1/λ = R_H × (1/n₁² − 1/n₂²), n₁ = 1, 2, …; n₂ = n₁ + 1, n₁ + 2, … (NCERT writes it in wavenumber, with R_H = 109,677 cm⁻¹ = 1.097 × 10⁷ m⁻¹). Series for n₁ = 1, 2, 3, 4, 5: Lyman (ultraviolet), Balmer (visible; the only lines of the hydrogen spectrum in the visible region), Paschen, Brackett and Pfund (infrared) (Table 2.3).
-- NCERT Class 11 Chemistry, Ch. 2, p. 45Hydrogen Spectrum
8 MCQs1 revision card9-step worked example
Source: NCERT Atoms and NucleiPYQ coverage: NEET 2023, 2024Official key: NTA-verifiedLast updated: 26 Sep 2026
Hydrogen Spectrum, explained for NEET
Write the Rydberg formula the wrong way round and the arithmetic hands you a negative wavelength. That is the whole of the trap on this topic, and it survives into the exam because a negative sign in an intermediate line is easy to drop on the way to a positive-looking answer.
The convention is fixed: 1/λ = R(1/n₁² − 1/n₂²) with n₂ > n₁, R = 1.097 × 10⁷ m⁻¹. Put the smaller-n term first inside the bracket. For emission, n₁ is the level the electron lands on and n₂ the level it left; for absorption the roles as initial and final swap, but the bracket does not — n₁ is always the lower level. A positive result is the check. If 1/λ comes out negative, you inverted the bracket, not the physics.
The lower level alone names the series. NCERT Class 11 Chemistry, Chapter 2, page 45 lists them: n₁ = 1 is Lyman (ultraviolet), n₁ = 2 is Balmer (the only series with lines in the visible), n₁ = 3 is Paschen, n₁ = 4 Brackett, n₁ = 5 Pfund — the last three all infrared. Change n₂ and you move along a series; change n₁ and you change series.
Two structural facts NEET likes. First, the series limit is n₂ → ∞, which kills the second term and leaves 1/λ = R/n₁² — the shortest wavelength the series can emit. Second, within a series the longest wavelength is the n₁+1 → n₁ line, because that is the smallest bracket. So each series occupies a band between its first line and its limit, and the Lyman limit (91.2 nm) sits far below the Balmer limit (364.6 nm).
Watch-out: the formula returns λ in vacuum, in metres, since R carries m⁻¹. Convert to nanometres or ångströms only at the end.
Can you answer these Hydrogen Spectrum MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The hydrogen spectral series whose lines lie in the visible region is the
Show answer and why every option is right or wrong
Answer: C. C is correct. The Balmer series has n₁ = 2, and its lines fall in the visible band; NCERT Class 11 Chemistry, Chapter 2, page 45 identifies it as the visible series.
Why A is wrong: A is wrong because the Lyman series has n₁ = 1 and lies entirely in the ultraviolet.
Why B is wrong: B is wrong because the Paschen series has n₁ = 3 and lies in the infrared.
Why D is wrong: D is wrong because the Brackett series has n₁ = 4 and lies further into the infrared than Paschen.
MCQ 2Easy RecallPractice
In the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²) as conventionally written for hydrogen, the integers satisfy
Show answer and why every option is right or wrong
Answer: A. A is correct. The convention fixes n₂ > n₁ and makes n₁ the lower level, so the bracket and hence 1/λ is positive (NCERT Class 11 Chemistry, Chapter 2, page 45).
Why B is wrong: B is wrong because it reverses the convention; writing the larger-n term first makes the bracket negative and λ negative, which is the standard inversion error.
Why C is wrong: C is wrong because n₁ is the lower level, which is the final level in emission and the initial level only in absorption — it is not always the initial level.
Why D is wrong: D is wrong because a modulus is not part of the formula; the ordering is a stated convention and a negative bracket signals a mistake rather than a sign to be discarded.
MCQ 3Easy RecallPractice
For a given hydrogen spectral series, the series limit corresponds to the transition in which
Show answer and why every option is right or wrong
Answer: B. B is correct. Letting n₂ → ∞ removes the 1/n₂² term, giving 1/λ = R/n₁², the shortest wavelength of that series (NCERT Class 11 Chemistry, Chapter 2, page 45).
Why A is wrong: A is wrong because n₂ = n₁ + 1 gives the smallest bracket and hence the longest wavelength in the series, which is the first line, not the limit.
Why C is wrong: C is wrong because n₁ fixes which series is being considered; letting it run to infinity does not define a limit within a series.
Why D is wrong: D is wrong because n₁ = n₂ makes the bracket zero, which corresponds to no transition at all.
MCQ 4Direct ApplicationPractice
A hydrogen atom emits a photon when its electron moves from the n = 5 level to the n = 3 level. Substituted correctly, the Rydberg expression for this line reads
Show answer and why every option is right or wrong
Answer: B. B is correct. The lower level is n₁ = 3 and the upper is n₂ = 5, and the smaller-n term goes first: 1/λ = R(1/3² − 1/5²), which is positive (NCERT Class 11 Chemistry, Chapter 2, page 45).
Why A is wrong: A is wrong because it puts the larger-n term first, giving a negative 1/λ and a negative wavelength — the n₁/n₂ inversion.
Why C is wrong: C is wrong because the quantum numbers appear as reciprocals of their squares, not of the numbers themselves.
Why D is wrong: D is wrong because the terms are subtracted, not added; adding them would give a wavelength shorter than the series limit, which no transition in this series can produce.
MCQ 5Direct ApplicationPractice
Taking R = 1.097 × 10⁷ m⁻¹, the series limit of the Lyman series has wavelength closest to
Show answer and why every option is right or wrong
Answer: C. C is correct. For the Lyman limit n₁ = 1 and n₂ → ∞, so 1/λ = R, giving λ = 1/(1.097 × 10⁷ m⁻¹) = 9.12 × 10⁻⁸ m (NCERT Class 11 Chemistry, Chapter 2, page 45).
Why A is wrong: A is wrong because 3.65 × 10⁻⁷ m is the Balmer limit, obtained with n₁ = 2 rather than n₁ = 1.
Why B is wrong: B is wrong because 1.22 × 10⁻⁷ m is the first Lyman line (2 → 1), the longest wavelength in the series, not its limit.
Why D is wrong: D is wrong because 8.20 × 10⁻⁷ m is the Paschen limit (n₁ = 3), 9/R; it belongs to a different series in the infrared.
MCQ 6Direct ApplicationPractice
A hydrogen line is measured at a wavelength of 4.86 × 10⁻⁷ m. Using R = 1.097 × 10⁷ m⁻¹, the transition responsible is
Show answer and why every option is right or wrong
Answer: B. B is correct. For 4 → 2, 1/λ = R(1/4 − 1/16) = 1.097 × 10⁷ × 0.1875 = 2.057 × 10⁶ m⁻¹, so λ = 4.86 × 10⁻⁷ m (NCERT Class 11 Chemistry, Chapter 2, page 45).
Why A is wrong: A is wrong because 3 → 2 gives R(1/4 − 1/9) = 1.524 × 10⁶ m⁻¹ and λ = 6.56 × 10⁻⁷ m, the red Balmer line, not this one.
Why C is wrong: C is wrong because 2 → 1 is a Lyman transition with λ = 1.22 × 10⁻⁷ m, in the ultraviolet.
Why D is wrong: D is wrong because 4 → 3 is a Paschen transition with λ = 1.87 × 10⁻⁶ m, in the infrared.
MCQ 7Concept TrapPractice
A student computes a hydrogen line for the transition n = 2 → n = 4 by writing 1/λ = R(1/4² − 1/2²) and obtains 1/λ = −2.057 × 10⁶ m⁻¹. The correct reading of this result is that
Show answer and why every option is right or wrong
Answer: C. C is correct. n₁ is the lower level regardless of direction, so the bracket is always (1/2² − 1/4²); the negative output is the inversion error and the wavelength is 4.86 × 10⁻⁷ m (NCERT Class 11 Chemistry, Chapter 2, page 45).
Why A is wrong: A is wrong because wavelength is a positive quantity by definition; a negative λ is a signal of an algebraic slip, not a label for absorption.
Why B is wrong: B is wrong because 2 → 4 is a perfectly ordinary absorption transition — the atom takes up the photon and the electron rises to the higher level.
Why D is wrong: D is wrong because the formula is not modified for absorption; the same bracket, smaller-n term first, serves both emission and absorption, and only the photon's role changes.
MCQ 8CalculationPractice
A hydrogen atom in the n = 4 level can de-excite to the ground state by various routes. Of all the photons that could be emitted in such de-excitations, the one with the longest wavelength arises from the transition
Show answer and why every option is right or wrong
Answer: D. D is correct. The longest wavelength means the smallest bracket; among the six possible transitions from n = 4 downward, 4 → 3 gives R(1/9 − 1/16) = 5.33 × 10⁵ m⁻¹, the smallest value, hence λ ≈ 1.87 × 10⁻⁶ m (NCERT Class 11 Chemistry, Chapter 2, page 45).
Why A is wrong: A is wrong because 4 → 1 has the largest bracket, R(1 − 1/16), and therefore the shortest wavelength of the set, not the longest.
Why B is wrong: B is wrong because 2 → 1 gives R(1 − 1/4) = 8.23 × 10⁶ m⁻¹, far larger than the 4 → 3 bracket, so its wavelength is much shorter.
Why C is wrong: C is wrong because 4 → 2 gives R(1/4 − 1/16) = 2.06 × 10⁶ m⁻¹, still nearly four times the 4 → 3 bracket; picking it usually comes from assuming the visible line must be the longest.
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Hydrogen Spectrum: quick recall before you leave
How do you solve a Hydrogen Spectrum question? A worked example
Pattern: P.PHY.U18.HYDROGEN_SPECTRUM_TRANSITION — identify the series and compute the wavelength from a stated transition.
- 1
Given.
A hydrogen atom undergoes the transition n = 3 → n = 2. Rydberg constant R = 1.097 × 10⁷ m⁻¹.
- 2
Required.
The series to which this line belongs, and its wavelength in nanometres.
- 3
Concept.
A transition between two Bohr levels emits a single photon whose wavelength follows the Rydberg relation. The lower level fixes the series; the upper level fixes which line within it. Here the electron ends at n = 2, so this is a Balmer line — the visible series.
- 4
Formula.
1/λ = R(1/n₁² − 1/n₂²), with n₁ the lower level and n₂ > n₁.
- 5
Substitution.
Emission from 3 down to 2 means n₁ = 2 (lower), n₂ = 3 (upper). Smaller-n term first:
1/λ = (1.097 × 10⁷ m⁻¹)(1/2² − 1/3²) - 6
Calculation.
1/2² − 1/3² = 1/4 − 1/9 = (9 − 4)/36 = 5/36 = 0.13889
1/λ = 1.097 × 10⁷ × 0.13889 = 1.5236 × 10⁶ m⁻¹
λ = 1/(1.5236 × 10⁶ m⁻¹) = 6.563 × 10⁻⁷ m
The integers 2 and 3, and the arithmetic constants 4, 9 and 36 that follow from them, are exact quantum numbers and exact counting values. They place no limit on significant figures. The precision is set entirely by R, quoted to four significant figures. - 7
Final answer.
λ = 6.563 × 10⁻⁷ m = 656.3 nm, a red line in the Balmer series.
- 8
Common trap.
Writing the bracket as (1/3² − 1/2²) because the electron starts at n = 3. That gives −1.5236 × 10⁶ m⁻¹ and a negative wavelength. Dropping the minus sign at that point happens to recover 656.3 nm here, which is why the habit survives — but the same slip on a comparison question ("which line is longer?") reverses the ranking and costs the mark. The fix is mechanical: identify the lower level, put its term first, and expect a positive number.
- 9
Similar NEET-style question.
A hydrogen atom in the n = 5 level drops to the n = 2 level. Taking R = 1.097 × 10⁷ m⁻¹, find the wavelength of the emitted photon and state whether it lies at a longer or shorter wavelength than the 656.3 nm line found above. *(Bracket: 1/4 − 1/25 = 0.21; λ ≈ 4.34 × 10⁻⁷ m = 434 nm — shorter, as the larger bracket requires.)*
What to remember before solving Hydrogen Spectrum questions
Which Hydrogen Spectrum formulas do you need for NEET?
1 formula — click to collapse
Rydberg formula (hydrogen spectrum)
Wavelengths of hydrogen spectral lines. R = 1.097e7 1/m. Lyman (n1=1, UV), Balmer (n1=2, visible), Paschen+ (IR).
| Symbol | Quantity | SI Unit |
|---|---|---|
| lambda | wavelength | m |
| R | Rydberg constant | 1/m |
| n1, n2 | integers, n2>n1 | - |
Valid when
- Hydrogen atom
- Single electron transition
Where do students lose marks on Hydrogen Spectrum?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
2 items — click to collapse
Category: Sign Convention
Student writes 1/λ = R(1/n_2² − 1/n_1²) (negative wavelength). Convention: n_2 > n_1, so 1/λ = R(1/n_1² − 1/n_2²) is positive.
When it triggers
Hydrogen spectrum problem with electronic transition.
How to avoid
n_1 is the LOWER (final for emission, initial for absorption) level. n_2 > n_1 is the HIGHER level. Bracket arrangement: smaller-n term first inside the parens. Result must be positive.
Root cause: sign error
Correction
1/λ = R(1/n_1² - 1/n_2²) with n_2 > n_1. Smaller-n term first. Result must be positive.
More in Atoms and Nuclei: 1 exam trap or mistake · 5 formulas · 4 question patterns from its other lessons.
Hydrogen Spectrum questions from past NEET papers
2 questions from NEET 2023, 2024. Answers verified against NTA official keys. — click to collapse
NEET 2024Revised key
1A-II, B-I, C-IV, D-III
2A-III, B-IV, C-II, D-I
3A-IV, B-III, C-I, D-II
4A-I, B-II, C-III, D-IV
NTA Answer: Option 2(revised_final)
NEET 2023
14λ
29λ
316λ
42λ
NTA Answer: Option 1(final)
How does NEET ask about Hydrogen Spectrum?
1 recurring pattern from past papers — click to collapse
Rydberg formula 1/λ = R(1/n_1² - 1/n_2²). Identify series and wavelength.
Multi StepMedium
Common distractors
swaps n1 n2
Inverts initial and final levels
Sources
NCERT refs: Class 11 Chemistry Chapter 2, p.45
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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