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At the start of the twentieth century, the atom was imagined as a diffuse sphere of positive charge with electrons embedded throughout — the so-called "plum pudding" model proposed by J. J. Thomson. It was a reasonable guess given the evidence available, but it was about to be demolished by one of the most important experiments in physics.
In 1909, Hans Geiger and Ernest Marsden, working under the direction of Ernest Rutherford at the University of Manchester, fired alpha particles at a thin gold foil and observed where they went. Alpha particles are helium-4 nuclei — positively charged, relatively massive, and fast-moving.
If Thomson’s model were correct, the alpha particles should have passed through the gold atoms with only slight deflections. The positive charge was supposed to be spread thinly across the whole atom, so there would be nothing dense enough to cause a large deflection.
Rutherford famously said it was "almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you."
Rutherford concluded that the atom must contain a tiny, dense, positively charged centre — the nucleus. The key features of the nuclear model are:
The plum pudding model predicts a maximum deflection angle that can be calculated from the electric field of a diffuse charge sphere. For gold atoms, this maximum angle is approximately 0.02° — far too small to account for the large-angle scattering Geiger and Marsden observed. Only a concentrated point-like charge can produce the strong Coulomb repulsion needed to turn an alpha particle through 90° or more.
Experiments using electron diffraction (high-energy electrons scattered off nuclei) have shown that the nuclear radius R depends on the nucleon number A according to the empirical formula:
R = R₀ A^(1/3)
where R₀ ≈ 1.2 – 1.4 fm (femtometres, 10⁻¹⁵ m).
This tells us something remarkable: nuclear volume is proportional to A (since V = 4/3 πR³ ∝ A), which means that every nucleon occupies roughly the same volume regardless of the size of the nucleus. This leads directly to the conclusion that nuclear density is approximately constant.
Using R₀ = 1.4 fm:
This is extraordinarily dense — about 10¹⁴ times denser than water. A teaspoon of nuclear matter would have a mass of roughly 5 billion tonnes. Neutron stars, which are essentially giant nuclei, have densities in this range.
Problem: Calculate the ratio of the nuclear radius of uranium-238 to carbon-12.
R_U / R_C = (A_U / A_C)^(1/3) = (238 / 12)^(1/3) = (19.83)^(1/3) = 2.71
So the uranium nucleus has a radius about 2.7 times that of carbon-12, despite having almost 20 times as many nucleons. This is a direct consequence of the cube-root relationship.
Problem: Using R₀ = 1.2 fm, calculate the nuclear density.
R = R₀ A^(1/3), so V = (4/3)πR³ = (4/3)πR₀³ A
Mass of nucleus = A × m_nucleon = A × 1.67 × 10⁻²⁷ kg
Density = mass / volume = (A × 1.67 × 10⁻²⁷) / ((4/3)π × (1.2 × 10⁻¹⁵)³ × A)
The A cancels:
ρ = 1.67 × 10⁻²⁷ / ((4/3)π × 1.728 × 10⁻⁴⁵) = 1.67 × 10⁻²⁷ / (7.24 × 10⁻⁴⁵) = 2.31 × 10¹⁷ kg m⁻³
Note that A cancels out entirely, proving that nuclear density is independent of nucleon number.
Every nucleus is characterised by two numbers:
The standard notation for a nuclide is:
ᴬ_Z X
For example, carbon-12 is ¹²₆C — it has 6 protons and 6 neutrons.
Isotopes are atoms of the same element (same Z) with different numbers of neutrons (different A). For example:
| Isotope | Protons | Neutrons | Nucleon Number |
|---|---|---|---|
| ¹²₆C | 6 | 6 | 12 |
| ¹³₆C | 6 | 7 | 13 |
| ¹⁴₆C | 6 | 8 | 14 |
All three are carbon (Z = 6), so they have the same electron configuration and almost identical chemical properties. However, they have different masses and different nuclear stability — ¹⁴C is radioactive and is the basis of carbon dating.
Two other terms are useful for classifying nuclides:
| Term | Same | Different | Example |
|---|---|---|---|
| Isotopes | Z (proton number) | A (nucleon number) | ¹²C, ¹⁴C |
| Isobars | A (nucleon number) | Z (proton number) | ⁴⁰K, ⁴⁰Ca |
| Isotones | N (neutron number) | Z (proton number) | ¹³C (N=7), ¹⁴N (N=7) |
If the nucleus contains multiple protons packed into a tiny space, why doesn’t the electrostatic repulsion between them blow the nucleus apart? The answer is the strong nuclear force — one of the four fundamental forces of nature.
| Property | Detail |
|---|---|
| Nature | Attractive between all nucleons (proton–proton, proton–neutron, neutron–neutron) |
| Range | Very short range — acts only up to about 3 fm; negligible beyond this |
| At very short range (< 0.5 fm) | Becomes repulsive, preventing nucleons from being squeezed into each other |
| Strength | Much stronger than the electromagnetic force at nuclear distances |
| Charge independence | Acts equally between all pairs of nucleons regardless of charge |
The balance between the attractive strong force and the repulsive electromagnetic force explains nuclear stability:
For a nucleus with Z protons, the total number of proton–proton repulsive pairs is Z(Z−1)/2, each exerting a long-range Coulomb force across the entire nucleus. The number of nearest-neighbour strong-force pairs grows roughly in proportion to A (each nucleon has a fixed number of nearest neighbours). So:
Eventually the quadratic term dominates, and the nucleus becomes unstable. This is why no stable nuclei exist above bismuth-209.
The nuclear atom model, established by Rutherford’s scattering experiment, revealed that atoms consist of a tiny, dense, positively charged nucleus surrounded by orbiting electrons. The nucleus has a radius given by R = R₀A^(1/3), leading to a remarkably constant nuclear density of approximately 10¹⁷ kg m⁻³. Nuclei are characterised by their proton number Z and nucleon number A, with isotopes being atoms of the same element with different neutron counts. The strong nuclear force holds nuclei together against electromagnetic repulsion, but its short range means that very large nuclei become unstable.
Edexcel 9PH0 specification Topic 11 — Nuclear and particle physics covers the evidence for the nuclear model of the atom from Rutherford's alpha-scattering experiment, the determination of nuclear size and density, the relationship R=R0A1/3, and the limitations of the Rutherford model in light of later quantum-mechanical descriptions (refer to the official specification document for exact wording). Material on atomic structure also threads back into Topic 8 — Nuclear radiation, where decay processes and the nucleon model build on the picture established here. Although introduced as a discrete topic, the nuclear atom is examined synoptically across both Paper 1 (mechanics, fields, particles) and Paper 2 (electric and magnetic fields, thermodynamics, nuclear radiation), often as the opening AO1 prompt of a longer multi-part question. The Edexcel formula booklet provides R=R0A1/3 and constants such as ε0 and e, but does not state the closest-approach formula explicitly — students must derive it from energy conservation.
Question (8 marks):
An alpha particle with initial kinetic energy Ek=5.5 MeV is fired head-on at a stationary gold nucleus (Z=79). Assume the gold nucleus is fixed and treat the interaction as purely electrostatic.
(a) Calculate the closest distance of approach d of the alpha particle to the centre of the gold nucleus. (5)
(b) Compare d with the accepted nuclear radius of gold (R≈7.0×10−15 m) and explain what your result implies about the experiment's ability to probe nuclear structure. (3)
Solution with mark scheme:
(a) Step 1 — set up energy conservation.
At the closest approach, the alpha particle is momentarily stationary; all of its initial kinetic energy has been converted to electrostatic potential energy:
Ek=4πε01d(2e)(Ze)
M1 — correct statement of energy conservation between initial KE and electrostatic PE at closest approach. The factor 2e comes from the alpha particle (charge +2e) and Ze from the gold nucleus.
Step 2 — convert energy to joules.
Ek=5.5 MeV=5.5×106×1.60×10−19 J=8.80×10−13 J.
M1 — correct unit conversion using 1 eV=1.60×10−19 J. Common error: candidates leave the energy in MeV and then divide by SI quantities, producing nonsense answers ten or twelve orders of magnitude wrong.
Step 3 — rearrange for d.
d=4πε01Ek2Ze2
M1 — correct rearrangement.
Step 4 — substitute values.
With 4πε01=8.99×109 N m2 C−2, Z=79, e=1.60×10−19 C:
d=8.80×10−13(8.99×109)(2)(79)(1.60×10−19)2
d=8.80×10−13(8.99×109)(2)(79)(2.56×10−38)
d≈4.13×10−14 m≈4.1×10−14 m
A1 — correct numerical answer to 2 s.f.
A1 — correct unit (metres) carried through, with reasonable significant figures (2 or 3 s.f.).
(b) Step 1 — compare magnitudes.
d≈4.1×10−14 m is approximately 5.9×R — that is, the alpha particle stops about six nuclear radii from the centre of the gold nucleus.
B1 — correct quantitative comparison.
Step 2 — interpret physically.
Because d>R, the alpha particle never penetrates the nucleus. This means the experiment can probe the upper limit of the nuclear radius (the alpha particle would need higher energy to "touch" the nucleus and reveal its true size).
B1 — explanation that d>R implies the alpha never reaches the nuclear surface.
Step 3 — extend.
Higher-energy alpha sources, or electron-scattering experiments at much higher energies, are required to map the nuclear charge distribution directly. The Rutherford geometry establishes only that the nucleus is no larger than d.
B1 — recognising the experiment yields an upper bound on R, motivating later electron-scattering work (Hofstadter, 1950s).
Total: 8 marks (M3 A2 B3, split as shown).
Question (6 marks): A student claims that "the nucleus contains nearly all the mass of the atom but occupies almost none of its volume."
(a) Using R=R0A1/3 with R0=1.2×10−15 m, estimate the density of nuclear matter for a generic nucleus of nucleon number A. Take the mass of one nucleon as 1.67×10−27 kg. (4)
(b) Use your answer to (a) to comment on the student's claim, given that the density of ordinary solid matter is around 104 kg m−3. (2)
Mark scheme decomposition by AO:
(a)
(b)
Total: 6 marks split AO1 = 2, AO2 = 2, AO3 = 2. Edexcel Paper 2 nuclear questions typically blend AO2 (analysis) with AO3 (real-world interpretation), reflecting the topic's role as a bridge between abstract calculation and physical reasoning.
Connects to:
Topic 8 — Nuclear radiation: the model of the nucleus as a tiny dense object surrounded by mostly empty space underpins every subsequent description of alpha, beta and gamma decay. The mean free path of a beta particle in air, the penetrating power of gamma rays, and the energy spectrum of decay products all depend on the geometry established here.
Topic 11 — Particle physics: the nuclear atom is the starting point for the standard model. Once Rutherford had established the nucleus, Chadwick's discovery of the neutron (1932), Yukawa's prediction of the pion, and modern QCD descriptions of quark confinement all built on the same scaffold.
GCSE / earlier work — Atomic structure and the periodic table: electron configuration, ionisation energies, and chemical bonding all assume the Rutherford-Bohr picture of nucleus + orbiting electrons. The "shells" of GCSE chemistry are the lowest-energy approximation of the quantum atom.
Topic 5 — Electric fields: the Coulomb potential V=4πε0rQ governs the alpha-scattering interaction. Closest-approach calculations are direct applications of electric potential energy, and synoptic Paper 2 questions often pair scattering with capacitance or field-strength items.
Topic 12 — Gravitational fields and astrophysics: neutron-star matter has densities comparable to nuclear matter (∼1017 kg m−3). The same R∝A1/3 relation that controls a uranium nucleus also describes, approximately, the radius of a neutron star treated as one giant nucleus held together by gravity rather than the strong force.
Nuclear-atom questions on 9PH0 split AO marks across all three objectives:
| AO | Typical share | Earned by |
|---|---|---|
| AO1 (knowledge / recall) | 35–45% | Stating the conclusions of Rutherford's experiment, recalling R=R0A1/3, defining nucleon number and proton number, identifying isotopes |
| AO2 (analysis / application) | 35–45% | Performing closest-approach calculations, deriving nuclear density, applying energy conservation to scattering geometries |
| AO3 (evaluation / problem-solving) | 15–25% | Comparing model predictions to experimental data, evaluating model limitations, linking nuclear-atom evidence to the development of quantum mechanics |
Examiner-rewarded phrasing: "by conservation of energy, the initial kinetic energy equals the electrostatic potential energy at closest approach"; "the alpha particle is momentarily at rest at distance d from the centre of the nucleus"; "this calculation gives an upper bound on the nuclear radius because the alpha particle does not penetrate the nucleus". Phrases that lose marks: writing "the alpha particle hits the nucleus" (it does not, in head-on Rutherford scattering); confusing "atom" and "nucleus"; treating the gold nucleus as moving (the question's "fixed" assumption is load-bearing for the energy-conservation step).
A specific Edexcel pattern to watch: questions that ask you to explain why most alpha particles pass straight through demand the explicit answer "the atom is mostly empty space" — not simply "alpha particles are small". The empty-space argument is the conclusion of the experiment; the size of the alpha is incidental.
Question: State the conclusions Rutherford drew from the gold-foil scattering experiment about the structure of the atom.
Grade C response (~150 words):
Rutherford concluded that atoms are mostly empty space, because most alpha particles passed through. He also said that there is a small dense centre to the atom called the nucleus, because some alpha particles bounced back. The nucleus contains all the positive charge of the atom.
Examiner commentary: 2/3 marks. The candidate captures two of the three conclusions correctly — the "mostly empty space" inference and the existence of a small dense nucleus. They miss the third standard mark for stating that the nucleus contains almost all the mass of the atom (or equivalently that electrons orbit the nucleus). The phrasing is loose ("a small dense centre") rather than the textbook "small, dense, positively charged nucleus", which would secure the second mark beyond doubt.
Grade A response (~190 words):*
Rutherford's interpretation of the Geiger-Marsden alpha-scattering data led to three conclusions about atomic structure. First, because the great majority of alpha particles passed through the gold foil with little or no deflection, the atom must be mostly empty space — the diffuse positive charge of Thomson's plum-pudding model could not account for this. Second, because a small fraction of alpha particles were deflected through large angles, and a very small number bounced almost straight back, the atom must contain a small region of concentrated positive charge — the nucleus — capable of producing the strong electrostatic repulsion needed for back-scattering. Third, because the nucleus must contain almost all the atomic mass (electrons being far too light to deflect alpha particles significantly), the picture that emerged was a tiny, dense, positively charged nucleus surrounded by orbiting electrons.
Examiner commentary: Full marks (3/3). All three standard mark-scheme points covered — empty space, small dense positively charged nucleus, mass concentration — each tied explicitly to the experimental observation that justifies it. The phrase "Geiger-Marsden" is unnecessary for the marks but signals subject-aware sophistication.
Question: A 4.0 MeV alpha particle is fired head-on at a stationary aluminium nucleus (Z=13). Calculate the closest approach distance and comment on whether the experiment probes the structure of the aluminium nucleus, given RAl≈3.6×10−15 m.
Grade B response (~210 words):
Ek=4.0×106×1.6×10−19=6.4×10−13 J.
Setting kinetic energy equal to electrostatic PE:
6.4×10−13=d(8.99×109)(2)(13)(1.6×10−19)2
Solving: d=6.4×10−13(8.99×109)(2)(13)(2.56×10−38)≈9.4×10−15 m.
This is bigger than the aluminium nuclear radius, so the alpha doesn't quite reach the nucleus.
Examiner commentary: 4/6. The numerical work is essentially correct and the candidate scores the M1, M1, A1 method-and-answer marks. They lose marks on the "comment" element. The expected response is a quantitative comparison (d≈2.6×RAl, so the alpha stops well outside the nuclear surface) plus an interpretive statement (the experiment yields an upper bound on RAl but cannot resolve internal structure). "Doesn't quite reach" is too vague to earn the AO3 marks. Grade B candidates often complete the calculation cleanly but truncate the discussion.
Grade A response (~270 words):*
By conservation of energy, the alpha particle's initial kinetic energy is fully converted to electrostatic potential energy at closest approach d:
Ek=4πε01d(2e)(Ze)
Converting Ek=4.0 MeV=4.0×106×1.60×10−19=6.40×10−13 J.
Rearranging for d and substituting Z=13, e=1.60×10−19 C, 4πε01=8.99×109 N m2 C−2:
d=6.40×10−13(8.99×109)(2)(13)(1.60×10−19)2≈9.36×10−15 m
So d≈9.4×10−15 m to 2 s.f.
Comparing with RAl≈3.6×10−15 m: d/RAl≈2.6. The alpha particle stops about 2.6 nuclear radii from the centre of the aluminium nucleus, so it does not penetrate the nuclear surface. This means the experiment can establish only an upper bound on the nuclear radius — it cannot resolve internal structure or charge distribution. To probe inside the nucleus, much higher-energy projectiles are needed (electron-scattering at hundreds of MeV, for example, accesses sub-femtometre detail).
Examiner commentary: Full marks (6/6). The energy-conservation equation is stated explicitly, the substitution is shown step-by-step, and the comment is both quantitative (the 2.6×R ratio) and interpretive (upper-bound logic, plus the synoptic link to electron-scattering). The "to probe inside…" sentence is the AO3 sophistication that distinguishes A* from A on Paper 2 nuclear items.
Question: Using the relation R=R0A1/3 with R0=1.2×10−15 m, derive an expression for the density of nuclear matter and show that it is approximately constant across all nuclei. Comment on the physical significance of this result, including a comparison with ordinary solid matter (ρ∼104 kg m−3) and one astrophysical implication.
Grade A response (~370 words):*
The volume of a nucleus of nucleon number A is
V=34πR3=34πR03A
since R3=R03A. The mass of the nucleus is approximately m=mnucA, where mnuc≈1.67×10−27 kg is the average nucleon mass.
Density:
ρ=Vm=34πR03AmnucA=34πR03mnuc
The factor A cancels — nuclear density is independent of nucleon number.
Substituting:
ρ=34π(1.2×10−15)31.67×10−27=7.24×10−451.67×10−27≈2.3×1017 kg m−3
So nuclear density is approximately 2×1017 kg m−3 for all nuclei.
Physical significance. The constancy of nuclear density implies that nucleons in a nucleus pack roughly like an incompressible fluid: adding more nucleons increases volume linearly, not by densifying. This is consistent with the short range and saturation of the strong nuclear force — each nucleon interacts only with its nearest neighbours, so the "binding density" stays uniform.
Comparison with ordinary matter. Ordinary solids have densities ∼104 kg m−3, so nuclear matter is about 1013 times denser. The atom is therefore mostly empty space; if all the nuclei in 1 cm³ of solid matter could be packed together with no electron clouds in between, that cube would weigh roughly 1011 kg.
Astrophysical implication. Neutron stars are essentially giant nuclei held together by gravity rather than the strong force, with densities of order 1017 kg m−3 — comparable to nuclear matter. This is why neutron stars have radii of ∼10 km despite containing more mass than the Sun: their constituent matter is at nuclear density.
Examiner commentary: Full marks (9/9). The derivation is complete, the cancellation of A is highlighted explicitly, the numerical density is calculated correctly, and three distinct interpretive layers (incompressibility, comparison with ordinary matter, neutron-star synoptic) earn the full AO3 allocation.
The errors that distinguish A from A* on the nuclear-atom topic:
"The atom is mostly empty space" interpreted as literal vacuum. The atom's volume is filled with electron probability density (in the quantum picture) and electromagnetic field. "Empty" here means "no massive matter" — not "nothing at all".
Confusing nuclear and atomic radii. The atom has radius ∼10−10 m; the nucleus is ∼10−15 m — five orders of magnitude smaller. Many candidates conflate these in calculations and write R≈10−10 m for a nucleus, throwing density estimates off by 1015.
Treating nuclear density as varying with A. A common slip is to compute ρ=M/V with M=Amnuc and V∝R03 (forgetting the A in volume). This gives a density that grows linearly with A — wrong by construction. The cancellation of A is the whole point.
Believing the Rutherford model is "wrong" rather than "incomplete". The nuclear-atom picture is correct as a coarse description; what is missing is the quantum-mechanical treatment of electron motion. A* candidates can articulate the precise sense in which Bohr and Schrödinger superseded Rutherford without contradicting his nuclear conclusion.
Misidentifying who discovered what. Rutherford's 1909–1911 work established the nuclear model. The proton was identified by Rutherford in 1917 as a separate experiment; the neutron was discovered by Chadwick in 1932. The electron was identified by Thomson in 1897. Mixing up these dates and discoverers loses straight recall marks.
Assuming alpha particles in scattering experiments penetrate the nucleus. They do not — at MeV energies the closest approach is many femtometres, well outside the nuclear surface. Penetration requires GeV-scale projectiles.
Thinking the strong nuclear force has infinite range. It is short-ranged (∼1 fm), saturating after one or two nucleon spacings. This is why nuclear density is constant: each nucleon interacts only with its nearest neighbours, not with the entire nucleus.
Three patterns repeatedly cost candidates marks on Paper 2 closest-approach items. They are all about consistent application of energy conservation.
This pattern is endemic to Paper 2 nuclear items: candidates know the physics, lose marks on dimensional or assumption-level care.
The nuclear atom points directly toward several undergraduate trajectories:
Oxbridge interview prompt: "If alpha particles in Rutherford's experiment never penetrated the gold nucleus, how do we know the nuclear radius is ∼10−15 m and not, say, 10−16 m? What experiment would resolve the difference?"
Edexcel A-Level Physics specifies a required practical (Core Practical) on simulating Rutherford scattering — typically using a marble rolling down a ramp toward a hidden hill (representing the nuclear potential) and recording its scattered trajectory. By varying the impact parameter and recording the angle of scattering, students reconstruct an effective "potential" and infer the scatterer's geometry, mirroring the logic of the original 1909 experiment.
A complementary practical uses an alpha source (americium-241, in a sealed and licensed school source), a Geiger-Müller tube, and a ratemeter to measure the count rate at varying angles from a thin metal foil. Although schools cannot reach the energies needed for genuine nuclear scattering, the apparatus illustrates the principles of detector geometry, count-rate statistics, and inverse-square dependence of intensity on distance. Students practise using a ratemeter to record stable averages, correcting for background radiation, and assessing uncertainty in count-rate measurements. The experiment also reinforces safe-handling protocols for sealed sources — distance, time, shielding — that recur throughout Topic 8.
This content is aligned with the Pearson Edexcel GCE A Level Physics (9PH0) specification, Paper 2 — Topic 11: Nuclear and particle physics, with synoptic threads into Topic 8: Nuclear radiation. For the most accurate and up-to-date information, please refer to the official Pearson Edexcel specification document.
graph TD
A["Pre-1909 model:<br/>Thomson plum pudding"] --> B["Rutherford-Geiger-Marsden<br/>alpha-scattering experiment"]
B --> C{"Observations"}
C -->|"Most α pass through"| D["Atom is<br/>mostly empty space"]
C -->|"Few α deflected > 90°"| E["Small region of<br/>concentrated charge"]
C -->|"Tiny number<br/>back-scattered"| F["Nucleus is<br/>very dense"]
D --> G["Nuclear model:<br/>tiny dense positive nucleus<br/>+ orbiting electrons"]
E --> G
F --> G
G --> H["R = R₀ A^(1/3)<br/>R₀ ≈ 1.2 fm"]
H --> I["Nuclear density<br/>ρ ≈ 2 × 10¹⁷ kg m⁻³<br/>(constant for all A)"]
G --> J["Closest approach<br/>d = 2Ze² / (4πε₀ E_k)"]
J --> K["Upper bound on R<br/>(α never penetrates)"]
G --> L["Superseded by quantum<br/>Bohr → Schrödinger<br/>(electrons as orbitals)"]
style G fill:#27ae60,color:#fff
style I fill:#3498db,color:#fff