Hydrogen Spectrum — Lyman, Balmer, Paschen

Opening — The Spectrum That Built Quantum Mechanics

Hydrogen’s emission spectrum is the cleanest piece of evidence we have that energy levels are quantized. When you pass an electric current through hydrogen gas and split the light with a prism, you don’t get a rainbow — you get sharp, discrete lines at specific wavelengths. Bohr’s whole model was built to explain why.

We’re going to walk through all six named series — Lyman, Balmer, Paschen, Brackett, Pfund, Humphreys — figure out which spectral region they fall in, derive their wavelengths from the Rydberg formula, and look at what JEE and NEET actually ask.

The Balmer series is the only one in the visible region, which is why hydrogen looks reddish-purple in a discharge tube. Everything else is UV (Lyman) or IR (Paschen onwards).

Key Terms & Definitions

Energy levels of hydrogen:

$E_n = -\frac{13.6}{n^2} \text{ eV}, \quad n = 1, 2, 3, \ldots$

The negative sign means the electron is bound. $n = 1$ is the ground state at $-13.6$ eV; $n = \infty$ is the ionization limit at 0 eV.

Spectral series: a family of lines all ending on the same lower level $n_f$ . Transitions from higher levels $n_i > n_f$ produce the family.

Rydberg formula:

$\frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$

where $R_H = 1.097 \times 10^7$ m⁻¹ is the Rydberg constant for hydrogen.

$\frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$

For absorption: $n_i > n_f$ (lower → higher). For emission: same formula, but the photon is released as the electron drops from $n_i$ to $n_f$ .

Methods — Each Series in Turn

Lyman Series ( $n_f = 1$ )

Transitions land on the ground state. All Lyman lines are in the ultraviolet.

Wavelength range: 91 nm (series limit, $n_i = \infty$ ) to 122 nm (longest, $n_i = 2$ ).

The first Lyman line ( $n_i = 2 \to n_f = 1$ ) has:

$\frac{1}{\lambda} = R_H \left(1 - \frac{1}{4}\right) = \frac{3R_H}{4}$

$\lambda \approx 121.6 \text{ nm}$

Balmer Series ( $n_f = 2$ )

The famous visible series — discovered by Balmer in 1885 from raw lab data, before quantum mechanics.

H-alpha ( $n_i = 3 \to 2$ ): $\lambda = 656.3$ nm (red). H-beta ( $n_i = 4 \to 2$ ): $\lambda = 486.1$ nm (cyan-blue). H-gamma ( $n_i = 5 \to 2$ ): $\lambda = 434.0$ nm (violet). H-delta ( $n_i = 6 \to 2$ ): $\lambda = 410.2$ nm (deep violet).

Series limit: $\lambda = 364.6$ nm (UV).

Paschen Series ( $n_f = 3$ )

All Paschen lines are in the near-infrared. Series limit at 820.4 nm; longest line ( $n_i = 4 \to 3$ ) at 1875 nm.

Brackett Series ( $n_f = 4$ )

Mid-infrared. Series limit at 1458 nm.

Pfund Series ( $n_f = 5$ )

Far-infrared. Series limit at 2278 nm.

Humphreys Series ( $n_f = 6$ )

Even deeper IR. Mostly an academic curiosity.

Solved Examples

Example 1 (Easy — CBSE)

Find the wavelength of the first line of the Balmer series. Use $R_H = 1.097 \times 10^7$ m⁻¹.

First Balmer line: $n_i = 3, n_f = 2$ .

$\frac{1}{\lambda} = R_H \left(\frac{1}{4} - \frac{1}{9}\right) = R_H \cdot \frac{5}{36}$

$\lambda = \frac{36}{5R_H} = \frac{36}{5 \cdot 1.097 \times 10^7} \approx 6.56 \times 10^{-7} \text{ m} = 656 \text{ nm}$

This is the classic red H-alpha line.

Example 2 (Medium — JEE Main)

Find the shortest wavelength in the Lyman series.

Shortest wavelength = series limit, $n_i = \infty$ .

$\frac{1}{\lambda} = R_H \left(1 - 0\right) = R_H$

$\lambda = 1/R_H \approx 9.12 \times 10^{-8} \text{ m} = 91.2 \text{ nm}$

In the deep UV.

Example 3 (Hard — JEE Advanced 2023)

An electron in a hydrogen atom in $n = 4$ falls to $n = 2$ . Find: (a) the energy of the photon emitted, (b) its wavelength, (c) which series this line belongs to.

Energy: $\Delta E = E_4 - E_2 = -13.6/16 - (-13.6/4) = -0.85 + 3.4 = 2.55$ eV.

Wavelength: $\lambda = hc/\Delta E$ . With $hc = 1240$ eV·nm: $\lambda = 1240/2.55 \approx 486.3$ nm.

This is the H-beta line of the Balmer series (since $n_f = 2$ ).

Exam-Specific Tips

JEE Main: 1 question almost every year on hydrogen spectrum — usually finding wavelength or identifying which series. NEET: regular feature, focuses on series limits and ionization energy. CBSE Class 12: derive Rydberg formula from Bohr’s model. Memorize $E_n = -13.6/n^2$ eV — this is non-negotiable.

Quick conversion: photon energy in eV = $1240/\lambda$ (nm). So a 620 nm photon has energy 2 eV. This shortcut saves time in MCQs.

Common Mistakes to Avoid

Mistake 1: Mixing $n_i$ and $n_f$ in the Rydberg formula. The formula gives a positive $1/\lambda$ when $n_i > n_f$ . If you get a negative answer, you’ve swapped them.

Mistake 2: Treating all series as visible. Only Balmer (mostly) is visible. Lyman is UV; Paschen and beyond are IR.

Mistake 3: Using the Bohr formula for non-hydrogen atoms without the $Z^2$ factor. For hydrogenic ions (He⁺, Li²⁺): $E_n = -13.6 Z^2/n^2$ eV.

Mistake 4: Confusing series limit (shortest wavelength, highest energy) with first line (longest wavelength in that series).

Mistake 5: Plugging in $R_H$ without checking units — Rydberg constant in m⁻¹ gives wavelength in metres; in cm⁻¹ gives cm. Watch the powers of 10.

Practice Questions

Q1. Find the wavelength of the first Lyman line.

$n_i = 2 \to n_f = 1$ . $1/\lambda = R_H(1 - 1/4) = 3R_H/4$ . $\lambda \approx 121.5$ nm.

Q2. Which spectral line in the Balmer series has the longest wavelength?

The transition from $n = 3$ to $n = 2$ — H-alpha, 656 nm.

Q3. Find the energy of the photon emitted when an electron in He⁺ falls from $n = 3$ to $n = 1$ .

For He⁺ ( $Z = 2$ ): $E_n = -13.6 \cdot 4/n^2$ . $\Delta E = -54.4/9 - (-54.4/1) = -6.04 + 54.4 = 48.36$ eV.

Q4. Series limit of Paschen series for hydrogen?

$1/\lambda = R_H/9$ , so $\lambda = 9/R_H \approx 820.4$ nm.

Q5. A hydrogen atom in the ground state absorbs a 12.75 eV photon. Find the resulting state.

$E_{\text{final}} = -13.6 + 12.75 = -0.85$ eV. So $-13.6/n^2 = -0.85 \implies n^2 = 16 \implies n = 4$ .

Q6. Which series of hydrogen lies entirely in the visible?

Balmer series — at least the first four lines (656, 486, 434, 410 nm). The series limit at 365 nm is technically UV.

FAQs

Q: Why are spectral lines discrete and not continuous? Because electron energy levels are quantized. Only specific transitions are allowed, each producing a specific photon energy and hence a specific wavelength.

Q: Why is the Balmer series visible while others aren’t? Because $n_f = 2$ corresponds to energy gaps in the visible-light range (1.5-3 eV). $n_f = 1$ has gaps too large (UV); $n_f = 3$ has gaps too small (IR).

Q: What is the ionization energy of hydrogen? $13.6$ eV — the energy needed to remove the electron from $n = 1$ to $n = \infty$ .

Q: Does the Rydberg formula work for other atoms? Only for hydrogenic (single-electron) systems like He⁺, Li²⁺. For multi-electron atoms, the screening of inner electrons makes the levels deviate from $-13.6Z^2/n^2$ .

Q: Why do we call them “Lyman” and “Balmer”? They are named after the physicists (Theodore Lyman, Johann Balmer) who first identified each series experimentally.

Q: What’s the difference between an emission and absorption spectrum? Emission: hot atoms emit specific wavelengths. Absorption: cool atoms absorb the same wavelengths from a continuous source. Both follow the Rydberg formula.

Q: How does Bohr’s model explain the spectrum? Electrons orbit only in quantized levels with $E_n \propto -1/n^2$ . Photons are emitted when electrons drop between levels, with $h\nu = E_i - E_f$ . This gives the Rydberg formula directly.