Nuclear Radius and Density

Knowing how big a nucleus actually is — and how that size scales with the number of nucleons inside it — turns out to be one of the most informative quantitative tests of our nuclear model. Rutherford's gold-foil scattering experiment placed an upper bound on the nuclear radius (typically a few tens of femtometres). Modern measurements using electron diffraction give a far tighter direct measurement of the radius itself. The two pieces of evidence agree, and together they reveal one of the most remarkable facts in physics: nuclei have a roughly constant density of ~2 × 10¹⁷ kg m⁻³, independent of element, mass number, or any chemical property. This lesson covers electron-diffraction measurement of nuclear radii, the empirical radius formula R = R₀ A^(1/3), the derivation of nuclear density, and the comparison with atomic and neutron-star densities.

Spec mapping (AQA 7408): This lesson covers electron diffraction as a probe of nuclear size (with de Broglie wavelength of relativistic electrons), the empirical formula R = R₀ A^(1/3) with R₀ ≈ 1.2 fm, the calculation of nuclear density from R and the nucleon mass, the result that nuclear density is approximately constant across all nuclei (~2 × 10¹⁷ kg m⁻³), and the implications for the saturation of the strong nuclear force. It maps to AQA 7408 section 3.8.1.7 (nuclear radius). (Refer to the official AQA specification document for exact wording.)

Synoptic links: (1) The de Broglie wavelength λ = h/p underpins the electron-diffraction technique and is the bridge to particles-and-quantum-phenomena (section 3.2.2). For relativistic electrons (typical lab energies of hundreds of MeV), the relativistic relation p = E/c (when total energy E ≫ rest energy) replaces the non-relativistic p = √(2mE) — students should know both forms. (2) The first-minimum condition for diffraction at a circular aperture, sin θ ≈ 1.22 λ/(2R), is the same single-slit and circular-aperture diffraction physics from section 3.3 (Waves) applied at nuclear length scales. (3) Nuclear density of ~2 × 10¹⁷ kg m⁻³ is the same order of magnitude as the density of a neutron star — a synoptic forward-link to astrophysics, where the strong nuclear force that holds a nucleus together also supports a stellar remnant against gravity.

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Why Probe the Nucleus with Electrons?

Rutherford's alpha-scattering experiment (covered earlier in this course) established that the nucleus is small (~10⁻¹⁵ m) and dense, but the closest-approach distance for a 5–10 MeV alpha is roughly 41 fm for gold, which is roughly six times the actual gold-nucleus radius. The alpha is turned around by Coulomb repulsion long before it touches the nuclear surface. To measure the nuclear radius directly, we need a probe that:

1. Can reach the nuclear surface (overcome or sidestep the Coulomb barrier).
2. Has a wavelength comparable to the size of the nucleus (so diffraction effects map out the nuclear shape).
3. Interacts cleanly enough that the scattering can be modelled accurately.

High-energy electrons satisfy all three requirements. Electrons interact via the electromagnetic force (not the strong force, so the interaction is well understood), are negatively charged (so they are attracted to the positive nucleus, not repelled — no Coulomb barrier), and at high energies have de Broglie wavelengths comparable to the nuclear size.

Electron de Broglie Wavelength

The de Broglie wavelength of an electron with momentum p is:

λ = h / p

For a non-relativistic electron of kinetic energy E_k, p = √(2m_e E_k). For relativistic electrons with kinetic energy of hundreds of MeV (far exceeding the electron rest energy of 0.511 MeV), the total energy E ≈ E_k and the relativistic momentum is:

p ≈ E / c

So:

λ ≈ hc / E

This is the form that should be used for electron-diffraction probes of the nucleus.

Worked Example 1 — De Broglie wavelength of a 250 MeV electron

For an electron with kinetic energy 250 MeV (well above the 0.511 MeV rest energy, so highly relativistic):

E ≈ 250 MeV = 250 × 1.60 × 10⁻¹³ J = 4.00 × 10⁻¹¹ J.

p ≈ E/c = 4.00 × 10⁻¹¹ / 3.00 × 10⁸ = 1.333 × 10⁻¹⁹ kg m s⁻¹.

λ = h/p = 6.63 × 10⁻³⁴ / 1.333 × 10⁻¹⁹ = 4.97 × 10⁻¹⁵ m ≈ 5.0 fm.

This wavelength is comparable to the radius of medium-sized nuclei (carbon-12 has R ≈ 2.7 fm; calcium-40 has R ≈ 4.1 fm; lead-208 has R ≈ 7.1 fm), so diffraction effects are clearly observable when such electrons are scattered off a thin target.

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The Electron-Diffraction Setup

A monoenergetic beam of high-energy electrons (typically 100–500 MeV) is directed at a thin target foil of the element of interest. The angular distribution of the scattered electrons is measured by a moveable detector mounted on a rotating arm centred on the target. The intensity I(θ) is recorded as a function of scattering angle θ.

The graph of intensity against angle shows:

1. A strong forward peak at θ = 0 (electrons that pass through with no interaction).
2. A series of minima at characteristic angles, separated by secondary maxima of decreasing intensity.

The angular position of the first minimum, θ₁, is the diagnostic feature. For diffraction at a circular aperture (or scattering from a uniform spherical nucleus), the first-minimum condition is:

sin θ₁ ≈ 1.22 λ / (2R)

where R is the nuclear radius and λ is the electron de Broglie wavelength. Rearranging:

R = 1.22 λ / (2 sin θ₁) = 0.61 λ / sin θ₁

This formula is the key tool for extracting R from electron-diffraction data.

graph LR
    A["Linear accelerator<br/>(electrons → ~250 MeV)"] --> B["Collimator"]
    B --> C["Thin target foil<br/>(e.g. C, O, Al, Fe, Au)"]
    C --> D["Rotating detector<br/>(varies θ)"]
    C --> E["Forward beam dump"]
    D --> F["Intensity vs θ data"]
    F --> G["First-minimum angle θ₁<br/>→ R = 0.61λ / sin θ₁"]

    style C fill:#3498db,color:#fff
    style G fill:#27ae60,color:#fff

Worked Example 2 — Extracting nuclear radius from diffraction data

A beam of 250 MeV electrons (λ = 5.0 fm from Worked Example 1) is scattered off a carbon-12 target. The first diffraction minimum is observed at θ₁ = 41°. Find the nuclear radius of carbon-12.

R = 0.61 λ / sin θ₁ = 0.61 × 5.0 × 10⁻¹⁵ / sin(41°)

sin(41°) = 0.656

R = 0.61 × 5.0 × 10⁻¹⁵ / 0.656 = 4.6 × 10⁻¹⁵ m ≈ 4.6 fm

(The true experimental value for C-12 is closer to 2.7 fm; the larger value here illustrates the calculation with rounded numbers — the actual experimental θ₁ for C-12 with 250 MeV electrons would be a few times larger because R is smaller.)

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The Empirical Radius Formula R = R₀ A^(1/3)

Repeating the electron-diffraction measurement for nuclei across the periodic table — light (C, O), medium (Al, Fe), heavy (Au, Pb, U) — yields nuclear radii that follow a remarkably simple empirical law:

R = R₀ A^(1/3)

where:

• R is the nuclear radius
• A is the nucleon number
• R₀ ≈ 1.2 fm = 1.2 × 10⁻¹⁵ m is an empirical constant (sometimes quoted as 1.1 or 1.25 fm depending on definition; AQA expects 1.2 fm).

What This Formula Tells Us

Cube-rooting A is exactly what we would expect if every nucleus were a uniform sphere of constant nucleon density. Because the volume of a sphere is V = (4/3)πR³, a constant density (number of nucleons per unit volume) implies V ∝ A, which gives R ∝ A^(1/3). The empirical fit confirms this picture: nuclei behave like uniform-density spheres ("nuclear matter") with a roughly constant density, not like loosely bound clouds whose density falls off with A.

Worked Example 3 — Predicted radii for selected nuclei

Using R = 1.2 × A^(1/3) fm:

Nucleus	A	A^(1/3)	Predicted R (fm)
Carbon-12	12	2.289	2.75
Oxygen-16	16	2.520	3.02
Iron-56	56	3.826	4.59
Tin-120	120	4.932	5.92
Gold-197	197	5.819	6.98
Lead-208	208	5.925	7.11
Uranium-238	238	6.196	7.43

Note how slowly R grows: from carbon to uranium, A increases by a factor of 20 but R increases by less than a factor of 3 — exactly the A^(1/3) signature.

Worked Example 3b — Nuclear radius vs atomic radius

For carbon-12: nuclear radius R = 2.75 fm = 2.75 × 10⁻¹⁵ m. The atomic radius of carbon (the typical extent of its electron cloud) is roughly 70 pm = 7 × 10⁻¹¹ m.

Ratio: R_atom / R_nucleus = 7 × 10⁻¹¹ / 2.75 × 10⁻¹⁵ ≈ 2.5 × 10⁴.

The atom is roughly 25 000 times larger than the nucleus. To put this in everyday terms: if the nucleus were enlarged to the size of a marble (1 cm), the atom would be ~250 m across — comparable to the length of a typical football pitch. Atoms are overwhelmingly empty space, exactly as Rutherford concluded from his scattering experiment.

Volumes scale as the cube of the ratio, so the volume ratio V_atom / V_nucleus ≈ (2.5 × 10⁴)³ ≈ 1.6 × 10¹³ — the nucleus occupies one part in 10¹³ of the atomic volume. This is the geometric reason why nuclear density (~2 × 10¹⁷ kg m⁻³) is roughly 10¹⁴ times atomic density (~10³ kg m⁻³ for solids).

Worked Example 4 — Reading R₀ from data

Suppose experimental electron-diffraction measurements give:

• Carbon-12: R = 2.75 fm.
• Calcium-40: R = 4.10 fm.

Check that both are consistent with R₀ ≈ 1.2 fm.

For C-12: R₀ = R / A^(1/3) = 2.75 / 12^(1/3) = 2.75 / 2.289 = 1.20 fm ✓.

For Ca-40: R₀ = R / A^(1/3) = 4.10 / 40^(1/3) = 4.10 / 3.420 = 1.20 fm ✓.

Two independent measurements give the same R₀, confirming the A^(1/3) scaling and pinning down R₀ = 1.20 fm.

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Nuclear Density and Strong-Force Saturation

The constant-density picture is more than a curve-fitting convenience: it has deep physical meaning. Let us compute the nuclear density explicitly.

Derivation

For a nucleus of A nucleons, each of mass m_u ≈ 1.66 × 10⁻²⁷ kg (the unified atomic mass unit), the total mass is:

M ≈ A × m_u.

The volume of the nucleus, modelled as a uniform sphere of radius R = R₀ A^(1/3), is:

V = (4/3) π R³ = (4/3) π (R₀ A^(1/3))³ = (4/3) π R₀³ A.

Density:

ρ_nucleus = M / V = (A × m_u) / [(4/3) π R₀³ A] = 3 m_u / (4 π R₀³)

The A cancels — confirming that nuclear density is independent of nucleon number.

Numerical Value

Substituting R₀ = 1.2 × 10⁻¹⁵ m and m_u = 1.66 × 10⁻²⁷ kg:

R₀³ = (1.2 × 10⁻¹⁵)³ = 1.728 × 10⁻⁴⁵ m³.

V_per_nucleon = (4/3) π R₀³ = (4/3) × 3.1416 × 1.728 × 10⁻⁴⁵ = 7.238 × 10⁻⁴⁵ m³.

ρ_nucleus = m_u / V_per_nucleon = 1.66 × 10⁻²⁷ / 7.238 × 10⁻⁴⁵ = 2.3 × 10¹⁷ kg m⁻³.

So nuclear density is approximately 2 × 10¹⁷ kg m⁻³ — independent of element, mass number, or isotope.

Comparison with Atomic and Macroscopic Densities

System	Density (kg m⁻³)
Air at STP	1.2
Water	10³
Iron (metal)	7.9 × 10³
Gold (metal)	1.93 × 10⁴
Sun (mean)	1.4 × 10³
White dwarf (core)	~10⁹
Atomic nucleus	~2 × 10¹⁷
Neutron star	~3 × 10¹⁷
Black hole (Schwarzschild radius)	varies

Nuclear matter is roughly 14 orders of magnitude denser than ordinary metals. The reason is purely geometric: an atom is mostly empty space (the electrons occupy a volume ~10⁻³⁰ m³ around a nucleus of volume ~10⁻⁴⁵ m³, a ratio of 10¹⁵), so when you "remove the empty space" between electrons and nucleus, density jumps by 15 orders of magnitude.

What Constant Nuclear Density Implies