The Bohr Model of Hydrogen

This is where the atomic thread of the optics track meets quantum mechanics, and the two must agree on who owns what. The microscopic derivation of the hydrogen spectrum — the proof that a normalizable radial wavefunction admits bound states only at $E_n = -13.6\ \mathrm{eV}/n^2$ , and that Rydberg's fitted constant is calculable from fundamental constants — is owned by aq-qm-hydrogen-energy-spectrum-rydberg, recalled here by name and not re-derived. The natural length that sets every atomic scale, the Bohr radius $a_0 = 0.529\ \text{Å}$ and its family of units, is owned by aq-qm-bohr-radius-atomic-scales. Quantum mechanics keeps the differential equation. What this article owns is the road not taken by Schrödinger: the semiclassical construction Niels Bohr wrote in 1913, a decade before there was a wave equation to solve — force balance on a Coulomb orbit, plus one bold quantization postulate — and the ledger of exactly what that construction gets miraculously right and catastrophically wrong. The audacity is that it works at all. Graft a single quantum rule, $L = n\hbar$ , onto a planet-and-sun picture of an electron circling a proton, and out fall the correct energy levels, the correct atomic size, and the Rydberg constant to a fraction of a percent — all from algebra a student can do in an afternoon. Bohr's model turned the hydrogen spectrum from a table of measured wavelengths into a theory, predicted the correspondence with classical orbital frequencies at large $n$ , and, once the finite proton mass was folded in, split hydrogen from deuterium precisely enough to discover a new isotope in the third decimal place of a wavelength. It also assigns the ground state an angular momentum it does not have, cannot compute a single line's brightness, and collapses the instant a second electron appears. Both halves of that ledger are the subject here.

Key takeaways

The Bohr model imposes one quantum rule — orbital angular momentum $L = m_e v r = n\hbar$ — on a classical Coulomb orbit, and that single postulate discretises both the radius and the energy.
Force balance plus quantization give $r_n = n^2 a_0$ and the energy ladder $E_n = -m_e e^4/(8\varepsilon_0^2 h^2 n^2) = -13.6057\ \mathrm{eV}/n^2$ , with ground-state orbital speed $v_1 = \alpha c = 2.188\times10^6\ \mathrm{m/s}$ (about $0.7\%$ of $c$ ).
The emission wavenumbers reproduce the Rydberg formula, and the Rydberg constant emerges built from $m_e, e, \varepsilon_0, \hbar, c$ — recalled as calculable from aq-qm-hydrogen-energy-spectrum-rydberg.
At large $n$ the photon frequency of an $n\to n-1$ jump approaches the classical orbital frequency (at $n = 100$ they agree to $1.5\%$ ) — Bohr's correspondence principle, the bridge back to classical radiation.
Folding in the finite nuclear mass ( $m_e\to\mu$ ) shifts $R_\infty$ to $R_{\mathrm H}$ and separates hydrogen from deuterium by $\Delta\lambda \approx 0.18\ \mathrm{nm}$ at H $\alpha$ — the split that discovered deuterium.
Key idea: one quantization postulate on a Coulomb orbit yields the exact hydrogen energies and atomic scales; the model is a triumph for $E_n$ yet wrong about angular momentum (the ground state has $\ell = 0$ , not $\hbar$ ), silent about intensities, and useless beyond one electron.

The core insight

Postulate that the electron's orbital angular momentum is quantized, $L = n\hbar$ , and impose it on a classical circular Coulomb orbit. Force balance then fixes the radii $r_n = n^2 a_0$ and the energies $E_n = -13.6\ \mathrm{eV}/n^2$ . The model reproduces the hydrogen spectrum and the Rydberg constant exactly for a two-body Coulomb system — a genuine success — while getting the ground-state angular momentum qualitatively wrong and failing for every atom with more than one electron.

Where this lives today

The Bohr model is superseded as physics but immortal as scaffolding: it is still the first quantitative atom every student meets, and its energy ladder $E_n = -13.6\ \mathrm{eV}/n^2$ remains exactly right for two-body Coulomb systems. Where its spirit genuinely lives on is at the precision frontier. Two-photon spectroscopy of hydrogen's $1S$ – $2S$ interval, its frequency counted against a cesium clock through an optical frequency comb, now pins the Rydberg constant to about two parts in $10^{12}$ — and the same hydrogen lines, cross-checked against muonic hydrogen (a Bohr atom shrunk $186$ -fold by a heavy orbiter), exposed and then helped settle the "proton radius puzzle." Bohr's reduced-mass correction, the one that split hydrogen from deuterium, is the working principle behind these isotope and exotic-atom comparisons. The correspondence principle, too, is a live instrument. Rydberg atoms excited to $n \sim 50$ – $100$ are giant, nearly classical objects whose adjacent levels are separated by microwave and radio frequencies — exactly the near-classical orbital frequencies the principle predicts — and radio astronomers routinely detect their recombination lines from interstellar clouds. Those same long-lived, strongly interacting Rydberg states are now the workhorses of neutral-atom quantum computing and quantum simulation, where the $n^2$ scaling of the orbit that Bohr derived in an afternoon sets the blockade radius that entangles the qubits.

The physical picture

Think of the electron as a planet and the proton as its sun, with the gravitational inverse-square pull replaced by the Coulomb inverse-square pull. Classically any orbit is allowed and, worse, an accelerating charge radiates — the electron should spiral into the nucleus in about sixteen picoseconds, and there would be no atoms. Bohr's leap was to forbid the spiral by decree: only orbits whose angular momentum is an integer multiple of $\hbar$ are permitted, and on those stationary states the electron, against all classical logic, does not radiate. Radiation happens only in the jump between two allowed orbits, carrying off the energy difference as a single photon.

Figure

The allowed orbits to scale — radii grow as n², with a Lyman jump to n=1 and a Balmer jump to n=2

An equivalent and more physical reading of the quantization rule, due to de Broglie a decade later, is that a stable orbit must hold a whole number of electron wavelengths: $2\pi r = n\lambda = n h/(m_e v)$ , which rearranges to exactly $m_e v r = n\hbar$ . The quantization is a standing-wave condition — the first hint that the electron is a wave, made rigorous only when Schrödinger replaced the orbit with a cloud. Every energy in the ladder is negative because the electron is bound; the zero of energy is a free electron at rest infinitely far away, so ionizing the ground state costs exactly $+13.6\ \mathrm{eV}$ .

Figure

The energy ladder with real values, and the transitions that end on n=1 (Lyman) and n=2 (Balmer)

The ladder's shape — levels crowding toward the top as $1/n^2$ — is the whole spectroscopic story in one picture. Transitions ending on $n=1$ are the ultraviolet Lyman series; on $n=2$ , the visible Balmer series that a schoolteacher found first because it lands in the eye's window; on $n=3$ and beyond, the infrared. Each family is a set of jumps sharing a floor, and the wavelengths follow from reading two rungs off the ladder and subtracting.

The core equations

Two ingredients start everything. The first is entirely classical: for a circular orbit the Coulomb attraction must supply exactly the centripetal force. Nothing quantum has entered yet — this is Newton with an electrostatic force law.

\frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2} = \frac{m_e v^2}{r} \qquad (1)

where: $e$ — elementary charge; $\varepsilon_0$ — vacuum permittivity; $m_e$ — electron mass (strictly the reduced mass $\mu$ , corrected below); $v$ — orbital speed; $r$ — orbital radius. Read alone, this relation permits any radius whatever — the classical catastrophe. The quantum content is the second ingredient, imposed by fiat. Bohr's postulate selects a discrete subset of those orbits by quantizing the orbital angular momentum. This is the one genuinely new physical assumption, and everything discrete in the atom descends from it.

L = m_e v r = n\hbar, \qquad n = 1, 2, 3, \dots \qquad (2)

where: $L$ — orbital angular momentum; $n$ — principal quantum number; $\hbar$ — reduced Planck constant. Eliminating $v$ between Equations (1) and (2) (Derivation A) collapses the continuum of orbits to a discrete ladder of radii, whose smallest rung defines the Bohr radius.

r_n = n^2 a_0, \qquad a_0 = \frac{4\pi\varepsilon_0\hbar^2}{m_e e^2} = 0.529177\ \text{Å}, \qquad v_n = \frac{\alpha c}{n} \qquad (3)

where: $a_0$ — Bohr radius (the $n=1$ orbit), the natural atomic length whose anatomy and family of scales are owned by aq-qm-bohr-radius-atomic-scales; $\alpha = e^2/(4\pi\varepsilon_0\hbar c)\approx1/137$ — the fine-structure constant; $v_1 = \alpha c = 2.188\times10^6\ \mathrm{m/s}$ — the ground-state speed, a clean $0.7\%$ of $c$ that justifies treating the atom non-relativistically. Feeding the quantized radius into the total energy gives the ladder.

E_n = -\frac{m_e e^4}{8\varepsilon_0^2 h^2}\frac{1}{n^2} = -\frac{13.6057\ \mathrm{eV}}{n^2} = -\frac{1}{2}m_e c^2\alpha^2\frac{1}{n^2} \qquad (4)

where: $E_n$ — total (kinetic plus potential) energy of level $n$ ; $h$ — Planck constant; $m_e c^2 = 511\ \mathrm{keV}$ — the electron rest energy. The last form is the most revealing: the binding energy is $\tfrac12\alpha^2$ of the rest energy, so atoms are "soft" ( $13.6\ \mathrm{eV}$ against half an MeV) precisely because $\alpha$ is small — and fine-structure corrections, entering at the next order in $\alpha^2$ , are guaranteed tiny. Photon emission between two rungs reproduces the empirical Rydberg formula, with the Rydberg constant now a computed quantity rather than a fitted one.

\frac{1}{\lambda} = R_\infty\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right), \qquad R_\infty = \frac{m_e e^4}{8\varepsilon_0^2 h^3 c} = 1.0973732\times10^7\ \mathrm{m^{-1}} \qquad (5)

where: $\lambda$ — emitted wavelength; $n_1 < n_2$ — lower and upper levels; $R_\infty$ — the Rydberg constant for an infinitely heavy nucleus. That $R_\infty$ is calculable from fundamental constants — the fact that turned Rydberg's fitted number into physics — is the result aq-qm-hydrogen-energy-spectrum-rydberg owns from the wave equation; here it drops out of the semiclassical orbits. The small deficit against the measured hydrogen value is repaired by the finite proton mass: $R_{\mathrm H} = R_\infty\,\mu/m_e = 1.0967758\times10^7\ \mathrm{m^{-1}}$ , with $\mu = m_e M/(m_e + M)$ .

Derivation AFrom two postulates to the hydrogen energy ladder

The goal is to show that combining one classical equation with one quantum postulate forces the entire discrete structure of hydrogen — the quantized radii $r_n = n^2 a_0$ and the energy ladder $E_n = -m_e e^4/(8\varepsilon_0^2 h^2 n^2)$ — with no further input. The strategy is to eliminate the orbital speed between force balance and quantization to get the radii, then use the virial relation to turn a radius into an energy. It is deliberately elementary: the whole point of Bohr's 1913 achievement is that a first-year mechanics toolkit reaches the right answer.

Step 1. From the quantization rule $m_e v r = n\hbar$ , solve for the speed, $v = n\hbar/(m_e r)$ , and substitute it into the force balance $e^2/(4\pi\varepsilon_0 r^2) = m_e v^2/r$ . Why: force balance alone has two unknowns, $r$ and $v$ ; the postulate supplies the second equation that pins them both.

Step 2. The substitution gives $e^2/(4\pi\varepsilon_0 r^2) = m_e/r\cdot(n\hbar/m_e r)^2 = n^2\hbar^2/(m_e r^3)$ . Solving for $r$ yields $r_n = 4\pi\varepsilon_0\hbar^2 n^2/(m_e e^2) = a_0 n^2$ , with the $n=1$ prefactor defining the Bohr radius. Why: the radius grows as $n^2$ because angular momentum rises linearly with $n$ while the Coulomb law ties speed to radius — the two constraints together stretch the orbit quadratically.

Step 3. For the energy, add kinetic and potential parts, $E = \tfrac12 m_e v^2 - e^2/(4\pi\varepsilon_0 r)$ . Force balance says $m_e v^2 = e^2/(4\pi\varepsilon_0 r)$ , i.e. the kinetic energy is exactly half the magnitude of the potential — the virial theorem for an inverse-square force — so $E = -\tfrac12\cdot e^2/(4\pi\varepsilon_0 r_n)$ . Why: the virial relation means the total energy is just minus the kinetic energy, collapsing two terms into one and fixing the sign as bound (negative).

Step 4. Insert $r_n = a_0 n^2$ : $E_n = -m_e e^4/(8\varepsilon_0^2 h^2)\cdot1/n^2$ , which evaluates to $-13.6057\ \mathrm{eV}/n^2$ . (Writing $\hbar = h/2\pi$ turns the natural prefactor $2(4\pi\varepsilon_0)^2\hbar^2$ into $8\varepsilon_0^2 h^2$ .) Why: the prefactor is a pure combination of fundamental constants, which is exactly why the Rydberg constant it contains could be predicted rather than fitted. *Solving Equations (1)–(2) symbolically gives $r_n = 4\pi\varepsilon_0\hbar^2 n^2/(m_e e^2)$ and confirms $E_n = -m_e e^4/(8\varepsilon_0^2 h^2 n^2)$ identically, evaluating to $-13.6057\ \mathrm{eV}$ and $a_0 = 0.529177\ \text{Å}$ .*

Go deeper · Derivation B · 2 minThe correspondence principle: quantum jumps that sound classical

Derivation BThe correspondence principle: quantum jumps that sound classical

The goal is to make good on Bohr's deepest claim — that quantum mechanics must reduce to classical physics where classical physics works, namely at large quantum numbers — by showing that the photon frequency of a jump between adjacent high- $n$ levels approaches the classical frequency at which the electron actually orbits. The strategy is to compute both frequencies and take their ratio as $n$ grows, then evaluate it at $n = 100$ to see how fast the two converge.

Step 1. The classical orbital frequency is $f_{\mathrm{orb}} = v_n/(2\pi r_n)$ . Using $v_n = \alpha c/n$ and $r_n = n^2 a_0$ , this is $f_{\mathrm{orb}} = \alpha c/(2\pi n^3 a_0)$ , which one can rewrite as $f_{\mathrm{orb}} = 2 c R_\infty/n^3$ . Why: the orbital frequency falls as $n^{-3}$ because the orbit's circumference grows as $n^2$ while the speed falls as $n^{-1}$ .

Step 2. The photon frequency for the jump $n\to n-1$ is $f_{\mathrm{photon}} = cR_\infty[1/(n-1)^2 - 1/n^2]$ . For large $n$ , expand the bracket: $1/(n-1)^2 - 1/n^2 \approx 2/n^3$ , so $f_{\mathrm{photon}}\to 2cR_\infty/n^3 = f_{\mathrm{orb}}$ . Why: adjacent high- $n$ levels are so close that a downward jump radiates at nearly the frequency of the orbit itself — the quantum ladder blurs into the continuous classical spiral.

Step 3. Quantify the approach at $n = 100$ . The exact bracket is $1/99^2 - 1/100^2 = 2.0304\times10^{-6}$ against the asymptotic $2/100^3 = 2.0000\times10^{-6}$ , a ratio $f_{\mathrm{photon}}/f_{\mathrm{orb}} = 1.0152$ . Why: the two frequencies already agree to $1.5\%$ at $n = 100$ , and the residual shrinks as $1/n$ , so by $n = 1000$ it is $0.15\%$ — the classical limit is reached smoothly, not abruptly.

Step 4. Read the consequence. It was this insistence that quantum results merge into classical ones at large $n$ — the correspondence principle — that let Bohr fix signs, select rules, and gauge his model's reach; and it is why microwave photons couple so fluently to the giant orbits of Rydberg atoms, whose adjacent levels are separated by exactly such near-classical frequencies. Why: the principle is not a limit to be taken once but a design constraint on the whole theory — any successor must reproduce classical radiation where the correspondence demands it. *The ratio $f_{\mathrm{photon}}/f_{\mathrm{orb}}$ evaluated at $n = 10, 100, 1000$ gives $1.173$ , $1.0152$ , $1.0015$ — convergence to unity as $1 + O(1/n)$ .*

Limiting cases

Case	Result	Meaning
large $n$ (correspondence)	$f_{\mathrm{photon}}(n\to n-1) \to f_{\mathrm{orb}}$	adjacent high- $n$ jumps radiate at the classical orbital frequency (agree to $1.5\%$ at $n=100$ )
finite nuclear mass	$\mu = m_e M/(m_e + M)$ , $R_{\mathrm H} = R_\infty\,\mu/m_e$	the proton recoils; the $0.05\%$ shift separates hydrogen from deuterium
hydrogenic ion (charge $Ze$ )	$E_n = -Z^2\cdot13.6\ \mathrm{eV}/n^2$ , $r_n = n^2 a_0/Z$	one-electron ions ( $\mathrm{He^+}$ , $\mathrm{Li^{2+}}$ ) scale as $Z^2$ ; He $^+$ ground state is $-54.4\ \mathrm{eV}$
wave-mechanical limit	$L = n\hbar \longrightarrow L = \sqrt{\ell(\ell+1)}\,\hbar$ , $\ell = 0,\dots,n-1$	Schrödinger keeps $E_n = -13.6\ \mathrm{eV}/n^2$ but replaces Bohr's angular-momentum rule

Common trapBohr's rule L = nℏ is the ground state's true angular momentum

It is not, and this is the model's deepest flaw. Bohr's postulate gives the $n=1$ ground state an orbital angular momentum $L = \hbar$ , but the Schrödinger solution shows the true $1s$ state has $\ell = 0$ , hence $L = 0$ — the electron cloud carries no orbital angular momentum at all. Bohr's $n$ is the principal quantum number, which sets the energy; it is not the angular-momentum quantum number, a role that belongs to $\ell$ , running from $0$ to $n-1$ . The coincidence that both schemes give the same energies makes the error easy to miss: for the pure Coulomb potential the energy happens to depend only on $n$ , so Bohr's mislabelled angular momentum never shows up in the spectrum he was fitting. It shows up everywhere else — in the shape of the ground state (spherical, not a ring), in the Zeeman pattern, in selection rules — which is why "keep the energies, discard the orbits" is the right way to carry the model forward.

Worked examples

1 · The Lyman-α line from the energy ladder (confidence builder)

Find the energy of the $n = 2 \to 1$ transition in hydrogen and the wavelength of the photon it emits (Lyman- $\alpha$ ). Compare the bare Bohr value with the measured one.

Step 1. Read two rungs off Equation (4): $E_1 = -13.6057\ \mathrm{eV}$ and $E_2 = -13.6057/4 = -3.4014\ \mathrm{eV}$ .

Step 2. The transition energy is the difference, $\Delta E = E_2 - E_1 = -3.4014 - (-13.6057) = \mathbf{10.204\ eV}$ — carried off as one ultraviolet photon.

Step 3. Convert to wavelength with $\lambda = hc/\Delta E = 1239.84\ \mathrm{eV\,nm}/10.204\ \mathrm{eV} = \mathbf{121.50\ nm}$ , deep in the vacuum ultraviolet where no glass transmits.

Step 4. The measured Lyman- $\alpha$ sits at $121.57\ \mathrm{nm}$ . The tiny gap is not error but physics: the infinite-mass Rydberg energy overstates the binding slightly, and correcting $m_e\to\mu$ (using $R_{\mathrm H}$ ) gives $\lambda = 4/(3R_{\mathrm H}) = 121.57\ \mathrm{nm}$ — dead on. $\Delta E = 10.20\ \mathrm{eV}$ ; $\lambda = 121.50\ \mathrm{nm}$ (bare), $121.57\ \mathrm{nm}$ (reduced-mass, measured).

2 · The hydrogen–deuterium isotope shift (exam level)

Deuterium's nucleus is a proton plus a neutron, roughly twice the mass. Predict how far its H $\alpha$ line ( $3\to2$ ) sits from ordinary hydrogen's, and say which way.

Step 1. The only thing that changes is the reduced mass. For hydrogen $M_{\mathrm H} = 1836.15\,m_e$ , giving $R_{\mathrm H} = R_\infty/(1 + m_e/M_{\mathrm H}) = 1.096776\times10^7\ \mathrm{m^{-1}}$ ; for deuterium $M_{\mathrm D} = 3670.48\,m_e$ , giving $R_{\mathrm D} = 1.097074\times10^7\ \mathrm{m^{-1}}$ — larger, because the heavier nucleus recoils less.

Step 2. Compute H $\alpha$ for each from $1/\lambda = R(1/4 - 1/9) = 5R/36$ . Hydrogen: $\lambda_{\mathrm H} = 36/(5R_{\mathrm H}) = \mathbf{656.47\ nm}$ (vacuum). Deuterium: $\lambda_{\mathrm D} = 36/(5R_{\mathrm D}) = \mathbf{656.29\ nm}$ (vacuum).

Step 3. The split is $\Delta\lambda = \lambda_{\mathrm H} - \lambda_{\mathrm D} = \mathbf{0.18\ nm}$ (about $1.8\ \text{Å}$ ), with deuterium on the blue side — its larger reduced mass deepens every level and shortens every wavelength.

Step 4. This is precisely the measurement Harold Urey made in 1931–32: he photographed faint satellite lines about $1.8\ \text{Å}$ to the blue of hydrogen's Balmer lines in evaporated liquid hydrogen, and a new isotope was announced from the third decimal place of a wavelength. $\Delta\lambda \approx 0.18\ \mathrm{nm}$ , deuterium blueward.

Physics of the portrait

The model is inseparable from the man who dared it — a physicist who was willing to postulate a rule he could not justify because it gave the right spectrum, and who spent the rest of his life on what such rules mean.

Origins · 1 minNiels Bohr and how this was discovered

Niels Bohr

1885–1962 · Copenhagen

Niels Bohr was a Danish physicist who, working in Manchester under Ernest Rutherford in 1912–13, took the newly discovered nuclear atom and asked what would stabilise it. His answer — the 1913 trilogy of papers "On the Constitution of Atoms and Molecules" — grafted quantization onto Rutherford's orbits and, at a stroke, computed the Rydberg constant and explained the hydrogen spectrum. He knew the model was a hybrid, classical orbits ruled by a non-classical postulate, and he made a virtue of the tension: his correspondence principle, that quantum results must merge into classical ones at large quantum numbers, guided a decade of atomic physics before wave mechanics arrived. Awarded the Nobel Prize in 1922, he built the Institute for Theoretical Physics in Copenhagen into the gravitational centre of the new quantum theory, where the Copenhagen interpretation and the idea of complementarity took shape — and where his long, famous argument with Einstein over the meaning of quantum mechanics played out. Half-Jewish by descent, he escaped occupied Denmark in 1943, contributed to and then argued against the atomic-bomb project, and spent his later years pressing for openness and international arms control. He died in Copenhagen in 1962, the semiclassical model that made his name long since superseded by the wave mechanics his institute helped create — yet still the first thing every physicist learns about the atom.

History

1885

— Balmer fits the four visible hydrogen lines with an integer formula

1888

— Rydberg generalises the pattern to wave-number term differences with a universal constant

1911

— Rutherford's scattering experiments establish the nuclear atom

1913

— Bohr's trilogy quantises the hydrogen orbits and computes the Rydberg constant

1914

— Franck and Hertz confirm discrete atomic energy levels by electron impact

1916

— Sommerfeld extends the orbits to ellipses and introduces the fine-structure constant

1926

— Pauli (matrix mechanics, submitted January) and Schrödinger (wave mechanics) each derive the hydrogen spectrum anew

1932

— Urey publishes the identification of deuterium through its isotope-shifted Balmer satellite lines, observed in late 1931

Going further

Texts. Foot, Atomic Physics, ch. 1 · Demtröder, Laser Spectroscopy, ch. 2 · Hecht, Optics, §3 (atomic emission) · Loudon, The Quantum Theory of Light, ch. 1 · Saleh & Teich, Fundamentals of Photonics, ch. 13. Key numerical anchors: $a_0 = 5.29\times10^{-11}\ \mathrm{m}$ , $E_1 = -13.6\ \mathrm{eV}$ , $R_\infty = 1.097\times10^7\ \mathrm{m^{-1}}$ , the Balmer wavelengths $656/486/434\ \mathrm{nm}$ , and $v_1 = \alpha c$ with $\alpha\approx1/137$ . The Schrödinger derivation of the spectrum and the Rydberg constant is recalled from aq-qm-hydrogen-energy-spectrum-rydberg; the Bohr radius and atomic-scale ladder from aq-qm-bohr-radius-atomic-scales — this article owns the semiclassical construction, its correspondence-principle check, and the spectroscopic numbers.

Prerequisites. The Hydrogen Spectrum and Spectral Series · Quantum Numbers and Orbital Angular Momentum. Leads to. Solution of the Schrödinger Equation for Hydrogen · Hydrogenic Wavefunctions and Radial Distributions · Fine Structure and Spin-Orbit Coupling · The Lamb Shift.

If Bohr's angular momentum is wrong, how does the model get the energies exactly right?

Because for the pure Coulomb potential the energy depends only on the principal quantum number $n$ , not on the angular momentum at all — the famous "accidental" degeneracy of hydrogen. Bohr labelled his single quantum number $n$ and attached to it both an energy and an angular momentum $n\hbar$ ; the energy assignment happens to match the exact answer, while the angular-momentum assignment does not. The full theory shows that a given $n$ contains many states of different $\ell$ (from $0$ to $n-1$ ), all sharing the same energy, so nothing in the spectrum Bohr was fitting could reveal his angular-momentum mistake. The error is real but invisible to the one measurement the model was built to explain — a piece of luck specific to the $1/r$ potential.

What is the de Broglie standing-wave version of the quantization rule?

De Broglie's 1924 insight recasts Bohr's ad-hoc postulate as a wave condition. If the electron has wavelength $\lambda = h/(m_e v)$ , then a stable orbit must contain a whole number of those wavelengths around its circumference: $2\pi r = n\lambda = n h/(m_e v)$ . Rearranging gives $m_e v r = n h/(2\pi) = n\hbar$ — exactly Bohr's quantization. The postulate that looked arbitrary in 1913 became, a decade later, the requirement that the electron wave close on itself without destructive interference. It is a genuine improvement in understanding, though still not the full story: the real standing waves are the three-dimensional solutions of the Schrödinger equation, not circles.

Why does the Bohr model fail completely for helium?

Because helium has two electrons, and the model has no machinery for the electron–electron repulsion. Bohr's construction relies on a single particle in a fixed central Coulomb field, where force balance and one quantization rule suffice. Add a second electron and each now moves in a field that is the sum of the nucleus's attraction and the other electron's shifting repulsion — a genuine three-body problem with no closed orbits to quantize. Early attempts to force helium into Bohr-Sommerfeld orbits gave badly wrong energies, and it was this failure, as much as any success, that showed a deeper theory was needed. The Schrödinger equation handles it only approximately too, but at least it provides a systematic framework (variational and self-consistent-field methods) where the Bohr model provides none.

What actually changed when Schrödinger replaced Bohr — if the energies are the same?

Almost everything except the energies. The Schrödinger solution keeps $E_n = -13.6\ \mathrm{eV}/n^2$ (until fine structure), but it replaces the sharp orbit with a probability cloud, corrects the angular momentum to $\sqrt{\ell(\ell+1)}\,\hbar$ with $\ell = 0,\dots,n-1$ (so the ground state is spherical with $L = 0$ ), introduces the degeneracy of $n^2$ orbital states per level, and — crucially — makes transition intensities and selection rules computable from wavefunction overlaps, which Bohr's model could never do. It also gives the right count of states in a magnetic field (the Zeeman effect) and the correct spatial distributions probed by everything from chemistry to scanning-probe imaging. Same ladder of energies, an entirely different atom underneath.

The reduced-mass correction is only 0.05%. Why does anyone care?

Because spectroscopy reads far more than five significant figures, and a $0.05\%$ shift is enormous at that precision. The correction $m_e\to\mu = m_e M/(m_e+M)$ is what separates $R_\infty$ (the idealised infinite-mass constant) from $R_{\mathrm H}$ (what a real hydrogen lamp emits), and it is what makes isotopes distinguishable: hydrogen and deuterium differ only in nuclear mass, so their entire spectral difference lives in this term. Urey discovered deuterium by exactly this route. The same $\mu$ turns hydrogen into positronium (equal masses, $\mu = m_e/2$ , every level at half depth) or muonic hydrogen (a heavy orbiter, levels $186\times$ deeper), each an incisive test of the theory. A correction you can ignore at three figures is a discovery engine at six.

Is the electron really orbiting the proton in the Bohr picture?

No — and this is the picture's most seductive lie. The Bohr model draws a point electron on a definite circular path with a definite speed at a definite radius, which is useful for getting the energy scales and impossible to reconcile with quantum mechanics: an electron in a definite orbit would have simultaneously definite position and momentum, violating the uncertainty principle. The true ground state is a static, spherical cloud with no net circulation and, as noted, zero orbital angular momentum. What the "orbit" gets right is the scale — the most probable radius of the $1s$ cloud really is close to $a_0$ — and the energy. What it gets wrong is the kinematics: nothing is going around. Keep the number $a_0$ , keep the energy $-13.6\ \mathrm{eV}$ , and drop the little planet.