# Atomic Models & Spectral Series > Pass a current through hydrogen gas and it emits light at a handful of razor-sharp wavelengths — > a barcode, not a rainbow. By 1885 the barcode had a formula; nobody knew why. This lesson follows > the atom from Thomson's plum pudding through Rutherford's nucleus to Bohr's quantized orbits: a > model built from one act of theft (Planck's $h$) that predicts the hydrogen spectrum to four > significant figures. It is gloriously, instructively wrong — and it left behind the quantum > numbers we still use. ## Learning Objectives After this lesson you will be able to: 1. Apply the **Rydberg formula** to the named hydrogen series and state the experimental facts of **emission** and **absorption line spectra**. 2. Explain why **Thomson's model** fails and how **Rutherford scattering** establishes the nuclear atom, computing the **distance of closest approach**. 3. Quantify the classical **radiation catastrophe**: estimate the spiral-in time of an orbiting electron. 4. Derive $r_n = n^2a_0$ and $E_n = -13.6\ \mathrm{eV}/n^2$ from **Bohr's postulates**, and obtain the **Rydberg constant** from fundamental constants. 5. Extend the model to **hydrogen-like ions** and apply the **reduced-mass** correction. 6. Explain how the **Franck–Hertz experiment** directly confirms discrete energy levels. --- ## Intuition A blackbody ([P.2.1](01-blackbody-radiation.md)) glows with a smooth continuum because it is dense matter — countless interacting charges. A *dilute gas* is different: excited atoms emit only discrete wavelengths (**emission lines**), and cold gas in front of a continuum source removes exactly the same wavelengths (**absorption lines**). Every element has its own pattern — atomic fingerprints sharp enough that helium was found in the Sun's spectrum before it was found on Earth. Lines scandalize classical physics twice over. First, a classical atom could oscillate — hence radiate — over a continuum of frequencies, not a barcode. Second, a classical atom shouldn't exist at all: an orbiting electron accelerates, an accelerating charge radiates, and the electron should spiral into the nucleus in picoseconds. Bohr's resolution: atoms possess a discrete set of allowed energies, and light is emitted only in jumps between them — each line is one jump, its frequency fixed by Planck's $E = h\nu$. --- ## Theory ### The empirical order: Balmer and Rydberg Hydrogen's four visible lines (656.3, 486.1, 434.0, 410.2 nm) fit Balmer's 1885 formula, later generalized by Rydberg to $$ \frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right), \qquad n_2 > n_1, \qquad R_H = 1.097\times10^{7}\ \mathrm{m^{-1}} . $$ Each choice of $n_1$ defines a **series**: | Series | $n_1$ | Region | First line | Series limit | |---|---|---|---|---| | Lyman | 1 | ultraviolet | 121.6 nm | 91.2 nm | | Balmer | 2 | visible | 656.3 nm | 364.6 nm | | Paschen | 3 | infrared | 1875 nm | 820.4 nm | | Brackett | 4 | infrared | 4051 nm | 1458 nm | The differences-of-inverse-squares structure (the *Ritz combination principle*: all line frequencies are differences of a fixed set of "terms") is the crucial clue — it smells like energy conservation between discrete levels, $h\nu = E_{n_2} - E_{n_1}$. ### Thomson's model and its failure Thomson (1904) pictured a sphere of uniform positive charge with embedded electrons — the "plum pudding." Inside a uniform charge sphere the restoring force is linear ($F \propto -r$, Gauss's law), so an electron oscillates at a *single* frequency — $\nu \sim 2.5\times10^{15}$ Hz for a hydrogen-sized sphere. One frequency and its harmonics cannot produce the Rydberg pattern of infinitely many lines converging to series limits. And the model fails quantitatively against scattering, as Rutherford showed. ### Rutherford scattering and the nuclear atom Geiger and Marsden (1909–13) fired $\alpha$ particles (charge $+2e$, kinetic energy $K \approx 7.7$ MeV) at thin gold foil. Most passed nearly straight through — but about 1 in 8000 scattered beyond $90^\circ$. **Why plum pudding cannot do this.** Spread gold's $Ze$ over a $10^{-10}$ m sphere and the maximum field is feeble: one atom deflects a 7.7 MeV $\alpha$ by at most $\sim10^{-2}$ degrees. Crossing $N\sim10^{4}$ atoms gives a random walk of r.m.s. angle $\sqrt N\times10^{-2}\,^\circ\approx1^\circ$; the probability that the walk exceeds $90^\circ$ is of order $e^{-(90/1)^2}\sim e^{-8100}$ — effectively zero, versus the observed $10^{-4}$. The deflection must occur in a **single violent encounter**: the positive charge is concentrated in a tiny, massive **nucleus**. With a point-charge Coulomb field, Rutherford's classical hyperbolic-orbit calculation reproduced the measured angular distribution ($\propto\sin^{-4}(\theta/2)$) exactly. **Distance of closest approach.** Head-on, the $\alpha$ stops where all kinetic energy has become Coulomb potential energy. With projectile charge $ze$ and target $Ze$: $$ K = \frac{1}{4\pi\varepsilon_0}\frac{zZe^2}{d} \quad\Longrightarrow\quad d = \frac{zZe^2}{4\pi\varepsilon_0 K} . $$ Using $e^2/4\pi\varepsilon_0 = 1.44\ \mathrm{eV\,nm}$, for gold ($Z=79$) at $K = 7.7$ MeV: $d = \frac{2\times79\times1.44}{7.7\times10^{6}}\ \mathrm{nm} = 3.0\times10^{-14}\ \mathrm m$. Scattering stayed perfectly Coulombic down to this distance, so the nucleus is smaller than $3\times10^{-14}$ m — over 3000 times smaller than the atom. Atoms are almost entirely empty space. ### The classical catastrophe: the atom should not exist A nuclear atom needs orbiting electrons, and an accelerating charge radiates with the Larmor power $$ P = \frac{e^2a^2}{6\pi\varepsilon_0c^3} . $$ For a circular orbit of radius $r$, $a = e^2/(4\pi\varepsilon_0mr^2)$ and $E(r) = -e^2/(8\pi\varepsilon_0 r)$. Setting $dE/dt = -P$ gives $r^2\,dr \propto -dt$; integrating from $a_0 = 5.29\times10^{-11}$ m to zero: $$ t_{spiral} = \frac{(4\pi\varepsilon_0)^2m^2c^3a_0^3}{4e^4} \approx 1.6\times10^{-11}\ \mathrm s . $$ Classically, every atom in your body collapses in sixteen picoseconds, emitting a continuous swan-song of rising frequency. The stability of matter is itself a quantum effect. ### The Bohr model Bohr (1913) cut the knot with three postulates: 1. **Stationary states.** The electron occupies certain allowed orbits *without radiating*, despite its acceleration. 2. **Quantized angular momentum.** Allowed orbits satisfy $L = mvr = n\hbar$, $n = 1,2,3,\dots$, with $\hbar \equiv h/2\pi = 1.055\times10^{-34}\ \mathrm{J\,s}$. 3. **Quantum jumps.** Radiation is emitted/absorbed only in transitions, with $h\nu = E_i - E_f$. **Orbits.** Coulomb attraction supplies the centripetal force: $$ \frac{mv^2}{r} = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2} \quad\Longrightarrow\quad mv^2 = \frac{e^2}{4\pi\varepsilon_0 r} . \tag{1} $$ Quantization gives $v = n\hbar/mr$; substituting into (1) and solving for $r$: $$ \boxed{\,r_n = \frac{4\pi\varepsilon_0\hbar^2}{me^2}\,n^2 = n^2a_0\,}, \qquad a_0 = 5.29\times10^{-11}\ \mathrm m = 0.529\ \text{Å}. $$ **Energies.** By (1), $K = \tfrac12mv^2 = -\tfrac12U$, so $$ E = K + U = \tfrac12U = -\frac{e^2}{8\pi\varepsilon_0r_n} = -\left[\frac{me^4}{2(4\pi\varepsilon_0)^2\hbar^2}\right]\frac{1}{n^2} \quad\Longrightarrow\quad \boxed{\,E_n = -\frac{13.6\ \mathrm{eV}}{n^2}\,}. $$ The ground state $E_1 = -13.6$ eV is precisely hydrogen's measured ionization energy. **The Rydberg constant, predicted.** A jump $n_2\to n_1$ emits $hc/\lambda = E_{n_2} - E_{n_1}$: $$ \frac{1}{\lambda} = \frac{me^4}{8\varepsilon_0^2h^3c}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right), \qquad R_\infty \equiv \frac{me^4}{8\varepsilon_0^2h^3c} = 1.0974\times10^{7}\ \mathrm{m^{-1}} . $$ This is the Rydberg formula with $R$ built from $m,e,\varepsilon_0,h,c$ alone — matching the spectroscopic $R_H = 1.0968\times10^{7}\ \mathrm{m^{-1}}$ to 0.05%. Lyman, Balmer, Paschen, Brackett are simply the jumps ending on $n_1 = 1,2,3,4$. **Hydrogen-like ions.** For one electron around charge $Ze$ (He$^+$, Li$^{2+}$, …), replace $e^2 \to Ze^2$: $r_n = n^2a_0/Z$ and $E_n = -13.6\,Z^2/n^2$ eV. **Reduced mass (brief).** Electron and nucleus orbit their common center of mass; replacing $m$ by $\mu = mM/(m+M)$ gives $R_H = R_\infty/(1+m/M)$. This 0.05% correction closes the gap with experiment exactly — and the hydrogen–deuterium line shift it predicts led to deuterium's discovery (E5). > **Caution.** Bohr's orbits **do not exist**. The model gets hydrogen's energies right from a > picture wrong in nearly every detail: the electron has no trajectory, the true ground state has > *zero* orbital angular momentum (not $\hbar$), and the model fails for every atom with two or > more electrons. What survives is the skeleton — discrete stationary states labeled by quantum > numbers, photons at $h\nu = \Delta E$. The honest treatment is > [P.6.2 The Radial Equation & the Hydrogen Atom](../c06-angular-momentum/02-radial-equation-hydrogen.md). ### Franck–Hertz: energy levels without light Franck and Hertz (1914) accelerated electrons through mercury vapor and measured the collector current as the voltage rose. The current climbed, then *dipped* sharply at 4.9 V, again at 9.8 V, at 14.7 V — every 4.9 V. Below 4.9 eV an electron can only collide elastically (the atom has no state to accept less); at 4.9 eV it can excite mercury's first level, losing its kinetic energy and failing to reach the collector; at 9.8 eV, twice. The tube glowed at $\lambda = hc/4.9\ \mathrm{eV} = 253$ nm — exactly mercury's known ultraviolet line. Discrete levels are real, and they are the same levels spectroscopy sees. ### Scorecard of the Bohr model | Successes | Failures | |---|---| | Hydrogen spectrum to 4 significant figures | Any atom with $\ge2$ electrons (helium: hopeless) | | $R_\infty$ from fundamental constants | No line **intensities** or transition rates | | Correct size scale $a_0$ and ionization energy | No mechanism: *why* don't stationary states radiate? | | Hydrogen-like ions with $Z^2$ scaling | Wrong angular momentum ($L=\hbar$ vs the true $L=0$ ground state) | | Reduced-mass isotope shifts (deuterium) | Fine structure, Zeeman anomalies, chemistry — silence | The model is a patch: quantum rules bolted onto classical orbits. The next lesson, [Correspondence & the Limits of the Old Quantum Theory](03-correspondence-and-limits.md), pushes the patching as far as it can go — and watches it fail. --- ## Worked Examples ### Example 1 — The Balmer series, computed For Balmer, $n_1 = 2$. First line ($n_2 = 3$, H$\alpha$): $\frac{1}{\lambda} = 1.097\times10^{7}(\tfrac14 - \tfrac19) = 1.524\times10^{6}\ \mathrm{m^{-1}}$, so $\lambda = 656.3$ nm (red). Next, $n_2 = 4$: $1/\lambda = 1.097\times10^{7}\times0.1875 = 2.057\times10^{6}$, $\lambda = 486.2$ nm (blue-green). Series limit ($n_2\to\infty$): $\lambda = 4/R_H = 364.6$ nm, beyond which lies the ionization continuum. All four visible hydrogen lines and their convergence point, from one constant. ### Example 2 — How small is the nucleus? A 7.7 MeV $\alpha$ ($z=2$) heads straight at a gold nucleus ($Z=79$): $$ d = \frac{zZe^2}{4\pi\varepsilon_0K} = \frac{2\times79\times1.44\ \mathrm{eV\,nm}}{7.7\times10^{6}\ \mathrm{eV}} = 2.96\times10^{-5}\ \mathrm{nm} \approx 3.0\times10^{-14}\ \mathrm m . $$ Since the data followed the pure-Coulomb prediction at all angles, the nuclear charge lies within $3.0\times10^{-14}$ m. Against the atomic radius $\sim10^{-10}$ m, the nucleus fills $\lesssim10^{-11}$ of the atom's volume: if the atom were a cathedral, the nucleus would be a housefly — a very dense housefly carrying 99.97% of the mass. --- ## Hands-on (Python) ```python import numpy as np import matplotlib.pyplot as plt from scipy.integrate import solve_ivp R_H = 1.097e7 # m^-1 # --- 1. Hydrogen spectral series: stem plot of wavelengths ---------------- series = {"Lyman": 1, "Balmer": 2, "Paschen": 3, "Brackett": 4} colors = ["purple", "tab:blue", "tab:red", "tab:brown"] fig, ax = plt.subplots(figsize=(9, 3.5)) for (name, n1), color in zip(series.items(), colors): n2 = np.arange(n1 + 1, n1 + 15) lam_nm = 1e9 / (R_H * (1 / n1**2 - 1 / n2**2)) ax.stem(lam_nm, np.ones_like(lam_nm), linefmt=color, markerfmt=" ", basefmt=" ", label=f"{name} (n1={n1})") ax.axvspan(400, 750, alpha=0.15, color="gold", label="visible band") ax.set_xscale("log"); ax.set_xlim(50, 4500); ax.set_yticks([]) ax.set_xlabel("wavelength (nm)"); ax.legend(); plt.tight_layout(); plt.show() # Expected: only Balmer lands in the visible band, lines at 656/486/434/410 nm # crowding toward the 364.6 nm series limit. # --- 2. Bohr energy-level diagrams for Z = 1 and Z = 2 -------------------- fig, axes = plt.subplots(1, 2, sharey=True, figsize=(7, 5)) for ax, Z in zip(axes, [1, 2]): for n in range(1, 8): E = -13.6 * Z**2 / n**2 # eV ax.hlines(E, 0, 1, color="k"); ax.text(1.03, E, f"n={n}", fontsize=8) ax.set_title(f"Z = {Z}"); ax.set_xticks([]) axes[0].set_ylabel("E (eV)"); plt.tight_layout(); plt.show() # Expected: He+ levels are 4x deeper -- ground state -54.4 eV vs -13.6 eV. # --- 3. Rutherford trajectories: alpha on gold ---------------------------- # Units: length in fm, energy in MeV, c = 1. k = zZ e^2/4pi eps0 = 2*79*1.44 MeV fm. k, K, m = 2 * 79 * 1.44, 7.7, 3727.0 # MeV fm, MeV, MeV/c^2 v0 = np.sqrt(2 * K / m) # in units of c def rhs(t, y): x, yy, vx, vy = y r3 = (x**2 + yy**2) ** 1.5 return [vx, vy, k * x / (m * r3), k * yy / (m * r3)] # repulsive Coulomb plt.figure(figsize=(7, 5)) for b in [2, 5, 10, 20, 40, 80]: # impact parameters (fm) sol = solve_ivp(rhs, [0, 25000], [-600, b, v0, 0], max_step=5.0, rtol=1e-9) plt.plot(sol.y[0], sol.y[1], label=f"b = {b} fm") theta = np.degrees(np.arctan2(sol.y[3][-1], sol.y[2][-1])) print(f"b = {b:3d} fm -> angle = {theta:6.1f} deg " f"(theory {np.degrees(2*np.arctan(k/(2*K*b))):6.1f})") plt.plot(0, 0, "ro", label="nucleus"); plt.xlabel("x (fm)"); plt.ylabel("y (fm)") plt.legend(fontsize=8); plt.show() # Expected: hyperbolic orbits; numerics match theta = 2 arctan(k / 2Kb): # b = 2 fm scatters ~164 deg (nearly backward), b = 80 fm only ~21 deg. ``` --- ## Exercises **E1 (easy).** Compute the longest and shortest wavelengths of the Lyman series. In what region do they lie?

Solution

Longest ($n_2=2$): $1/\lambda = R_H(1-\tfrac14)$, $\lambda = 121.5$ nm. Shortest (series limit): $\lambda = 1/R_H = 91.2$ nm. Both ultraviolet — which is why Lyman found his series decades after Balmer's visible one.

**E2 (easy).** What minimum photon energy ionizes hydrogen from the $n=2$ state, and what wavelength is that?

Solution

$E = 0 - E_2 = 13.6/4 = 3.4$ eV; $\lambda = hc/E = 1240\ \mathrm{eV\,nm}/3.4 = 365$ nm — precisely the Balmer series limit, as it must be (a jump from $n=2$ to the continuum edge).

**E3 (medium).** Show the Bohr orbital speed is $v_n = \alpha c/n$ with $\alpha = e^2/4\pi\varepsilon_0\hbar c \approx 1/137$, and evaluate $v_1$. Was Bohr justified in ignoring relativity?

Solution

$v_n = \frac{n\hbar}{mr_n} = \frac{n\hbar}{m}\cdot\frac{me^2}{4\pi\varepsilon_0\hbar^2n^2} = \frac{e^2}{4\pi\varepsilon_0\hbar}\cdot\frac1n = \frac{\alpha c}{n}$; $v_1 = c/137 = 2.19\times10^{6}\ \mathrm{m\,s^{-1}}$, so $(v/c)^2 \approx 5\times10^{-5}$. Relativistic corrections are $\sim10^{-4}$ eV — negligible at Rydberg accuracy, but real: they appear as *fine structure* (next lesson, Sommerfeld).

**E4 (medium).** A 5.5 MeV $\alpha$ scatters off aluminum ($Z=13$). Find the head-on distance of closest approach, compare with the nuclear radius $R \approx 1.2A^{1/3}\ \mathrm{fm} \approx 3.6$ fm ($A=27$), and predict what the data show.

Solution

$d = \frac{2\times13\times1.44\ \mathrm{eV\,nm}}{5.5\times10^{6}\ \mathrm{eV}} = 6.8\times10^{-6}\ \mathrm{nm} = 6.8$ fm — only about twice the nuclear radius. In the most violent (large-angle) collisions the $\alpha$ begins to feel the nuclear force and finite nuclear size, so the cross-section *deviates* from the Rutherford formula. Such deviations were later used to measure nuclear radii.

**E5 (hard).** Deuterium was discovered (Urey, 1931) through a tiny Balmer-line shift. Using $R_M = R_\infty/(1+m_e/M)$, derive $\lambda_H - \lambda_D$ for a given transition and evaluate it for H$\alpha$ (656.3 nm). Take $M_D \approx 2M_H$, $m_e/M_H = 1/1836$.

Solution

For a fixed transition $1/\lambda \propto R_M$, so $\lambda \propto 1 + m_e/M$ and $$ \frac{\lambda_H-\lambda_D}{\lambda} \approx \frac{m_e}{M_H} - \frac{m_e}{M_D} = \frac{m_e}{M_H}\left(1-\frac12\right) = \frac{1}{3672}. $$ $\Delta\lambda = 656.3/3672 = 0.179$ nm. Urey resolved this 1.8 Å shift as a faint satellite of H$\alpha$ in evaporation-enriched hydrogen — a Nobel Prize riding on the last decimal of the reduced mass.

--- ## Checkpoint 1. Write the Rydberg formula and identify $n_1$ for the Lyman, Balmer, Paschen, and Brackett series. 2. What single observation killed the plum-pudding model, and why can't multiple small deflections explain it? 3. State Bohr's three postulates and identify which directly contradicts classical electrodynamics. 4. Sketch the derivation chain from $L = n\hbar$ + Coulomb force to $E_n = -13.6\ \mathrm{eV}/n^2$. 5. Why are the Franck–Hertz dips spaced by 4.9 V, and what light does the tube emit?

Answers

1. $1/\lambda = R_H(1/n_1^2 - 1/n_2^2)$, $R_H = 1.097\times10^7\ \mathrm{m^{-1}}$; Lyman $n_1=1$ (UV), Balmer $n_1=2$ (visible), Paschen $n_1=3$ (IR), Brackett $n_1=4$ (IR). 2. Large-angle ($>90^\circ$) $\alpha$ scattering at rate $\sim10^{-4}$. Multiple scattering off diffuse charge is a random walk with r.m.s. $\sim1^\circ$; reaching $90^\circ$ has probability $\sim e^{-8100}$ — the deflection must be a single encounter with concentrated charge. 3. (i) Non-radiating stationary states; (ii) $L = n\hbar$; (iii) $h\nu = E_i - E_f$. Postulate (i) is the direct contradiction: classically an accelerating charge must radiate (Larmor) and spiral in within $\sim1.6\times10^{-11}$ s. 4. Coulomb = centripetal gives $mv^2 = e^2/4\pi\varepsilon_0r$; with $v = n\hbar/mr$ this yields $r_n = n^2a_0$; then $E = \tfrac12U = -e^2/8\pi\varepsilon_0r_n = -13.6\ \mathrm{eV}/n^2$. 5. Mercury's first excited state lies 4.9 eV up: each time an electron accumulates 4.9 eV it can lose it all in one inelastic collision, so the current dips at every multiple of 4.9 V. De-exciting atoms emit $\lambda = 1240/4.9 \approx 253$ nm — mercury's UV line.

--- ## Further Reading - **[ER]** Eisberg & Resnick, §4.1–4.8 — Thomson, Rutherford scattering (with the full cross-section derivation), and the Bohr model. - **[ER]** Eisberg & Resnick, §4.9–4.10 — Franck–Hertz and the finite-nuclear-mass correction. - **[Gri]** Griffiths & Schroeter, §4.2 — the true hydrogen atom, for comparison once you reach [P.6.2](../c06-angular-momentum/02-radial-equation-hydrogen.md). --- ← Prev: [Blackbody Radiation & Planck's Law](01-blackbody-radiation.md) · Up: [Pre-Term](../README.md) · Next: [Correspondence & the Limits of the Old Quantum Theory](03-correspondence-and-limits.md) →