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Nuclear radius and density

How do you measure something a hundred thousand times smaller than the atom it sits in? Twice over: once by seeing how close an alpha can climb, once by diffracting electrons off it. The answers agree on a startling law: every nucleus, light or heavy, is packed to the same density.

Year 13AQA 3.8.1.5OCR A 6.4.1

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.

WHAT YOU PROBABLY THINK

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 = Qq4πε0rminNOT ON THE DATA SHEET — LEARN IT
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 estimate of nuclear size is made.

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.

Electron diffraction, the sharper ruler

The sharper ruler is the electron. Electrons are leptons, blind to the strong force, so they probe charge alone; and by wave-particle duality a beam of them diffracts off a nucleus exactly as light diffracts through a slit. To make the de Broglie wavelength λ = h/mv 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 is the single-slit story replayed: a strong central maximum, a first minimum at an angle set by λ and the nuclear diameter, then faint ripples. Measure the angle of that minimum and the radius follows. The results are the trusted ones: a few femtometres, 10-15 m, 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 does not grow in proportion to nucleon number A, but to its cube root:

R = R0A1/3ON YOUR 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 femtometresA⅓
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 order-of-magnitude radius estimates one keystroke long: gold, with A = 197, has R = 1.2 × 1971/3 ≈ 7 fm.

One density for every nucleus

Cube the radius law and something remarkable drops out. Nuclear volume is (4/3)πR3 = (4/3)πR03A, directly proportional to A: 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:

ρ = 3u4πR03NOT ON THE 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.

Every nucleus, hydrogen to uranium, shares one density near 2.3 × 1017 kg m-3, which finishes the myth above: heavier nuclei are bigger, never denser. Set it against ordinary matter and the scattering lesson closes its loop: solid gold manages 1.9 × 104 kg m-3, some 1013 times less, because an atom is a vast emptiness with its mass gathered in one dense point.

THE EXAM BIT

  • 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.
  • Charges in Coulomb's equation are 2e and Ze, not 2 and Z. Forgetting one factor of e, or squaring the wrong bracket, is the standard slip.
  • State why electrons: no strong force to muddy the probing, and a de Broglie wavelength that can be made femtometre-sized. Both halves carry credit.
  • Closest approach overestimates; electron diffraction is the reliable determination. Comparison questions want that verdict with the reason.
  • For density, show the cancellation: volume proportional to A, mass proportional to A, so the ratio is constant. Quote around 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: A cancels, so every nucleus shares this density.

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

So every nucleus, light or heavy, shares one density near 2 × 10¹⁷ kg m⁻³.

No animated video for this topic yet; these notes stand alone. InkPhysics on YouTube.