Hydrogen Spectrum

8 MCQs2 revision cards9-step worked example

Source: NCERT Structure of AtomPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

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 18):

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 → ∞.

Practice 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?

MCQ 2Easy RecallPractice

Which spectral series of hydrogen falls in the visible region?

MCQ 3Easy RecallPractice

The series limit (shortest wavelength) of the Balmer series of hydrogen corresponds to a transition from:

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)

MCQ 5Direct ApplicationPractice

The wavelength of the first line of the Lyman series for He⁺ (Z = 2) is:

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:

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:

MCQ 8CalculationPractice

The ratio of energies of the first lines of the Lyman and Balmer series for hydrogen is:

Quick recall before you leave

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] ---

Before solving, remember these

Formula

Rydberg formula for hydrogen spectrum

1/λ = R_H × (1/n₁² - 1/n₂²), where R_H = 1.097 × 10⁷ m⁻¹. Series: Lyman (n₁=1, UV), Balmer (n₁=2, visible), Paschen (n₁=3, IR).

-- NCERT Class 11 Chemistry, Ch. 2, p. 18

Formulas

Bohr energy (hydrogen-like)

E_{n} = - 13.6 \frac{Z ^{2}}{n ^{2}} eV

Energy of nth orbit. Negative (bound). Ground state H: -13.6 eV.

Symbol	Quantity	SI Unit
E_n	orbit energy	eV
Z	nuclear charge	-
n	principal	-

Valid when

Hydrogen-like atom
Non-relativistic

Bohr radius (hydrogen-like)

r_{n} = (0.529 \overset{˚}{A}) \frac{n ^{2}}{Z}

Radius of nth Bohr orbit for hydrogen-like atom of nuclear charge Z.

Symbol	Quantity	SI Unit
n	principal quantum number	-
Z	nuclear charge	-
r_n	orbit radius	Å

Valid when

Hydrogen-like (one-electron) atom
Non-relativistic

de Broglie wavelength

λ = h / (m v)

Wavelength associated with moving particle of momentum mv.

Symbol	Quantity	SI Unit
h	Planck 6.626e-34	J*s
m	mass	kg
v	velocity	m/s

Valid when

Non-relativistic

Heisenberg uncertainty

Δ x Δ p \geq \frac{h}{4 π}

Position and momentum cannot both be known with arbitrary precision.

Symbol	Quantity	SI Unit
Δx	position uncertainty	m
Δp	momentum uncertainty	kg*m/s

Valid when

Quantum scale; meaningful only when Δx, Δp comparable to atomic dimensions

Rydberg formula (H spectrum)

\frac{1}{λ} = R_{H} Z^{2} (1/ n_{1}^{2} - 1/ n_{2}^{2})

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

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Trap

Cr / Cu electronic configuration anomaly

Category: Inorganic Exception

Student writes Cr as [Ar]3d⁴4s² (expected) instead of actual [Ar]3d⁵4s¹. Same for Cu: actual [Ar]3d¹⁰4s¹ (one e⁻ promoted from 4s to 3d).

When it triggers

Question asks for ground-state electronic configuration of Cr (Z=24) or Cu (Z=29).

How to avoid

Half-filled (d⁵) and fully filled (d¹⁰) configurations have extra stability from exchange energy and symmetry. Cr and Cu adopt these configurations by promoting one 4s electron.

Trap

Hydrogen-like Z² factor

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.

Mistake

Writes Cr as [Ar]3d⁴4s² (expected) instead of actual [Ar]3d⁵4s¹. Same for Cu.

Root cause: concept gap

Correction

Half-filled (d⁵) and full-filled (d¹⁰) configurations have extra exchange-energy stability. Cr and Cu adopt these by promoting one 4s electron.

Mistake

Drops Z² factor when applying Bohr formulas to He⁺ or Li²⁺.

Root cause: formula misuse

Correction

Always include Z². E_n = -13.6 × Z²/n². For He+: 4× more energetic than H. For Li²⁺: 9×.

Past Year Questions

9 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.

NEET 2025

Higher yield of NO in - 40 - N 2 (g ) + O 2 (g ) 2 N O (g ) can be obtained at [H of the reaction = +180.7 kJ mol–1] A. Higher temperature B. Lower temperature C. Higher concentration of N 2 D. Higher concentration of O 2 Choose the correct answer from the options given below :

1A, C, D only

2A, D only

3B, C only

4B, C, D only

NTA Answer: Option 1(final)

NEET 2024Revised key

Match List I with List II. List I List II (Molecule) (Number and types of bond/s between two carbon atoms) A. ethane I. one -bond and two -bonds B. ethene II. two -bonds C. carbon molecule, C III. one -bonds 2 D. ethyne IV. one -bond and one -bond Choose the correct answer from the options given below:

1A-I, B-IV, C-II, D-III

2A-IV, B-III, C-II, D-I

3A-III, B-IV, C-II, D-I

4A-III, B-IV, C-I, D-II

NTA Answer: Option 3(revised_final)

NEET 2024Revised key

Match List I with List II List I List II (Quantum Number) (Information provided) A. m I. Shape of orbital l B. m II. Size of orbital s C. l III. Orientation of orbital D. n IV. Orientation of spin of electron Choose the correct answer from the options given below :

1A-I, B-III, C-II, D-IV

2A-III, B-IV, C-I, D-II

3A-III, B-IV, C-II, D-I

4A-II, B-I, C-IV, D-III

NTA Answer: Option 2(revised_final)

NEET 2024Revised key

Given below are two statements : Statement I : [Co(NH ) ]3+ is a homoleptic complex whereas [Co(NH ) Cl ]+ is a heteroleptic complex. 3 6 3 4 2 Statement II : Complex [Co(NH ) ]3+ has only one kind of ligands but [Co(NH ) Cl ]+ has more than one kind 3 6 3 4 2 of ligands. In the light of the above statements, choose the correct answer from the options given below.

1Both Statement I and Statement II are true

2Both Statement I and Statement II are false

3Statement I is true but Statement II is false

4Statement I is false but Statement II is true

NTA Answer: Option 1(revised_final)

NEET 2023

Select the correct statements from the following A. Atoms of all elements are composed of two fundamental particles. B. The mass of the electron is 9.10939 × 10–31 kg. C. All the isotopes of a given element show same chemical properties: D. Protons and electrons are collectively known as nucleons. E. Dalton’s atomic theory, regarded the atom as an ultimate particles of matter Choose the correct answer from the options given below

1C, D and E only

2A and E only

3B, C and E only

4A, B and C only

NTA Answer: Option 3(final)

NEET 2023

On balancing the given redox reaction, c aCrO2 bSO2(aq)cH(aq) 2aCr3(aq)bSO2(aq) H O(l) 2 7 3 4 2 2 the coefficients a, b and c are found to be, respectively-

13, 8, 1

21, 8, 3

38, 1, 3

41, 3, 8

NTA Answer: Option 4(final)

NEET 2022

Match List-I with List-II List-I List-II (a) Li (i) absorbent for carbon dioxide (b) Na (ii) electrochemical cells (c) KOH (iii) coolant in fast breeder reactors (d) Cs (iv) photoelectric cell Choose the correct answer from the options given below :

1(a) - (ii), (b) – (iii), (c) – (i), (d) – (iv)

2(a) - (iv), (b) – (i), (c) – (iii), (d) – (ii)

3(a) - (iii), (b) – (iv), (c) – (ii), (d) – (i)

4(a) - (i), (b) – (iii), (c) – (iv), (d) – (ii)

NTA Answer: Option 1(final)

NEET 2021

Statement I : Acid strength increases in the order given as HF << HCl << HBr << HI. Statement II : As the size of the elements F, Cl, Br, I increases down the group, the bond strength of HF, HCl, HBr and HI decreases and so the acid strength increases. In the light of the above statements, choose the correct answer from the options given below.

1Statement I is incorrect but Statement II is true

2Both statement I and Statement II are true

3Both Statement I and Statement II are false

4Statement I : correct but statement II is false

NTA Answer: Option 2(final)

NEET 2021

From the following pairs of ions which one is not an iso-electronic pair? Fe2+, Mn2+

1O2–, F–

2Na+, Mg2+

3Mn2+, Fe3+

4Answer

NTA Answer: Option 1(final)

How NEET usually asks this

Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.

Write electronic configuration, especially handling Cr/Cu anomalies and ions.

InterpretationMedium

Common distractors

misses cr cu anomaly

Writes 3d⁴4s² instead of 3d⁵4s¹

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

Apply Δx·Δp ≥ h/(4π) to find minimum uncertainty given the other.

Direct ApplicationEasy

Common distractors

uses h instead of h over 4pi

Drops 4π factor

Given orbital (e.g. 3p_z) or electron, identify n, l, m_l, m_s. Or check forbidden quantum number combinations.

InterpretationEasy

Common distractors

uses wrong l range

Uses l ≤ n instead of l < n

Sources

NCERT refs: Class 11 Chemistry Chapter 2, p.18

Test yourself on this topic with real past-paper questions:

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Hydrogen Spectrum

Lesson

Practice MCQs

Quick recall before you leave

Rydberg formula

Number of spectral lines

Worked Example

Before solving, remember these

Formulas

Valid when

Valid when

Valid when

Valid when

Valid when

Exam Traps & Common Mistakes

When it triggers

How to avoid

When it triggers

How to avoid

Correction

Correction

Past Year Questions

How NEET usually asks this

Common distractors

Common distractors

Common distractors

Common distractors

Sources

Free NEET study resources