Hydrogen spectrum — Balmer, Lyman, Paschen series and wavelength calculation

Question

Using the Rydberg formula, calculate the wavelength of the first line of the Balmer series for hydrogen. Also, find the shortest wavelength (series limit) of the Lyman series. Identify which series falls in the visible, UV, and infrared regions.

(NCERT Class 12 — Atoms)

Solution — Step by Step

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

where $R = 1.097 \times 10^7$ m $^{-1}$ is the Rydberg constant, $n_1$ is the lower level, and $n_2$ is the upper level ( $n_2 > n_1$ ).

Balmer series: transitions to $n_1 = 2$ . First line: $n_2 = 3$ .

\frac{1}{\lambda} = R\left(\frac{1}{4} - \frac{1}{9}\right) = R \times \frac{9 - 4}{36} = \frac{5R}{36}

\lambda = \frac{36}{5R} = \frac{36}{5 \times 1.097 \times 10^7} = \frac{36}{5.485 \times 10^7}

\lambda = 6.56 \times 10^{-7} \text{ m} = \mathbf{656 \text{ nm (red light)}}

This is the H-alpha line — the famous red line in hydrogen’s spectrum.

Lyman series: transitions to $n_1 = 1$ . Series limit: $n_2 \to \infty$ .

\frac{1}{\lambda} = R\left(\frac{1}{1} - \frac{1}{\infty}\right) = R

\lambda = \frac{1}{R} = \frac{1}{1.097 \times 10^7} = \mathbf{91.2 \text{ nm (deep UV)}}

Series	Transitions to $n$	Region	Wavelength Range
Lyman	$n = 1$	Ultraviolet	91-122 nm
Balmer	$n = 2$	Visible	365-656 nm
Paschen	$n = 3$	Infrared	820-1875 nm
Brackett	$n = 4$	Infrared	1458-4051 nm
Pfund	$n = 5$	Far infrared	2279+ nm

Only the Balmer series falls in the visible region. The first four lines are: H-alpha (656 nm, red), H-beta (486 nm, blue-green), H-gamma (434 nm, violet), H-delta (410 nm, violet).

Why This Works

Bohr’s model tells us that hydrogen’s electron can only occupy discrete energy levels $E_n = -13.6/n^2$ eV. When the electron jumps from a higher level to a lower one, it emits a photon with energy $E = E_{n_2} - E_{n_1}$ .

The Rydberg formula translates this energy difference into a wavelength. Each series corresponds to a fixed lower level: Lyman ( $n=1$ ), Balmer ( $n=2$ ), Paschen ( $n=3$ ), etc.

The series limit is the shortest wavelength (highest energy) in each series — it corresponds to ionisation from that level ( $n_2 \to \infty$ ).

Alternative Method

Instead of the wavelength formula, use energies directly: $E = 13.6\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ eV, then convert to wavelength using $\lambda = hc/E$ . For the Balmer first line: $E = 13.6(1/4 - 1/9) = 13.6 \times 5/36 = 1.89$ eV, $\lambda = 1240/1.89 = 656$ nm. The factor $1240$ eV·nm is worth memorising.

For NEET and JEE, memorise: Lyman = UV, Balmer = visible, Paschen and beyond = IR. Also remember $1240$ eV·nm — it converts between energy (eV) and wavelength (nm) instantly: $\lambda = 1240/E$ .

Common Mistake

Students often get the subtraction backwards in the Rydberg formula, writing $\frac{1}{n_2^2} - \frac{1}{n_1^2}$ instead of $\frac{1}{n_1^2} - \frac{1}{n_2^2}$ . This gives a negative $1/\lambda$ , which is physically meaningless. The smaller $n$ always goes first because the lower energy level has larger magnitude. Check: the result must give a positive wavelength.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next