Nuclear Radius

Explore how nuclear radii are measured, the relationship R = R₀A^(1/3), and why nuclear density is remarkably constant.

AQA A-Level Physics · Section 8 · Nuclear Physics

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Use R = R₀A^(1/3) with R₀ = 1.2 fm

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Describe electron diffraction as evidence for nuclear radius

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Explain diffraction minima in electron scattering

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Calculate nuclear density (~10¹⁷ kg/m³)

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Show that nuclear density is approximately constant for all nuclei

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Convert between femtometres and metres

Nuclear Radius Formula

The radius R of a nucleus depends on its nucleon number (mass number) A:

R = R₀ A^(1/3)

Where:

R = nuclear radius (m)
R₀ = 1.2 × 10⁻¹⁵ m = 1.2 fm (femtometres)
A = nucleon number (mass number)
1 fm = 10⁻¹⁵ m (femtometre)

This relationship implies that nuclei behave like incompressible spheres with constant nucleon density — the volume is proportional to A, and since V ∝ R³ ∝ A, density is constant for all nuclei.

Nucleus	A	R / fm
Hydrogen-1 (proton)	1	1.2
Carbon-12	12	1.2 × 12^(1/3) ≈ 2.75
Iron-56	56	1.2 × 56^(1/3) ≈ 4.60
Lead-208	208	1.2 × 208^(1/3) ≈ 7.11

Electron Diffraction Evidence

High-energy electrons are used to probe nuclear radii because:

Electrons are point-like (no internal structure) — they probe the nucleus cleanly
At energies of ~100–1000 MeV, de Broglie wavelength λ is comparable to nuclear size (~fm)
Electrons interact via the electromagnetic force only (unlike neutrons or protons), making results easier to interpret

Electrons are scattered by the charge distribution of the nucleus (protons). The intensity pattern shows a central maximum and diffraction minima — similar to single-slit diffraction of light.

sin θ_min ≈ 0.61 λ / R

By measuring the angle θ of the first diffraction minimum and knowing the wavelength λ of the electrons (from de Broglie: λ = h/p), the nuclear radius R can be calculated.

Interpreting the Diffraction Pattern

The electron scattering experiment (Hofstadter, 1950s) showed that:

Nuclei have a roughly uniform charge density inside and a sharp boundary — like a "hard sphere"
The diffraction minima occur at predictable angles determined by nuclear size
Plotting ln R vs ln A gives a straight line with gradient 1/3, confirming R ∝ A^(1/3)

The first diffraction minimum in electron scattering from a nucleus (treated as a uniform sphere) occurs at sin θ = 1.22λ/(2R) for a sphere of radius R. This is analogous to the Rayleigh criterion for a circular aperture.

Do not confuse electron diffraction (used to find nuclear radius) with alpha particle scattering (Rutherford's experiment, which found the existence of the nucleus but not its precise radius).

Nuclear Density

The volume of a nucleus is:

V = (4/3)πR³ = (4/3)π(R₀A^(1/3))³ = (4/3)πR₀³ A

The mass of the nucleus ≈ A × m_u (where m_u = 1.661 × 10⁻²⁷ kg, the atomic mass unit):

ρ = mass / volume = (A × m_u) / ((4/3)πR₀³ A) = m_u / ((4/3)πR₀³)

The A cancels — nuclear density is the same for ALL nuclei:

ρ ≈ 1.661 × 10⁻²⁷ / ((4/3)π(1.2 × 10⁻¹⁵)³) ≈ 2.3 × 10¹⁷ kg/m³

This is roughly 10¹⁴ times the density of ordinary matter — equivalent to the entire human population compressed to the size of a sugar cube!

Significance of Constant Nuclear Density

The constant density has profound implications:

Nucleons are packed together uniformly — no "compressible" interior
The strong nuclear force must be saturating — each nucleon only interacts with its nearest neighbours, not all other nucleons
This is analogous to a liquid drop — basis of the liquid drop model of the nucleus

Neutron stars (collapsed stellar cores held up by neutron degeneracy pressure) have a density close to nuclear density (~10¹⁷ kg/m³) — essentially a giant nucleus of neutrons tens of kilometres across.

Example 1: Calculating nuclear radius

Calculate the radius of a gold-197 nucleus (A = 197).

1 Use R = R₀A^(1/3): R = 1.2 × 10⁻¹⁵ × (197)^(1/3)

2 (197)^(1/3) = ∛197 ≈ 5.818

3 R = 1.2 × 10⁻¹⁵ × 5.818

R = 6.98 × 10⁻¹⁵ m ≈ 7.0 fm

Example 2: Finding A from nuclear radius

A nucleus has radius 3.46 fm. Identify the nucleus by finding its nucleon number.

1 R = R₀A^(1/3) → A^(1/3) = R/R₀ = 3.46/1.2 = 2.883

2 A = (2.883)³ = 24.0

A = 24 → This is most likely Magnesium-24 (²⁴Mg) or Carbon-24, etc.

Example 3: Calculating nuclear density

Calculate the density of an oxygen-16 nucleus (A = 16, R₀ = 1.2 fm, m_u = 1.661 × 10⁻²⁷ kg).

1 R = 1.2 × 10⁻¹⁵ × (16)^(1/3) = 1.2 × 10⁻¹⁵ × 2.520 = 3.024 × 10⁻¹⁵ m

2 V = (4/3)π R³ = (4/3)π(3.024 × 10⁻¹⁵)³ = 1.155 × 10⁻⁴³ m³

3 Mass ≈ 16 × 1.661 × 10⁻²⁷ = 2.658 × 10⁻²⁶ kg

4 ρ = 2.658 × 10⁻²⁶ / 1.155 × 10⁻⁴³

ρ = 2.30 × 10¹⁷ kg/m³ — confirming the universal nuclear density

Example 4: Electron diffraction angle

Electrons with de Broglie wavelength 1.5 × 10⁻¹⁵ m are scattered by a nucleus of radius 4.0 fm. At what angle does the first diffraction minimum occur?

1 Use sin θ_min ≈ 0.61λ/R: sin θ = 0.61 × 1.5 × 10⁻¹⁵ / (4.0 × 10⁻¹⁵)

2 sin θ = 0.915 / 4.0 = 0.2288

3 θ = arcsin(0.229)

θ = 13.2°

Q1. The nuclear radius formula is R = R₀A^(1/3). What does the exponent 1/3 tell us about the relationship between volume and nucleon number?

Q2. Why are high-energy electrons (rather than visible light) used to probe nuclear structure?

Q3. A nucleus has A = 27. What is its radius? (R₀ = 1.2 fm)

Q4. Why is nuclear density the same for all nuclei?

Q5. Two nuclei have nucleon numbers A and 8A. What is the ratio of their radii?

Challenge 1. A student fires electrons with kinetic energy 500 MeV at a nucleus of A = 64. (a) Calculate the de Broglie wavelength (use E = pc for ultra-relativistic electrons, h = 6.63 × 10⁻³⁴ J s, 1 MeV = 1.6 × 10⁻¹³ J). (b) Calculate the nuclear radius. (c) Find the angle of the first diffraction minimum.

Challenge 2. A neutron star has a density of 2.0 × 10¹⁷ kg/m³ and mass 2.0 × 10³⁰ kg. Show that it is approximately nuclear-scale dense, and calculate its radius in km.

Challenge 3. By considering R = R₀A^(1/3) and the mass of a nucleus ≈ Am_u, derive an expression for nuclear density and show it does not depend on A. Calculate ρ numerically using R₀ = 1.2 × 10⁻¹⁵ m and m_u = 1.661 × 10⁻²⁷ kg.