Atoms

By the early 1900s, J. J. Thomson had proposed that an atom is a sphere of positive charge with electrons embedded in it, like seeds in a watermelon (the "plum-pudding" model). To test this, Hans Geiger and Ernest Marsden, working under Ernest Rutherford, fired a narrow beam of fast, energetic alpha (α) particles (helium nuclei, charge +2e) at a very thin gold foil and observed how the particles scattered using a rotatable zinc-sulphide detecting screen.

Three observations stood out. Most α-particles passed straight through the foil with little or no deflection. A small fraction were deflected through moderate angles. Most strikingly, about 1 in every 8000 α-particles bounced back through angles greater than 90°, some almost straight back towards the source. A spread-out positive charge (as in Thomson's model) could never produce such large single-collision deflections, since it exerts only a weak, gradually varying force on a passing α-particle.

Rutherford concluded that the positive charge and almost the entire mass of the atom must be concentrated in an extremely small central region called the nucleus, with electrons revolving around it at comparatively large distances, held by Coulomb attraction. Since most α-particles pass through undeflected, an atom is mostly empty space; the nucleus occupies a radius of about 10^-14 m compared with the atom's overall radius of about 10^-10 m — the nucleus is roughly 10,000 times smaller than the atom.

For a head-on (zero impact parameter) collision, the α-particle momentarily comes to rest at the point of closest approach, where its entire initial kinetic energy K has converted into electrostatic potential energy. Equating these gives the distance of closest approach r₀, an experimental upper bound on the size of the nucleus. For an off-centre collision with impact parameter b, the scattering angle θ decreases as b increases — small b (near head-on) gives large θ, and large b gives only a small deflection.

Worked example: An α-particle of kinetic energy 5.5 MeV is scattered head-on by a gold nucleus (Z = 79). Find the distance of closest approach.
K = 5.5 MeV = 5.5 × 10⁶ × 1.6 × 10⁻¹⁹ J = 8.8 × 10⁻¹³ J.
r₀ = (1/4πε₀) × (2Ze²)/K = (9 × 10⁹ × 2 × 79 × (1.6 × 10⁻¹⁹)²)/(8.8 × 10⁻¹³)
r₀ = (9 × 10⁹ × 158 × 2.56 × 10⁻³⁸)/(8.8 × 10⁻¹³) ≈ 3.64 × 10⁻²⁶/8.8 × 10⁻¹³ ≈ 4.14 × 10⁻¹⁴ m.
This is about 41 femtometres, confirming that gold's positive charge is confined to a region roughly 10,000 times smaller than the atom itself.

When a gas is excited (for example, by an electric discharge) and its light is passed through a prism or grating, it does not give a continuous rainbow but a series of sharp, bright lines at specific wavelengths — an emission line spectrum. The reverse experiment, passing white light through a cool gas, produces dark lines at the very same wavelengths — an absorption spectrum. Every element has its own unique pattern of lines, which acts like a fingerprint and can be used to identify elements even in distant stars.

Hydrogen, having only one electron, has the simplest and most systematic spectrum. In 1885, Johann Balmer found an empirical formula that fit the four then-known visible hydrogen lines. This was later generalised by Johannes Rydberg into a single formula covering all the observed series of hydrogen lines, well before any theoretical model could explain why it worked.

Each series corresponds to transitions ending on the same lower level n₁, while the upper level n₂ takes successive integer values greater than n₁: the Lyman series (n₁ = 1, ultraviolet), Balmer series (n₁ = 2, partly visible), Paschen series (n₁ = 3, infrared), Brackett series (n₁ = 4, infrared) and Pfund series (n₁ = 5, infrared). Within each series, the lines crowd together and converge to a shortest-wavelength series limit as n₂ → ∞.

Worked example: Find the series limit (shortest wavelength) of the Balmer series.
Series limit corresponds to n₁ = 2, n₂ → ∞, so 1/n₂² → 0.
1/λ = R(1/2² − 0) = R/4 = 1.097 × 10⁷/4 = 2.7425 × 10⁶ m⁻¹.
λ = 1/(2.7425 × 10⁶) ≈ 3.646 × 10⁻⁷ m = 364.6 nm.
All Balmer lines therefore lie between 364.6 nm and the first (longest-wavelength) member of the series.

Rutherford's planetary model correctly located the nucleus but classical electromagnetism predicted disaster: an electron orbiting (i.e. accelerating) should continuously radiate energy, spiral inward, and collapse into the nucleus in a fraction of a second — and it should emit a continuous spectrum as it does so. Neither is observed; atoms are stable and emit only discrete lines. In 1913, Niels Bohr resolved this by proposing that, for the hydrogen atom, classical mechanics is supplemented by quantum conditions, expressed as three postulates.

Postulate 1 (stationary orbits): An electron can revolve only in certain special orbits, called stationary states, without radiating energy, even though it is accelerating. In these orbits, the Coulomb force of attraction on the electron provides exactly the centripetal force needed for circular motion.

Postulate 2 (quantisation of angular momentum): Only those orbits are allowed for which the orbital angular momentum is an integral multiple of h/2π (i.e. of ħ). This is an ad hoc quantum rule with no classical justification at the time; it was justified later by de Broglie's matter-wave hypothesis.

Postulate 3 (Bohr frequency condition): An atom can emit or absorb radiation only when an electron jumps from one stationary orbit to another. If the electron falls from a higher-energy orbit Eᵢ to a lower-energy orbit E_f, a photon of frequency ν is emitted such that the photon energy equals the energy difference between the two orbits. The same relation governs absorption when the electron jumps up.

Together these three postulates explain both why atoms are stable (electrons don't radiate while in a stationary orbit) and why atomic spectra are discrete lines rather than a continuum (only specific energy jumps, hence specific photon energies/wavelengths, are allowed).

For an electron of charge −e moving in a circular orbit of radius r around a nucleus of charge +Ze, the electrostatic force of attraction supplies the centripetal force: (1/4πε₀)(Ze²/r²) = mv²/r, which gives v² = Ze²/(4πε₀mr). Combining this with Bohr's quantisation condition mvr = nh/2π and eliminating v algebraically gives the radius of the nth allowed orbit.

For hydrogen (Z = 1), the smallest orbit (n = 1) has radius a₀ = 0.529 × 10⁻¹⁰ m, called the Bohr radius. Since rₙ ∝ n², orbits expand rapidly for higher n. Substituting rₙ back into v = nh/(2πmrₙ) gives the orbital speed vₙ, which for hydrogen's ground state works out to about 2.19 × 10⁶ m/s (roughly c/137) and decreases as 1/n for higher orbits.

The total energy of the electron in the nth orbit is the sum of kinetic energy (½mv²) and electrostatic potential energy (−(1/4πε₀)(Ze²/r)); working through the algebra gives Eₙ = −13.6 Z²/n² eV for hydrogen-like atoms. The negative sign shows the electron is bound to the nucleus; as n → ∞, Eₙ → 0, corresponding to a free (ionised) electron. The ground state energy E₁ = −13.6 eV is therefore also the ionisation energy of hydrogen — the minimum energy needed to remove the electron completely.

Worked example: Find the radius and energy of the n = 2 orbit of a hydrogen atom.
r₂ = n²a₀ = 2² × 0.529 Å = 4 × 0.529 Å = 2.116 Å = 2.116 × 10⁻¹⁰ m.
E₂ = −13.6/2² eV = −13.6/4 eV = −3.4 eV.
So an electron lifted from the ground state (−13.6 eV) to n = 2 (−3.4 eV) has absorbed 10.2 eV of energy.

Bohr's third postulate says a photon is emitted when an electron falls from an orbit n₂ to a lower orbit n₁, with photon energy hc/λ = Eₙ₂ − Eₙ₁. Substituting the orbital energy formula Eₙ = −13.6Z²/n² eV derived from the first two postulates gives 1/λ = RZ²(1/n₁² − 1/n₂²), which is exactly the empirical Rydberg formula found decades earlier — and it even predicts the correct numerical value of R purely from fundamental constants (e, m, ε₀, h, c). This was one of the great triumphs of Bohr's theory.

Every observed hydrogen series now has a clean explanation as transitions ending on a common lower level: Lyman (n₁=1), Balmer (n₁=2), Paschen (n₁=3), Brackett (n₁=4) and Pfund (n₁=5), with the upper level n₂ taking all integer values greater than n₁. An atom initially excited to level n can cascade down to the ground state through several possible intermediate steps, so more than one wavelength can be emitted from a single excited sample; the total number of distinct spectral lines possible is n(n−1)/2.

Worked example 1: Find the wavelength of the Hα line (Balmer series, n₂ = 3 → n₁ = 2).
1/λ = R(1/2² − 1/3²) = 1.097 × 10⁷ × (0.25 − 0.1111) = 1.097 × 10⁷ × 0.1389 ≈ 1.524 × 10⁶ m⁻¹.
λ = 1/(1.524 × 10⁶) ≈ 6.563 × 10⁻⁷ m = 656.3 nm, the familiar red Balmer line, matching observation.

Worked example 2: If a hydrogen atom is excited to n = 4, how many spectral lines can it emit while returning to the ground state?
N = n(n−1)/2 = 4 × 3/2 = 6 possible transitions (4→3, 4→2, 4→1, 3→2, 3→1, 2→1), so up to 6 distinct wavelengths can appear in the spectrum.

Bohr's angular-momentum condition (mvr = nh/2π) worked brilliantly but had no physical justification — it was simply assumed to fit the data. In 1924, Louis de Broglie proposed that every moving particle has an associated matter wave of wavelength λ = h/(mv). Applying this to the orbiting electron gives a natural explanation: a stationary orbit is stable precisely when the electron's wave forms a standing wave that closes smoothly on itself around the orbit, i.e. the orbit's circumference must contain a whole number of de Broglie wavelengths.

This standing-wave condition is written 2πrₙ = nλ. Substituting the de Broglie wavelength λ = h/(mv) immediately gives 2πrₙ = nh/(mv), which rearranges to mvrₙ = nh/2π — exactly Bohr's second postulate, now derived rather than assumed. If the orbit did not contain a whole number of wavelengths, the electron wave would interfere destructively with itself over successive revolutions and the orbit could not persist.

Worked example: Verify the standing-wave condition for the n = 1 (ground-state) orbit of hydrogen.
r₁ = 0.529 × 10⁻¹⁰ m, so circumference = 2πr₁ = 2π × 0.529 × 10⁻¹⁰ ≈ 3.324 × 10⁻¹⁰ m.
Using v₁ = 2.19 × 10⁶ m/s: λ = h/(mv₁) = (6.626 × 10⁻³⁴)/(9.11 × 10⁻³¹ × 2.19 × 10⁶) ≈ 3.32 × 10⁻¹⁰ m.
The circumference equals almost exactly one de Broglie wavelength (n = 1), confirming the standing-wave picture.

Despite this success, the Bohr model has real limitations. It gives accurate results only for hydrogen and other single-electron (hydrogen-like) systems such as He⁺ and Li²⁺; it cannot correctly predict the spectra of multi-electron atoms, where electron-electron repulsion matters. It also cannot explain the relative intensities of different spectral lines, the fine structure (small splitting) of spectral lines, or the splitting of lines in electric fields (Stark effect) and magnetic fields (Zeeman effect). Being a semi-classical patchwork of classical orbits plus an imposed quantum rule, the Bohr model was eventually superseded by full quantum mechanics based on the Schrödinger equation, which treats the electron entirely as a wave (via a probability amplitude) rather than a particle in a definite orbit.