Rutherford scattering and nuclear radiusAQA A-Level Physics: Mind map
Rutherford scattering
Changing models
Nuclear size
Scattering evidence
R = R₀A^(1/3)ρ ≈ 2.3 × 10¹⁷R ≈ 1–8 fm
Closest approach
Electron diffraction
Radius and density
Exam questions on Rutherford scattering and nuclear radius
- In an alpha particle scattering experiment, a narrow beam of alpha particles from a radioactive source is fired at a thin gold foil inside an evacuated chamber. A detector counts the alpha particles scattered through different angles. Almost all the particles pass through the foil, but about 1 in 8000 is scattered through an angle greater than 90°.The gold foil used is only a few hundred atoms thick. Explain why a much thicker foil would make the results difficult to interpret.2 marks
- An alpha particle of kinetic energy 7.7 MeV is fired directly towards the centre of a stationary gold nucleus, which has proton number 79. The alpha particle is brought momentarily to rest at a distance r from the centre of the nucleus before it is repelled. The gold nucleus can be assumed to remain at rest.The distance of closest approach for this alpha particle is 3.0 × 10⁻¹⁴ m. Explain why this gives only an upper limit for the radius of the gold nucleus.2 marks
- A beam of electrons of kinetic energy 420 MeV is directed at a thin target of carbon-12 nuclei. The scattered electrons are detected at different angles and the intensity shows a diffraction pattern, with the first minimum at 40° to the original direction. For these electrons, use E = pc. The first minimum occurs when sin θ = 1.22λ/D, where λ is the electron wavelength and D is the diameter of the nucleus.Calculate the wavelength of the electrons.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).