The Rydberg formula

Result

The Rydberg formula

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Statement

For a hydrogen atom, the wavelength \(\lambda\) of light emitted or absorbed as the electron transitions between two allowed energy levels \(n_1\) and \(n_2\) (integers, \(n_2>n_1\ge1\)) satisfies \(\dfrac{1}{\lambda}=R_H\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)\), where \(R_H\approx1.097\times10^7\,\text{m}^{-1}\) is the Rydberg constant for hydrogen. For a hydrogen-like ion of nuclear charge \(Z\), the same formula holds with \(R_H\) replaced by \(Z^2R_H\).

Why it matters

This is the formula chemistry inherited before it had any theoretical explanation: Johannes Rydberg fit it empirically to hydrogen's known spectral lines in 1888, twenty-five years before Bohr's model gave a mechanism for why it should hold at all. It is the working tool spectroscopists actually use — given any two energy levels, it predicts the exact wavelength of the transition between them, and conversely, measuring a spectral line's wavelength pins down which two levels are involved.

Its historical role goes beyond hydrogen. The pattern Rydberg found — a formula built from the difference of two reciprocal-square terms — turned out to describe every one-electron system, and inspired the search for similar regularities in the far more complicated many-electron spectra of the rest of the periodic table, seeding what became the systematic classification of atomic spectra used throughout the rest of this unit.

Hypotheses

One-electron atom or ion (built on the Bohr model's own restriction).For any atom with two or more electrons, the energy levels are not simply \(-13.6Z^2/n^2\,\text{eV}\) (electron–electron repulsion breaks the clean \(1/n^2\) dependence, splitting a single \(n\) into multiple sub-levels of different energy depending on the orbital shape), so the two-term reciprocal-square difference formula does not hold; multi-electron spectra require an empirical Rydberg-Ritz correction with an extra, element-specific "quantum defect" added to \(n\). Transition is between two definite, allowed energy levels (an electric-dipole-allowed transition).Not every pair of levels gives an observed line: quantum-mechanical selection rules (governed by the orbital angular momentum quantum number \(l\), covered in this unit's quantum-numbers-and-orbitals result) forbid many transitions the bare energy-difference formula would otherwise predict as allowed; the Rydberg formula gives the correct wavelength if the transition occurs, but says nothing about whether it does.

Proof

E_{n_1} = -\frac{13.6\,Z^2}{n_1^2}\text{ eV}, \qquad E_{n_2} = -\frac{13.6\,Z^2}{n_2^2}\text{ eV}

The two energy levels involved in the transition, from the Bohr model result (this unit, previous page), applied to a hydrogen-like system of nuclear charge \(Z\). A

\Delta E = E_{n_2}-E_{n_1} = -13.6Z^2\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)\text{ eV} = 13.6Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\text{ eV}

Take \(n_2>n_1\), so the electron falls from the higher level \(n_2\) to the lower level \(n_1\), releasing energy as a photon; \(\Delta E\) here is defined as the (positive) energy of that emitted photon, hence the sign flip on subtracting. A

\Delta E = \frac{hc}{\lambda} \quad\Longrightarrow\quad \frac{1}{\lambda} = \frac{\Delta E}{hc}

The Planck–Einstein relation for a photon's energy in terms of its wavelength; \(h\) is Planck's constant, \(c\) the speed of light. Solve for \(1/\lambda\) directly. A

\frac{1}{\lambda} = \frac{13.6Z^2\,\text{eV}}{hc}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \equiv Z^2R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)

Substitute Step 2 into Step 3 and define \(R_H\equiv\dfrac{13.6\,\text{eV}}{hc}\), collecting every physical constant (electron mass, charge, \(\varepsilon_0\), \(h\), \(c\)) into the single number \(R_H\approx1.097\times10^7\,\text{m}^{-1}\) that Rydberg originally fit purely from spectral-line measurements, decades before any of these individual constants entered the story. B

Result

\frac{1}{\lambda} = Z^2R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right), \qquad R_H\approx1.097\times10^7\,\text{m}^{-1}

Reading. Every spectral line of a one-electron atom or ion is fixed by just two integers, the initial and final principal quantum numbers; the same single constant \(R_H\) (rescaled by \(Z^2\) for ions) governs every line of every one-electron species.

Scope. Exact for hydrogen and hydrogen-like ions (He\(^+\), Li\(^{2+}\), …). For multi-electron atoms, only an approximate, element-specific analogue (the Rydberg–Ritz formula, with an empirical quantum defect) holds, and even that fails to capture fine structure and other multi-electron effects.

Corollaries & converses

Fixing \(n_1=1\) and varying \(n_2=2,3,4,\dots\) generates the Lyman series (ultraviolet); \(n_1=2\) generates the Balmer series (visible, historically the first observed); \(n_1=3\) the Paschen series (infrared) — three named series, one formula.
As \(n_2\to\infty\) for fixed \(n_1\), \(1/\lambda\to Z^2R_H/n_1^2\), the series limit — the shortest wavelength (highest energy) line in that series, corresponding to ionisation from level \(n_1\).
Converse: given a measured wavelength that fits the formula for some integers \(n_1,n_2\), one can work backward to identify the transition — the basis of atomic emission spectroscopy as an identification technique, since every element's (multi-electron, Rydberg-Ritz-corrected) spectral "fingerprint" is unique.

Fails without

Drop the one-electron restriction: sodium's characteristic yellow doublet (589.0 and 589.6 nm, the "sodium D lines") comes from a single-electron transition in the outermost electron, but the two very close wavelengths cannot be reproduced by simple integers \(n_1,n_2\) in the bare formula — that near-degeneracy splitting is a spin–orbit coupling effect entirely outside the model this formula is built on.
Drop the requirement that the transition is dipole-allowed: the formula would predict a line for, say, an \(n=2,l=0\) to \(n=1,l=0\) transition (both \(s\)-orbitals) at the correct energy, but this specific transition is forbidden by the electric-dipole selection rule \(\Delta l=\pm1\) and is not observed at meaningful intensity; the wavelength the formula predicts is real, but no line appears there in an ordinary emission spectrum.

Common errors

Swapping \(n_1\) and \(n_2\) and getting a negative \(1/\lambda\). By convention \(n_1\) is always the smaller (lower-energy) level; \(n_2>n_1\) always, whether the transition is emission (\(n_2\to n_1\)) or absorption (\(n_1\to n_2\)) — the wavelength magnitude is the same either way, only the direction of energy flow differs.
Forgetting to square \(Z\) for a hydrogen-like ion, applying \(R_H\) unscaled to He\(^+\) or Li\(^{2+}\) spectral lines.
Using the Rydberg formula directly on a multi-electron atom's spectrum without the Rydberg–Ritz quantum-defect correction, then being confused when the predicted lines do not match measurement.

Discussion

Rydberg's 1888 formula was itself a generalisation of an earlier, narrower pattern: Johann Balmer had noticed in 1885 that the four visible hydrogen lines then known fit \(\lambda=364.56\,\text{nm}\times\dfrac{n^2}{n^2-4}\) for \(n=3,4,5,6\) — a numerological curiosity with no physical grounding. Rydberg recast Balmer's formula in reciprocal-wavelength form and showed it generalised to other known series (in alkali-metal spectra) with a second free integer in place of Balmer's fixed \(4=2^2\), arriving at the two-integer form used today. Neither Balmer nor Rydberg had any explanation for why the pattern held; that had to wait for Bohr's model twenty-five years later.

The Rydberg constant itself is one of the most precisely measured quantities in physics — known today to better than one part in \(10^{12}\) via laser spectroscopy of hydrogen — making the Bohr-derived expression for it (Proof, Step 4) a genuine, exacting test of the underlying atomic constants \(m_e,e,\varepsilon_0,h\) that enter it.

A small but real correction (below the precision quoted on this page): the Bohr/Rydberg derivation uses the electron mass \(m_e\) alone, treating the nucleus as infinitely heavy and fixed. The correct two-body treatment replaces \(m_e\) with the reduced mass \(\mu=m_em_N/(m_e+m_N)\) of the electron–nucleus system, shifting \(R_H\) by about \(1\) part in \(1800\) (the electron-to-proton mass ratio) between ordinary hydrogen and, say, deuterium — a shift large enough to have let Harold Urey discover deuterium spectroscopically in 1931, purely from this tiny wavelength difference.

Common misconception: that the Rydberg formula and the Bohr model are the same result stated two ways. The Rydberg formula is older, purely empirical, and correct as a curve fit; the Bohr model is a physical mechanism that happens to reproduce it exactly for one-electron systems. A theory can match the Rydberg formula's numbers (as Bohr's does) without the formula itself explaining why — that explanatory gap is precisely what Bohr's postulate closed.

Worked examples

\text{Balmer-}\alpha\ (\text{H-}\alpha): n_1=2,\ n_2=3 \ \Longrightarrow\ \frac{1}{\lambda}=R_H\left(\frac14-\frac19\right)=R_H\cdot\frac{5}{36}

The first line of the Balmer series, the red line at the heart of the visible hydrogen spectrum, historically the first hydrogen line pattern noticed by Balmer. A

\lambda = \frac{36}{5R_H} = \frac{36}{5\times1.097\times10^7\,\text{m}^{-1}} \approx 656.1\,\text{nm}

Direct evaluation. \(656.1\,\text{nm}\) is the well-known deep-red H-\(\alpha\) line, used throughout astronomy to identify hydrogen gas (e.g. emission nebulae, solar prominences); the small gap to the precisely measured \(656.28\,\text{nm}\) is the reduced-mass correction discussed below. A

\lambda_{\text{H-}\alpha}\approx656.1\,\text{nm}\ \ (\text{electron mass; precisely measured value: }656.28\,\text{nm})

Reading. One line, two small integers, and a single constant reproduce a wavelength close to its measured value; the residual \(\sim\!0.2\,\text{nm}\) gap is accounted for below, not a failure of the formula.

Scope. The same two-step calculation, for any \(n_1,n_2\) and any hydrogen-like \(Z\), gives every line this system can emit or absorb.

Problems

Find the wavelength of the Lyman-series limit (the shortest-wavelength line of the Lyman series, \(n_1=1\), \(n_2\to\infty\)) for hydrogen.
Solution
As \(n_2\to\infty\), \(1/n_2^2\to0\), so \(1/\lambda\to R_H/n_1^2=R_H\) (since \(n_1=1\)). Then \(\lambda=1/R_H=1/(1.097\times10^7\,\text{m}^{-1})\approx91.1\,\text{nm}\) — the shortest wavelength hydrogen can emit in the Lyman series, corresponding to an electron falling from the ionisation threshold (\(n=\infty\), \(E=0\)) directly to the ground state.
A spectral line at \(\lambda=102.6\,\text{nm}\) is observed. Identify the transition in hydrogen that produces it (assume it is a Lyman-series line, \(n_1=1\)).
Solution
\(1/\lambda=1/(102.6\times10^{-9}\,\text{m})\approx9.75\times10^6\,\text{m}^{-1}\). Setting \(9.75\times10^6=1.097\times10^7(1-1/n_2^2)\) gives \(1-1/n_2^2\approx0.889\), so \(1/n_2^2\approx0.111\), \(n_2^2\approx9\), \(n_2=3\). This is the Lyman-\(\beta\) line, the \(n=3\to n=1\) transition.
Use the formula to find the ratio of the Lyman-\(\alpha\) (\(n_1=1,n_2=2\)) to Balmer-\(\alpha\) (\(n_1=2,n_2=3\)) wavelengths, and check the result against the two computed values on this page (\(121.5\,\text{nm}\) and \(656.1\,\text{nm}\)).
Solution
\(1/\lambda_{\text{Ly-}\alpha}=R_H(1-\tfrac14)=\tfrac34R_H\); \(1/\lambda_{\text{H-}\alpha}=R_H(\tfrac14-\tfrac19)=\tfrac{5}{36}R_H\). Ratio: \(\dfrac{\lambda_{\text{Ly-}\alpha}}{\lambda_{\text{H-}\alpha}}=\dfrac{1/\lambda_{\text{H-}\alpha}}{1/\lambda_{\text{Ly-}\alpha}}=\dfrac{5/36}{3/4}=\dfrac{5}{27}\approx0.185\). Checking against the computed values: \(121.5/656.1\approx0.185\) — matches, confirming both worked examples are mutually consistent.
Explain, using the Hypotheses above, why the Rydberg formula (unmodified) cannot be used to predict the wavelength of the sodium D line at \(589\,\text{nm}\), even though sodium's outer electron is, loosely, "one electron."
Solution
Sodium has ten inner (core) electrons in addition to its one valence electron; although the valence electron is often described loosely as "hydrogen-like," it moves in the field of the nucleus screened by the ten core electrons, not the bare nuclear charge \(Z=11\) the formula assumes. The effective nuclear charge felt by the valence electron is far smaller and depends on the orbital shape (through the quantum defect), which the bare Rydberg formula has no term for — exactly the one-electron-atom hypothesis violation flagged above, requiring the Rydberg–Ritz correction instead.