Chapter 12: Atoms

The internal architecture of the atom remained a mystery until the dawn of the twentieth century. Cathode-ray experiments showed the electron exists, but how is it arranged within neutral matter? This chapter traces the elegant chain of reasoning — from Thomson's plum-pudding, through Rutherford's gold-foil scattering, to Bohr's semi-classical quantization — that decoded the hydrogen atom and produced the Rydberg formula from first principles. We will end with de Broglie's beautiful reinterpretation of Bohr's quantization condition as a standing-wave requirement, and the limitations that made full quantum mechanics inevitable.

Concept Map

Cathode rays → electron (Thomson 1897)
        │
        ├── Thomson model (plum-pudding) ─ fails spectra
        │
        ├── α-scattering (Geiger-Marsden) → nucleus
        │       │
        │       └── Rutherford model ─ fails stability
        │
        └── Bohr model (postulates)
                │
                ├── quantized r_n, v_n, E_n
                ├── Hydrogen spectrum (Lyman, Balmer, Paschen, …)
                ├── Rydberg formula
                ├── de Broglie standing-wave 2πr = nλ
                └── Limitations → wave mechanics

12.1 Thomson's Plum-Pudding Model

Definition

Following his 1897 discovery of the electron, J. J. Thomson proposed that an atom is a sphere of uniform positive charge of radius $\sim 10^{-10}\ \text{m}$ in which negatively charged electrons are embedded like raisins in a pudding (or seeds in a watermelon). The atom as a whole is neutral, with total positive charge $+Ze$ balanced by $Z$ electrons.

Salient features:

• The positive charge density $\rho_+ = \dfrac{3Ze}{4\pi R^3}$ is uniform inside the sphere.
• Electrons reside at positions where the electrostatic force on them vanishes (equilibrium).
• For a single electron displaced by $r < R$ from the centre, the restoring force is $F = -\dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze^2}{R^3}r$ — i.e. simple-harmonic with frequency $\nu = \dfrac{1}{2\pi}\sqrt{\dfrac{Ze^2}{4\pi\varepsilon_0 m R^3}}$.

Derivation — oscillation frequency in Thomson's atom

Inside a uniformly charged sphere the electric field at distance $r$ from the centre is

The force on a displaced electron of charge $-e$ is therefore

which is Hooke-like with spring constant $k = \dfrac{Ze^2}{4\pi\varepsilon_0 R^3}$. Newton's second law gives $m\ddot r + kr = 0$, so

Worked Example

For hydrogen ($Z=1$) with $R = 1\ \text{Å} = 10^{-10}\ \text{m}$, compute $\nu$.

This corresponds to $\lambda = c/\nu \approx 120\ \text{nm}$, well in the ultraviolet — but the model predicts only this single frequency, not the rich line spectrum of hydrogen.

Pitfalls

• Thomson's atom predicts a continuous oscillatory emission at a single $\nu$, not the discrete Lyman/Balmer lines.
• It fails dramatically to explain the large-angle deflections observed in α-scattering — soft positive jelly cannot back-scatter a 5 MeV α.
• Modern reading: the positive charge in real atoms is concentrated in a nucleus $10^{4}$ times smaller than the atom.

12.2 Rutherford's α-Scattering Experiment

Definition

Geiger and Marsden (1909–1911), under Rutherford's direction, bombarded a thin gold foil ($\sim 10^{-7}\ \text{m}$) with α-particles ($Z=2$, $E\approx 5.5\ \text{MeV}$) from a radium source. A fluorescent ZnS screen detected scattered α's. Key observations:

Rutherford concluded that all of the positive charge $+Ze$ and nearly all of the mass of the atom are concentrated in a nucleus of radius $\sim 10^{-15}\ \text{m}$, with electrons orbiting at $\sim 10^{-10}\ \text{m}$ — an atomic radius $10^5$ times the nuclear radius.

Derivation — distance of closest approach $r_0$

A head-on α-particle with kinetic energy $K$ approaches a target nucleus of charge $+Ze$. At closest approach, the entire $K$ has been converted to electrostatic PE:

For $K = 5.5\ \text{MeV}$ α on gold ($Z=79$):

Thus the gold nucleus is at most $\sim 4\times 10^{-14}\ \text{m}$ in size — an upper bound, since actual nuclear radii are smaller still.

Derivation — impact parameter and scattering angle

The Rutherford formula relating impact parameter $b$ to scattering angle $\theta$ (for a Coulomb potential, derived from conservation of energy + angular momentum, hyperbolic trajectory) is

The differential cross-section (number scattered into solid angle $d\Omega$) is

This $\sin^{-4}(\theta/2)$ dependence was confirmed beautifully by Geiger and Marsden — the experimental signature of a point Coulomb scatterer.

Worked Example

A $7.7\ \text{MeV}$ α-particle is scattered by a copper nucleus ($Z=29$). Find $r_0$ for a head-on collision.

Pitfalls

• $r_0$ depends only on $K$ (not on α-mass) — confusing this with kinematic distances is common.
• The factor of $2$ in the numerator (charge of α) is often dropped — always write the product of charges explicitly.
• Scattering angle distribution is $\sin^{-4}(\theta/2)$, not $\sin^{-4}\theta$.

12.3 Drawbacks of Rutherford's Model

Definition

Rutherford's planetary atom faces two fatal failures within classical electrodynamics:

1. Stability problem. An electron in circular orbit is accelerating (centripetal $a = v^2/r$). Maxwell's electromagnetism requires accelerating charges to radiate at the orbital frequency, losing energy. The electron should spiral into the nucleus in $\sim 10^{-8}\ \text{s}$. But atoms are stable for $> 10^{17}\ \text{s}$.

2. Spectral problem. As the orbit shrinks, the orbital frequency $\nu$ varies continuously, so the radiated frequency would form a continuous spectrum. Hydrogen, however, shows sharp discrete lines (Lyman, Balmer, …) at very specific frequencies.

Derivation — classical collapse time of hydrogen

The Larmor power radiated by an accelerating non-relativistic charge is $P = \dfrac{q^2 a^2}{6\pi\varepsilon_0 c^3}$. For a circular orbit $a = \dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{mr^2}$. The total energy is $E = -\dfrac{1}{8\pi\varepsilon_0}\dfrac{e^2}{r}$, so $dE/dr = +\dfrac{1}{8\pi\varepsilon_0}\dfrac{e^2}{r^2}$. Setting $\dot E = -P$ and integrating from $r_0 = 0.5\ \text{Å}$ to $0$ yields a fall time

where $r_e = e^2/(4\pi\varepsilon_0 m c^2) \approx 2.8\times 10^{-15}\ \text{m}$ is the classical electron radius. Nanoseconds — yet atoms are eternal. Classical physics has clearly broken down.

Pitfalls

• Note Rutherford did not fail to explain the gold-foil data — those he explained perfectly. The failures are in atomic stability and spectra.
• Do not confuse "orbital frequency" with "radiated frequency": classically they are equal; quantum-mechanically they are not.

12.4 Bohr's Postulates

Definition

Niels Bohr (1913) resolved both crises with three bold postulates that hybridize classical mechanics with quantum hypotheses:

1. Stationary orbits. Electrons revolve in certain allowed (stationary) circular orbits in which they do not radiate, despite the acceleration.
2. Quantization of angular momentum. The allowed orbits are those for which the angular momentum is an integer multiple of $\hbar = h/2\pi$:

3. Frequency condition. Radiation is emitted only when the electron jumps from a higher orbit of energy $E_i$ to a lower of energy $E_f$, with frequency

The first postulate breaks classical electrodynamics ad hoc; the second introduces $h$; the third — the Bohr frequency rule — connects to the photon picture.

Pitfalls

• The angular-momentum quantization is $n\hbar$, not $nh$.
• Bohr's postulates apply only to hydrogenic (one-electron) atoms: H, He$^+$, Li$^{2+}$, Be$^{3+}$, …
• The "stationary orbit" is a misnomer — the electron does move, but it does not radiate.

12.5 Bohr Radius $r_n$ — Derivation

Definition

For a hydrogen-like atom of nuclear charge $+Ze$, the electron of mass $m$ and charge $-e$ moves in a circular orbit of radius $r$ with speed $v$ under Coulomb attraction.

Derivation

The centripetal force is provided by Coulomb's law:

Bohr's quantization condition:

From (2): $v = \dfrac{nh}{2\pi m r}$. Substitute into (1):

The constant $a_0$ is the Bohr radius. The ground-state hydrogen radius is $r_1 = a_0 = 0.529\ \text{Å}$, $r_2 = 4a_0 = 2.12\ \text{Å}$, $r_3 = 9a_0$, etc. — orbits grow as $n^2$.

Worked Example

Find the radius of the third orbit of singly-ionized helium (He$^+$, $Z=2$).

Pitfalls

• $r_n \propto n^2/Z$, not $n/Z$. Many students forget the square.
• For multi-electron atoms, Bohr's formula does not apply; use effective nuclear charge approaches instead.

12.6 Orbital Velocity $v_n$ — Derivation

Definition

The speed of the electron in the $n^{\text{th}}$ orbit.

Derivation

From (2): $v_n = \dfrac{nh}{2\pi m r_n}$. Substitute $r_n = \dfrac{n^2 h^2 \varepsilon_0}{\pi m Z e^2}$:

Here $\alpha$ is the fine-structure constant. The ground-state hydrogen electron moves at $v_1 \approx c/137 \approx 2.19\times 10^{6}\ \text{m/s}$ — non-relativistic, justifying our use of $\tfrac{1}{2}mv^2$ for kinetic energy.

Worked Example

The velocity of the electron in the second Bohr orbit of Li$^{2+}$ ($Z=3$, $n=2$) is

Pitfalls

• $v_n \propto Z/n$ — inversely proportional to $n$, not $n^2$.
• Time period: $T_n = \dfrac{2\pi r_n}{v_n}\propto \dfrac{n^3}{Z^2}$.
• Orbital current $I_n = \dfrac{e}{T_n} \propto \dfrac{Z^2}{n^3}$.

12.7 Energy $E_n$ — Derivation

Definition

The total mechanical energy (kinetic + potential) of the electron in the $n^{\text{th}}$ stationary orbit.

Derivation

Kinetic energy: from (1), $\dfrac{1}{2}mv^2 = \dfrac{1}{2}\cdot\dfrac{Ze^2}{4\pi\varepsilon_0 r}$.

Potential energy (Coulomb, with zero at infinity): $U = -\dfrac{Ze^2}{4\pi\varepsilon_0 r}$.

Total: $E = K + U = \dfrac{Ze^2}{8\pi\varepsilon_0 r} - \dfrac{Ze^2}{4\pi\varepsilon_0 r} = -\dfrac{Ze^2}{8\pi\varepsilon_0 r}$.

Substitute $r_n$:

Key relations:

• $K_n = -E_n = +13.6\,Z^2/n^2\ \text{eV}$ (virial theorem).
• $U_n = 2E_n = -27.2\,Z^2/n^2\ \text{eV}$.
• $E_n / K_n = -1$, $U_n / K_n = -2$ (Coulomb virial).

Worked Example

Energy of the $n=2$ level of hydrogen:

Pitfalls

• Sign is negative for bound states; $E\to 0$ as $n\to \infty$.
• $E\propto Z^2/n^2$ — note the $Z^2$ scaling. He$^+$ ground state is $-54.4\ \text{eV}$, not $-27.2$.
• Total energy $= -K$, not $+K$ (sign trap).

12.8 Energy Level Diagram of Hydrogen

Definition

A vertical-axis plot of $E_n$ vs. $n$ shows the discrete bound states of hydrogen.

• Ionization energy of hydrogen (from ground state to $n=\infty$): $E_\infty - E_1 = 0 - (-13.6) = 13.6\ \text{eV}$.
• First excitation energy (from $n=1$ to $n=2$): $E_2 - E_1 = -3.4 - (-13.6) = 10.2\ \text{eV}$.
• Second excitation energy ($1\to 3$): $-1.51 - (-13.6) = 12.09\ \text{eV}$.

Worked Example

What minimum kinetic energy must an electron have to excite a ground-state H atom to $n=3$ in a collision?

The atom needs $12.09\ \text{eV}$. So $K_{\text{min}} = 12.09\ \text{eV}$ (the projectile electron need not stop — only transfer this energy).

Pitfalls

• Excitation energy is referenced from the ground state, not from $n=0$.
• For He$^+$, ionization energy $= 13.6 \times Z^2 = 54.4\ \text{eV}$.

12.9 Hydrogen Spectrum — Series

Definition

When excited hydrogen atoms relax, photons are emitted in discrete frequencies grouped into spectral series, each named after its discoverer and characterized by the lower level $n_1$:

Balmer's visible H$_\alpha$ (red), H$_\beta$ (cyan, $486\ \text{nm}$), H$_\gamma$ (blue, $434\ \text{nm}$), H$_\delta$ (violet, $410\ \text{nm}$) are the familiar lines that gave Balmer his empirical 1885 formula.

Pitfalls

• Lyman is UV (because $1\to 2$ requires $10.2\ \text{eV}$).
• Balmer is the only visible series for hydrogen.
• Wavelength formula uses $1/n_1^2 - 1/n_2^2$ — always $n_2 > n_1$, so the bracket is positive.

12.10 Rydberg Formula

Definition

The wave-number $\bar\nu = 1/\lambda$ of an emitted/absorbed line in the hydrogen-like spectrum:

where the Rydberg constant is

Derivation

From Bohr's frequency postulate $h\nu = E_{n_2} - E_{n_1}$ (where $E_{n_2} > E_{n_1}$ for emission going from $n_2 \to n_1$):

Divide by $hc$ and use $1/\lambda = \nu/c$:

Numerically, $R = 1.097\times 10^7\ \text{m}^{-1}$ corresponds to the photon energy $Rhc = 13.6\ \text{eV}$ for $n_1=1, n_2=\infty$ — the ionization energy, as expected.

Worked Example

Find the wavelength of H$_\alpha$ ($3\to 2$, hydrogen, $Z=1$).

so $\lambda = 656.3\ \text{nm}$ — the iconic red line of the Balmer series.

Worked Example — Series Limit

The Lyman series limit ($n_2\to\infty$) lies at

Photons shorter than $91.2\ \text{nm}$ will ionize ground-state hydrogen.

Pitfalls

• The bracket is $1/n_1^2 - 1/n_2^2$, not $1/n_2^2 - 1/n_1^2$ (which would give a negative wavelength).
• $R$ has units of $\text{m}^{-1}$; for energies use $Rhc = 13.6\ \text{eV}$.
• For deuterium, replace $m$ by reduced mass $\mu$ — gives slightly different $R_D$ (this isotope effect was historically critical).

12.11 de Broglie's Explanation of Bohr Quantization

Definition

In 1924 Louis de Broglie hypothesised that every particle of momentum $p = mv$ has an associated matter wave of wavelength

He then showed that Bohr's mysterious quantization condition is equivalent to demanding that the electron's matter wave form a standing wave around the orbit — i.e., the orbit's circumference is an integer number of de Broglie wavelengths.

Derivation

Standing-wave condition on an orbit of radius $r_n$:

Rearrange:

which is exactly Bohr's angular-momentum quantization (postulate 2). de Broglie thus reduced an ad hoc postulate to a wave-physics requirement — a step that helped motivate Schrödinger's full quantum theory in 1926.

Worked Example

In the ground state ($n=1$) of hydrogen, $v_1 = 2.19\times 10^6\ \text{m/s}$, so

Check: $2\pi r_1 = 2\pi\cdot 0.529\ \text{Å} = 3.33\ \text{Å}$. The orbit fits exactly one wavelength.

Pitfalls

• The standing-wave picture is a visualisation — modern QM does not treat the electron as a literal wave on a circle. It is the probability amplitude that is wave-like.
• $n=1$ contains one full wavelength, $n=2$ has two, etc. — not "half" wavelengths.

12.12 Limitations of Bohr's Model

Definition

Despite explaining hydrogen's spectrum to four significant figures, Bohr's model fails in several systematic ways:

1. Multi-electron atoms. Bohr's formulas were derived for one electron. They cannot account for helium, lithium, or any heavier atom — electron-electron repulsion is absent.
2. Fine structure. High-resolution spectroscopy reveals each Balmer line is actually a closely-spaced doublet. Bohr predicts a single line.
3. Hyperfine and Zeeman effects. Splitting under nuclear-spin coupling and external magnetic fields is unexplained.
4. Relative intensities of spectral lines. Bohr predicts which lines occur but not their relative brightness — only QM transition probabilities (Einstein $A$, $B$ coefficients) achieve this.
5. Wave-particle duality is ignored. The electron is treated as a particle in a definite orbit — yet the uncertainty principle $\Delta x\,\Delta p \ge \hbar/2$ forbids simultaneous knowledge of orbit radius and momentum.
6. No mechanism for radiation. Bohr postulates that an electron in a stationary orbit does not radiate, but offers no derivation; full QED supplies one.
7. Chemical bonding. Bohr's circular orbits cannot explain molecular formation (covalent bonds, hybridization, etc.).

Modern resolution: Schrödinger's wave equation. The "orbit" is replaced by a probabilistic orbital (1s, 2s, 2p, …), with the principal quantum number $n$, orbital angular-momentum quantum number $\ell$, magnetic quantum number $m_\ell$, and spin quantum number $m_s$.

Pitfalls

• Bohr's model is not wrong about hydrogen energies; the formula $E_n = -13.6/n^2$ eV is exact within non-relativistic QM. It is the interpretation (definite orbits) that fails.
• It explains $\sim 99\%$ of hydrogenic transitions but $\sim 0\%$ of multi-electron chemistry.

Solved Problems

Problem 1 — Closest approach for a non-head-on collision

An α-particle of $K = 4.0\ \text{MeV}$ is incident on a copper foil ($Z=29$). Find the closest approach distance for a head-on collision and discuss what happens for a non-head-on impact.

Solution. For head-on:

For non-head-on, conservation of angular momentum forces a finite "perihelion" distance $r_{\min} > r_0$. The trajectory is a hyperbola; $r_{\min}$ depends on impact parameter $b$ via the Rutherford geometry.

Problem 2 — Bohr radius of doubly-ionized lithium

Compute $r_1$ for Li$^{2+}$ ($Z=3$).

Solution. $r_1(\text{Li}^{2+}) = a_0/Z = 0.529/3 = 0.176\ \text{Å}$.

Problem 3 — Photon emitted when H electron jumps $4\to 2$

Find $\lambda$ and identify the colour.

Solution.

$\lambda = 486\ \text{nm}$ — H$_\beta$, cyan.

Problem 4 — Total energy lost when H atom ionizes

How much energy is required to ionize a hydrogen atom from the $n=3$ state?

Solution. $E_\infty - E_3 = 0 - (-1.51) = 1.51\ \text{eV}$.

Problem 5 — Maximum number of spectral lines

If a hydrogen atom is excited to $n = 5$, what is the maximum number of distinct spectral lines that can be emitted as it returns to the ground state?

Solution. Number of lines $= \dfrac{n(n-1)}{2} = \dfrac{5\cdot 4}{2} = 10$ lines.

Problem 6 — Series identification

A hydrogen sample emits light at $\lambda = 1216\ \text{Å}$. Identify the transition.

Solution. $\dfrac{1}{1216\times 10^{-10}} = 8.22\times 10^6\ \text{m}^{-1}$. Try $n_1 = 1$ (Lyman):

It is the Lyman-α line ($2\to 1$).

Problem 7 — Photon momentum and recoil

A hydrogen atom at rest emits a photon during $n=2\to n=1$ transition. Find the photon energy, photon momentum, and recoil velocity of the atom.

Solution. Photon energy $E = 10.2\ \text{eV} = 1.63\times 10^{-18}\ \text{J}$. Momentum $p = E/c = 5.44\times 10^{-27}\ \text{kg m/s}$. Atom mass $M_H = 1.67\times 10^{-27}\ \text{kg}$, so recoil $v = p/M_H \approx 3.26\ \text{m/s}$. (Small but non-zero; the photon energy in the atom's rest frame is slightly less than $10.2\ \text{eV}$ — the rest is recoil KE.)

JEE/NEET Edge Cases

1. Reduced mass correction. Replace $m_e$ by $\mu = \dfrac{m_e M}{m_e + M}$ for a finite nuclear mass. This explains the small wavelength shift between H and D (deuterium) spectra.
2. Relativistic correction at high $Z$. For heavy hydrogenic ions (e.g., U$^{91+}$), $v\sim Zc/137$ becomes comparable to $c$, breaking Bohr's non-relativistic derivation.
3. Magnetic moment of the orbiting electron.

The Bohr magneton $\mu_B$ is the natural unit of atomic magnetism. 4. Frank-Hertz experiment. Provided direct evidence for discrete atomic energy levels: electrons accelerated through Hg vapour at $4.9\ \text{V}$ lose exactly $4.9\ \text{eV}$ in inelastic collisions. 5. Selection rules (beyond Bohr; from QM): $\Delta\ell = \pm 1$, $\Delta m_\ell = 0, \pm 1$. Bohr's model has no $\ell$ and thus no selection rule. 6. Continuous absorption above ionization. For $\lambda < 91.2\ \text{nm}$, hydrogen absorbs continuously (photoionization). The spectrum is line-like below threshold and continuous above. 7. Numerical shortcut. Use $E_n = -13.6\,Z^2/n^2\ \text{eV}$ and $r_n = 0.529\,n^2/Z\ \text{Å}$ — memorize these.

Quick Recap

• Thomson: uniform positive sphere with embedded electrons — fails spectra and scattering.
• Rutherford: small heavy nucleus; $r_0 = (1/4\pi\varepsilon_0)(2Ze^2)/K$ — fails stability and discrete spectra.
• Bohr postulates: stationary orbits, $L = n\hbar$, $h\nu = \Delta E$.
• $r_n = n^2 a_0/Z$, $v_n = Z\alpha c/n$, $E_n = -13.6\,Z^2/n^2\ \text{eV}$.
• Rydberg: $1/\lambda = R Z^2(1/n_1^2 - 1/n_2^2)$ with $R = 1.097\times 10^7\ \text{m}^{-1}$.
• Series: Lyman (UV, $n_1=1$), Balmer (visible, $n_1=2$), Paschen/Brackett/Pfund (IR, $n_1=3,4,5$).
• de Broglie: $2\pi r_n = n\lambda_n$ — standing wave on orbit.
• Bohr fails for multi-electron atoms, fine structure, intensities, chemistry.

Formula Sheet