You are viewing a free preview of this lesson.
Subscribe to unlock all 10 lessons in this course and every other course on LearningBro.
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.