Nuclear Size, Shape, and Charge Distributions

The size of a nucleus has no single definition: the charge density, the nuclear-force density, and the matter density each fall off over a surface of finite width, and each probe measures its own radius. The sharpest picture comes from firing electrons at the nucleus. Electrons feel only the electromagnetic interaction, whose form is known exactly, so the deflection pattern maps the charge distribution without the theoretical uncertainty that clouds strong-interaction probes.¹ To resolve a structure of size $R$ the probe wavelength must satisfy $\lambda \lesssim R$. For $R\approx 5\mathrm{fm}$ the de Broglie relation $\lambda =h/p$ requires momenta $pc\gtrsim 250\mathrm{MeV}$, so the electron is ultrarelativistic and $E\approx pc$.

Elastic scattering and the Born approximation

A fast electron scattering elastically off a static charge distribution transfers momentum but no energy. Write the incident and outgoing wavevectors as ${\vec{k}}_i$ and ${\vec{k}}_f$ with $|{\vec{k}}_i|=|{\vec{k}}_f|=k$; the momentum transfer is

where $\theta$ is the scattering angle. In the first Born approximation the scattering amplitude is the Fourier transform of the interaction potential $V(\vec{r})$,

The electron interacts with the nuclear charge through the electrostatic potential $\varphi (\vec{r})$, which is tied to the charge density $\rho (\vec{r})$ by Poisson's equation ${\nabla }^2\varphi =-\rho /{ϵ}_0$. Splitting $V=-e\varphi$ and integrating by parts twice converts the transform of $\varphi$ into the transform of $\rho$, and the $1/q^2$ that appears carries the point-charge (Rutherford) amplitude. The result factorizes:

From form factor to charge density

For a spherically symmetric density the angular part of the Fourier integral is elementary. Writing $\vec{q}\cdot \vec{r}=qr\cos \alpha$ and integrating over the solid angle of $\vec{r}$,

The transform is invertible: measuring $F(q)$ over a wide range of $q$ and inverting gives $\rho (r)$ directly. Two limits carry most of the physics.

Small $q$ (the mean-square radius). Expanding $\sin qr/qr=1-(qr{)}^2/6+\cdots$,

The initial slope of $F$ against $q^2$ measures the mean-square charge radius with no assumption about the shape of $\rho$:

Large $q$ (diffraction minima). At high momentum transfer the wave scattered from the near edge and the far edge of the nucleus interfere, and $|F(q){|}^2$ develops diffraction zeros exactly as light does at a circular aperture. For a uniformly charged sphere of radius $R$ the integral gives

whose first zero is at the root of $\tan qR=qR$, namely $qR\approx 4.493$. The angular position of the first minimum therefore fixes the radius:

A real nucleus has a diffuse surface, so its minima are filled in rather than reaching zero, but their spacing still tracks $1/R$.

The Woods-Saxon charge distribution

Inverting many nuclei's form factors gives a density that is flat in the interior and falls smoothly to zero over a surface layer. The standard two-parameter fit is the Woods-Saxon (Fermi) form,

with $c$ the half-density radius where $\rho =\frac{1}{2}{\rho }_0$ and $a$ the diffuseness. Fits across the chart give

independent of $A$ for the diffuseness. The skin thickness $t$, the distance over which $\rho$ drops from $90\%$ to $10\%$ of ${\rho }_0$, follows from the logistic profile:

The central density ${\rho }_0$ is nearly the same for all but the lightest nuclei, the hallmark of saturation: nucleons pack to a fixed number density ${\rho }_0/e\approx 0.16nucleons/f{\mathrm{m}}^3$. For a distribution that is close to a uniform sphere the mean-square and half-density radii are related by $⟨r^2⟩=\frac{3}{5}{c}^2$ plus a small surface correction, giving the root-mean-square charge radius

Radii from mirror nuclei, muons, and isotope shifts

Electron scattering is the cleanest measurement but not the only one, and three independent methods agree.

• Mirror-nucleus Coulomb energies. A pair of mirror nuclei, $(Z,N)$ and $(N,Z)$ with $N=Z\pm 1$, would be identical if the nuclear force were exactly charge-symmetric; they differ only by the Coulomb energy of one extra proton and the neutron-proton mass difference. For a uniformly charged sphere the electrostatic self-energy is $E_C=\frac{3}{5}(Z^2{e}^2/4\pi {ϵ}_{0}R)$, so the Coulomb-energy difference between mirror partners is$$\Delta E_C=\frac{3}{5}\frac{e^2}{4\pi {ϵ}_{0}R}[Z^2-(Z-1{)}^2]=\frac{3}{5}\frac{e^2}{4\pi {ϵ}_{0}R}(2Z-1).$$
  Measuring $\Delta E_C$ from the beta-decay endpoint of the pair fixes $R$, and the values follow $R=R_0{A}^{1/3}$ with $R_0\approx 1.2\mathrm{fm}$. This measures the radius of the nuclear-force (proton) distribution, slightly larger than the charge radius because the proton itself has finite size.
• Muonic atoms. A muon captured into an atomic orbit has a Bohr radius smaller than the electron's by the mass ratio, $a_{\mu }=a_0(m_e/m_{\mu })=a_0/206.8$. For medium and heavy elements the muon's $1s$ orbit lies partly inside the nucleus, so its binding energy is strongly reduced by the finite charge distribution. The muonic $2p\to 1s$ X-ray energy shifts by an amount proportional to $⟨r^2⟩$, giving the most precise single-nucleus charge radii available.
• Optical isotope shifts. Comparing the same electronic transition in two isotopes, the spectral line shifts by a field shift proportional to the change in mean-square radius, $\delta {\nu }_{\text{FS}}\propto |\psi (0){|}^2\delta ⟨r^2⟩$, on top of a mass shift from the finite nuclear mass. Separating the two yields $\delta ⟨r^2⟩$ along an isotopic chain, tracking how the radius grows as neutrons are added.

Deformation and the second moment of the shape

The form factor also reports departures from spherical symmetry. Expanding $\rho$ in multipoles, the monopole term gives the radius while the quadrupole term encodes the deformation. A permanently deformed nucleus, such as the rare-earth species with $Z=57$ to $71$, has an elongated charge cloud whose mean-square radius depends on orientation. Parametrizing the surface as

with ${\beta }_2$ the quadrupole deformation, expands the charge distribution along the symmetry axis for ${\beta }_2>0$ (prolate, watermelon) and flattens it for ${\beta }_2<0$ (oblate). The corresponding intrinsic electric quadrupole moment is the static shape measure taken up in the nuclear moments lesson; here the point is that elastic electron scattering off an oriented or high-spin target already carries the signature of that deformation in the angular distribution.

Every method returns the same coefficient to within its systematic error, and the constancy of $R_0$ across five orders of magnitude in $A$ is the strongest single piece of evidence that nuclear matter is incompressible at a fixed density set by the saturation of the nuclear force. The next lesson turns from the charge distribution to the mass, where the binding energy that holds this incompressible drop together is read off from precise mass measurements.