3.8.1NuclearStretch

Rutherford Scattering and Nuclear Radius

Before 1911, physicists pictured the atom as a uniform "plum pudding" of positive charge with electrons scattered through it. A single experiment — firing alpha particles at gold foil, thinner than this page — completely overturned that model, revealing that almost all of an atom’s mass and all of its positive charge is crammed into a nucleus ten thousand times smaller than the atom itself.

40 min Video by Science Shorts 3.8.1 Radioactivity

Nuclear Radius & Alpha Scattering — A-level PhysicsWatch the full walkthrough before the notes below.

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What you'll be able to do

Describe the alpha particle scattering experiment and its key observations
Explain how the results provide evidence for a small, dense, positively charged nucleus
Estimate nuclear radius using the closest-approach method
Use the nuclear radius equation $R = R_{0} A^{1/3}$
Explain why nuclear density is approximately constant for all nuclei

The alpha particle scattering experiment

Geiger and Marsden, working under Rutherford, fired a beam of alpha particles at an extremely thin sheet of gold foil and observed where they ended up on a surrounding fluorescent screen. The vast majority of alpha particles passed straight through with little or no deflection — but a small fraction were deflected through large angles, and a tiny number bounced back almost the way they came, at angles greater than 90°.

These observations were completely inconsistent with the "plum pudding" model, in which positive charge (and mass) was spread thinly and evenly through the whole atom — such a diffuse charge distribution could never exert enough force to reverse the path of a fast, relatively heavy alpha particle. Rutherford concluded that an atom’s positive charge (and almost all its mass) must be concentrated in a tiny, dense $nucleus$ : most alpha particles pass through the mostly-empty space around it undeflected, but the rare few travelling almost directly at a nucleus are repelled strongly enough to scatter back.

Tip — Learn the three observations as a linked set: (1) most pass straight through → atom is mostly empty space; (2) some deflect at large angles → a concentrated positive charge exists; (3) a very few bounce back → that charge (and mass) is small and dense, not spread out.

Estimating nuclear radius: closest approach

An alpha particle fired directly at a nucleus is repelled by the Coulomb force and gradually slows down as it approaches, converting kinetic energy into electric potential energy. At the point of closest approach, the alpha particle is momentarily at rest (all its initial kinetic energy has become potential energy), before being repelled back the way it came. Setting the alpha particle’s initial kinetic energy equal to the electric potential energy at this point gives an estimate — an upper limit — for the nuclear radius.

E_{k} = \frac{Qq}{4 π ε _{0} r _{m i n}}

Closest-approach method: initial KE converts entirely to electric PE at the nucleus.

1Convert energy to joules:

E_{k} = 6.0 \times 1.60 \times 1 0^{- 13} = 9.6 \times 1 0^{- 13}

2Rearrange

E_{k} = \frac{Qq}{4 π ε _{0} r _{m i n}}

r_{m i n} = \frac{Qq}{4 π ε _{0} E _{k}}

3Charges:

Q = 2 e = 3.2 \times 1 0^{- 19}

q = 79 e = 1.264 \times 1 0^{- 17}

r_{m i n} = \frac{3.2 \times 1 0 ^{- 19} \times 1.264 \times 1 0 ^{- 17}}{4 π \times 8.85 \times 1 0 ^{- 12} \times 9.6 \times 1 0 ^{- 13}}

Answer

r_{m i n} \approx 3.8 \times 1 0^{- 14}

m — the right order of magnitude for a nuclear radius (a few tens of femtometres).

Tip — This method gives an upper limit on nuclear radius (the alpha particle turns around before reaching it exactly if the calculated distance is still larger than the true nuclear surface) — using higher-energy alpha particles, which approach closer, refines the estimate.

Nuclear radius and nuclear density

More detailed scattering experiments (also using high-energy electrons) show that nuclear radius depends on the nucleon number, following $R = R_{0} A^{1/3}$ , where $R_{0} \approx 1.2$ fm is approximately the same for all nuclei. This cube-root relationship means nuclear $volume$ ( $\propto R^{3}$ ) is directly proportional to $A$ — exactly what you’d expect if every nucleon takes up roughly the same amount of space and nucleons are simply packed together, rather than being squeezed more tightly in heavier nuclei.

Because both mass and volume are proportional to $A$ , $nuclear density$ is essentially the same for every nucleus, regardless of size — and it is enormous, roughly $1 0^{17}$ kg m⁻³, far denser than any ordinary matter.

R = R_{0} A^{1/3}

Nuclear radius; R₀ ≈ 1.2 × 10⁻¹⁵ m (1.2 fm), roughly constant for all nuclei.

1Ratio of radii:

\frac{R _{F e}}{R _{H e}} = (\frac{A _{F e}}{A _{H e}})^{1/3} = (\frac{56}{4})^{1/3} = 1 4^{1/3}

1 4^{1/3} \approx 2.41

AnswerThe iron-56 nucleus has a radius about 2.4 times larger than helium-4, despite having 14 times the mass — consistent with roughly constant nuclear density.

Tip — The A^(1/3) relationship is the single most exam-tested fact in this lesson — memorise it precisely, including that it is a CUBE root, not a square root.

Equation recap

E_{k} = \frac{Qq}{4 π ε _{0} r _{m i n}}

Closest-approach estimate of nuclear radius.

R = R_{0} A^{1/3}

Nuclear radius (R₀ ≈ 1.2 fm).

Common mistakes to avoid

Thinking the scattering experiment showed most alpha particles were deflected.

The vast MAJORITY passed straight through with little deflection — it is the rare large-angle and backscattered particles that provided the key evidence for a small, dense nucleus.

Using R = R₀A² or R₀A instead of the cube root relationship.

Nuclear radius follows R = R₀A^(1/3) — a cube root, consistent with nuclear volume (not radius) being directly proportional to A.

Assuming heavier nuclei are much denser than lighter ones, since they have more mass.

Because both mass AND volume scale with A, nuclear density is approximately the SAME for all nuclei — heavier nuclei are simply bigger, not denser.

Key takeaways

Rutherford scattering (mostly undeflected, some large-angle, rare backscattered alpha particles) is evidence for a small, dense, positively charged nucleus.
The closest-approach method estimates nuclear radius by equating an alpha particle’s kinetic energy to electric potential energy at minimum separation.
Nuclear radius: R = R₀A^(1/3), with R₀ ≈ 1.2 fm approximately constant for all nuclei.
Because volume ∝ A (from the cube-root radius relationship) and mass ∝ A, nuclear density (~10¹⁷ kg m⁻³) is approximately the same for every nucleus.

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Rutherford Scattering and Nuclear Radius

40 min Video by Science Shorts 3.8.1 Radioactivity