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How do we measure the size of a nucleus, and why is nuclear density the same for all atoms?

Estimating nuclear radius from closest approach of alpha particles and from electron diffraction, the dependence of radius on nucleon number, and the constancy of nuclear density.

A focused answer to AQA A-Level Physics 3.8.1.6, covering estimates of nuclear radius from alpha particle closest approach and electron diffraction, the relationship R proportional to the cube root of A, and the constant density of nuclear matter.

Generated by Claude Opus 4.89 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. Distance of closest approach
  3. Electron diffraction
  4. Radius and nucleon number
  5. Constant nuclear density
  6. Try this

What this dot point is asking

AQA specification point 3.8.1.6 wants you to describe how the nuclear radius is estimated from the closest approach of alpha particles and from electron diffraction, use the relationship between radius and nucleon number, and show that nuclear density is constant.

Distance of closest approach

When an alpha particle is fired head-on at a nucleus, it slows and stops at the point where all its kinetic energy has become electric potential energy. Equating the two gives an upper estimate of the nuclear radius.

This gives only an upper limit, because the alpha particle is repelled before it actually touches the nucleus, and is affected by the strong nuclear force at very close range.

Electron diffraction

The electrons must be very high energy so that their de Broglie wavelength (λ=hp\lambda = \dfrac{h}{p}) is of the order of femtometres, matching the nuclear size needed for diffraction.

Radius and nucleon number

Taking logarithms gives ln⁡R=ln⁡r0+13ln⁡A\ln R = \ln r_0 + \tfrac{1}{3}\ln A, so the experimental confirmation of the cube-root law is a straight line of gradient one third, with the intercept giving r0r_0.

Constant nuclear density

This density is enormous, about 101410^{14} times that of water, and is the same density found in neutron stars, which are essentially giant nuclei held together by gravity.

Try this

Q1. State how the nuclear radius depends on nucleon number. [1 mark]

  • Cue. R=r0A1/3R = r_0 A^{1/3}, so radius is proportional to the cube root of nucleon number.

Q2. Explain why nuclear density is approximately the same for all nuclei. [2 marks]

  • Cue. Volume is proportional to AA (since R∝A1/3R \propto A^{1/3}) and mass is proportional to AA, so density stays constant.

Q3. State why electron diffraction gives a more accurate nuclear radius than the alpha closest-approach method. [1 mark]

  • Cue. Electrons are not affected by the strong nuclear force.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20194 marksThe nuclear radius is related to nucleon number by R=r0A1/3R = r_0 A^{1/3} with r0=1.05 fmr_0 = 1.05 \text{ fm}. Calculate the radius of a gold-197 nucleus, and find the ratio of the radius of a gold-197 nucleus to that of a carbon-12 nucleus.
Show worked answer →

For gold-197: R=(1.05×10−15)(197)1/3R = (1.05 \times 10^{-15})(197)^{1/3}. The cube root of 197197 is 5.825.82, so R=(1.05×10−15)(5.82)=6.1×10−15 mR = (1.05 \times 10^{-15})(5.82) = 6.1 \times 10^{-15} \text{ m}.

The ratio depends only on the nucleon numbers: RAuRC=(19712)1/3=(16.4)1/3=2.5\dfrac{R_{\text{Au}}}{R_{\text{C}}} = \left(\dfrac{197}{12}\right)^{1/3} = (16.4)^{1/3} = 2.5.

Markers reward the cube root for the radius, the correct value in metres, and using the ratio of nucleon numbers (cube rooted) for the comparison.

AQA 20214 marksShow that the density of a nucleus is independent of its nucleon number, and estimate the density of nuclear matter. Take r0=1.05 fmr_0 = 1.05 \text{ fm} and the mass of a nucleon as 1.67×10−27 kg1.67 \times 10^{-27} \text{ kg}.
Show worked answer →

The nuclear mass is approximately m=Amnucleonm = A m_{\text{nucleon}} and the volume is V=43πR3=43π(r0A1/3)3=43πr03AV = \tfrac{4}{3}\pi R^3 = \tfrac{4}{3}\pi (r_0 A^{1/3})^3 = \tfrac{4}{3}\pi r_0^3 A.

Density =mV=Amnucleon43πr03A=mnucleon43πr03= \dfrac{m}{V} = \dfrac{A m_{\text{nucleon}}}{\tfrac{4}{3}\pi r_0^3 A} = \dfrac{m_{\text{nucleon}}}{\tfrac{4}{3}\pi r_0^3}, with AA cancelling, so density is independent of AA.

Substituting: ρ=1.67×10−2743π(1.05×10−15)3=3.4×1017 kg m−3\rho = \dfrac{1.67 \times 10^{-27}}{\tfrac{4}{3}\pi (1.05 \times 10^{-15})^3} = 3.4 \times 10^{17} \text{ kg m}^{-3}.

Markers reward cancelling AA between mass and volume, and a value of order 1017 kg m−310^{17} \text{ kg m}^{-3}.

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