Energy level diagram of hydrogen atom — transitions and spectral series

Question

Draw the energy level diagram of hydrogen. Identify the Lyman, Balmer, Paschen, Brackett, and Pfund series. Which series falls in the visible region? Calculate the wavelength of the first line of the Balmer series.

(JEE Main 2024 tested the maximum number of spectral lines; NEET asks series identification)

Solution — Step by Step

The energy of the $n^\text{th}$ level: $E_n = -\frac{13.6}{n^2}$ eV

$n = 1$ : $E_1 = -13.6$ eV (ground state)
$n = 2$ : $E_2 = -3.4$ eV
$n = 3$ : $E_3 = -1.51$ eV
$n = \infty$ : $E_\infty = 0$ eV (ionisation)

The negative sign means the electron is bound. More negative = more tightly bound.

Each series consists of transitions that end at the same lower level:

Series	Final level ( $n_f$ )	Region
Lyman	1	Ultraviolet
Balmer	2	Visible + near UV
Paschen	3	Infrared
Brackett	4	Infrared
Pfund	5	Far infrared

The first line of the Balmer series: transition from $n = 3$ to $n = 2$ .

Using Rydberg’s formula:

\frac{1}{\lambda} = R\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right) = 1.097 \times 10^7 \left(\frac{1}{4} - \frac{1}{9}\right)

\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{5}{36} = 1.524 \times 10^6 \text{ m}^{-1}

\lambda = 656 \text{ nm}

This is the H-alpha line — a prominent red line in the hydrogen spectrum.

If an electron is in the $n^\text{th}$ level, the maximum number of spectral lines it can emit while cascading down:

N = \frac{n(n-1)}{2}

For $n = 4$ : $N = 4 \times 3/2 = 6$ lines. This formula is a JEE Main favourite.

flowchart TD
    A["n=∞ (0 eV, ionisation)"] --> B["n=5 (−0.54 eV)"]
    B --> C["n=4 (−0.85 eV)"]
    C --> D["n=3 (−1.51 eV)"]
    D --> E["n=2 (−3.4 eV)"]
    E --> F["n=1 (−13.6 eV)"]
    D -->|"Paschen series (IR)"| D
    E -->|"Balmer series (Visible)"| E
    F -->|"Lyman series (UV)"| F
    style F fill:#ff6b6b,stroke:#333
    style E fill:#90EE90,stroke:#333

Why This Works

Bohr’s model explains why hydrogen has discrete energy levels — the electron can only occupy specific orbits where the angular momentum is quantised ( $L = n\hbar$ ). Transitions between these levels emit or absorb photons with energy exactly equal to the energy difference: $E = h\nu = E_{n_i} - E_{n_f}$ .

The Balmer series falls in the visible range because the energy differences for transitions to $n = 2$ correspond to photon wavelengths between 400-700 nm.

Alternative Method

For quick calculations, use the energy difference method instead of Rydberg’s formula. For the first Balmer line: $\Delta E = E_3 - E_2 = -1.51 - (-3.4) = 1.89$ eV. Then $\lambda = \frac{hc}{E} = \frac{1240 \text{ eV}\cdot\text{nm}}{1.89 \text{ eV}} = 656$ nm. The shortcut $hc = 1240$ eV $\cdot$ nm saves significant calculation time.

Common Mistake

Students confuse the series limit with the first line. The series limit of the Balmer series is the transition from $n = \infty$ to $n = 2$ (shortest wavelength, highest energy in the series). The first line is from $n = 3$ to $n = 2$ (longest wavelength, lowest energy in the series). “First line” means the lowest energy transition within the series — the one involving adjacent levels.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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