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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

  1. 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
  2. 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
  3. 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
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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).