AQA · A-Level · Physics
Course version: 7408 · Version from Sept 2016
Estimate of radius from closest approach of alpha particles and…
Full specification wording
Nuclear radius (A-level only): Estimate of radius from closest approach of alpha particles and determination of radius from electron diffraction.
Rutherford alpha scattering gives an upper estimate of nuclear radius: an alpha particle of charge 2e and kinetic energy Ek fired head-on at a nucleus of charge Ze stops at the distance of closest…
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- 4 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
Rutherford alpha scattering gives an upper estimate of nuclear radius: an alpha particle of charge 2e and kinetic energy Ek fired head-on at a nucleus of charge Ze stops at the distance of closest approach d, where Coulomb potential energy equals Ek, so d = 2Ze²/(4πε₀Ek). This treats the nucleus and alpha as point charges and ignores the strong force, so it overestimates the radius. Electron diffraction is more precise: electrons accelerated through a potential difference gain de Broglie wavelength λ = h/p, and the first minimum in intensity at angle θ satisfies sinθ ≈ 1.22λ/(2R), giving R. Higher electron energies probe the charge distribution more sharply.
What earns the marks
- States that for a head-on alpha collision all initial kinetic energy becomes electric potential energy at the distance of closest approach.
- Uses Coulomb potential energy U = (1/4πε₀)(q₁q₂/d) with q₁ = 2e for the alpha and q₂ = Ze for the nucleus.
- Rearranges to d = 2Ze²/(4πε₀Ek) and explains that d is an upper estimate because the alpha never actually touches the nuclear surface.
- Explains that electron diffraction determines radius from the angular position of the first minimum in the scattered electron intensity pattern.
- Uses the de Broglie relation λ = h/p, with p from accelerating voltage via eV = ½mv², to find the electron wavelength.
- Applies sinθ ≈ 1.22λ/(2R) or equivalent geometry to calculate R from the measured first-minimum angle.
Common mistakes and how to put them right
Using the alpha particle charge as e instead of 2e.
Put it right by remembering an alpha is a helium nucleus with charge 2e.
Forgetting to convert kinetic energy from MeV or keV into joules before substituting.
Put it right by multiplying by 1.60 × 10⁻¹³ J per MeV.
Treating the closest-approach distance as the exact nuclear radius.
Put it right by stating it is an upper limit because the alpha is repelled before reaching the surface.
Exam tips
- Write the energy-balance equation first, then substitute values, so the method is clear even if arithmetic slips.
- When using electron diffraction, calculate the electron momentum from the accelerating voltage before finding λ.
- State explicitly why the alpha-scattering result is an estimate, as this is often worth a marking point.
Try the lesson
The start of the Nuclear radius (A-level only): Estimate of radius from closest approach of alpha particles and determination of radius from electron diffraction. lesson, made from checked teaching notes for this specification point.
Have a go first
In the alpha-particle scattering experiment, what does the closest approach of a head-on alpha particle to a nucleus allow you to estimate?
Listen
Rutherford alpha scattering gives an upper estimate of nuclear radius: an alpha particle of charge 2e and kinetic energy Ek fired head-on at a nucleus of charge Ze stops at the distance of closest approach d, where Coulomb potential energy equals Ek, so d = 2Ze²/(4πε₀Ek). This treats the nucleus and alpha as point charges and ignores the strong force, so it overestimates the radius. Electron diffraction is more precise: electrons accelerated through a potential difference gain de Broglie wavelength λ = h/p, and the first minimum in intensity at angle θ satisfies sinθ ≈ 1.22λ/(2R), giving R. Higher electron energies probe the charge distribution more sharply. A strong answer earns marks for these points. States that for a head-on alpha collision all initial kinetic energy becomes electric potential energy at the distance of closest approach. Uses Coulomb potential energy U = (1/4πε₀)(q₁q₂/d) with q₁ = 2e for the alpha and q₂ = Ze for the nucleus. Rearranges to d = 2Ze²/(4πε₀Ek) and explains that d is an upper estimate because the alpha never actually touches the nuclear surface. Explains that electron diffraction determines radius from the angular position of the first minimum in the scattered electron intensity pattern. A common slip: Using the alpha particle charge as e instead of 2e. Instead, remembering an alpha is a helium nucleus with charge 2e. Exam tip: Write the energy-balance equation first, then substitute values, so the method is clear even if arithmetic slips.
Key words
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distance of closest approach
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upper estimate
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de Broglie wavelength
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first minimum
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See every definition
- distance of closest approach
- The distance d at which a head-on alpha particle stops, where Coulomb potential energy equals its initial kinetic energy Ek.
- upper estimate
- A value that overestimates the radius because the point-charge and strong-force approximations are ignored and the alpha never touches the nuclear surface.
- de Broglie wavelength
- The wavelength λ = h/p of an electron, where p is its momentum gained by acceleration through a potential difference.
- first minimum
- The angle θ in the scattered electron intensity pattern that satisfies sinθ ≈ 1.22λ/(2R) and gives the radius R.
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