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Nuclear radius and density

Nuclear radii can be estimated from the closest approach of an alpha particle and measured by diffracting electrons off the nucleus. The two methods agree, and they give R = R₀A^(1/3), from which it follows that medium and heavy nuclei share very nearly one density.

Year 13AQA 3.8.1.5OCR A 6.4.1(c), 6.4.1(f), 6.4.1(g)EDEXCEL 136

Builds on Rutherford scattering and the nuclear atom and Coulomb's law and electric field strength and Wave-particle duality.

IN THIS TOPIC

  • Estimate a nuclear radius from the closest approach of an alpha particle using energy conservation.
  • Describe radius determination by electron diffraction and sketch the intensity against angle graph.
  • Use R = R₀A^(1/3) and show that it makes nuclear density the same for every nucleus.

COMMON MISCONCEPTION

Bigger nuclei are packed tighter, so heavy elements have denser nuclei.

How close can an alpha get?

Fire an alpha particle straight at a nucleus and it climbs the electrical hill of the repulsion, trading kinetic energy for electrical potential energy until, for an instant, it stops. At that closest approach every joule is electrical, so with the Coulomb potential energy from the electric fields unit,

Ek=Qq4πε0rminE_{k} = \frac{Qq}{4πε_{0}r_{min}}NOT ON THE AQA DATA SHEET: LEARN IT
Closest approach: buying distance with energy (animated figure)the alpha spends its KE climbing, then rolls backclosest approachKEPEtotal
FIG. 1The alpha runs at the nucleus and climbs the Coulomb hill, its kinetic bar draining into the potential bar exactly as fast as it climbs. At the dashed line the kinetic bar hits zero, the alpha stops for an instant, and everything runs backwards. Equating the initial KE to the potential energy at that stop is how the closest-approach limit on nuclear size is made: an upper bound rather than a measured radius, because the alpha stops at the electrical wall, still short of the surface.

Rearrange for rmin and put numbers in for a 5.0 MeV alpha meeting gold. The charges are 2e and 79e, and Ek = 5.0 × 106 × 1.60 × 10-19 = 8.0 × 10-13 J, giving rmin = 8.99 × 109 × 158 × (1.60 × 10-19)2 / (8.0 × 10-13) = 4.5 × 10-14 m.

Treat that number as an upper estimate. The alpha stops short of the surface, and it never probes inside; the true radius must be smaller. A sharper ruler is needed.

GUIDED PRACTICE

Closest approach to silver

A 4.8 MeV alpha particle heads straight at a silver nucleus (Z = 47). Using energy conservation, find the distance of closest approach.

Show the working

At the stop, all the kinetic energy has become electrical potential energy, so Ek = (1/4πε0) × 2e × 47e/r.

r = 8.99 × 109 × 2 × 47 × (1.60 × 10−19)2/(4.8 × 106 × 1.60 × 10−19) = 2.8 × 10−14 m.

An upper estimate, as always with this method. The alpha stopped at the electrical wall, still short of the nuclear surface itself.

Electron diffraction, the sharper ruler

The sharper ruler is the electron. Electrons are leptons and feel no strong force, so they probe charge alone; and by wave-particle duality a beam of them diffracts off a nucleus much as light diffracts at a small circular aperture; the nucleus is a round charge distribution, not a literal slit, but the pattern is the same in kind. To make the de Broglie wavelength λ = h/p comparable to a nucleus, the electrons must carry energies of hundreds of MeV.

Electron diffraction by a nucleus: intensity against angle shows a central maximum, a first minimum that fixes the radius, and only weak ripples beyondfirst minimumintensityanglethe minimum's angle fixes the radius
FIG. 2Intensity against angle for electrons diffracted by a nucleus: the angle of the first minimum fixes the radius.

The pattern has the single-slit shape. A strong central maximum, a first minimum at an angle set by λ and the nuclear size, then faint ripples. For the round charge distribution the first minimum sits at sinθ≈1.22λ2R\sin θ ≈ \frac{1.22λ}{2R}, the circular-aperture result with 2R for the diameter, so measure that angle and the radius follows. These are the reliable results, a few femtometres or 10-15 m, set against 4.5 × 10-14 m from closest approach and around 10-10 m for the atom. The nucleus is smaller than its atom by a factor of ten thousand or more.

One rule for every radius

Measure many nuclides this way and a single pattern emerges. Radius grows as the cube root of nucleon number A, never in proportion to A itself.

R=R0A1/3R = R_{0}A^{1/3}ON THE AQA DATA SHEET
Nuclear radius against the cube root of nucleon number: a straight line through the origin whose gradient is R nought, about 1.2 femtometrescarbon-12iron-56gold-197gradient R₀ = 1.2 fmR in femtometrescube root of A
FIG. 3Measured radii against the cube root of A: a straight line through the origin with gradient R₀.

Plot R against A1/3 and the data fall on a straight line through the origin, whose gradient is the constant R0, about 1.2 fm from electron diffraction. That is how the equation is derived from experiment, and it makes an order-of-magnitude radius one keystroke long. Gold, with A = 197, gives R = 1.2 × 1971/3 ≈ 7 fm.

WORKED EXAMPLE

Lead against aluminium, no constants needed

How many times larger is a lead-208 nucleus than an aluminium-26 nucleus?

Take the ratio so R0 cancels, giving RPb/RAl = (208/26)1/3.

208/26 = 8, and the cube root of 8 is exactly 2, so lead's nucleus has twice the radius.

Eight times the nucleons, twice the radius. The cube-root law at its clearest, and a reminder that nuclear volume simply counts nucleons.

One density for every nucleus

Cube the radius law and a simple result follows. Nuclear volume is (4/3)πR3 = (4/3)πR03A, directly proportional to A, so volume simply counts nucleons, as if they were marbles packed shoulder to shoulder. Mass is close to Au, with u the atomic mass unit, so in the density m/V the nucleon number cancels away.

ρ=3u4πR03ρ = \frac{3u}{4πR_{0}^{3}}NOT ON THE AQA DATA SHEET: LEARN IT
Nuclear density is the same for every nucleus, about ten to the thirteen times denser than the solid gold the nuclei sit insidecarbonirongold2.3 × 10¹⁷ kg m⁻³ eachordinary solid goldabout 10¹³ times less dense19 300 kg m⁻³
FIG. 4Carbon, iron, gold: one nuclear density, about ten to the thirteen times that of the solid gold around it.

On this model every nucleus, hydrogen to uranium, shares one density near 2.3 × 1017 kg m-3, and that settles the misconception above. Heavier nuclei are bigger, not denser. The radius law itself is an approximation at its best for medium and heavy nuclei, and the lightest nuclei stray furthest from it, so read the constant density as a broad experimental rule rather than an exact theorem. Set the figure against ordinary matter and the comparison with the scattering lesson follows. Solid gold manages 1.9 × 104 kg m-3, some 1013 times less, because an atom is a vast emptiness with all its mass gathered into one dense point.

ASSESSMENT FOCUS

  • Closest approach is an energy argument, and saying so scores. Initial kinetic energy equals electrical potential energy at the stop. Set them equal before any algebra begins.
  • Charges in Coulomb's equation are 2e and Ze, never 2 and Z. Answers usually go wrong by dropping one factor of e, or by squaring the wrong bracket.
  • Say why electrons are the probe of choice. They feel no strong force to muddy the measurement, and their de Broglie wavelength can be made femtometre-sized. Both halves carry credit.
  • Closest approach overestimates; electron diffraction is the reliable determination. Comparison questions want that verdict with its reason attached.
  • For density, show the cancellation. Volume is proportional to A, mass is proportional to A, so the ratio is a constant. Quote about 2 × 10¹⁷ kg m⁻³ to one significant figure.

CHECK YOURSELF

Take R₀ = 1.2 fm. Find the radius of an iron-56 nucleus, and then its density, given the atomic mass unit u = 1.661 × 10⁻²⁷ kg. Comment on how the density would differ for gold-197.

Show a hint

Radius first from the cube-root law; then mass over the volume of a sphere.

Show the answer

R = R0A1/3 = 1.2 × 561/3 = 1.2 × 3.83 = 4.6 fm.

Mass = 56u = 9.3 × 10-26 kg; volume = (4/3)π(4.59 × 10-15)3 = 4.1 × 10-43 m3.

ρ = 9.3 × 10-26 / (4.1 × 10-43) = 2.3 × 1017 kg m-3, and gold's comes out the same, because A cancels and the radius law gives each of them this same density.

R = R₀A^(1/3), so nuclear volume simply counts nucleons.

On the radius law, medium and heavy nuclei share one density near 2 × 10¹⁷ kg m⁻³; the lightest stray furthest from it.

Closest approach only ever overestimates; electron diffraction measures.

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  • Estimate a nuclear radius from the closest approach of an alpha particle using energy conservation.
  • Describe radius determination by electron diffraction and sketch the intensity against angle graph.
  • Use R = R₀A^(1/3) and show that it makes nuclear density the same for every nucleus.

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