Nuclear constituents, size, and density

A nucleus with nucleon number A holds Z protons and N = A − Z neutrons: ¹⁹⁷Au has Z = 79 and N = 118. Proton and neutron each have a mass close to 1 u and the electron only 1/1823 u, so a nuclear mass is close to A u, and A is the natural variable for size. Before the neutron was found, A protons plus A − Z electrons looked economical. Confinement rules it out. An electron held within R ≈ 7 fm needs momentum of order ħ/R, so pc ≈ 197.3 MeV fm ÷ 7.0 fm = 28 MeV, 55 times its rest energy. Its kinetic energy √((pc)² + (mc²)²) − mc² is then about 28 MeV, well above the 16 MeV of Coulomb attraction at gold's surface and far above the energies, typically around 1 MeV, of the electrons that beta decay actually emits. A nucleon with the same momentum has K = (pc)²/2mc² = 0.42 MeV: the classical limit applies because pc ≪ 939 MeV, and modest binding holds it. Spin agrees: 14 protons and 7 electrons would give ¹⁴N a half-integer spin, but it is measured as 1.

Work in the centre-of-mass frame, where an alpha of charge ze approaches a nucleus of charge Ze with speed v₀, kinetic energy K_cm = KM/(M + m) and impact parameter b. At closest approach its velocity is perpendicular to r, so conservation of angular momentum gives vr = v₀b, and conservation of energy gives K_cm = K_cm b²/r² + zZe²/(4πε₀r). Dividing by K_cm and writing d = zZe²/(4πε₀K_cm) turns this into r² − dr − b² = 0, whose positive root is rₘᵢₙ = (d/2)[1 + √(1 + (2b/d)²)]. The hyperbolic orbit ties b to the deflection, b = (d/2)cot(θ/2), and because 1 + cot²x = 1/sin²x, rₘᵢₙ = (d/2)[1 + 1/sin(θ/2)]. Check the limits: θ = 180° needs b = 0 and returns rₘᵢₙ = d, the head-on result; θ → 0 needs b → ∞ and sends rₘᵢₙ to infinity. For 5.5 MeV alphas on gold, K_cm = 5.39 MeV and d = 42.2 fm, so even those counted at 150° turned back 43 fm from the centre, six times gold's 7.0 fm radius.

Rutherford's cross-section, dσ/dΩ = (d/4)²/sin⁴(θ/2), assumes two point charges and nothing else. Coulomb's law carries no length of its own, so the only length in the prediction is d, set by the beam energy. Agreement at angle θ therefore certifies only that nothing but the point-charge field acted outside rₘᵢₙ(θ): gold fitting at 5.5 MeV out to 150° places its charge within 43 fm, an upper bound and no more. Raise K and every rₘᵢₙ shrinks as 1/K until, at some energy and angle, the count falls below the prediction as alphas are drawn out of the elastic beam into reactions. Scans at different energies and angles put that onset at one value of rₘᵢₙ, and different targets put it at a distance that grows with A¹⁄³. A fixed distance, independent of the energy and angle that bring the alpha to it, is something pure Coulomb scattering cannot produce: it is the range of a second force. The law also fails at the opposite extreme, for another reason: a 5.5 MeV alpha passing gold with b beyond a₀/Z ≈ 670 fm, deflected by less than about 4°, meets a charge partly screened by the inner electrons.

Electrons feel no strong force, so they report the charge distribution alone, but only when their wavelength is comparable to it. For λ = 3.0 fm they need pc = hc/λ = 1239.8 MeV fm ÷ 3.0 fm = 413 MeV, about 800 times the electron's rest energy, so E ≈ pc; the first-year λ = h/√(2mK) would give 60 fm at that energy and is simply wrong here. The scattered amplitude is the Fourier transform of the charge density, evaluated at the momentum transfer q = (2p/ħ)sin(θ/2), and dividing the measured cross-section by the point-charge (Mott) prediction leaves the squared form factor |F(q)|². A uniform sphere of radius R has F(q) = 3[sin(qR) − qR cos(qR)]/(qR)³, which first vanishes at qR = 4.493, the first root of tan x = x. So 400 MeV electrons on a 5.0 fm sphere show a minimum where sin(θ/2) = 4.493 × 197.3 MeV fm ÷ (2 × 400 MeV × 5.0 fm) = 0.222, at θ = 25.6°. Real minima are shallow dips, because the edge is diffuse; fits report an rms radius, and a uniform sphere with the same rms radius has R = √(5/3) rᵣₘₛ.

Across nuclei from carbon to uranium the equivalent radii fall on R = R₀A¹⁄³ with R₀ ≈ 1.2 fm. Cube it and the volume, (4/3)πR₀³A, is proportional to A, so the number density n₀ = 3/(4πR₀³) = 0.138 fm⁻³ is the same for every nucleus; multiplying by the nucleon mass, 1.67 × 10⁻²⁷ kg, gives ρ ≈ 2.3 × 10¹⁷ kg m⁻³. The mean spacing n₀⁻¹⁄³ = 1.9 fm is little more than twice the proton's 0.84 fm charge radius, so the nucleons nearly touch. Atoms do the opposite: their radii stay between about 0.05 and 0.3 nm from hydrogen to uranium, because every electron is pulled by the whole nuclear charge through a long-range force, and added charge pulls the inner shells in. A nucleus instead grows like a drop of liquid, each added nucleon bringing its own volume. That is saturation: a nucleon is bound only to its near neighbours, so the range of the attraction is about the spacing, 1–2 fm, the same scale as the reach found where Rutherford's law fails.

R₀ is a fitted number, not a constant of nature. Electrons measure the charge and give about 1.2 fm; alphas and neutrons feel the strong force and report larger interaction radii, because the reach is included. Nor is the edge sharp: the density falls from 90% to 10% of its central value over about 2.3 fm, which is most of the radius of a light nucleus, so R₀A¹⁄³ is unreliable below A ≈ 12 and fails for halo nuclei such as ¹¹Li, whose two outermost neutrons spread far beyond it. Many heavy nuclei are not spheres at all. Above all, this is inference, not theory. The scattering data fix the distance at which a new force switches on and show that it is short-ranged; that 79 protons stay together in gold shows it is attractive and stronger than their mutual repulsion. None of it gives the force's shape, its repulsive core or its spin dependence, and none of it yields a binding energy: that takes the mass measurements of the next topic.