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Hydrogen spectrum questions appear in both physics and chemistry, and they reward a student who can move between three forms of the same idea: energy levels, photon energies and wavelengths. This post shows how they connect, works the standard question types, and checks the answers against real measured lines.
Where does this appear in the syllabus?
The JEE (Main) 2026 syllabus lists "Bohr model, energy levels, hydrogen spectrum" in physics and "spectrum of the hydrogen atom, Bohr model of a hydrogen atom" in chemistry. The NEET (UG) 2026 syllabus has the same physics line, and JEE Advanced 2026 lists "Bohr's theory of hydrogen-like atoms". The CBSE 2025-26 curriculum asks for hydrogen line spectra as a qualitative treatment, so board questions lean on energy levels while entrance tests push the calculations. The named series and the Rydberg formula are in the NCERT Class 11 chemistry chapter on atomic structure. Check your own exam year's syllabus.
How do the energy levels give the spectrum?
Niels Bohr's 1913 paper proposed stationary orbits and photon emission between them; he received the 1922 Nobel Prize in Physics "for his services in the investigation of the structure of atoms and of the radiation emanating from them". Our post on Bohr's model of the atom covers what it got right and what replaced it.
From the NCERT physics chapter on atoms:
Eₙ = −13.6/n² eV, so E₁ = −13.6 eV, E₂ = −3.40 eV, E₃ = −1.51 eV.
The ground state needs 13.6 eV to ionise. Raising it from n = 1 to n = 2 needs 10.2 eV.
A jump from nᵢ to a lower n_f emits a photon with hν = Eᵢ − E_f.
Turn that into wavelength and you get the Rydberg formula: 1/λ = R(1/n₁² − 1/n₂²), with n₂ > n₁. NCERT Class 11 gives R = 109,677 cm⁻¹ for hydrogen and names the series by their lower level:
Lyman (n₁ = 1): ultraviolet
Balmer (n₁ = 2): visible
Paschen (n₁ = 3): infrared
Brackett (n₁ = 4) and Pfund (n₁ = 5): infrared
Worked examples
Example 1: the red Balmer line. Find the wavelength for n = 3 to n = 2.
1/λ = 109,677 × (1/4 − 1/9) = 109,677 × 5/36 = 15,233 cm⁻¹.
λ = 1/15,233 cm = 6.565 × 10⁻⁵ cm = 656.5 nm.
Check with energies: E₃ − E₂ = 3.40 − 1.51 ≈ 1.89 eV, and λ = 1240/1.89 ≈ 656 nm.
The NIST table of strong hydrogen lines lists this line at 6562.85 Å (656.3 nm) measured in air. The small gap is mostly because light travels slightly slower in air than in vacuum, which shortens the wavelength.
Example 2: first Lyman line. For n = 2 to 1, 1/λ = R × 3/4 = 82,258 cm⁻¹, so λ = 121.6 nm. NIST lists 1215.67 Å, a close match.
Example 3: series limit. The shortest Balmer wavelength comes from n₂ → ∞: 1/λ = R/4 = 27,419 cm⁻¹, so λ = 364.7 nm. For Lyman, 1/λ = R and λ = 91.2 nm.
Example 4: counting lines. Atoms are excited to n = 4. How many emission lines are possible? Any pair of levels gives a line, so the count is n(n − 1)/2 = 4 × 3/2 = 6: three Lyman, two Balmer, one Paschen.
Example 5: hydrogen-like ions. For He⁺ (Z = 2), energies scale by Z², so Eₙ = −54.4/n² eV and every wavelength is a quarter of hydrogen's for the same jump. The He⁺ 3 → 2 line is 656.5/4 ≈ 164 nm.
Common mistakes and how to avoid them
Dropping the minus sign. Bound energies are negative. A photon's energy is the positive difference.
Swapping n₁ and n₂. n₁ is always the lower level, so the bracket stays positive.
Mixing up series and region. Lyman is ultraviolet, not visible. Only Balmer has visible lines.
Using the series limit as the longest wavelength. The limit (n₂ → ∞) is the shortest; the first line (n₂ = n₁ + 1) is the longest.
Counting lines for one atom. A single atom can emit at most n − 1 lines in one cascade; a sample of many atoms can show all n(n − 1)/2. Read which the question asks.
Forgetting Z². For He⁺ or Li²⁺, multiply energies by Z².
Our post on quantum numbers and atomic structure picks up where Bohr's model stops.
When your answer differs from the key by a factor such as 4 or 9, the error is usually in one line of the set-up. The Fermi Problem Library has a step-by-step solver that explains the reasoning, so you can compare each line with your own.
A practice routine for this chapter
Day 1: draw the energy level diagram for n = 1 to 5 with values, and mark one line from each series.
Day 2: ten wavelength calculations, half with the Rydberg formula and half with λ = 1240/E.
Day 3: five series-limit and longest-wavelength questions.
Day 4: five line-counting and hydrogen-like ion questions.
Day 5: a mixed timed set of ten from previous papers, checking each answer against the region it should fall in.
Today, calculate the second Balmer line (n = 4 to 2) and check that it is near 486 nm, the blue-green line in NIST's table.
