JEE Mains Physics · Atoms

Spectral Series and Wavelength Ratios

Each hydrogen series is every jump down to one fixed lower level; 1/λ = RZ²(1/n_f² − 1/n_i²) gives every line, and the ratio of two lines needs only the brackets.

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Why this matters

Twenty-three PYQs, thirteen of them multiple choice, and three from 2026. Eleven are about the series as a whole: which region a series lies in, its longest line and its limit, often scaled from one given wavelength, and one asks you to name lines on an energy-level diagram. Twelve take the ratio of two lines: two lines of one series, lines from two series, two frequencies in one ion, a line against its own series limit, or three levels whose lines combine.

Concept 1 of 2: Hydrogen spectral series, first lines and series limits

Fix the lower level and let the upper level run up the ladder: that family of lines is a series. The first line comes from the next level up, the smallest jump, so it has the least energy and the LONGEST wavelength. The series limit comes from n = ∞, the biggest jump into that level, so it has the SHORTEST wavelength. Every wavelength is a simple fraction times 1/R, so one given wavelength fixes all the others.

Definition

Rydberg formula: $\dfrac{1}{\lambda} = RZ^2\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right)$ , with $R = 1.097 \times 10^{7}\ \text{m}^{-1}$ and $1/R \approx 91.2$ nm.
First member (longest λ): $n_f + 1 \to n_f$ .
Series limit (shortest λ): $\infty \to n_f$ , $\dfrac{1}{\lambda} = \dfrac{R}{n_f^2}$ , so $\lambda = \dfrac{n_f^2}{R}$ .
The kth member counts up from the longest line: it comes from $n_f + k$ . The third Balmer member is 5 → 2.
Scaling from one given line: write the given wavelength as a multiple of 1/R, find 1/R, then build the line you need.

Series	Lower level	Region	Longest line (first member)	Shortest line (series limit)
Lyman	1	Ultraviolet	2 → 1: $4/3R$ , about 122 nm	∞ → 1: $1/R$ , about 91 nm
Balmer	2	Visible	3 → 2: $36/5R$ , about 656 nm	∞ → 2: $4/R$ , about 365 nm Balmer's first lines are visible, but its limit, 365 nm, is just into the ultraviolet.
Paschen	3	Infrared	4 → 3: $144/7R$ , about 1875 nm	∞ → 3: $9/R$ , about 820 nm
Brackett	4	Infrared	5 → 4: $400/9R$ , about 4050 nm	∞ → 4: $16/R$ , about 1458 nm
Pfund	5	Far infrared	6 → 5: $900/11R$ , about 7460 nm	∞ → 5: $25/R$ , about 2280 nm

For a hydrogen-like ion divide every wavelength by Z².

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 2 · Q7Moderate

Example 1 · Atoms · Spectral Series and Wavelength Ratios

The smallest wavelength of Lyman series is 91 nm. The difference between the largest wavelengths of Paschen and Balmer series is nearly

\_\_\_\_

nm.

The series limit is the shortest wavelength

The limit comes from n = ∞, the biggest jump into that level, so it carries the most energy and the shortest wavelength. The first member is the longest.

Each series has its own lower level

Only Lyman ends on n = 1. The Balmer limit is 4/R, not 1/R, and the Paschen limit is 9/R. Fix n_f from the series name before substituting.

Count members up from the longest line

The first Balmer member is 3 → 2, the second 4 → 2, the third 5 → 2. Counting down from the series limit gives the wrong jump.

Drill 10 more on hydrogen spectral series, first lines and series limits

Concept 2 of 2: Ratios of two spectral lines

Write the Rydberg formula for each line and divide. R cancels, and Z cancels too when both lines come from one ion. What is left is a ratio of two brackets. Energy, frequency and photon momentum are all proportional to the bracket, so their ratio is the bracket ratio. Wavelength is its inverse, so the wavelength ratio is the bracket ratio turned upside down.

Definition

$E$ , $\nu$ and $p = h/\lambda$ are all proportional to $Z^2\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right)$ ; λ is proportional to its inverse.
Brackets to know: 2 → 1: $\tfrac{3}{4}$ ; 3 → 1: $\tfrac{8}{9}$ ; 4 → 1: $\tfrac{15}{16}$ ; 3 → 2: $\tfrac{5}{36}$ ; 4 → 2: $\tfrac{3}{16}$ ; 4 → 3: $\tfrac{7}{144}$ .
A line against its own series limit: divide its bracket by $\dfrac{1}{n_f^2}$ .
Three levels A, B, C (A highest): energies add, $E_{AC} = E_{AB} + E_{BC}$ , so $\dfrac{1}{\lambda_{AC}} = \dfrac{1}{\lambda_{AB}} + \dfrac{1}{\lambda_{BC}}$ .
Same jump in two ions: $\lambda \propto \dfrac{1}{Z^2}$ .

Rydberg formula and combining lines

\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right), \qquad \frac{1}{\lambda_{AC}} = \frac{1}{\lambda_{AB}} + \frac{1}{\lambda_{BC}}

Worked example

In hydrogen, the 4 → 3 line has wavelength λ₀. Find the wavelength of the 3 → 1 line in terms of λ₀, and the ratio of their frequencies.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 26 July 2022 · Q29Moderate

Example 2 · Atoms · Spectral Series and Wavelength Ratios

In a hydrogen spectrum,

\lambda

be the wavelength of first transition line of Lyman series. The wavelength difference will be "

a\lambda

" between the wavelength of

3^{\text{rd~}}

transition line of Paschen series and that of

2^{\text{nd~}}

transition line of Balmer Series where

a =

Turn the bracket ratio upside down for wavelength

The bracket is 1/λ. A larger bracket means a shorter wavelength, so the wavelength ratio is the inverse of the bracket ratio. Energy, frequency and momentum ratios are not inverted.

Energies add; wavelengths do not

For levels A > B > C, E_AC = E_AB + E_BC. Adding the wavelengths instead gives a longer wavelength for the bigger jump, which is impossible.

Z cancels only within one ion

Two lines of the same ion share Z², so it cancels. The same jump in two different ions scales as Z², and forgetting it is off by a factor of 4 or 9.

Drill 11 more on ratios of two spectral lines

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)

Ratios of two spectral lines
Rydberg formula and combining lines
$\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right), \qquad \frac{1}{\lambda_{AC}} = \frac{1}{\lambda_{AB}} + \frac{1}{\lambda_{BC}}$

Reference tables (1)

Hydrogen spectral series, first lines and series limits5 rows

Series	Lower level	Region	Longest line (first member)	Shortest line (series limit)
Lyman	1	Ultraviolet	2 → 1: $4/3R$ , about 122 nm	∞ → 1: $1/R$ , about 91 nm
Balmer	2	Visible	3 → 2: $36/5R$ , about 656 nm	∞ → 2: $4/R$ , about 365 nm Balmer's first lines are visible, but its limit, 365 nm, is just into the ultraviolet.
Paschen	3	Infrared	4 → 3: $144/7R$ , about 1875 nm	∞ → 3: $9/R$ , about 820 nm
Brackett	4	Infrared	5 → 4: $400/9R$ , about 4050 nm	∞ → 4: $16/R$ , about 1458 nm
Pfund	5	Far infrared	6 → 5: $900/11R$ , about 7460 nm	∞ → 5: $25/R$ , about 2280 nm

For a hydrogen-like ion divide every wavelength by Z².

Watch out for (6)

The series limit is the shortest wavelength
→ Hydrogen spectral series, first lines and series limits
Each series has its own lower level
→ Hydrogen spectral series, first lines and series limits
Count members up from the longest line
→ Hydrogen spectral series, first lines and series limits
Turn the bracket ratio upside down for wavelength
→ Ratios of two spectral lines
Energies add; wavelengths do not
→ Ratios of two spectral lines
Z cancels only within one ion
→ Ratios of two spectral lines

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