# 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 a formula that Balmer, a Basel
schoolteacher, found in 1885 by pure numerology: $\lambda = B\,n^2/(n^2 - 4)$ with
$n = 3, 4, 5, \dots$ and $B = 364.56$ nm. The case $n = 3$ gives the red line, and the lines crowd
toward the **series limit** $\lambda \to B$ as $n \to \infty$. Inverting it, $1/\lambda = (4/B)\big(\tfrac14 - \tfrac1{n^2}\big)$, which suggested
working with wavenumbers. Rydberg (1888) did, and generalized Balmer's case $n_1 = 2$ 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 |
| Pfund | 5 | infrared | 7460 nm | 2279 nm |
Comparing the two forms, $R_H = 4/B \approx 1.097\times10^7\ \mathrm{m^{-1}}$. (The residual
difference of a few parts in $10^4$ comes mostly from Balmer's wavelengths having been measured in
air rather than vacuum.)
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.
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).
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).
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.
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.
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.