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Physics · Atoms

Line Spectra of the Hydrogen Atom

When an electric discharge passes through hydrogen gas, the light it emits is not a smooth band of colour but a set of sharp, isolated lines at fixed wavelengths. NCERT §12.5 shows how Bohr's third postulate turns these lines into arithmetic: every line is one electron falling between two discrete energy levels. This subtopic covers the Rydberg formula, the five named series, the meaning of a series limit, and the algebra NEET tests almost every year.

What a Line Spectrum Is

Atoms of most elements are stable and emit a characteristic spectrum. As NCERT records, the emission spectrum of atomic hydrogen consists of bright lines on a dark background, while passing white light through the gas produces dark absorption lines at exactly the same wavelengths. This set of isolated parallel lines is called a line spectrum, and because it is unique to each element it serves as a fingerprint for identifying a gas.

The decisive clue came in 1885, when Johann Jakob Balmer found a simple empirical formula for the wavelengths of a group of visible hydrogen lines. He had no physical model — only the numbers. The challenge for any theory of the atom was to derive Balmer's pattern, and it was this challenge that Bohr's model met in 1913.

Figure 1 · Visible Balmer lines 656.3 nm Hα · n=3→2 486.1 nm Hβ · n=4→2 434.1 nm Hγ · n=5→2 410.2 nm · Hδ · n=6→2 violet red

The four longest-wavelength Balmer lines, with the wavelengths matched in NEET 2024. Hα (red, 656.3 nm) is the first line; the lines crowd toward the violet end as the upper level rises.

The Rydberg Formula

Balmer's pattern was later generalised into a single relation, the Rydberg formula, which gives the wave number (the reciprocal of the wavelength) of every line in the hydrogen spectrum:

$$\frac{1}{\lambda} = R\left(\frac{1}{n_1^{2}} - \frac{1}{n_2^{2}}\right), \qquad n_2 > n_1$$

Here $n_1$ is the lower orbit, $n_2$ the higher orbit, and $R = 1.097 \times 10^{7}\ \text{m}^{-1}$ is the Rydberg constant. The integer $n_1$ is fixed for a given series; running $n_2$ over every integer greater than $n_1$ generates all the lines of that series. The quantity $1/\lambda$ is the wave number, with SI unit $\text{m}^{-1}$, and it is the form NEET most often asks you to evaluate.

The Five Spectral Series

Each value of $n_1$ collects the lines that end on the same lower orbit, and each such family was discovered separately and named after its discoverer. NIOS §24.3 lists all five, together with the spectral region in which they fall.

SeriesLower orbit $n_1$Upper orbits $n_2$RegionDiscovery
Lymann₁ = 12, 3, 4, …Ultraviolet1906
Balmern₁ = 23, 4, 5, …Visible1885
Paschenn₁ = 34, 5, 6, …Near infrared1908
Brackettn₁ = 45, 6, 7, …Mid infraredinfrared
Pfundn₁ = 56, 7, 8, …Far infraredinfrared

Only the Balmer series, with $n_1 = 2$, lies in the visible region — these are the lines the eye sees through a discharge tube. The Lyman series ($n_1 = 1$) ends on the ground state and carries the largest energy gaps, placing it firmly in the ultraviolet. Paschen, Brackett and Pfund all end on orbits beyond the second and so emit lower-energy infrared photons.

NEET Trap

Match the series to its region and lower orbit before you compute

Two confusions cost marks every year. First, mismatching the series to its region: Lyman → ultraviolet (n₁ = 1), Balmer → visible (n₁ = 2), the rest → infrared. Second, mixing up which line is which: the shortest wavelength of a series is its series limit (set n₂ = ∞), while the largest wavelength is its first line (the n₂ = n₁ + 1 transition).

Fix n₁ from the series name first; only then decide whether the question wants the first line (longest λ) or the series limit (shortest λ).

Series Limit and First Line

Within any one series the lines are not evenly spaced. As $n_2$ increases, the term $1/n_2^{2}$ shrinks toward zero, so the lines bunch closer and closer together until they converge on a single shortest-wavelength boundary called the series limit. Setting $n_2 = \infty$ in the Rydberg formula gives the limit directly:

$$\frac{1}{\lambda_{\text{limit}}} = \frac{R}{n_1^{2}}$$

Physically the series limit is the photon emitted when an electron from the ionised state (energy $0\ \text{eV}$, $n_2 = \infty$) drops straight to the level $n_1$. The first line of the series is the opposite extreme: the small-gap $n_2 = n_1 + 1 \to n_1$ transition, which has the smallest wave number and therefore the longest wavelength.

Figure 2 · Energy levels and transitions n=∞ (0 eV) n=4 (−0.85 eV) n=3 (−1.51 eV) n=2 (−3.40 eV) n=1 (−13.6 eV) Lyman (UV) Balmer (visible) Paschen (IR)

Downward transitions grouped by their final orbit. Energies $E_n = -13.6/n^{2}\ \text{eV}$ from NCERT Eq. (12.10). Within each group the longest arrow (smallest gap) is the first line; the shortest-wavelength line comes from n=∞.

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Build the foundation

Every line here is a transition between Bohr's quantised orbits. Revise where the levels $E_n = -13.6/n^2$ come from in Bohr Model of the Hydrogen Atom.

How Bohr Explains the Lines

The classical Rutherford picture predicted a continuous spectrum, because a spiralling electron would radiate at smoothly changing frequencies. Bohr resolved the contradiction with his third postulate: an atom emits radiation only when an electron makes a transition from a higher stationary state $n_i$ to a lower one $n_f$, and the energy difference is carried away by a single photon of frequency $\nu$,

$$h\nu = E_{n_i} - E_{n_f}$$

Because the allowed energies are discrete, $E_n = -13.6/n^{2}\ \text{eV}$, only specific energy differences are possible, so light is radiated at discrete frequencies — sharp lines, not a continuous band. Substituting the level energies into $h\nu = E_{n_i} - E_{n_f}$ reproduces the Rydberg formula exactly, with $R$ expressed in terms of fundamental constants. This derivation, NCERT notes, was a brilliant achievement that earned Bohr the 1922 Nobel Prize and predicted series later confirmed by experiment.

Worked Examples

Example 1 · NIOS-style

An electron in hydrogen falls from $n = 3$ to $n = 2$. Taking $R = 1.097 \times 10^{7}\ \text{m}^{-1}$, find the wavelength and identify the series.

Use $\dfrac{1}{\lambda} = R\left(\dfrac{1}{2^2} - \dfrac{1}{3^2}\right) = R\left(\dfrac{1}{4} - \dfrac{1}{9}\right) = R \cdot \dfrac{5}{36}$.

$\dfrac{1}{\lambda} = 1.097 \times 10^{7} \times 0.1389 \approx 1.524 \times 10^{6}\ \text{m}^{-1}$, so $\lambda \approx 6.56 \times 10^{-7}\ \text{m} = 6563\ \text{Å}$.

The final orbit is $n_1 = 2$, the wavelength is visible — this is the first line of the Balmer series (the red Hα line), matching the NIOS answer of 6563 Å.

Example 2 · Series limit

Find the wave number of the series limit (last line) of the Balmer series, given $R = 10^{7}\ \text{m}^{-1}$.

For the series limit set $n_2 = \infty$, so $\dfrac{1}{\lambda} = R\left(\dfrac{1}{2^2} - 0\right) = \dfrac{R}{4}$.

$\dfrac{1}{\lambda} = \dfrac{10^{7}}{4} = 0.25 \times 10^{7}\ \text{m}^{-1}$. This is the shortest-wavelength line of the Balmer series.

Quick Recap

Hydrogen line spectra in one glance

  • Rydberg formula: $\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1^{2}} - \dfrac{1}{n_2^{2}}\right)$, with $R = 1.097 \times 10^{7}\ \text{m}^{-1}$.
  • Series by lower orbit: Lyman ($n_1=1$, UV), Balmer ($n_1=2$, visible), Paschen ($n_1=3$), Brackett ($n_1=4$), Pfund ($n_1=5$) — last three infrared.
  • Series limit = shortest wavelength, from $n_2 = \infty$, giving $1/\lambda = R/n_1^{2}$.
  • First line = longest wavelength, from $n_2 = n_1 + 1$.
  • Bohr's third postulate $h\nu = E_{n_i} - E_{n_f}$ with $E_n = -13.6/n^{2}\ \text{eV}$ turns discrete levels into discrete lines.

NEET PYQ Snapshot — Line Spectra of the Hydrogen Atom

Real NEET questions on the Rydberg formula and the spectral series, with worked answers.

NEET 2023

In the hydrogen spectrum, the shortest wavelength in the Balmer series is $\lambda$. The shortest wavelength in the Brackett series is:

  1. $16\lambda$
  2. $2\lambda$
  3. $4\lambda$
  4. $9\lambda$
Answer: (3) 4λ

Shortest wavelength = series limit ($n_2 = \infty$). Balmer: $1/\lambda = R/2^2 = R/4$. Brackett: $1/\lambda' = R/4^2 = R/16$. Dividing, $\lambda'/\lambda = 16/4 = 4$, so $\lambda' = 4\lambda$.

NEET 2017

The ratio of wavelengths of the last line of the Balmer series and the last line of the Lyman series is:

  1. 0.5
  2. 2
  3. 1
  4. 4
Answer: (4) 4

"Last line" = series limit ($n_2 = \infty$). Lyman: $1/\lambda_1 = R/1^2 = R$, so $\lambda_1 = 1/R$. Balmer: $1/\lambda_2 = R/2^2 = R/4$, so $\lambda_2 = 4/R$. Therefore $\lambda_2/\lambda_1 = 4$.

NEET 2016

Given the Rydberg constant as $10^{7}\ \text{m}^{-1}$, the wave number of the last line of the Balmer series in the hydrogen spectrum will be:

  1. $0.5 \times 10^{7}\ \text{m}^{-1}$
  2. $0.25 \times 10^{7}\ \text{m}^{-1}$
  3. $2.5 \times 10^{7}\ \text{m}^{-1}$
  4. $0.025 \times 10^{4}\ \text{m}^{-1}$
Answer: (2) 0.25 × 10⁷ m⁻¹

Last line of Balmer = series limit: $\bar{\nu} = R\left(\dfrac{1}{2^2} - \dfrac{1}{\infty}\right) = \dfrac{R}{4} = \dfrac{10^{7}}{4} = 0.25 \times 10^{7}\ \text{m}^{-1}$.

NEET 2024

Match the spectral lines of hydrogen (transitions to $n_1 = 2$) with their wavelengths. A: $n_2=3\to2$; B: $n_2=4\to2$; C: $n_2=5\to2$; D: $n_2=6\to2$. Wavelengths (nm): I. 410.2, II. 434.1, III. 656.3, IV. 486.1.

  1. A-II, B-I, C-IV, D-III
  2. A-III, B-IV, C-II, D-I
  3. A-IV, B-III, C-I, D-II
  4. A-I, B-II, C-III, D-IV
Answer: (2) A-III, B-IV, C-II, D-I

These are Balmer lines. The smallest gap ($n_2=3$) gives the longest wavelength, so A → 656.3 nm (III). As $n_2$ rises the wavelength falls: B → 486.1 (IV), C → 434.1 (II), D → 410.2 (I).

FAQs — Line Spectra of the Hydrogen Atom

Quick answers to the doubts that surface most often in this subtopic.

What is the Rydberg formula for the hydrogen spectrum?

The Rydberg formula gives the wave number of each emitted line as 1/λ = R(1/n₁² − 1/n₂²), where n₁ is the lower orbit and n₂ the higher orbit (n₂ > n₁), and R = 1.097 × 10⁷ m⁻¹ is the Rydberg constant. Fixing n₁ defines a spectral series; varying n₂ over the integers greater than n₁ generates every line within that series.

Which hydrogen series lies in the visible region?

The Balmer series (n₁ = 2), discovered by Johann Balmer in 1885, lies in the visible region. Its first line is the red Hα at 656.3 nm (n₂ = 3); successive lines at 486.1 nm, 434.1 nm and 410.2 nm move into the blue–violet. The Lyman series lies in the ultraviolet, while Paschen, Brackett and Pfund lie in the infrared.

What is the series limit of a hydrogen spectral series?

The series limit is the shortest-wavelength line of a series, obtained by setting n₂ = ∞ in the Rydberg formula so 1/λ = R/n₁². It marks the photon emitted when an electron from the ionised state (energy 0 eV) falls directly to the level n₁. Beyond the limit the lines crowd together into a continuum.

How does Bohr's model explain the line spectrum of hydrogen?

Bohr's third postulate states that when an electron falls from a higher level n₂ to a lower level n₁, the energy difference is carried away by a single photon of frequency ν such that hν = E(n₂) − E(n₁). Because the energies E(n) = −13.6/n² eV are discrete, only certain frequencies appear, producing sharp lines rather than a continuous band.

Which line of a series has the longest wavelength?

The first line of a series — the transition from n₂ = n₁ + 1 down to n₁ — has the smallest energy gap and therefore the longest wavelength. The shortest wavelength is the series limit at n₂ = ∞. So within any series, the first line is the longest and the series limit is the shortest.

Why is the hydrogen spectrum a line spectrum and not continuous?

Because the electron in hydrogen can occupy only discrete energy levels, the energy released in a transition can take only specific values. Each allowed energy difference corresponds to one photon frequency, so the emitted light appears as a set of isolated bright lines on a dark background — a fingerprint unique to hydrogen.

Related Topics
Atoms — Chapter Hub All subtopics, formulas and PYQs for the Atoms chapter. Rutherford Model The nuclear atom and why it failed to explain line spectra. Atomic Spectra Emission and absorption spectra as fingerprints of atoms. Bohr Model of Hydrogen Quantised orbits and the energy levels behind every line. De Broglie & Bohr's Postulate The wave picture that justifies quantised angular momentum.