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Exam level8 past questions

Hydrogen spectrum and spectral series

Understand the origin, calculation, and interpretation of spectral lines, series limits, wavelength and frequency relations, and transitions in hydrogen and hydrogen-like ions.

Why this shows up in the exam

You will be asked to solve problems on spectral line wavelengths, frequencies, and transitions in NEET.

How NEET tests this

Numerical · 7 QsApplicationDirect recall

Learn the idea

Spectral lines arise when an electron jumps between two quantised orbits; each series groups all transitions that end on the same lower orbit, and the longest wavelength in a series comes from the smallest jump (n=lower+1 → lower).

🧠 Memory hook: Lyman lands on 1, Balmer balances on 2, Paschen parties on 3 – think L=1, B=2, P=3, and the longest wave is the smallest step up.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • Rydberg formula: 1/λ = R Z²(1/n₁² – 1/n₂²) (n₂>n₁)
  • Series names: Lyman (n₁=1), Balmer (n₁=2), Paschen (n₁=3) etc.
  • Energy of a level: Eₙ = –13.6 Z²/n² eV
  • λ ν = c and ν = c/λ
  • Longest wavelength in a series: transition from n = n₁+1 to n₁
  • For hydrogen‑like ions wavelength ∝ 1/Z²

How to approach it

  1. 1Identify the lower level n₁ that defines the series
  2. 2Write the transition that gives the longest wavelength: n₂ = n₁+1
  3. 3Plug n₁ and n₂ into the Rydberg formula (include Z² if ion is hydrogen‑like)
  4. 4If a ratio of wavelengths is required, use 1/λ expressions and simplify; cancel R and any common factors

Worked example — watch it click

Ratio of longest wave lengths corresponding to Lyman and Balmer series in hydrogen spectrum is:

  • ✅5/27
  • B)3/23
  • C)7/29
  • D)9/31

The concept behind this problem

The example asks for the ratio of the longest wavelengths of Lyman (2→1) and Balmer (3→2); using the Rydberg formula for those specific n values directly tests the idea that longest wavelength comes from the nearest upper level.

Step by step

  1. 1Longest wavelength in Lyman (2→1): 1/λ_L = R(1/1 - 1/4) = 3R/4.
  2. 2Longest wavelength in Balmer (3→2): 1/λ_B = R(1/4 - 1/9) = 5R/36.
  3. 3Ratio: λ_L/λ_B = (5R/36)/(3R/4) = (5/36) × (4/3) = 20/108 = 5/27.

Watch out

Students often invert the ratio, writing 27/5 instead of the correct 5/27.

Common slip-ups that cost marks

  • •Swapping n₁ and n₂ gives a negative 1/λ
  • •For hydrogen‑like ions forgetting the Z² factor in the Rydberg constant
  • •Mixing up longest (small Δn) with shortest (large Δn) wavelength

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Practise it

These are real questions from past NEET papers that test this exact idea.

Question 1 of 8NEET 2016

If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength λ. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be:

Push further

More challenging

2 harder questions built from the past papers above — a step up in difficulty, with distractors designed so you can't get there by elimination. Written and checked by our reviewers, not from a real paper.

Question 1 of 2

A photon with a wavelength of 486 nm is emitted when an electron in a hydrogen atom transitions from a higher energy level to the n = 2 level. What was the initial energy level (nᵢ) of the electron? (Given h = 6.625 × 10⁻³⁴ Js, c = 3 × 10⁸ m/s, and ionization energy of H = 2.18 × 10⁻¹⁸ J/atom)

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