The Bohr Energy Spectrum and Rydberg Constant

Before anyone knew what an atom was, Johannes Rydberg in Lund was doing something strange with spectral tables: instead of cataloguing wavelengths, he inverted them. Written as wave numbers — reciprocal wavelengths — every line of every series he examined became a difference of two terms, and the terms of hydrogen were breathtakingly simple: a single constant divided by the square of an integer. He published the formula in 1888. Nobody, including Rydberg, could say why it worked, or why his constant $R$ came out the same for series after series. It was a law waiting twenty-five years for a theory. The previous article supplied the theory: the Coulomb radial equation tolerates bound states only at $E_n = -13.6\ \mathrm{eV}/n^2$ . This article unpacks that spectrum — what the $13.6$ is made of, why it can be written three equivalent ways (one of them, $\tfrac12\alpha^2 m_ec^2$ , explains at a glance why atoms are "soft"), and how the ladder of levels becomes the ladder of lines: Lyman in the ultraviolet, Balmer in the visible, Paschen beyond. Rydberg's fitted constant becomes a calculable number, $R_\infty = 1.0974\times10^7\ \mathrm{m^{-1}}$ — today the most precisely known constant in physics. Then comes the fine print that turned into discoveries. The electron does not orbit a fixed point; the proton recoils, and honesty about the two-body problem replaces $m_e$ by the reduced mass $\mu$ . That one-part-in-two-thousand correction split hydrogen from deuterium on a photographic plate — a Nobel Prize hiding in the third decimal place.

Key takeaways

The hydrogen bound-state ladder is $E_n = -E_{\mathrm{Ry}}/n^2$ with $E_{\mathrm{Ry}} = 13.6057\ \mathrm{eV}$ : binding energy, ionization cost, and photon bookkeeping all read off one formula.
The Rydberg energy wears three equivalent outfits: $\dfrac{m_e e^4}{2(4\pi\varepsilon_0)^2\hbar^2} = \dfrac{e^2}{8\pi\varepsilon_0 a_0} = \dfrac{1}{2}\alpha^2 m_e c^2$ — the last one says atomic physics runs at $\alpha^2 \sim 10^{-4}$ of rest-mass energies.
Lines are differences: $1/\lambda = R\left(1/n_f^2 - 1/n_i^2\right)$ , with $R_\infty = E_{\mathrm{Ry}}/hc = 1.09737\times10^7\ \mathrm{m^{-1}}$ ; the choice of $n_f$ names the series (Lyman, Balmer, Paschen…).
Levels crowd toward the ionization edge as $n^{-2}$ , with spacing shrinking as $2E_{\mathrm{Ry}}/n^3$ — the corridor where quantum transitions merge into classical orbit frequencies.
The honest two-body problem replaces $m_e$ by $\mu = m_e M/(m_e + M)$ : hydrogen's constant drops to $R_{\mathrm H} = 1.09678\times10^7\ \mathrm{m^{-1}}$ , deuterium's sits $0.027\%$ higher — a measurable $1.79\ \mathrm{\mathring{A}}$ split at H $\alpha$ that discovered deuterium.
Key idea: one calculable constant converts integer arithmetic into every line of the hydrogen spectrum — and its small corrections are precision physics.

The core insight

For the attractive Coulomb potential the requirement that the radial wavefunction be normalizable (decaying at infinity, regular at the origin) quantizes the energy as $E_n=-E_{\mathrm{Ry}}/n^2$ . The energy depends only on $n=n_r+\ell+1$ , so different $\ell$ at fixed $n$ are degenerate — a hallmark of the pure $1/r$ potential.

Where this lives today

The Rydberg constant is the precision anchor of modern physics: two-photon spectroscopy of hydrogen's 1S–2S interval, compared against cesium clocks, pins $R_\infty$ to about $2$ parts in $10^{12}$ . That precision is a lever — the same measurements, cross-checked between ordinary and muonic hydrogen, exposed and then resolved the "proton radius puzzle" of the 2010s, revising the proton's charge radius by four percent. When CODATA adjusts the fundamental constants, hydrogen's spectrum sits near the top of the evidence pile. In astronomy the ladder works at both ends. Balmer lines are the workhorse tracers of star formation and galactic redshifts — H $\alpha$ surveys map ionized gas across the sky, with every observed wavelength just equation (1) plus Doppler shift. At the other end, radio telescopes log recombination lines between rungs like $n = 110 \to 109$ , photons from atoms so distended they exist only in interstellar near-vacuum. And the eV-scale arithmetic anchors chemistry itself: $13.6\ \mathrm{eV} = 1312\ \mathrm{kJ/mol}$ is the first entry in every ionization-energy table, the benchmark against which all other elements' electrons are judged.

The physical picture

Rydberg's move — think in wave numbers, and think in differences — is the picture to internalize. Draw the levels as a ladder of terms $T_n = R/n^2$ hanging below the ionization edge. A photon's inverse wavelength is never a term; it is always a gap between two terms. Fix the lower rung and sweep the upper one, and you generate a whole series of lines converging to a limit (the lower rung's depth); change the lower rung and the entire series relocates wholesale to a different waveband. That is why hydrogen's barcode comes in families: Lyman ( $n_f = 1$ ) crammed in the far ultraviolet, Balmer ( $n_f = 2$ ) straddling the visible, Paschen ( $n_f = 3$ ) in the infrared. The combination principle — any line's wave number is the difference of two others' — stops being a spectroscopic curiosity and becomes obvious bookkeeping. (Its dynamical face, that only differences of level energies are observable frequencies, is the Bohr-frequency doctrine of Spectral Decomposition and Time Evolution.)

Figure

Terms and differences — Rydberg's bookkeeping made visible

The second picture is about scale. Why electron-volts? Why visible light? The answer hides in the fine-structure constant: binding energies are $\tfrac12\alpha^2 m_ec^2$ , and $\alpha^2 \approx 5\times10^{-5}$ knocks the electron's half-MeV rest energy down to $13.6\ \mathrm{eV}$ . Atoms are soft because $\alpha$ is small: chemistry happens at billionths of nuclear energy scales, and the light atoms emit lands, not coincidentally, where thermal sources shine and photographic plates respond. The same number budgets the corrections: relativity enters at relative order $\alpha^2 \sim 10^{-4}$ , which is exactly where the clean $-1/n^2$ ladder starts to fray into fine structure. And a third, quieter picture: the nucleus moves. The electron and proton waltz about their common center of mass, so the "electron mass" in every formula is really the reduced mass of the pair. Make the nucleus heavier — deuterium instead of hydrogen — and every level deepens by a hundredth of a percent, every line shifts a fraction of an ångström toward the blue. Small numbers, but spectroscopes read five decimal places; this is where the spectrum becomes a scale for weighing nuclei.

The math: one constant, three faces, every line

The starting point is inherited whole from The Hydrogen Atom: Coulomb Radial Equation — termination of the radial series forced $E_n = -E_{\mathrm{Ry}}/n^2$ , and we do not re-derive it. This article's business is the constant out front. Write it explicitly:

E_n = -\frac{m_e e^4}{2(4\pi\varepsilon_0)^2\hbar^2}\,\frac{1}{n^2} = -\frac{13.6057\ \mathrm{eV}}{n^2}, \qquad n = 1, 2, 3, \dots \qquad (1)

where: $m_e$ — electron mass (upgraded to the reduced mass below); the prefactor $E_{\mathrm{Ry}}$ — the Rydberg energy. Read it three ways: $|E_n|$ is the binding energy of level $n$ , the ionization cost from level $n$ , and (in differences) the photon budget for every transition. The ground state binds by $13.6057\ \mathrm{eV}$ — equivalently $1312\ \mathrm{kJ/mol}$ , a number every chemistry table lists as hydrogen's ionization energy. The same constant wears three outfits, and each teaches something different:

E_{\mathrm{Ry}} = \frac{m_e e^4}{2(4\pi\varepsilon_0)^2\hbar^2} = \frac{e^2}{8\pi\varepsilon_0 a_0} = \frac{1}{2}\alpha^2 m_e c^2 \qquad (2)

where: $a_0$ — the Bohr radius; $\alpha = e^2/4\pi\varepsilon_0\hbar c \approx 1/137.036$ — the fine-structure constant. The first form is what the derivation delivers; the second says the binding energy is half the Coulomb energy at one Bohr radius (a virial-theorem fingerprint); the third prices atomic physics against relativity — $\alpha^2/2 \approx 2.66\times10^{-5}$ of the electron's rest energy, which is why a non-relativistic equation gets five digits right. The anatomy of $a_0$ and the full ladder of atomic scales built from these constants belong to The Bohr Radius and Characteristic Atomic Scales.

Derivation AFrom the energy ladder to the Rydberg constant

The goal is the working formula of spectroscopy: the wave number of any hydrogen line as integer arithmetic times one measurable constant, and the proof that the three faces of $E_{\mathrm{Ry}}$ in equation (2) are the same number. The strategy is bookkeeping, done carefully — energy conservation for the photon, then two substitutions. It is short because the hard work (the spectrum itself) was already done by the radial equation; what this card adds is the interface to experiment, which is wave numbers, not energies. Watch how naturally the "difference of two terms" structure Rydberg guessed in 1888 falls out.

Step 1. A photon carries the energy difference. For a downward jump $n_i \to n_f$ :

h\nu = E_{n_i} - E_{n_f} = E_{\mathrm{Ry}}\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right).

Why: emission requires $n_i > n_f$ , making the bracket positive; absorption runs the same arithmetic upward.

Step 2. Convert to the spectroscopist's currency. Divide by $hc$ to get inverse wavelength:

\frac{1}{\lambda} = R_\infty\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right), \qquad R_\infty = \frac{E_{\mathrm{Ry}}}{hc} = 1.09737\times10^7\ \mathrm{m^{-1}}.

Why: this is exactly Rydberg's 1888 form — a difference of terms $T_n = R_\infty/n^2$ — with his fitted constant now computed from $m_e, e, \hbar, c$ , $\varepsilon_0$ . The subscript $\infty$ flags the idealization of an infinitely heavy, motionless nucleus, repaired in the next card.

Step 3. Prove the three faces. Substitute $a_0 = 4\pi\varepsilon_0\hbar^2/m_ee^2$ into $e^2/8\pi\varepsilon_0 a_0$ : the $a_0$ upstairs cancels one power of $e^2$ and delivers $m_ee^4/2(4\pi\varepsilon_0)^2\hbar^2$ — face one equals face two. Then insert $\alpha = e^2/4\pi\varepsilon_0\hbar c$ into $\tfrac12\alpha^2 m_ec^2$ : the two powers of $c$ cancel and the same expression appears — face three. Why: the algebra is two lines each, but the payoff is conceptual: one face comes from the differential equation, one from electrostatics at the natural length, one from special relativity's price list. *Verified: all three faces evaluate to $13.6057\ \mathrm{eV}$ from CODATA inputs (agreement to $10^{-6}$ relative), and their symbolic equality is confirmed with sympy; $R_\infty = E_{\mathrm{Ry}}/hc = 1.0973731\times10^7\ \mathrm{m^{-1}}$ ; matches Griffiths §4.2 and the CODATA value.*

The series taxonomy is now mechanical. Fix $n_f$ and read off the family, every number below computed with the hydrogen-corrected constant $R_{\mathrm H}$ of the next card (vacuum wavelengths):

series	$n_f$	first line	series limit	waveband
Lyman	1	Ly $\alpha$ ( $2\to1$ ): $121.57\ \mathrm{nm}$	$91.18\ \mathrm{nm}$	far ultraviolet
Balmer	2	H $\alpha$ ( $3\to2$ ): $656.47\ \mathrm{nm}$	$364.71\ \mathrm{nm}$	visible (the barcode)
Paschen	3	( $4\to3$ ): $1875.6\ \mathrm{nm}$	$820.4\ \mathrm{nm}$	infrared
Brackett	4	( $5\to4$ ): $4.05\ \mathrm{\mu m}$	$1.46\ \mathrm{\mu m}$	infrared

Only Balmer's family lands in the eye's window — its photon energies run $1.89$ to $3.40\ \mathrm{eV}$ — which is why a nineteenth-century schoolteacher found those four lines and not Lyman's. (One convention flag: these are vacuum wavelengths; in air the readings sit about $0.18\ \mathrm{nm}$ lower at H $\alpha$ , $656.28\ \mathrm{nm}$ — the number most discharge-tube tables print.)

Go deeper · Derivation B · 1 minTwo bodies, one effective electron — the reduced mass

Derivation BTwo bodies, one effective electron — the reduced mass

The goal is to repair the one dishonest assumption so far: that the proton stands still. The strategy is the classic two-body separation, done at the operator level — split the Hamiltonian into center-of-mass and relative pieces, and watch the entire hydrogen solution survive with one substitution, $m_e \to \mu$ . This matters because the correction, though only a twentieth of a percent, is the leverage for comparing atoms that differ only in their nucleus: hydrogen against deuterium, ordinary hydrogen against muonic. Watch which mass appears where — the trap is assuming "heavier nucleus" changes nothing.

Step 1. Write the honest two-body Hamiltonian and change coordinates to $\mathbf{R} = (m_e\mathbf{r}_e + M\mathbf{r}_N)/(m_e + M)$ and $\mathbf{r} = \mathbf{r}_e - \mathbf{r}_N$ :

H = \frac{\mathbf{p}_e^2}{2m_e} + \frac{\mathbf{p}_N^2}{2M} + V(|\mathbf{r}_e - \mathbf{r}_N|) = \frac{\mathbf{P}^2}{2(m_e + M)} + \frac{\mathbf{p}^2}{2\mu} + V(r), \qquad \mu = \frac{m_e M}{m_e + M}.

Why: the kinetic identity is algebra (verified below); $[\mathbf{R}, \mathbf{P}]$ and $[\mathbf{r}, \mathbf{p}]$ each satisfy canonical commutators and commute with each other, so the two problems separate exactly.

Step 2. Discard the free center-of-mass motion and recognize the relative problem: it is the hydrogen problem already solved, with $\mu$ in place of $m_e$ . Every result inherits the swap:

E_n = -\frac{E_{\mathrm{Ry}}}{n^2}\,\frac{\mu}{m_e}, \qquad R_M = \frac{R_\infty}{1 + m_e/M}.

Why: $E_{\mathrm{Ry}} \propto m$ , so the whole ladder scales by $\mu/m_e = 1/(1 + m_e/M) < 1$ — a lighter effective particle binds slightly less.

Step 3. Put in numbers. For hydrogen $M = 1836.15\,m_e$ : $R_{\mathrm H} = R_\infty/1.000545 = 1.09678\times10^7\ \mathrm{m^{-1}}$ . For deuterium $M = 3670.48\,m_e$ : $R_{\mathrm D} = 1.09707\times10^7\ \mathrm{m^{-1}}$ — larger, because the heavier nucleus recoils less. Every deuterium line sits at slightly shorter wavelength: at H $\alpha$ , $656.47$ vs $656.29\ \mathrm{nm}$ , a split of $1.79\ \mathrm{\mathring{A}}$ . Why: this is the split Harold Urey resolved in 1932 — he photographed faint satellite lines exactly $1.8\ \mathrm{\mathring{A}}$ blueward of hydrogen's Balmer lines in evaporated liquid hydrogen, and deuterium was discovered inside a reduced-mass correction. *Verified: the kinetic-energy identity checked symbolically; $\mu \to m_e$ as $M \to \infty$ and $\mu = m_e/2$ for equal masses (sympy); $R_{\mathrm H}, R_{\mathrm D}$ and the $1.786\ \mathrm{\mathring{A}}$ H $\alpha$ split recomputed from mass ratios $1836.153$ and $3670.483$ ; matches Griffiths §5.2 problem set and Bethe & Salpeter.*

The reduced mass is also a dial. Turn it hard and you get new atoms wholesale: positronium (an electron bound to a positron, $M = m_e$ ) has $\mu = m_e/2$ exactly, so its Rydberg is half of hydrogen's — ground state $-6.80\ \mathrm{eV}$ , every line at double the wavelength. Muonic hydrogen (a muon, $m_\mu = 206.77\,m_e$ , bound to a proton) has $\mu = 185.84\,m_e$ : its ladder is $186\times$ deeper — a $2.53\ \mathrm{keV}$ ground state, X-ray territory — and its orbit $186\times$ smaller, about $285\ \mathrm{fm}$ . Those are mass-scalings; the equally powerful charge-scaling dial ( $Z^2$ , with its own stellar detective story) is the business of Hydrogenic Ions and Scaling with Nuclear Charge, which borrows this card's $\mu$ machinery whenever an exotic atom turns both dials at once.

Figure

One ladder, three atoms — the reduced mass as a dial

One more structural fact costs nothing now: the level spacing. Between adjacent rungs, $E_{n+1} - E_n \approx 2E_{\mathrm{Ry}}/n^3$ for large $n$ — the ladder crowds toward the edge faster than the levels approach it. That $n^{-3}$ is the corridor to classical physics: the photon frequency of an $n \to n-1$ hop approaches exactly the orbital frequency of a classical electron on the $n$ -th orbit (at $n = 50$ they differ by $3\%$ , vanishing as $n$ grows). Quantum jumps between high rungs sound like the classical orbit — the correspondence limit, and the reason microwave photons talk so fluently to Rydberg states.

Limiting cases

Case	Result	Meaning
ground state	$E_1 = -13.6057\ \mathrm{eV}$	hydrogen's ionization energy; $1312\ \mathrm{kJ/mol}$ in chemistry's units
$n \to \infty$	$E_n \to 0^-$ , spacing $\sim 2E_{\mathrm{Ry}}/n^3$	levels crowd toward the continuum; transition frequencies merge into classical orbit frequencies
$M \to \infty$	$\mu \to m_e$ , $R \to R_\infty$	the textbook idealization; real atoms sit a part-in- $M/m_e$ below it
equal masses	$\mu = m_e/2$	positronium: the whole spectrum at exactly half depth, $E_1 = -6.80\ \mathrm{eV}$
beyond leading order	shifts $\sim \alpha^2 E_n \sim 10^{-4}E_n$	fine structure splits the $\ell$ -degenerate rungs — the ladder's first crack (APRX chapter)

Common trap“The Rydberg is 13.6” — which Rydberg, and 13.6 what?

Two different objects share the name. The Rydberg energy $E_{\mathrm{Ry}} = 13.6057\ \mathrm{eV}$ is an energy; the Rydberg constant $R_\infty = 1.0974\times10^7\ \mathrm{m^{-1}}$ is an inverse wavelength. They are related by $E_{\mathrm{Ry}} = hcR_\infty$ , and mixing them up scrambles every spectroscopy calculation by a factor of $hc$ . Second ambiguity: $R_\infty$ versus $R_{\mathrm H}$ — the infinite-nucleus idealization versus hydrogen's actual constant, $0.054\%$ apart. Use $R_\infty$ for the pure theory number, $R_{\mathrm H}$ the moment you compare with a hydrogen lamp; at four significant figures the difference is already visible ( $656.11$ vs $656.47\ \mathrm{nm}$ for H $\alpha$ in vacuum). Third: $n$ is the principal quantum number, not the radial node count $n_r$ — they satisfy $n = n_r + \ell + 1$ , bookkeeping owned by the radial-equation article.

Worked examples

1 · Auditing the Balmer series (confidence builder)

Predict all four visible hydrogen lines from $R_{\mathrm H}$ , and use the fourth decimal place to tell hydrogen from deuterium.

Step 1. Balmer means $n_f = 2$ : $\dfrac{1}{\lambda} = R_{\mathrm H}\left(\dfrac{1}{4} - \dfrac{1}{n_i^2}\right)$ with $R_{\mathrm H} = 1.09678\times10^7\ \mathrm{m^{-1}}$ .

Step 2. Crank $n_i = 3, 4, 5, 6$ : $\lambda = \mathbf{656.47},\ \mathbf{486.27},\ \mathbf{434.17},\ \mathbf{410.29\ nm}$ (vacuum) — red, blue-green, and the violet pair. Modern vacuum measurements agree to about a part in $10^5$ , where fine structure (not an error in $R_{\mathrm H}$ ) takes over.

Step 3. The series limit: $n_i \to \infty$ gives $\lambda = 4/R_{\mathrm H} = \mathbf{364.7\ nm}$ — beyond it, a continuum from ionizing captures. Balmer's 1885 formula $\lambda = B\,n^2/(n^2 - 4)$ is this arithmetic in disguise: his fitted $B$ is the series limit, $4/R_{\mathrm H}$ .

Step 4. Now deuterium. $R_{\mathrm D} = 1.09707\times10^7\ \mathrm{m^{-1}}$ shifts H $\alpha$ to $656.29\ \mathrm{nm}$ : a split of $\Delta\lambda = \mathbf{1.79\ \mathring{A}}$ , deuterium on the blue side. Urey's 1932 plates showed precisely these faint satellites — a new isotope announced by the third decimal of a wavelength.

2 · A hydrogen atom the size of a virus (exam level)

Radio astronomers detect hydrogen recombination lines from atoms at $n = 50$ . Characterize the state: binding energy, transition frequency to $n = 49$ , physical size, and why such atoms cannot survive in a lab beaker.

Step 1. Binding: $E_{50} = -13.6057/2500 = \mathbf{-5.44\ meV}$ — four thousandths of a percent of the ground state's grip.

Step 2. The $50 \to 49$ hop: $\Delta E = 13.6057\left(\dfrac{1}{49^2} - \dfrac{1}{50^2}\right) = \mathbf{0.224\ meV}$ , i.e. $\nu = \Delta E/h = \mathbf{54.3\ GHz}$ — a radio photon, exactly the regime of the H $50\alpha$ recombination lines mapped in interstellar clouds. (The crowding estimate $2E_{\mathrm{Ry}}/n^3 = 0.218\ \mathrm{meV}$ undershoots by $3\%$ — the $n^{-3}$ law is exact only asymptotically.)

Step 3. Size: $\langle r \rangle \approx n^2 a_0 = 2500\,a_0 = \mathbf{132\ nm}$ — the atom is the size of a small virus, about 2,500 times its ground-state self.

Step 4. Fragility: room-temperature blackbody photons carry $kT = 25.9\ \mathrm{meV}$ , five times the binding energy — ambient radiation ionizes the atom almost on contact. Such states live only in the near-vacuum of interstellar space or in cryogenic, shielded traps — which is exactly where Rydberg-atom quantum computing keeps them.

Origins · 1 minJohannes Robert Rydberg and how this was discovered

Johannes Robert Rydberg

1854–1919 · Lund

Rydberg never measured a spectrum in his life; he mined other people's tables. Working through Thalén's and Ångström's wavelength catalogues at Lund in the 1880s, he made one move nobody else had committed to: he worked with the reciprocal of wavelength, and looked for structure in differences. In 1888–1890 it paid off spectacularly — series after series, element after element, fit $n = n_0 - N_0/(m + \mu')^2$ , and the constant $N_0$ was the same for every element, a universality nobody had any right to expect. Balmer's celebrated hydrogen formula turned out to be one special case of Rydberg's scheme. What he could not do was explain any of it: the constant sat in his papers as a brute empirical fact for twenty-five years, until Bohr's 1913 model — and then honestly, the radial equation — computed it from $e$ , $m_e$ , $\hbar$ , and $c$ . Rydberg lived just long enough to see the derivation; he died in 1919, having spent his last years in poor health, repeatedly passed over for the recognition his constant would earn. The constant outlived every atomic model proposed during his lifetime, and is now the most precisely determined quantity in physics — twelve digits, anchored in the very hydrogen lines he never needed to look at.

History

1868

— Ångström publishes precision wavelengths of four visible hydrogen lines, the raw material of everything that follows.

1885

— Balmer fits them with integers: λ = B n²/(n² − 4).

1888

— Rydberg generalizes to wave-number term differences with one universal constant across elements.

1908

— Ritz elevates the pattern to the combination principle: every line is a difference of two terms.

1913

— Bohr computes Rydberg's constant from fundamental constants — the fitted number becomes theory.

1932

— Urey resolves deuterium's Balmer satellites, 1.8 Å blueward: an isotope discovered by reduced mass.

Going further

Texts. Griffiths & Schroeter, Introduction to Quantum Mechanics, 3rd ed., §4.2 · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., §3.7 · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 13 · Cohen-Tannoudji, Diu & Laloë, Vol. 1, Ch. VII · Bethe & Salpeter, Quantum Mechanics of One- and Two-Electron Atoms.

Provenance. Energies and constants are standard CODATA/textbook values: $E_{\mathrm{Ry}}=13.6057\,\mathrm{eV}$ , $R_\infty=1.0973731\times10^7\,\mathrm{m^{-1}}$ , $a_0=0.52918\,\text{\AA}$ , $\alpha^{-1}=137.036$ . A verifier should: (1) confirm $E_{\mathrm{Ry}}=\tfrac12\alpha^2 m_e c^2$ numerically using $m_ec^2=0.511\,\mathrm{MeV}$ ; (2) reproduce the series-termination condition $\rho_0=2n$ from the recursion in Griffiths §4.2; (3) verify the SO(4) Casimir relation against Shankar Ch. 13 / Pauli's 1926 derivation; (4) check Balmer/Lyman wavelengths from the Rydberg formula. References: Griffiths 3rd ed., Sakurai 2nd ed., Shankar 2nd ed., Cohen-Tannoudji Vol. 1, Bethe & Salpeter. (Editorial note: item (2) is executed at "The Hydrogen Atom: Coulomb Radial Equation", which owns the termination derivation; item (3) at "Accidental Degeneracy and the SO(4) Symmetry" — this article consumes both by citation.)

Prerequisites. The hydrogen Coulomb radial equation · the radial equation · boundary conditions and normalization of radial wavefunctions · effective potential and the centrifugal barrier · the Bohr radius and characteristic atomic scales. Leads to. Hydrogenic wavefunctions and quantum numbers · degeneracy and the magnetic quantum number · accidental degeneracy and SO(4) · the Runge–Lenz vector · hydrogenic ions and Z-scaling.

When do I use R_∞ and when R_H?

$R_\infty$ is the theoretical constant for a fixed (infinitely heavy) nucleus — use it when quoting fundamental-constant physics or building the $Z$ -scaled formulas. $R_{\mathrm H} = R_\infty/(1 + m_e/m_p)$ is what a hydrogen lamp actually emits — use it the moment you compare with measured wavelengths. The difference is $0.054\%$ : invisible at three significant figures, decisive at four. Deuterium, tritium, and every hydrogenic ion each get their own $R_M$ by the same one-line recipe.

Why did Rydberg work with 1/λ instead of wavelength or frequency?

Because the physics is additive in wave numbers. Energy conservation makes photon energies differences of level energies, and $1/\lambda = E/hc$ inherits that: every line is a difference of two terms, and series relationships leap off the page (Lyman line minus Lyman line equals a Balmer line). In wavelength the same structure is a mess of reciprocals — Balmer's $n^2/(n^2-4)$ contortion is what the pattern looks like in the wrong variable. Frequency would work as well; wave numbers were what spectroscopes calibrated best in 1888.

Is the 13.6 eV exact?

No — it is the leading term of a systematically improvable ladder. The number $13.6057\ \mathrm{eV}$ already includes nothing beyond Coulomb attraction and non-relativistic quantum mechanics with an infinitely heavy proton. Honesty adds, in order: the reduced mass (one part in $1836$ ), fine structure (relativity plus spin–orbit, order $\alpha^2 \sim 10^{-4}$ relative), the Lamb shift (QED, order $\alpha^3$ with logarithms), and hyperfine structure (nuclear spin). Each correction is a later article; the point here is that each is small because $\alpha$ is small — the ladder converges fast.

Can an electron jump several levels at once, or only to the next rung?

Any $n_i \to n_f$ is energetically allowed, and the spectrum proves it — Lyman- $\beta$ is a $3 \to 1$ triple-rung drop. What transitions prefer is governed not by energy but by angular-momentum selection rules ( $\Delta\ell = \pm1$ for one-photon electric-dipole emission), which is time-dependent physics belonging to a later chapter. Energy conservation writes the wavelength menu; the selection rules decide how brightly each item is ordered.

Why do sodium and the other alkalis have series like hydrogen's but shifted?

Because their single valence electron sees a hydrogen-like $-1/r$ tail outside a screening core, its terms fit $R/(n - \delta_\ell)^2$ with an $\ell$ -dependent quantum defect $\delta_\ell$ — Rydberg discovered exactly this form. The defect measures how much each orbit penetrates the core (large for $s$ , small for $d$ ), and it is the cleanest experimental display of the fact that the $\ell$ -degeneracy is special to pure $1/r$ . The screening story is taken up at Hydrogenic Ions and Scaling with Nuclear Charge.

How can R_∞ possibly be known to twelve digits when λ measurements are less precise?

Because modern determinations do not measure wavelengths with rulers — they measure frequencies against atomic clocks. Two-photon spectroscopy of the 1S–2S interval (a transition naturally a thousand times narrower than ordinary lines, and Doppler-free in the two-photon geometry) gives a frequency to fifteen digits; theory expresses that interval as a known multiple of $cR_\infty$ plus calculable corrections. Inverting yields $R_\infty$ at parts in $10^{12}$ , limited mostly by how well the proton's charge radius is known — which is why muonic hydrogen mattered so much.

Positronium has μ = m_e/2 exactly. Is its spectrum then exactly half of hydrogen's?

The Coulomb ladder is exactly halved — $E_n = -6.80/n^2\ \mathrm{eV}$ , all wavelengths doubled, and unlike hydrogen there is no further reduced-mass correction to argue about since $\mu = m_e/2$ is exact. But positronium adds physics hydrogen lacks: the electron and positron annihilate (the atom lives $\sim0.1$ – $142\ \mathrm{ns}$ depending on spin state), and spin–spin plus annihilation interactions split its levels differently than hydrogen's fine structure. Half is the right first answer, and only the first.