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.

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Intuition

A blackbody (P.2.1) 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 $E = h\nu$.

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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.

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, pushes the patching as far as it can go — and watches it fail.

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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.

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Hands-on (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.

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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.

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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.

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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.

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