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 MCQs5 revision cards9-step worked example
Source: NCERT Structure of AtomOfficial key: NTA-verifiedLast updated: 24 Sep 2026
Hydrogen Spectrum, explained for NEET
The trap that costs marks: You apply the Rydberg formula correctly for hydrogen — then a question asks for He⁺ (Z = 2). You forget the Z² factor. The wavelength you calculate is 4× too large. That's one mark gone to a single missing term.
The Rydberg formula governs the emission spectrum of hydrogen and hydrogen-like species (NCERT Class 11 Chemistry Chapter 2, page 46):
1/λ = R_H × Z² × (1/n₁² − 1/n₂²), where n₂ > n₁
R_H = 1.097 × 10⁷ m⁻¹. For hydrogen, Z = 1. For He⁺, Z = 2. For Li²⁺, Z = 3.
Spectral series by n₁ value:
- Lyman (n₁ = 1): UV region
- Balmer (n₁ = 2): visible region
- Paschen (n₁ = 3): IR region
- Brackett (n₁ = 4), Pfund (n₁ = 5): far IR
The series limit (shortest wavelength, highest energy) for any series occurs when n₂ → ∞, reducing the bracket to 1/n₁².
NEET connection: Questions typically give you a transition (e.g., n = 5 → n = 2) and ask for wavelength, or give wavelength and ask you to identify the transition. The Z² trap surfaces whenever the species is not neutral hydrogen.
Watch-out: When a question says "first line of Balmer series," it means n₂ = 3 → n₁ = 2 (the lowest-energy transition in that series), NOT n₂ = 1. "Last line" or "series limit" means n₂ → ∞.
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 first line of the Lyman series of hydrogen corresponds to which transition?
Show answer and why every option is right or wrong
Answer: C. The first line of any series corresponds to the transition from the immediately higher level to n₁. For Lyman, n₁ = 1, so the first line is n = 2 → n = 1 (NCERT Class 11 Chemistry Chapter 2, page 46).
Why A is wrong: A is the second line of the Lyman series (n = 3 → n = 1), not the first.
Why B is wrong: B describes absorption (lower → higher), not an emission line of the series.
Why D is wrong: D is the series limit (n = ∞ → n = 1), the highest-energy line, not the first line.
MCQ 2Easy RecallPractice
Which spectral series of hydrogen falls in the visible region?
Show answer and why every option is right or wrong
Answer: B. The Balmer series (n₁ = 2) produces wavelengths in the visible region (400–700 nm). Lyman is UV; Paschen and Brackett are IR (NCERT Class 11 Chemistry Chapter 2, page 46).
Why A is wrong: A (Lyman) lies in the ultraviolet region, not visible.
Why C is wrong: C (Paschen) lies in the infrared region.
Why D is wrong: D (Brackett) lies in the far infrared region.
MCQ 3Easy RecallPractice
The series limit (shortest wavelength) of the Balmer series of hydrogen corresponds to a transition from:
Show answer and why every option is right or wrong
Answer: B. The series limit occurs when n₂ → ∞, giving the maximum value of (1/n₁² − 1/n₂²) = 1/n₁². For Balmer, n₁ = 2, so the limit is n = ∞ → n = 2 (NCERT Class 11 Chemistry Chapter 2, page 46).
Why A is wrong: A is the first line of the Balmer series (longest wavelength in the series), not the series limit.
Why C is wrong: C is the series limit of the Lyman series (n₁ = 1), not Balmer.
Why D is wrong: D is the first line of the Lyman series, unrelated to Balmer.
MCQ 4Direct ApplicationPractice
Calculate the wavelength of light emitted when an electron in a hydrogen atom transitions from n = 3 to n = 2. (R_H = 1.097 × 10⁷ m⁻¹)
Show answer and why every option is right or wrong
Answer: A. 1/λ = R_H(1/2² − 1/3²) = 1.097 × 10⁷ × (1/4 − 1/9) = 1.097 × 10⁷ × 5/36 = 1.524 × 10⁶ m⁻¹. λ = 6.56 × 10⁻⁷ m = 656 nm (NCERT Class 11 Chemistry Chapter 2, page 46).
Why B is wrong: B (486 nm) corresponds to n = 4 → n = 2 (second line of Balmer). Confusing the transition number with n₂ = 4 instead of 3.
Why C is wrong: C (434 nm) corresponds to n = 5 → n = 2. Wrong initial level selected.
Why D is wrong: D (122 nm) corresponds to n = 2 → n = 1 (Lyman series). Wrong series entirely — used n₁ = 1 instead of n₁ = 2.
MCQ 5Direct ApplicationPractice
The wavelength of the first line of the Lyman series for He⁺ (Z = 2) is:
Show answer and why every option is right or wrong
Answer: D. 1/λ = R_H × Z² × (1/1² − 1/2²) = 1.097 × 10⁷ × 4 × 3/4 = 3.291 × 10⁷ m⁻¹. λ = 3.04 × 10⁻⁸ m = 30.4 nm (NCERT Class 11 Chemistry Chapter 2, page 46). The Z² = 4 factor shortens the wavelength by a factor of 4 compared to hydrogen.
Why A is wrong: A (243.2 nm) divides by Z instead of multiplying by Z² — inverts the effect of nuclear charge.
Why B is wrong: B (121.6 nm) is the hydrogen Lyman first line — you forgot the Z² factor entirely (trap: hydrogen-like Z² factor).
Why C is wrong: C (60.8 nm) uses Z = 2 instead of Z² = 4. The formula requires Z², not Z.
MCQ 6Direct ApplicationPractice
For a hydrogen atom, if the wavelength of the series limit of the Lyman series is λ₁ and of the Balmer series is λ₂, then the ratio λ₁/λ₂ is:
Show answer and why every option is right or wrong
Answer: A. Series limit: 1/λ = R_H × (1/n₁²). For Lyman: 1/λ₁ = R_H/1² = R_H. For Balmer: 1/λ₂ = R_H/2² = R_H/4. Therefore λ₁ = 1/R_H and λ₂ = 4/R_H. Ratio λ₁/λ₂ = (1/R_H)/(4/R_H) = 1/4 (NCERT Class 11 Chemistry Chapter 2, page 46).
Why B is wrong: B (4) is the inverted ratio λ₂/λ₁. Confusing which series has shorter wavelength — Lyman is higher energy, therefore shorter λ.
Why C is wrong: C (1/2) uses n₁ ratio directly (1/2) instead of n₁² ratio (1/4). The formula has 1/n₁², not 1/n₁.
Why D is wrong: D (2) is the inverse of option C — same conceptual error (linear instead of quadratic).
MCQ 7CalculationPractice
An electron in a hydrogen atom jumps from the 4th orbit to the 2nd orbit. The number of spectral lines possible in the emission spectrum of this atom (considering all possible downward transitions from n = 4) is:
Show answer and why every option is right or wrong
Answer: A. From n = 4, the possible downward transitions are: 4→3, 4→2, 4→1, 3→2, 3→1, 2→1. Total = n(n−1)/2 = 4×3/2 = 6 lines (NCERT Class 11 Chemistry Chapter 2, page 46). The formula n(n−1)/2 counts all transitions when all intermediate levels can be populated.
Why B is wrong: B (4) likely counts n − 1 = 3 transitions from n = 4 plus one extra, but misses the full cascade from all intermediate levels.
Why C is wrong: C (3) counts only the direct transitions from n = 4 (4→3, 4→2, 4→1), ignoring subsequent emissions from populated intermediate levels.
Why D is wrong: D (10) uses n(n+1)/2 = 10, which is the wrong formula. The correct formula for spectral lines is n(n−1)/2.
MCQ 8CalculationPractice
The ratio of the energy of the first line of the Lyman series to that of the first line of the Balmer series, for hydrogen, is:
Show answer and why every option is right or wrong
Answer: A. A is correct. The energy of a line goes as the bracket (1/n₁² − 1/n₂²). First Lyman line, n = 2 → 1: 1 − 1/4 = 3/4. First Balmer line, n = 3 → 2: 1/4 − 1/9 = 5/36. Ratio = (3/4)/(5/36) = (3/4)(36/5) = 27/5. In electron-volts: 10.2 eV and 1.89 eV (NCERT Class 11 Chemistry Chapter 2, page 46).
Why B is wrong: B (5/27) is the ratio the wrong way up. The Lyman line is the more energetic of the two — ultraviolet against visible — so the ratio must be greater than 1.
Why C is wrong: C (4) is the ratio of the SERIES LIMITS, n = ∞ → 1 against n = ∞ → 2, which is 1 : 1/4. The first line of a series is the lowest-energy transition in it, not the highest, so the upper level has to be the next one up, not infinity.
Why D is wrong: D (3/4) is the Lyman bracket, 1 − 1/4, on its own. It has not been divided by the Balmer bracket, 5/36, so it is not a ratio of the two lines at all.
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Hydrogen Spectrum: quick recall before you leave
How do you solve a Hydrogen Spectrum question? A worked example
Pattern: Bohr energy transition — compute photon wavelength using Rydberg formula for a hydrogen-like ion.
- 1
Given
A He⁺ ion (Z = 2) has an electron that transitions from n = 4 to n = 2. R_H = 1.097 × 10⁷ m⁻¹.
- 2
Required
Wavelength of the emitted photon.
- 3
Concept
The Rydberg formula relates electronic transitions to emitted wavelength. For hydrogen-like species, the Z² factor must be included.
- 4
Formula
1/λ = R_H × Z² × (1/n₁² − 1/n₂²)
- 5
Substitution
1/λ = 1.097 × 10⁷ × (2)² × (1/2² − 1/4²)
1/λ = 1.097 × 10⁷ × 4 × (1/4 − 1/16)
1/λ = 1.097 × 10⁷ × 4 × (4/16 − 1/16)
1/λ = 1.097 × 10⁷ × 4 × 3/16 - 6
Calculation
1/λ = 1.097 × 10⁷ × 12/16 = 1.097 × 10⁷ × 0.75 = 8.228 × 10⁶ m⁻¹
λ = 1/(8.228 × 10⁶) = 1.215 × 10⁻⁷ m = 121.5 nm
Note: Z = 2, n₁ = 2, n₂ = 4 are exact integers and do not limit significant figures. R_H (4 sig figs) governs precision. - 7
Final answer
λ = 121.5 nm (UV region — this is the Balmer-equivalent series for He⁺, but shifted into UV by the Z² = 4 factor).
- 8
Common trap
If you forget Z²: 1/λ = 1.097 × 10⁷ × (1/4 − 1/16) = 1.097 × 10⁷ × 3/16 = 2.057 × 10⁶ m⁻¹ → λ = 486 nm. This is the hydrogen Balmer line (n = 4 → 2), exactly 4× too large. The Z² factor shortens wavelength by Z² for He⁺.
- 9
Similar NEET-style question
Li²⁺ (Z = 3) undergoes a transition from n = 3 to n = 1. Calculate the wavelength of emitted radiation. [Answer: 1/λ = R_H × 9 × (1 − 1/9) = R_H × 9 × 8/9 = 8R_H → λ = 11.4 nm]
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What to remember before solving Hydrogen Spectrum questions
Which Hydrogen Spectrum formulas do you need for NEET?
Rydberg formula (H spectrum)
Spectral wavelengths of hydrogen-like atoms. Lyman (n1=1, UV), Balmer (n1=2, visible), Paschen (n1=3, IR).
| Symbol | Quantity | SI Unit |
|---|---|---|
| lambda | wavelength | m |
| R_H | Rydberg 1.097e7 | 1/m |
| Z | nuclear charge | - |
| n1, n2 | integers, n2>n1 | - |
Valid when
- One-electron atom
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.
Category: Similar Terms
Student forgets Z² scaling when applying Bohr formulas to He⁺ (Z=2) or Li²⁺ (Z=3).
When it triggers
Question involves hydrogen-like ion (He+, Li2+, etc.).
How to avoid
E_n = -13.6 × Z²/n² eV. r_n = (0.529/Z) × n² Å. He+: 4× more bound than H. Li²⁺: 9× more bound. Always include Z².
Root cause: formula misuse
Correction
Always include Z². E_n = -13.6 × Z²/n². For He+: 4× more energetic than H. For Li²⁺: 9×.
More in Structure of Atom: 2 exam traps and mistakes · 4 formulas · 2 question patterns from its other lessons.
Hydrogen Spectrum questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
How does NEET ask about Hydrogen Spectrum?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Bohr energy difference between two levels; compute photon energy/wavelength using Rydberg or E_n formulas.
Multi StepMedium
Common distractors
forgets z squared
Drops Z² for hydrogen-like
Sources
NCERT refs: Class 11 Chemistry Chapter 2, p.46
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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