Energy level diagram of hydrogen atom — transitions and spectral series

Question

Draw the energy level diagram for the hydrogen atom. Which transitions produce the Lyman, Balmer, Paschen, Brackett, and Pfund series?

Solution — Step by Step

The energy of the $n$ -th level in hydrogen is:

E_n = -\frac{13.6}{n^2} \text{ eV}

Level ( $n$ )	Energy (eV)
1 (ground)	$-13.6$
2	$-3.4$
3	$-1.51$
4	$-0.85$
5	$-0.544$
$\infty$	$0$ (ionisation)

The negative sign means the electron is bound. The gap between levels decreases as $n$ increases — levels crowd together near the ionisation limit.

When an electron transitions from a higher level ( $n_2$ ) to a lower level ( $n_1$ ), it emits a photon. The series are named by the lower level:

Lyman series: $n_1 = 1$ (transitions to ground state) — UV region
Balmer series: $n_1 = 2$ — Visible region (this is the one we can see)
Paschen series: $n_1 = 3$ — Infrared
Brackett series: $n_1 = 4$ — Infrared
Pfund series: $n_1 = 5$ — Far infrared

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

where $R = 1.097 \times 10^7 \text{ m}^{-1}$ is the Rydberg constant.

For the first line of Balmer series ( $n_2 = 3 \to n_1 = 2$ ):

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

\lambda = \frac{36}{5R} = 656 \text{ nm (red light)}

graph TD
    A["n=5, E=-0.54 eV"] --> E5[Pfund: to n=5]
    B["n=4, E=-0.85 eV"] --> E4[Brackett: to n=4]
    C["n=3, E=-1.51 eV"] --> E3[Paschen: to n=3, IR]
    D["n=2, E=-3.4 eV"] --> E2[Balmer: to n=2, Visible]
    E["n=1, E=-13.6 eV"] --> E1[Lyman: to n=1, UV]
    A -->|drop to n=1| E1
    B -->|drop to n=2| E2
    C -->|drop to n=3| E3

Why This Works

The energy levels follow from solving the Schrodinger equation for the hydrogen atom (or, at the Bohr model level, from balancing centripetal force with electrostatic attraction while quantising angular momentum).

Key patterns to remember:

Shortest wavelength (series limit) in each series: $n_2 \to \infty$
Longest wavelength (first line): $n_2 = n_1 + 1$
Maximum number of spectral lines from level $n$ : $\frac{n(n-1)}{2}$

For NEET, questions typically ask for the total number of lines when an electron is in level $n$ , or for the wavelength of a specific transition using the Rydberg formula.

Alternative Method

For quick calculations, memorise these ratios for the Balmer series:

$H_\alpha$ (3 $\to$ 2): $\lambda = 656$ nm (red)
$H_\beta$ (4 $\to$ 2): $\lambda = 486$ nm (blue-green)
$H_\gamma$ (5 $\to$ 2): $\lambda = 434$ nm (violet)
Series limit ( $\infty \to$ 2): $\lambda = 365$ nm

For Lyman series limit: $\lambda = 91.2$ nm (UV)

Common Mistake

Students often confuse which series falls in which region. The key: only the Balmer series is in the visible region. Lyman is UV (higher energy, shorter wavelength), and everything from Paschen onward is infrared. NEET 2023 asked “which series of hydrogen spectrum falls in the visible region?” — the answer is Balmer ( $n_1 = 2$ ), not Lyman.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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