ADV-2026-P2-CHE-Q10 · Virtual Atomic Spectroscopy Lab

01Original question — verbatim

X^a+ and Y^b+ are hydrogen-like species. The wavelength of light absorbed during the transition between the states with principal quantum numbers n = 1 and n = 2 of X^a+ is λ. The wavelength of light absorbed during the transition between the states with principal quantum numbers n = 2 and n = 4 of Y^b+ is 9λ. The lowest possible value of (a + b) is ______.

Section 3 — non-negative integer answer (+4 for the correct value, 0 otherwise, no negative marking). There are no printed options. What there are are four decisions you have to get right on the way, and every one of them is a place students lose the mark. The lab settles all four by measurement.

Claim A

"Hydrogen-like" means the species has exactly one proton.
Claim B

The transition in Y, at 9λ, involves more energy than the transition in X, at λ.
Claim C

λ_Y = 9λ_X forces Z_X : Z_Y = 3 : 2, so the smallest atomic numbers are 3 and 2.
Claim D

For a hydrogen-like ion the charge on the ion equals its atomic number.

02Bench A — the ion foundry

Before you can measure anything you have to know what is in the tube. Add protons to the nucleus, strip electrons off the outside, and watch the label change. One rule decides whether the Bohr formula is allowed to run at all.

Assemble the species

Protons in the nucleus, Z1

hydrogen

Electrons left on it1

Predict, then build

Claim A says a hydrogen-like species has exactly one proton. Before you build anything — is that what the term means?

Commit to an answer, then build two hydrogen-like species with different Z.

Read-out

protons Z

electrons

charge

species

Hydrogen-like species built

Build two different hydrogen-like species and compare how many protons each one has.

claim A disproved

03Bench B — the absorption spectrometer

A photon only gets absorbed if its energy matches a gap in the ladder exactly. Tune the source, fire it at the sample, and watch what happens to the electron. Anything that is not absorbed carries straight on to the detector.

Source → collimator → sample chamber → detector. The packet drawn on the beam has the wavelength you tuned; short wavelength means high energy.

The sample and the transition

Sample: hydrogen-like ion with Z1

Start level n_i1

Finish level n_f2

Photon energy from the source10.20 eV

Absorption needs an exact match. Try missing it on purpose once.

Energy ladder

ΔE needed

—

wavelength

—

region

—

electron now at

n = 1

Pick a transition, tune the source and fire.

Every absorbed photon leaves a dark line at its own wavelength. This strip is the spectrum you are building.

Predict, then fire

Claim B: the Y photon has nine times the wavelength of the X photon. Does that mean it carries more energy?

Load X, absorb it, then load Y and absorb that. Compare the two energies.

Record both lines — X at Z = 3 and Y at Z = 2 — and the comparison makes itself.

claim B disproved

Why Z matters so much: ΔE against Z

ΔE for n_i → n_f as set your current sample

04Bench C — the ratio hunter

Now both ions at once, each locked to the transition the question gives it. You are not told the atomic numbers — you are told one number, that λ_Y is nine times λ_X. Turn the two dials until the apparatus says nine.

Two samples, two fixed transitions

X^a+ — atomic number Z_X, transition n = 1 → 2 1

Y^b+ — atomic number Z_Y, transition n = 2 → 4 1

Both ladders have exactly the same shape. Only the scale changes — and it changes as Z².

The measurement

λ_X

—

λ_Y

—

—λ_Y ÷ λ_X — the question says this must be 9

Turn the dials.

λ_Y/λ_X at your Z_Y the target, 9

Pairs that give exactly 9

Nothing found yet.

claim C confirmed

05Bench D — the charge counter

The question does not ask for atomic numbers. It asks for a and b, the charges. That is one subtraction away — and it is the subtraction most of the marks are lost on.

Count what is actually there

Claim D: for a hydrogen-like ion the charge equals the atomic number. Ion X has Z = 3 — is it X³⁺?

Count both ions and read the tally.

Tally

Count ion X, then ion Y.

Is that really the lowest?

Any pair (Z_X, Z_Y) = (3m, 2m) satisfies the ratio. Slide m and watch the total.

multiplier m1

claim D disproved

06Evidence board

Everything below came out of a bench you ran. The traps are the wrong turns this question is built to catch — open one only after you have formed your own view.

Lab notebook

Written as you work. Only what you actually demonstrated appears here.

Nothing recorded yet — work through the four benches above.

Common traps

Predict, then check

07Step-by-step solution

The solution is still locked

Test all four statements on the bench above and the derivation opens by itself.
Statements tested: 0 of 4 — build an ion, take both spectra, lock the ratio at 9 and count the charges

What "hydrogen-like" actually fixes

A hydrogen-like species has exactly one electron — the number of protons is free. H, He⁺, Li²⁺, Be³⁺ are all hydrogen-like. That single electron is the whole reason the Bohr result is allowed to run:

E_n = −13.6 Z² / n² eV · 1/λ = R Z² (1/n₁² − 1/n₂²)

It also fixes the bookkeeping you need at the end: one electron and Z protons means the charge is Z − 1, so Z = a + 1. Claims (A) and (D) both die here.
Write down both transitions

X: n = 1 → 2 1/λ = R Z_X² (1/1² − 1/2²) = R Z_X² · 3/4

Y: n = 2 → 4 1/(9λ) = R Z_Y² (1/2² − 1/4²) = R Z_Y² · 3/16

The two geometric factors are 3/4 and 3/16 — the 3's are about to cancel, which is why the numbers come out clean.
Divide, and the constants disappear

λ_Y/λ_X = [Z_X² · 3/4] ÷ [Z_Y² · 3/16] = 4 Z_X²/Z_Y² = 9

Z_X²/Z_Y² = 9/4 ⟹ Z_X : Z_Y = 3 : 2

R never had to be known, and neither did λ. That is what makes this a ratio question rather than a calculation. Claim (C) is correct.
Smallest integers that fit

Z must be a whole number, so (Z_X, Z_Y) = (3m, 2m) for m = 1, 2, 3 … The smallest is:

Z_X = 3 → Li · Z_Y = 2 → He
Atomic number is not the charge

Each ion keeps one electron, so subtract it:

a = Z_X − 1 = 3 − 1 = 2 ⟹ X^a+ = Li²⁺

b = Z_Y − 1 = 2 − 1 = 1 ⟹ Y^b+ = He⁺

Stopping at Z_X + Z_Y = 5 is the single commonest way to lose this mark.
Check that m = 1 really is the minimum

a + b = (3m − 1) + (2m − 1) = 5m − 2

That increases with m — 3, 8, 13, 18 … — so m = 1 gives the lowest value. Both ions are also genuinely positive there (a = 2, b = 1), which the notation X^a+, Y^b+ requires.
Answer

a + b = 2 + 1 = 3

Marking: +4 for 3, 0 otherwise. No negative marking, so there is never a reason to leave this blank.

✓ Independent verification

Derived before anything was looked up. The ratio, the pair (3, 2) and the value 3 were fixed from the Bohr result alone, and only then compared with a published solution.
The geometric factors computed as exact fractions, not decimals: 1 − 1/4 = 3/4 and 1/4 − 1/16 = 3/16, whose quotient is exactly 4. So the condition is 4Z_X²/Z_Y² = 9 with no rounding anywhere.
Brute force over every integer pair with 1 ≤ Z ≤ 60: the pairs that give a ratio of exactly 9.000000 are (3,2), (6,4), (9,6), (12,8) … — every one in the ratio 3 : 2, and none outside it. The smallest is (3, 2).
Two independent numerical routes agree: Bohr energies with the Rydberg energy 13.6057 eV give λ_X = 13.5003 nm and λ_Y = 121.5023 nm; the Rydberg constant 1.0973732 × 10⁷ m⁻¹ gives the same two numbers. Their quotient is 9.0000000000.
ΔE/Z² checked to be constant across Z = 1, 2, 3, 4, 6 at 2.5511 eV for n = 2 → 4, confirming the Z² scaling the whole argument rests on.
The direction of the ratio checked: reversing it to Z_X : Z_Y = 2 : 3 gives 1.7778, and two hydrogens give exactly 4 — neither is 9.
The traps reproduced deliberately: Z_X + Z_Y = 5 is a different number from a + b = 3, and a + b = 5m − 2 was confirmed strictly increasing.
Answer key cross-check: a published worked solution for this paper reports the same intermediate values — Z_X = 3, Z_Y = 2, a = 2, b = 1 — and the same answer, 3. IIT Roorkee's official key was not retrievable at the time of writing.
Headless-browser run of this page: every control driven, every drawing checked, clean console, and zero horizontal overflow at 390 px and 360 px.

08Concepts and classification

Relations used

hydrogen-like ⟺ exactly one electron ⟹ charge = Z − 1

E_n = −13.6 Z²/n² eV (one-electron ions only)

ΔE = 13.6 Z² (1/n₁² − 1/n₂²) eV

1/λ = R Z² (1/n₁² − 1/n₂²), R = 1.097 × 10⁷ m⁻¹

λ = hc/ΔE ⟹ longer wavelength means less energy

n = 1 → 2 factor = 3/4 · n = 2 → 4 factor = 3/16 · their ratio = 4

4 Z_X²/Z_Y² = 9 ⟹ Z_X : Z_Y = 3 : 2

a + b = (3m − 1) + (2m − 1) = 5m − 2, minimum at m = 1

orbit radius r_n = a₀ n²/Z — bigger n, smaller Z, wider orbit

Classification

Simulation ID	ADV-2026-P2-CHE-Q10
Exam / Year / Paper	JEE Advanced · 2026 · Paper 2
Subject	Chemistry (Physical)
Chapter	Atomic Structure
Topic	Bohr model of hydrogen-like species and the Rydberg equation
Subtopics	Hydrogen-like species · Energy levels and the Z² scaling · Absorption and emission transitions · Rydberg equation · Ratio of wavelengths · Atomic number versus ionic charge
Question type	Non-negative integer (+4 / 0, no negative marking)
Difficulty	Moderate
Answer	— run the experiment

Teacher notes

Objective. Get students to treat "hydrogen-like" as a statement about electrons, and to finish the question rather than stopping at the atomic numbers.
Teaching order that works. Do bench D first. Ask the class what the charge on a hydrogen-like lithium is before any spectroscopy — the wrong answer comes fast and is memorable.
Expected wrong answers. 5 by far (Z_X + Z_Y, forgetting the electron); then 2 (using the ratio 3:2 as the answer itself); occasionally 8 (taking m = 2).
The step that decides it. 9 = 4Z_X²/Z_Y², not Z_X²/Z_Y². Students who drop the factor of 4 get Z_X:Z_Y = 3:1 and then a + b = 4.
Discussion question. "The Y photon has nine times the wavelength. Is it a bigger photon?" Longer wavelength, lower energy — worth the two minutes.
Board demonstration. Draw the ladder for Z = 1 and Z = 2 side by side to scale. Identical shape, four times the size. Everything else follows from that picture.

SPECTRUM
SOLVED

Four benches run · both ions identified from their light alone

a + b = 3

01Original question — verbatim

02Bench A — the ion foundry

Assemble the species

Predict, then build

Read-out

Hydrogen-like species built

03Bench B — the absorption spectrometer

The sample and the transition

Energy ladder

Predict, then fire

Why Z matters so much: ΔE against Z

04Bench C — the ratio hunter

Two samples, two fixed transitions

The measurement

Pairs that give exactly 9

05Bench D — the charge counter

Count what is actually there

Tally

Is that really the lowest?

06Evidence board

Lab notebook

Common traps

Predict, then check

07Step-by-step solution

The solution is still locked

What "hydrogen-like" actually fixes

Write down both transitions

Divide, and the constants disappear

Smallest integers that fit

Atomic number is not the charge

Check that m = 1 really is the minimum

Answer

✓ Independent verification

08Concepts and classification

Relations used

Classification

Teacher notes