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Alpha-particle scattering is the classic “how science works” story in nuclear physics, and nuclear radius is where it becomes quantitative. Expect recall MCQs on what each observation shows, a 6-mark description of the experiment, closest-approach calculations that reuse electric potential energy, and graph questions on R=r0A1/3 that finish with nuclear density.
By the end you’ll be able to
Describe the Rutherford scattering experiment and explain how its observations led to the nuclear model
Estimate the distance of closest approach by equating kinetic energy to electric potential energy
Describe how high-energy electron diffraction measures nuclear radius, including the first-minimum condition
Use R = r₀A^(1/3) (including log-graph analysis) and show that nuclear density is approximately constant
What the exam asks
Describe and explain the Geiger–Marsden experiment, linking each observation to the nuclear model. This is a common 6-mark extended response. Contrast it with the “plum pudding” model.
Calculate the distance of closest approach, and explain why it is only an upper limit for the nuclear radius.
Electron diffraction: why high-energy electrons are used, their wavelength, and R from the first minimum (AQA, OCR B).
Use R=r0A, including a log–log graph, and show that nuclear density is constant (AQA, OCR A, CCEA).
vii.Check your understanding
3 questions on Rutherford scattering and nuclear size. Every option is explained once you answer.
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PromptCard 1 of 3
What are the three key observations of alpha scattering, and what does each show?
1/3
Core ideas
The experiment
Geiger and Marsden, working with Rutherford (1909–1911), fired a narrow beam of alpha particles from a radioactive source at a very thin gold foil in an evacuated chamber. A zinc sulfide screen viewed through a microscope detected the scattered alpha particles as tiny flashes, and could be moved to count them at different scattering angles.
Design choice
Reason
Vacuum
Alpha particles travel only a few cm in air; air molecules would absorb and scatter them
Very thin foil (a few hundred nm)
Each alpha particle is scattered by at most one nucleus, so each deflection reveals a single encounter
Gold
Can be beaten into extremely thin foil; its large nuclear charge (Z=79) gives strong deflections; its massive nucleus hardly recoils
Collimated, single-energy beam
So every alpha particle starts with the same direction and kinetic energy
Observations and conclusions
Observation
Conclusion
Almost all alpha particles passed straight through or were deflected by very small angles
Most of the atom is empty space; the nucleus is tiny compared with the atom
A small fraction were deflected through large angles
The positive charge is concentrated in a very small region, producing an intense electric field
A very few (of the order of 1 in 10⁴) were deflected by more than 90∘, some back towards the source
The nucleus contains most of the mass as well as the positive charge; a light or spread-out charge could not turn a fast alpha particle round
In the plum pudding model (Thomson), positive charge was spread through the whole atom. The field anywhere would be weak, so every deflection would be tiny, and back-scattering would be impossible. The number scattered at each angle matched Rutherford’s calculation for an inverse-square Coulomb repulsion from a point charge. That fit is what made the nuclear model convincing.
The paths are hyperbolas curving away from the nucleus, because the force is repulsive. The closer an alpha particle’s line of approach passes the nucleus, the larger its deflection. A head-on alpha particle stops and returns along its own path.
Distance of closest approach
In a head-on approach, all the alpha particle’s kinetic energy becomes electric potential energy at the closest point:
Ek=4πε0rQq⇒r=4πε0Ek(2e)(Ze)
Convert MeV to joules (1MeV=1.60×10−13 J), and use 4πε01=8.99×109N m2C−2. For a 5 MeV alpha particle and gold, r≈4.5×10−14 m. This is an upper limit for the nuclear radius: the alpha particle is turned back before it reaches the nucleus, so the nucleus must be smaller than r. The distance r is inversely proportional to Ek and proportional to Z.
Measuring the radius with electron diffraction
High-energy electrons diffract around a nucleus, much as light diffracts around a small obstacle. Electrons are used because they are leptons. They do not feel the strong nuclear force, so they probe the charge distribution cleanly. Alpha particles and neutrons interact through the strong force, which complicates the scattering.
The de Broglie wavelength must be comparable with the nuclear diameter (about 10−15 m), so the electrons need energies of hundreds of MeV. At these energies E≫mec2 (0.511 MeV), so p≈cE and
λ=ph≈Ehc
The intensity of the scattered electrons falls with angle and shows a first minimum at angle θ, given (on AQA and in the question) by
sinθ≈R0.61λ
The minimum does not fall to zero, because the nucleus has no sharp edge.
Nuclear radius and density
Experiments show that
R=r0A1/3,r0≈1.2fm
where A is the nucleon number. Values between about 1.1 and 1.4 fm appear in questions, and the question always gives one. Two graphs test this:
R against A1/3: a straight line through the origin with gradient r0.
lnR against lnA: a straight line with gradient 31 and intercept lnr0, since lnR=31lnA+lnr.
Since the volume V=34πR3=34πr03A is proportional to A, the density
ρ=34πr03AAu=4πr033u≈2×1017kg m−3
is the same for all nuclei. Nucleons pack together at a constant density, like the molecules of a liquid, because the strong force is short-range. Solid matter has a density of about 104kg m−3, roughly 1013 times less, which confirms again that atoms are mostly empty space.
Worked examples
Exam technique
6-mark description: apparatus → observations → conclusions, pairing each observation with its conclusion (“most went straight through, so most of the atom is empty space”). Give the reasons for the vacuum and the thin foil.
Closest approach: write the energy equation first, with charges 2eandZe, and convert MeV to J.
Ratios:r∝EkZ and R∝A1/3. Use proportionality instead of recalculating.
Log graphs: write lnR=31lnA+lnr first. If the axis does not start at zero, find the intercept from .
Density “show that”: keep A as a symbol so that it visibly cancels, and give one more significant figure than the value quoted in the question.
Common mistakes
Quick recap
Most alpha particles pass straight through (atom mostly empty space); a very few bounce back (small, massive, positive nucleus).
Vacuum: alpha particles have a short range in air. Thin gold foil: single scattering events.
Closest approach: Ek=4πε0r(2e)(Ze) gives an upper limit for R, about 10−14 m.
Electron diffraction: leptons feel no strong force; λ≈Ehc; first minimum at sinθ≈.
R=r0A1/3 with r fm; against has gradient .
Nuclear density 4πr033u≈ is the same for every nucleus.