 1/a − (1/2nπ) sin(2nπ/a). The first term is the classical value and the second is a wave term that shrinks as 1/n. This general form was asked in UPSC Civil Services 2022, Chemistry Paper I, Q2(b), for 20 marks.

Which book should I read for the particle in a box for UPSC Chemistry?

Any one standard book is enough. Atkins' Physical Chemistry (9th edition), Section 8.1, has the worked example and the self-test that match the UPSC questions. Puri, Sharma and Pathania or McQuarrie's Quantum Chemistry cover the same derivation.

How often is the particle in a box asked in UPSC chemistry papers?

From 2021 to 2025 it appeared in 13 of 15 papers across Civil Services Paper I, Combined Geo-Scientist Mains Paper II and IFoS Paper I: 3 times as this probability question and 10 times as related questions on energy levels, degeneracy, emitted light or the wavefunction.

Read the complete answers

CSE 2025 Q2(a) solution CSE 2022 Q2(b) solution All atomic structure questions Paper I book

By VKC, your mentor. Integrated M.Sc. Chemistry, IIT Roorkee. CSIR NET Chemical Sciences, All India Rank 4. New posts are announced first on Telegram.