The line spectra of the hydrogen atom

What are the line spectra of the hydrogen atom?

Scope: the basic one-photon Bohr transition model, neglecting nuclear recoil, fine structure and line broadening. Quantum selection rules and state populations determine which lines are present and their strengths. Wavelength formulas use vacuum wavelengths; rounded quoted visible wavelengths are adequate at the precision used here.

In plain words: the electron in hydrogen can only stand on certain steps of an energy staircase. Each time it jumps down a step it throws out one packet of light whose energy is the height of that jump. Different jumps give different colours, so the light comes out as a few sharp lines.

Photon emitted in a transition: hν = E_i − E_f

• By Bohr’s third postulate, a jump from a higher state n_i to a lower state n_f (n_f < n_i) emits a photon carrying the energy difference (NCERT Eq. (12.11), pages 300–301):

• Since n_i and n_f are whole numbers, light is radiated only at certain discrete frequencies. These spectral lines are the emission lines (NCERT page 301).
• With E_n = −13.6 eV/n² (NCERT Eq. (12.10), page 299), the photon energy is (reasoned):

• As a wavelength. Beyond the NCERT Physics text: dividing by hc gives the Rydberg formula, which NCERT Class 11 Chemistry gives with 109 677 cm⁻¹ for hydrogen (Unit 2, page 45). With an infinitely heavy nucleus, R∞ = me⁴/(8ε₀²h³c). For hydrogen, replacing m by the electron–proton reduced mass gives R_H ≈ 1.09678 × 10⁷ m⁻¹; both round to R ≈ 1.097 × 10⁷ m⁻¹ at the precision of these examples. For n_f = 2 this is Balmer’s formula of 1885 (NCERT page 291).

• Spectral series: the lines that end on the same level form a series. NCERT Figure 12.5 (page 297) names the Lyman, Balmer and Paschen series and marks the wavelengths below except 486, 434 and 410 nm. Beyond the NCERT Physics text: the final levels and the regions are from NCERT Class 11 Chemistry’s table of the series (Unit 2, Table 2.3, page 45), and those three wavelengths from the Rydberg formula.
  • n_f = 1, the Lyman series, in the ultraviolet: from 122 nm down to 91 nm.
  • n_f = 2, the Balmer series, spanning visible and near-ultraviolet wavelengths: its first four prominent lines, 656, 486, 434 and 410 nm, are visible; the rest crowd down to 365 nm.
  • n_f = 3, the Paschen series, in the infrared: from 1875 nm down to 820 nm. The Brackett (n_f = 4) and Pfund (n_f = 5) series lie further into the infrared.
• NCERT slip (page 297): NCERT Figure 12.5 misspells the visible series as Balmar. It is the Balmer series, as NCERT page 291 spells the name.
• In each series the longest wavelength comes from the smallest jump (n_i = n_f + 1), and the shortest, the series limit, from n_i → ∞ (beyond the NCERT text).
• In the simplified single-photon cascade model, without re-excitation, an atom descending from level n emits at most n − 1 photons. An ensemble with all relevant downward transitions accessible has up to n(n − 1)/2 level-pair transitions between n and the ground state; this is the usual exam line count, before selection rules and unresolved or coincident frequencies (beyond the NCERT text).
• Hydrogen-like ions (beyond the NCERT text): for an ion with one electron and nuclear charge +Ze, such as He⁺ (Z = 2), every level and photon energy is Z² times hydrogen’s, so each wavelength is divided by Z². The 4 → 2 line of He⁺ has the wavelength of hydrogen’s 2 → 1 line, 122 nm.
• Bohr received the Nobel Prize in Physics in 1922 (NCERT page 301).

Absorption as an upward transition

• For a bound-to-bound transition an atom can absorb a photon whose energy matches the level gap within its line width, provided the lower state is populated and the transition is allowed (NCERT page 301). The atom absorbs radiation of the same frequencies that it emits (NCERT Summary point 6(c), page 303):

• So if light with a continuous range of frequencies passes through a rarefied gas, and the background outshines the gas’s own emission into the sightline, the absorbed frequencies show up as dark absorption lines in the continuous spectrum (NCERT page 301).
• At room temperature most hydrogen atoms are in the ground state (NCERT page 300), so cool hydrogen absorbs mainly the lines that start from n = 1, the Lyman lines in the ultraviolet (reasoned).
• A photon of 13.6 eV or more can free the electron from the ground state: above 0 eV the energies form a continuum, so it need not match exactly, and the extra appears as the electron’s kinetic energy (reasoned from NCERT page 300).
• An atom can also be excited by collisions, for example with electrons (NCERT page 300). Unlike a photon, an electron can hand over only part of its energy, so a 12.5 eV electron can lift a hydrogen atom from n = 1 to n = 2 or n = 3, but not to n = 4, which needs 12.75 eV (reasoned; see Example 2).

Solved examples

Common mistakes to avoid

Quick revision summary

• A jump from n_i down to n_f emits one photon with hν = E_i − E_f; whole-number levels give discrete lines.
• For hydrogen, hν = 13.6(1/n_f² − 1/n_i²) eV, or 1/λ = R(1/n_f² − 1/n_i²) with R ≈ 1.097 × 10⁷ m⁻¹ (the 1/λ form is beyond the NCERT Physics text).
• Series (final levels and regions beyond the NCERT Physics text; NCERT Class 11 Chemistry, Table 2.3): Lyman to n = 1 (ultraviolet), Balmer to n = 2 (visible and near ultraviolet), Paschen to n = 3 (infrared).
• Absorption is the upward jump, E_i + hν = E_f; the atom absorbs the frequencies it can emit, and cool hydrogen absorbs the Lyman lines.
• Electron collisions can excite atoms by giving up part of their energy; photons must match a gap exactly, unless they ionise.

Questions students ask

How do you find the number of spectral lines emitted by hydrogen atoms in the nth level?

In the usual level-pair counting model, the maximum number of possible transitions for an ensemble initially at level n and cascading to the ground state is n(n − 1)/2, the number of different pairs of levels, since every downward jump between two levels gives one line; if the atoms fall no lower than level n₁, it is (n − n₁)(n − n₁ + 1)/2 (reasoned; beyond the NCERT text). For a gas of atoms excited to n = 7 that is 7 × 6/2 = 21 lines: 6 Lyman lines ending on n = 1, 5 Balmer lines ending on n = 2 and 10 more ending on levels 3 to 6. A single atom makes one jump at a time, so it gives at most n − 1 of them (worked example).

Why does the Rydberg constant have two values?

The Rydberg constant seems to have two values because it is quoted in two units: NCERT Class 11 Chemistry gives it as 109 677 cm^⁻¹ (page 45), a wavenumber that gives 1/λ for a line, and as R_H = 2.18 × 10^⁻¹⁸ J (page 47), the same constant multiplied by hc to make it an energy, 13.6 eV (how they match). There is also a small real difference: 109 677 cm^⁻¹ is hydrogen’s own value, about 0.05% below the R = 1.09737 × 10^⁷ m^⁻¹ of an infinitely heavy nucleus, because the proton is only about 1836 times as heavy as the electron (beyond the NCERT text; NIST, CODATA 2022).

What is the difference between the Balmer formula and the Rydberg formula?

Balmer’s formula of 1885 was an empirical fit to one group of hydrogen’s lines (NCERT page 291), the series NCERT Figure 12.5 marks from 656 to 365 nm, while the Rydberg formula, 1/λ = R(1/n_f² − 1/n_i²), covers every series of hydrogen, with Balmer’s formula as its case n_f = 2 (beyond the NCERT text). Balmer wrote his as λ = Bn^²/(n^² − 4), with B = 4/R, and the Rydberg form extends the relation to the other series. Both were found by fitting measured wavelengths; Bohr’s model of 1913 was the first to explain them (NCERT pages 298 and 301).

Are the Lyman, Balmer and Paschen series in the JEE Main and NEET syllabus?

Yes, the spectral series of hydrogen are examinable in JEE Main and NEET (UG) 2026, although neither syllabus names them: both list the “hydrogen spectrum” in Physics Unit 18, Atoms and Nuclei, and the spectrum of the hydrogen atom in Chemistry Unit 2, Atomic Structure. NCERT Class 12 Physics Part II (2026–27) has no section on the series and names the Lyman, Balmer and Paschen series only in its Figure 12.5 (page 297). NCERT Class 11 Chemistry still teaches all five series and the Rydberg formula (page 45), as does the chemistry page on the line spectrum of hydrogen.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Find the wavelength of the first line of the Lyman series, from n = 2 to n = 1.

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   hν = 13.6 × (1 − 1/4) = 10.2 eV, so λ ≈ 1240/10.2 ≈ 122 nm, in the ultraviolet (reasoned from NCERT Eqs. (12.10)–(12.11)).

2. Q2. Find the energy and wavelength of the photon emitted when hydrogen goes from n = 4 to n = 2.

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   hν = 13.6 × (1/4 − 1/16) = 2.55 eV, so λ ≈ 1240/2.55 ≈ 486 nm, the blue-green Balmer line (reasoned from NCERT Eqs. (12.10)–(12.11)).

3. Q3. Hydrogen atoms are excited to n = 4. How many different lines can the gas emit as they return to the ground state?

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   n(n − 1)/2 = 4 × 3/2 = 6 lines: 4→3, 4→2, 4→1, 3→2, 3→1 and 2→1 (beyond the NCERT text).