JEE Advanced 2026 Paper 2 — Full Mock Test (54 Questions, 180 Marks)
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This MCQ module is based on: Line Spectra Hydrogen
TOPIC 15 OF 27
Line Spectra Hydrogen
🎓 Class 12
Physics
CBSE
Theory
Ch 12 – Atoms
⏱ ~14 min
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Line Spectra Hydrogen
12.10 Line Spectra — Atomic Fingerprints
Heat any element to incandescence — or pass an electric discharge through a tube of its gas — and it glows with a unique pattern of sharp spectral lines. Sodium glows yellow at 589 nm; hydrogen produces four prominent visible lines at 656 nm (red), 486 nm (blue-green), 434 nm (violet) and 410 nm (deep violet). These line spectra are fingerprints of each element, and they had baffled physicists for half a century.
Why lines and not a continuum? A glowing solid emits a continuous black-body spectrum (all wavelengths). A low-density gas of separated atoms emits only at discrete wavelengths — those allowed by Bohr's frequency condition hν = Ei − Ef.
12.11 Rydberg's Formula (1888)
Long before Bohr, the Swedish school-teacher Johannes Rydberg had found, by patient curve-fitting, an empirical formula that fits all observed wavelengths in the hydrogen spectrum:
\(\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_f^{2}} - \dfrac{1}{n_i^{2}}\right)\quad(n_i > n_f)\)
Here \(R = 1.097 \times 10^{7}\) m−1 is the Rydberg constant. \(n_i\) is the principal quantum number of the upper (initial) level and \(n_f\) of the lower (final) level. Bohr's brilliance was to derive this formula from first principles:
\[\frac{1}{\lambda} = \frac{m_e e^{4}}{8\varepsilon_0^{2}h^{3}c}\left(\frac{1}{n_f^{2}} - \frac{1}{n_i^{2}}\right) = R\left(\frac{1}{n_f^{2}} - \frac{1}{n_i^{2}}\right)\]Plugging in the constants gives R = 1.097 × 10⁷ m⁻¹ — agreeing with Rydberg's empirical value to better than one part in 10,000. Triumph!
12.12 The Five Spectral Series of Hydrogen
Each value of \(n_f\) (the final level) defines a complete series of lines. As \(n_i\) ranges from \(n_f+1, n_f+2, \dots, \infty\), an infinite ladder of lines is generated, converging to a series limit at \(n_i \to \infty\).
| Series name | nf | ni values | Wavelength range | Region | Discoverer / year |
|---|---|---|---|---|---|
| Lyman | 1 | 2,3,4,… | 122 – 91 nm | Ultraviolet | Lyman, 1906 |
| Balmer | 2 | 3,4,5,… | 656 – 365 nm | Visible (4 lines visible) | Balmer, 1885 |
| Paschen | 3 | 4,5,6,… | 1875 – 820 nm | Near infrared | Paschen, 1908 |
| Brackett | 4 | 5,6,7,… | 4050 – 1460 nm | Mid infrared | Brackett, 1922 |
| Pfund | 5 | 6,7,8,… | 7460 – 2280 nm | Far infrared | Pfund, 1924 |
The Balmer series — first to be discovered
In 1885, Swiss school-teacher Johann Balmer noticed that the four visible hydrogen lines fit:
\[\lambda = 364.6\,\text{nm}\,\frac{n^2}{n^2 - 4}\]This is exactly the Rydberg formula with \(n_f = 2\) and \(n_i = n\). The four visible lines:
- Hα (n=3 → 2): 656.3 nm — the famous red Balmer-α.
- Hβ (n=4 → 2): 486.1 nm — blue-green.
- Hγ (n=5 → 2): 434.0 nm — violet.
- Hδ (n=6 → 2): 410.2 nm — deep violet.
12.13 Series Limits
For each series, the highest-frequency (shortest-wavelength) line corresponds to \(n_i = \infty\), giving:
\[\frac{1}{\lambda_{\text{lim}}} = \frac{R}{n_f^2}\]| Series | nf | Limit λ (nm) | Limit energy (eV) |
|---|---|---|---|
| Lyman | 1 | 91.2 | 13.60 |
| Balmer | 2 | 364.6 | 3.40 |
| Paschen | 3 | 820.4 | 1.51 |
| Brackett | 4 | 1458.7 | 0.85 |
| Pfund | 5 | 2278.8 | 0.54 |
12.14 Emission vs Absorption Spectra
If the gas is hot, its atoms are continuously excited and de-excited; we see an emission spectrum — bright coloured lines on a dark background.
If white light passes through a cool sample of the same gas, atoms in the gas absorb exactly those photons that match their allowed transitions. The transmitted light shows an absorption spectrum — dark lines on the rainbow background, at exactly the same wavelengths as the emission lines. The Sun's spectrum, when carefully examined, shows hundreds of such Fraunhofer absorption lines from atoms in its cooler outer atmosphere.
Activity 12.3 — Calculate Balmer-α from Bohr
Predict the wavelength of the Hα red line (n=3 → n=2) using the Rydberg formula and compare with the laboratory value of 656.3 nm.
Hint: 1/λ = R(1/2² − 1/3²) = R(5/36).
1/λ = (1.097×10⁷)(5/36) = 1.524×10⁶ m⁻¹.
λ = 1/1.524×10⁶ = 6.563×10⁻⁷ m = 656.3 nm. ✓ Bohr's model is exact for hydrogen!
λ = 1/1.524×10⁶ = 6.563×10⁻⁷ m = 656.3 nm. ✓ Bohr's model is exact for hydrogen!
Interactive — Spectral Series Generator
Pick a final level nf (this defines the series) and a starting level ni. Read off the photon wavelength and watch a coloured marker appear on the EM spectrum strip.
3
Wavelength λ
656.3 nm
656.3 nm
Photon energy
1.89 eV
1.89 eV
Worked Examples
Example 1 — Wavelength of the H-β line
Compute the wavelength of the Balmer-β line (n=4 → n=2) using R = 1.097×10⁷ m⁻¹.
1/λ = R(1/4 − 1/16) = R × 3/16 = (1.097×10⁷)(0.1875) = 2.057×10⁶ m⁻¹.
λ = 4.86×10⁻⁷ m = 486 nm — blue-green.
λ = 4.86×10⁻⁷ m = 486 nm — blue-green.
Example 2 — Series limit of Balmer
Find the shortest wavelength in the Balmer series.
n_i = ∞: 1/λ_lim = R(1/4 − 0) = R/4.
λ_lim = 4/R = 4/(1.097×10⁷) = 3.646×10⁻⁷ m = 364.6 nm — just inside the UV. Beyond this wavelength, the Balmer continuum begins.
λ_lim = 4/R = 4/(1.097×10⁷) = 3.646×10⁻⁷ m = 364.6 nm — just inside the UV. Beyond this wavelength, the Balmer continuum begins.
Example 3 — Identifying a series
An emission line of hydrogen at 1875 nm is observed. Identify the series and the transition.
1/λ = 1/1.875×10⁻⁶ = 5.33×10⁵ m⁻¹.
1/λ = R(1/n_f² − 1/n_i²) ⟹ (5.33×10⁵)/(1.097×10⁷) = 0.0486 = 1/n_f² − 1/n_i².
Try n_f = 3: 1/9 = 0.1111. So 1/n_i² = 0.1111 − 0.0486 = 0.0625 = 1/16, giving n_i = 4.
This is the Paschen-α line (n=4 → n=3).
1/λ = R(1/n_f² − 1/n_i²) ⟹ (5.33×10⁵)/(1.097×10⁷) = 0.0486 = 1/n_f² − 1/n_i².
Try n_f = 3: 1/9 = 0.1111. So 1/n_i² = 0.1111 − 0.0486 = 0.0625 = 1/16, giving n_i = 4.
This is the Paschen-α line (n=4 → n=3).
Example 4 — H-α photon energy
What is the energy of an Hα photon (656.3 nm)?
E = hc/λ = 1240/656.3 nm·eV/nm = 1.89 eV. Equivalently, E_3 − E_2 = −1.51 − (−3.40) = 1.89 eV. ✓
Competency-Based Questions
Q1. The series of hydrogen lines that lies entirely in the visible region is:
(b) Balmer (transitions to n=2; first four lines are visible).
Q2. The shortest wavelength in the Lyman series is approximately:
(a) 91 nm — the Lyman series limit (n=∞ → 1).
Q3. (Short Answer) State Rydberg's formula and define each symbol.
1/λ = R(1/n_f² − 1/n_i²). Here λ is the wavelength of emitted/absorbed light, R = 1.097×10⁷ m⁻¹ is the Rydberg constant, n_i is the upper level and n_f is the lower level (n_i > n_f).
Q4. (Fill in the blank) The transition n=2 → n=1 emits a photon belonging to the ______ series.
Lyman (final level n_f = 1).
Q5. (HOT) Which has a longer wavelength: the first line of the Balmer series or the first line of the Paschen series? Justify briefly.
Paschen-α (n=4→3), λ = 1875 nm. Balmer-α (n=3→2), λ = 656 nm. Paschen has the longer wavelength because the energy gap is smaller for transitions between higher orbits.
Assertion–Reason Questions
Options: (A) Both true, R correct explanation of A. (B) Both true, R not the correct explanation. (C) A true, R false. (D) A false, R true.
Assertion: Hydrogen produces a discrete line spectrum, not a continuous one.
Reason: The energies of the bound states of hydrogen are discrete.
(A) Both correct and the reason explains the assertion. Discrete energy levels imply discrete photon energies.
Assertion: The Balmer series of hydrogen lies entirely in the visible region.
Reason: The Balmer series limit is at 364.6 nm, on the very edge of the visible.
(C) Assertion is partly true (only the first four lines are visible; higher members are in the near-UV). Reason is true. Best answer (C).
Assertion: The wavelengths of the emission and absorption spectra of a sample of hydrogen at the same wavelengths.
Reason: Both involve the same set of energy-level differences in hydrogen.
(A) Both correct and the reason explains the assertion. The same atom can either emit or absorb the photons of these specific energies.
Frequently Asked Questions - Line Spectra Hydrogen
What is the main concept covered in Line Spectra Hydrogen?
In NCERT Class 12 Physics Chapter 12 (Atoms), "Line Spectra Hydrogen" covers the core principles and equations students need for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Line Spectra Hydrogen useful in real-life applications?
Real-life applications of "Line Spectra Hydrogen" from NCERT Class 12 Physics Chapter 12 include electronics, communication systems, medical imaging, solar energy, semiconductor devices, and modern technology. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Line Spectra Hydrogen?
Key formulas in "Line Spectra Hydrogen" (NCERT Class 12 Physics Chapter 12 Atoms) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 12?
NCERT Class 12 Physics Chapter 12 (Atoms) is structured so each part builds on the previous one. "Line Spectra Hydrogen" connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Line Spectra Hydrogen?
CBSE board questions from "Line Spectra Hydrogen" typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Line Spectra Hydrogen" lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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