JEE Mains Chemistry · Structure of Atom

Hydrogen Spectrum and Spectral Series

When the electron of a hydrogen-like species drops between orbits it emits one line; the Rydberg equation gives each line's wavenumber, and the lines group into series by the orbit they land on.

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Why this matters

Eighteen PYQs, thirteen of them multiple choice, and eight from 2026 — the page most tilted to recent papers. Ten use the Rydberg equation with Z² to find a line's energy, wavenumber or wavelength, or to compare two lines; eight name a series and its region, count lines, or test line spectra and Moseley's law. Two ideas cover the page.

Concept 1 of 2: Rydberg equation for hydrogen-like species

A line's energy is the gap between two Bohr levels. Since

E_n\propto Z^2/n^2

, the wavenumber is

RZ^2

times the difference of

1/n^2

for the two levels. Most questions compare two lines, so

R

cancels and only the brackets matter. Two lines from different species match when

Z^2

times the bracket is the same.

Definition

$\bar\nu=\dfrac{1}{\lambda}=RZ^2\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)$ , $n_1<n_2$ .
$R=1.097\times10^{7}\ \mathrm{m^{-1}}=109677\ \mathrm{cm^{-1}}$ .
Energy of the line: $E=hc\bar\nu$ , so $E\propto\bar\nu\propto1/\lambda$ .
The k-th line of a series with lower level $n_1$ comes from $n_2=n_1+k$ .
The first line has the lowest energy and the longest wavelength. The series limit ( $n_2=\infty$ ) has the highest energy and the shortest wavelength.

Rydberg equation

\bar\nu=RZ^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)

Worked example

Find the ratio of the wavelength of the first Lyman line to that of the first Balmer line of hydrogen.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 2 · Q39Moderate

Example 1 · Structure of Atom · Hydrogen Spectrum and Spectral Series

The energy of first (lowest) Balmer line of H atom is x J . The energy (in J) of second Balmer line of H atom is :

Count lines from the landing level

The third Paschen line is

6\to3

, not

3\to\ldots

5\to3

. The k-th line of a series starts from

n_1+k

Longest wavelength is the first line

The first line of a series has the smallest energy gap, so the longest wavelength. The shortest wavelength is the series limit, from

n_2=\infty

Drill 9 more on rydberg equation for hydrogen-like species

Concept 2 of 2: Spectral series, line counts and other spectra

Lines that land on the same lower orbit form one series. Lyman lands on

n=1

and is ultraviolet; Balmer lands on

n=2

and is the only visible series; the rest are infrared. Many atoms together can show every possible drop, but a single electron takes one path down.

Definition

Many atoms falling from $n_2$ to $n_1$ : number of lines $=\dfrac{\Delta n(\Delta n+1)}{2}$ , $\Delta n=n_2-n_1$ . From $n$ to the ground state this is $\dfrac{n(n-1)}{2}$ .
One electron falling from $n_2$ to $n_1$ : at most $n_2-n_1$ lines.
An emission spectrum of a gas-phase atom has bright lines, not a continuous spread. An absorption spectrum is its photographic negative: dark lines on a bright background. Line spectra identify elements; helium was found in the sun this way.
Moseley's law: $\sqrt{\nu}=a(Z-b)$ for X-ray lines. $\sqrt\nu$ against atomic number is a straight line. $\nu$ against $Z$ is a parabola, and atomic mass gives no straight line.

Series	Lands on (n₁)	First line	Series limit	Region
Lyman	1	2 → 1, about 122 nm	∞ → 1, about 91 nm	UltravioletQ
Balmer	2	3 → 2, about 656 nm	∞ → 2, about 365 nm	Visible
Paschen	3	4 → 3, about 1875 nm	∞ → 3, about 821 nm	InfraredQ With R rounded to 10⁵ cm⁻¹, the Paschen limit ∞ → 3 is 9/R = 900 nm, the infrared used for heat therapy.
Brackett	4	5 → 4, about 4052 nm	∞ → 4, about 1459 nm	Infrared
Pfund	5	6 → 5, about 7460 nm	∞ → 5, about 2280 nm	Infrared

Wavelengths are for hydrogen, from

1/R=91.2\ \mathrm{nm}

. For a hydrogen-like ion divide by

Z^2

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 1 · Q41Moderate

Example 2 · Structure of Atom · Hydrogen Spectrum and Spectral Series

The wave numbers of three spectral lines of H atom are considered. Identify the set of spectral lines belonging to Balmer series. (

R =

Rydberg constant)

One electron is not many atoms

\dfrac{n(n-1)}{2}

counts every drop a large sample can make. A single electron cascading from

n=5

to the ground state gives at most 4 lines.

Moseley used atomic number, not mass

\sqrt\nu

is linear in

Z

. A statement that

\sqrt\nu

against atomic mass, or

\nu

against

Z

, is a straight line is false.

Drill 7 more on spectral series, line counts and other spectra

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Rydberg equation for hydrogen-like species
Rydberg equation
$\bar\nu=RZ^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

Reference tables (1)

Spectral series, line counts and other spectra5 rows

Series	Lands on (n₁)	First line	Series limit	Region
Lyman	1	2 → 1, about 122 nm	∞ → 1, about 91 nm	UltravioletQ
Balmer	2	3 → 2, about 656 nm	∞ → 2, about 365 nm	Visible
Paschen	3	4 → 3, about 1875 nm	∞ → 3, about 821 nm	InfraredQ With R rounded to 10⁵ cm⁻¹, the Paschen limit ∞ → 3 is 9/R = 900 nm, the infrared used for heat therapy.
Brackett	4	5 → 4, about 4052 nm	∞ → 4, about 1459 nm	Infrared
Pfund	5	6 → 5, about 7460 nm	∞ → 5, about 2280 nm	Infrared

Wavelengths are for hydrogen, from

1/R=91.2\ \mathrm{nm}

. For a hydrogen-like ion divide by

Z^2

Watch out for (4)

Count lines from the landing level
→ Rydberg equation for hydrogen-like species
Longest wavelength is the first line
→ Rydberg equation for hydrogen-like species
One electron is not many atoms
→ Spectral series, line counts and other spectra
Moseley used atomic number, not mass
→ Spectral series, line counts and other spectra

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