Atomic Structure and the Rutherford Scattering Experiment

The story of atomic structure is one of the most important narratives in physics. Over roughly a century, physicists moved from thinking of atoms as indivisible solid spheres to our modern picture of a tiny, dense, positively charged nucleus surrounded by orbiting electrons. This lesson covers the key experiments and models that built our understanding, the notation used to describe nuclei, and the concept of specific charge. These ideas form the foundation of AQA A-Level Physics sections 3.2.1 and 3.8.1.

Early Models of the Atom

The Plum Pudding Model (Thomson, 1897)

After J.J. Thomson discovered the electron in 1897, he proposed a model of the atom in which negative electrons were embedded in a uniform sphere of positive charge — like plums in a pudding (or raisins in a Christmas pudding). Key features:

The atom was imagined as a sphere of uniformly distributed positive charge, roughly 10⁻¹⁰ m in diameter.
Negatively charged electrons were dotted throughout this positive sphere.
The overall atom was electrically neutral.
There was no concentrated nucleus in this model.

This model was widely accepted for over a decade, until a crucial experiment proved it wrong.

The Rutherford-Geiger-Marsden Experiment (1909–1911)

Experimental Setup

Ernest Rutherford, along with Hans Geiger and Ernest Marsden, directed a beam of alpha particles (helium-4 nuclei, charge +2e, mass ≈ 4 u) at a very thin gold foil (only a few hundred atoms thick). A zinc sulfide screen surrounding the foil was used to detect the scattered alpha particles — each particle produced a tiny flash of light (scintillation) when it struck the screen.

Observations

Observation	Proportion of alpha particles
Passed straight through with little or no deflection	The vast majority (> 99%)
Deflected through small angles (< 10°)	A small fraction
Deflected through large angles (> 90°)	Approximately 1 in 8000
Deflected back towards the source (> 150°)	Very rare (approximately 1 in 20 000)

Conclusions — The Nuclear Model

Rutherford analysed the results and concluded:

Most of the atom is empty space — because the vast majority of alpha particles passed straight through the foil with no deflection.
The positive charge and nearly all the mass are concentrated in a very small, dense region — the nucleus — because some alpha particles were deflected through large angles, and a few even bounced back. Only a very concentrated positive charge could exert a Coulomb repulsion strong enough to reverse the direction of a fast-moving alpha particle.
The nucleus is very small compared to the atom — the nuclear radius is approximately 10⁻¹⁵ m (1 femtometre, fm), while the atomic radius is approximately 10⁻¹⁰ m. This means the nucleus is roughly 100 000 times smaller than the atom.
Electrons orbit the nucleus at relatively large distances, occupying the otherwise empty space.

Key Point: The fact that very few alpha particles were deflected through large angles shows that the nucleus is extremely small. If the positive charge were spread out (as in the plum pudding model), the maximum deflection would be very small — large-angle scattering would be impossible.

Why the Plum Pudding Model Failed

In Thomson's model, the positive charge is spread over the full atomic volume (~10⁻¹⁰ m). An alpha particle passing through such a diffuse charge distribution would experience only weak electrostatic forces and would never be deflected through large angles. The observation of back-scattering was, in Rutherford's own words, "as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you."

Nuclear Notation

Every nuclide (a specific nuclear species) is described using the notation:

ᴬ_Z X

where:

X is the chemical symbol of the element
Z is the proton number (also called atomic number) — the number of protons in the nucleus
A is the nucleon number (also called mass number) — the total number of protons and neutrons (nucleons) in the nucleus

The number of neutrons is therefore: N = A − Z

Examples

Nuclide	Symbol	Protons (Z)	Neutrons (N)	Nucleons (A)
Hydrogen-1	¹₁H	1	0	1
Carbon-12	¹²₆C	6	6	12
Carbon-14	¹⁴₆C	6	8	14
Uranium-235	²³⁵₉₂U	92	143	235
Uranium-238	²³⁸₉₂U	92	146	238

Isotopes

Isotopes are atoms of the same element (same number of protons, Z) that have different numbers of neutrons (different A).

Isotopes have identical chemical properties (because chemistry depends on electron configuration, which is determined by Z).
Isotopes have different physical properties — for example, different masses and different nuclear stability.
Some isotopes are radioactive (radioisotopes) — their nuclei are unstable and decay by emitting radiation.

Example: Carbon has three naturally occurring isotopes: ¹²₆C (stable, 98.9% abundance), ¹³₆C (stable, 1.1% abundance), and ¹⁴₆C (radioactive, trace amounts — used in carbon dating).

Specific Charge

The specific charge of a particle is defined as the ratio of its charge to its mass:

Specific charge = Q / m

The units are C kg⁻¹.

This quantity is important because it determines how a particle behaves in electric and magnetic fields. Particles with a larger specific charge are deflected more in a given field.

Worked Example 1 — Specific charge of a proton

A proton has charge Q = +1.60 × 10⁻¹⁹ C and mass m = 1.673 × 10⁻²⁷ kg.

Specific charge = Q / m = 1.60 × 10⁻¹⁹ / 1.673 × 10⁻²⁷

Specific charge = 9.56 × 10⁷ C kg⁻¹

Worked Example 2 — Specific charge of a nucleus

Find the specific charge of a carbon-12 nucleus (⁶ protons, 6 neutrons).

Charge: Q = 6 × 1.60 × 10⁻¹⁹ = 9.60 × 10⁻¹⁹ C

Mass: m = 12 × 1.661 × 10⁻²⁷ = 1.993 × 10⁻²⁶ kg

Specific charge = 9.60 × 10⁻¹⁹ / 1.993 × 10⁻²⁶

Specific charge = 4.82 × 10⁷ C kg⁻¹

Worked Example 3 — Specific charge of an electron

An electron has charge Q = −1.60 × 10⁻¹⁹ C (magnitude 1.60 × 10⁻¹⁹ C) and mass m = 9.11 × 10⁻³¹ kg.

Specific charge = 1.60 × 10⁻¹⁹ / 9.11 × 10⁻³¹

Specific charge = 1.76 × 10¹¹ C kg⁻¹

Note that the specific charge of an electron is about 1836 times larger than that of a proton. This is why electrons are deflected much more than protons in electric and magnetic fields.

Exam Tip: When calculating specific charge, use the magnitude of the charge (ignore the sign) unless the question specifically asks for the sign. Always state the units C kg⁻¹. Questions often ask you to compare the specific charges of different particles — the electron always has by far the largest value.

Estimating Nuclear Size from Scattering

The closest approach distance of an alpha particle to a nucleus can be estimated using energy conservation. At the point of closest approach, all the kinetic energy of the alpha particle has been converted to electrical potential energy:

½mv² = kQα Qnucleus / r

where k = 1/(4πε₀) = 8.99 × 10⁹ N m² C⁻², Qα = 2e, and Qnucleus = Ze.

Rearranging: r = kQα Qnucleus / Eₖ

Worked Example 4 — Closest approach to a gold nucleus

An alpha particle with kinetic energy 7.7 MeV is fired at a gold-197 nucleus (Z = 79). Find the distance of closest approach.

Convert energy: Eₖ = 7.7 × 10⁶ × 1.60 × 10⁻¹⁹ = 1.232 × 10⁻¹² J

Charges: Qα = 2 × 1.60 × 10⁻¹⁹ = 3.20 × 10⁻¹⁹ C; Q_Au = 79 × 1.60 × 10⁻¹⁹ = 1.264 × 10⁻¹⁷ C

r = (8.99 × 10⁹ × 3.20 × 10⁻¹⁹ × 1.264 × 10⁻¹⁷) / 1.232 × 10⁻¹²

r = (3.636 × 10⁻²⁶) / (1.232 × 10⁻¹²)

r ≈ 2.95 × 10⁻¹⁴ m ≈ 29.5 fm

This is an upper limit on the nuclear radius. The actual nuclear radius of gold is about 7 fm, so the alpha particle does not quite reach the nuclear surface at this energy.

Common Misconception: Students sometimes think the closest approach distance equals the nuclear radius. It is actually an upper bound — the alpha particle is turned around by the Coulomb repulsion before it reaches the nuclear surface. Higher energy alpha particles get closer and give a tighter upper bound.