Hydrogen spectral series

Heat hydrogen gas until it glows and, instead of a smooth rainbow, you get a handful of sharp, isolated lines — a fingerprint unique to hydrogen. Bohr's model explains exactly why: an electron falling from a higher energy level $n_i$ to a lower one $n_f$ emits a photon of one precise energy, and every landing level $n_f$ produces its own family of lines, named Lyman, Balmer, Paschen, Brackett and Pfund. A single formula, the Rydberg formula, predicts every wavelength in every series to four significant figures, and it was one of the first great triumphs of quantum theory when it matched decades-old spectroscopic data almost perfectly. This article builds that formula from Bohr's energy levels, works through the classic Balmer red line and the Lyman series limit, and flags the mix-ups examiners love to test.

The core insight

Each hydrogen spectral series is the set of photon wavelengths from transitions $n_i \to n_f$ landing on a fixed lower level $n_f$ , and every wavelength obeys the Rydberg formula $1/\lambda = R(1/n_f^2 - 1/n_i^2)$ .

The core idea

A hydrogen atom's electron can only occupy discrete energy levels $E_n = -13.6/n^2$ eV. When it drops from an initial level $n_i$ to a final level $n_f < n_i$ , it emits a single photon carrying away exactly the energy difference $\Delta E = E_{n_i} - E_{n_f}$ . Because $\Delta E = hc/\lambda$ , this energy gap fixes the photon's wavelength precisely — no continuous range is possible, which is why the emission spectrum is a set of sharp lines rather than a smear. Fixing the landing level $n_f$ and letting $n_i$ run over $n_f+1, n_f+2, n_f+3,\dots$ generates one whole family of lines — a series. There are five series inside the syllabus, and each sits in a different region of the electromagnetic spectrum:

\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right)

where: $\lambda$ is the emitted wavelength (report in nm, as the syllabus specifies), $R = 1.097\times10^{7}\ \text{m}^{-1}$ is the Rydberg constant for hydrogen, $n_f$ is the fixed final (lower) level that names the series, and $n_i > n_f$ is the initial (higher) level, taking values $n_f+1, n_f+2, \dots$ for successive lines.

Lyman ( $n_f=1$ ): every line ultraviolet.
Balmer ( $n_f=2$ ): the only series with lines in the visible band — this is the historical reason it was discovered first, before Bohr's theory existed.
Paschen, Brackett, Pfund ( $n_f = 3, 4, 5$ ): all infrared.

Within a series, as $n_i \to \infty$ the wavelength shrinks toward a minimum called the series limit, $1/\lambda_{\text{limit}} = R/n_f^2$ , corresponding physically to the electron being ejected entirely (ionisation from level $n_f$ ). At the other end, the transition $n_i = n_f+1 \to n_f$ has the smallest energy gap and so gives the series' longest-wavelength, lowest-energy line.

Figure

Energy level diagram of hydrogen showing downward arrows from higher levels to n equals one forming the Lyman series in ultraviolet to n equals two forming the Balmer series in visible light and to n equals three forming the Paschen series in infrared

Why the formula follows from Bohr's model

Derivation ADeriving the Rydberg formula from the Bohr energy levels

The goal is to show that the Rydberg formula is not a separate empirical rule but a direct algebraic consequence of Bohr's postulate that hydrogen's electron energies are quantised as $E_n = -13.6/n^2$ eV combined with the photon frequency condition, so that any student who accepts Bohr's energy-level formula can derive the wavelength law for every spectral series without any new physical assumption.

Step 1. Write the photon energy released in a transition from $n_i$ to $n_f$ (with $n_i > n_f$ ) as the difference of the two Bohr energy levels: $\Delta E = E_{n_i} - E_{n_f} = -\dfrac{13.6}{n_i^2} - \left(-\dfrac{13.6}{n_f^2}\right) = 13.6\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right)$ eV. Since $n_i > n_f$ , the bracket is positive, so $\Delta E$ is positive — energy is released, as expected for emission.

Step 2. Convert this energy into a wavelength using the photon relation $\Delta E = h\nu = hc/\lambda$ , so $\dfrac{1}{\lambda} = \dfrac{\Delta E}{hc}$ . Substituting the Step 1 expression gives $\dfrac{1}{\lambda} = \dfrac{13.6\ \text{eV}}{hc}\left(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}\right)$ .

Step 3. Identify the constant in front as the Rydberg constant: $R \equiv \dfrac{13.6\ \text{eV}}{hc}$ . Working this out numerically using $h = 6.63\times10^{-34}\ \text{J s}$ , $c = 3.0\times10^{8}\ \text{m/s}$ and $13.6\ \text{eV} = 2.18\times10^{-18}\ \text{J}$ gives $R = 1.097\times10^{7}\ \text{m}^{-1}$ , matching the value measured spectroscopically decades before Bohr's theory existed.

Step 4. Read off the result: $\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}\right)$ , the Rydberg formula, valid for any pair $n_i > n_f$ . Fixing $n_f$ and varying $n_i$ produces one named series; this agreement between a constant derived from atomic theory and a constant measured from spectral lines was strong evidence for Bohr's model.

Worked example

Calculate the wavelength of the $H_\alpha$ line of the Balmer series (transition $n_i = 3 \to n_f = 2$ ), and find the shortest wavelength (series limit) of the Lyman series.

Step 1. For $H_\alpha$ , apply the Rydberg formula with $n_f = 2$ , $n_i = 3$ : $\dfrac{1}{\lambda} = R\left(\dfrac{1}{2^2}-\dfrac{1}{3^2}\right) = 1.097\times10^7\left(\dfrac{1}{4}-\dfrac{1}{9}\right) = 1.097\times10^7\times\dfrac{5}{36} = 1.524\times10^6\ \text{m}^{-1}$ .

Step 2. Invert: $\lambda = \dfrac{1}{1.524\times10^6} = 6.56\times10^{-7}\ \text{m} = 656\ \text{nm}$ — the well-known red line of the Balmer series.

Step 3. For the Lyman series limit, set $n_f = 1$ and let $n_i \to \infty$ , so $1/n_i^2 \to 0$ : $\dfrac{1}{\lambda} = R\left(\dfrac{1}{1^2}-0\right) = 1.097\times10^7\ \text{m}^{-1}$ .

Step 4. Invert: $\lambda = \dfrac{1}{1.097\times10^7} = 9.12\times10^{-8}\ \text{m} = 91.2\ \text{nm}$ . Answer: the Balmer $H_\alpha$ line is at $656\ \text{nm}$ (red, visible), and the Lyman series limit is at $91.2\ \text{nm}$ (ultraviolet), the shortest wavelength hydrogen can emit onto the ground state.

Competitive stretch

Find the longest wavelength line of the Paschen series ( $n_f=3$ , smallest gap is $n_i=4$ ). $\dfrac{1}{\lambda} = R\left(\dfrac{1}{9}-\dfrac{1}{16}\right) = 1.097\times10^7\times\dfrac{7}{144} = 5.33\times10^5\ \text{m}^{-1}$ , so $\lambda = 1.876\times10^{-6}\ \text{m} = 1876\ \text{nm}$ , firmly in the infrared — consistent with Paschen being an infrared-only series.

Common traps

Swapping $n_i$ and $n_f$ — the smaller quantum number is always the fixed final level naming the series, and $n_i$ must be strictly greater than $n_f$ .
Assuming every series has visible lines — only Balmer does; Lyman is entirely UV and Paschen/Brackett/Pfund are entirely infrared.
Confusing the series limit (shortest wavelength, $n_i \to \infty$ ) with the longest-wavelength line (smallest gap, $n_i = n_f+1$ ) — they are opposite ends of the same series.
Quoting wavelengths in ångströms out of habit instead of nanometres, as the syllabus requires.

Ask the tutor: Want to practise identifying which series a given wavelength belongs to, or work through more Rydberg-formula problems step by step?

Why does fixing n_f produce a whole series instead of a single line?

Because $n_f$ only fixes the level the electron lands on, not the level it starts from. Any higher level $n_i = n_f+1, n_f+2, n_f+3, \dots$ can supply the electron, and each starting point gives a different energy gap and therefore a different wavelength. All of these transitions share the same final level, so they are grouped together as one named series — Lyman for $n_f=1$ , Balmer for $n_f=2$ , and so on — even though every individual line within it has its own distinct wavelength.

Why is the Balmer series the only one with visible lines?

Visible light spans roughly $400$ to $700\ \text{nm}$ . Transitions landing on $n_f=1$ (Lyman) release more energy than any transition landing on $n_f=2$ , pushing every Lyman wavelength below $122\ \text{nm}$ , deep in the ultraviolet. Transitions landing on $n_f=3$ or higher release less energy than Balmer transitions, pushing those wavelengths above $700\ \text{nm}$ into the infrared. Balmer, sitting in between, is the one series whose energy gaps happen to correspond to visible-light photon energies — which is also why it was the first series ever observed, in the nineteenth century, well before Bohr's theory.

Does the same Rydberg constant R work for every series and every line?

Yes. $R = 1.097\times10^{7}\ \text{m}^{-1}$ is a single fixed constant that applies to every transition in every series of the hydrogen atom — only $n_f$ and $n_i$ change from line to line and series to series. This is what makes the formula so powerful: once you know $R$ , you can predict the wavelength of any hydrogen line, in any series, without needing separate formulas or separate measured constants for Lyman versus Balmer versus Paschen.