---
title: Nuclear Size, Shape, and Charge Distributions
module: Nuclear Properties
moduleNumber: 1
lessonNumber: 2
order: 102
summary: >
  Elastic electron scattering resolves the nucleus by its de Broglie wavelength.
  The measured cross section is the Mott point-charge cross section modulated by a
  form factor, and that form factor is the Fourier transform of the charge density.
  Diffraction minima fix the radius, the small-angle slope fixes the mean-square
  radius, and the fitted Woods-Saxon profile gives a central density and a skin
  thickness. Mirror-nucleus Coulomb energies, muonic-atom X-rays, and optical
  isotope shifts give independent radii that all track R = R0 A^(1/3).
topics: [Nuclear Properties]
sources:
  - book: Krane
    ref: "Ch. 3 — Nuclear Properties; §3.1 The Nuclear Radius, §3.4 The Distribution of Nuclear Charge"
  - book: Povh, Rith, Scholz, Zetsche
    ref: "Ch. 5 — Geometric Shapes of Nuclei; Form Factors and elastic electron scattering"
  - book: Wong
    ref: "Ch. 4 — Nuclear Collective Motion (electron scattering context)"
draft: false
---

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.[^krane-radius] 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

$$
\vec q = \vec k_i - \vec k_f, \qquad q = |\vec q| = 2k\sin\frac{\theta}{2},
$$

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

$$
f(\vec q) = -\frac{2m_e}{\hbar^2}\,\frac{1}{q}\int_0^\infty r\,V(r)\sin(qr)\,\d r
= -\frac{m_e}{2\pi\hbar^2}\int V(\vec r)\,e^{i\vec q\cdot\vec r}\,\d^3 r .
$$

The electron interacts with the nuclear charge through the electrostatic potential
$\phi(\vec r)$, which is tied to the charge density $\rho(\vec r)$ by Poisson's
equation $\nabla^2\phi = -\rho/\epsilon_0$. Splitting $V = -e\phi$ and integrating by
parts twice converts the transform of $\phi$ into the transform of $\rho$, and the
$1/q^2$ that appears carries the point-charge (Rutherford) amplitude. The result
factorizes:

$$
\left(\frac{\d\sigma}{\d\Omega}\right) = \left(\frac{\d\sigma}{\d\Omega}\right)_{\!\text{point}}\bigl|F(\vec q)\bigr|^2 .
$$

> **Definition (Nuclear form factor).** For a charge density $\rho(\vec r)$ normalized
> to the total charge $\int\rho\,\d^3 r = Ze$, the form factor is the normalized Fourier
> transform
> $$
> F(\vec q) = \frac{1}{Ze}\int \rho(\vec r)\,e^{i\vec q\cdot\vec r}\,\d^3 r .
> $$
> A point charge has $\rho = Ze\,\delta^3(\vec r)$ and $F \equiv 1$; any extended
> distribution has $|F| < 1$ away from $q = 0$, and the falloff of $F$ with $q$ measures
> the spatial extent of $\rho$.

$$
% caption: An electron of wavevector k_i scatters through angle theta to k_f; the
% momentum transfer q closes the vector triangle and sets the spatial scale 1/q that
% the measurement resolves.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  % target
  \draw[black] (0,0) circle (0.35);
  \node[anchor=north, black!70] at (0,-0.45) {nucleus};
  % incident ray
  \draw[very thick, ->] (-3.4,0) -- (-0.4,0);
  \node[black, anchor=south] at (-2.2,0.06) {$k_i$};
  \node[black, anchor=north] at (-2.6,-0.05) {incident electron};
  % scattered ray at 40 deg
  \draw[very thick, ->] (0.35,0.23) -- (2.75,1.85);
  \node[black, anchor=south east] at (2.6,1.7) {$k_f$};
  \node[black, anchor=west] at (2.7,1.9) {detector};
  % continuation of incident direction (dashed) to show angle
  \draw[black, dashed] (0.4,0) -- (3.2,0);
  % angle arc (scattering angle theta, labelled in caption)
  \draw[black] (1.3,0) arc (0:34:1.3);
  \node[black!70, font=\scriptsize] at (1.7,0.4) {angle};
  % momentum-transfer triangle inset
  \begin{scope}[xshift=0.2cm, yshift=-2.3cm]
    \draw[thick, ->] (0,0) -- (2.2,0) node[midway, below, black] {$k_i$};
    \draw[thick, ->] (0,0) -- (1.82,1.28) node[midway, above left, black] {$k_f$};
    \draw[acc, thick, ->] (1.82,1.28) -- (2.2,0);
    \node[acc, anchor=west] at (2.05,0.7) {$q$};
    \node[black, anchor=north] at (1.1,-0.5) {momentum transfer};
  \end{scope}
\end{tikzpicture}
$$

## 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 $\vec r$,

$$
F(q) = \frac{4\pi}{Ze}\int_0^\infty \rho(r)\,\frac{\sin qr}{qr}\,r^2\,\d r .
$$

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 $\sin qr/qr = 1 - (qr)^2/6 + \cdots$,

$$
F(q) = 1 - \frac{q^2}{6}\langle r^2\rangle + \frac{q^4}{120}\langle r^4\rangle - \cdots,
\qquad \langle r^2\rangle = \frac{1}{Ze}\int r^2\rho(\vec r)\,\d^3 r .
$$

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

$$
\langle r^2\rangle = -6\left.\frac{\d F}{\d(q^2)}\right|_{q^2=0}.
$$

**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

$$
F(q) = \frac{3}{(qR)^3}\bigl[\sin qR - qR\cos qR\bigr],
$$

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:

$$
R \approx \frac{4.49}{q_{\min}} = \frac{4.49\,\hbar c}{2E\sin(\theta_{\min}/2)}.
$$

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

$$
% caption: The elastic cross section falls steeply like the point-charge law and is
% modulated by diffraction minima whose spacing measures the radius; a diffuse surface
% fills the zeros to shallow dips.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (8.0,0) node[right, black!70] {$q$};
  \draw[->, black] (0,-0.2) -- (0,4.4) node[above, black!70] {log cross section};
  % point-charge falling reference (straight declining line on log axis)
  \draw[black, dashed] (0.3,4.1) -- (7.6,0.6);
  \node[black, anchor=south west, font=\scriptsize] at (5.3,1.6) {point charge};
  % measured curve: falling with diffraction dips (sharp for uniform sphere)
  \draw[acc, very thick]
    (0.3,4.05)
    .. controls (1.1,3.2) and (1.5,2.6) .. (1.9,2.0)
    .. controls (2.1,1.6) and (2.25,0.9) .. (2.35,0.9)
    .. controls (2.5,1.4) and (2.9,2.1) .. (3.3,2.05)
    .. controls (3.55,1.6) and (3.75,0.6) .. (3.9,0.6)
    .. controls (4.1,1.1) and (4.5,1.6) .. (4.9,1.5)
    .. controls (5.2,1.1) and (5.4,0.4) .. (5.55,0.4)
    .. controls (5.8,0.8) and (6.2,1.1) .. (6.6,1.0);
  % minima markers
  \draw[black, dashed] (2.35,0) -- (2.35,0.9);
  \draw[black, dashed] (3.9,0) -- (3.9,0.6);
  \node[black!70, anchor=north, font=\scriptsize] at (2.35,-0.05) {1st min};
  \node[black!70, anchor=north, font=\scriptsize] at (3.9,-0.05) {second min};
  \node[acc, anchor=west] at (0.5,3.7) {measured};
\end{tikzpicture}
$$

## 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,

$$
\rho(r) = \frac{\rho_0}{1 + \exp\!\bigl[(r - c)/a\bigr]},
$$

with $c$ the **half-density radius** where $\rho = \tfrac12\rho_0$ and $a$ the diffuseness.
Fits across the chart give

$$
c \approx 1.07\,A^{1/3}\ \mathrm{fm}, \qquad a \approx 0.54\ \mathrm{fm},
$$

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:

$$
t = a\Bigl[\ln 9 - \ln\tfrac19\Bigr] = 4a\ln 3 \approx 4.4\,a \approx 2.3\ \mathrm{fm}.
$$

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 $\rho_0/e \approx
0.16\ \mathrm{nucleons/fm^3}$. For a distribution that is close to a uniform sphere the
mean-square and half-density radii are related by $\langle r^2\rangle = \tfrac35 c^2$
plus a small surface correction, giving the root-mean-square charge radius

$$
\sqrt{\langle r^2\rangle} \approx 0.94\,A^{1/3}\ \mathrm{fm}.
$$

$$
% caption: The Woods-Saxon charge density is flat at rho0 in the interior, passes
% through half its central value at the radius c, and falls over a surface of
% diffuseness a; the 90-to-10 percent width is the skin thickness t.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (7.4,0) node[right, black!70] {$r$};
  \draw[->, black] (0,0) -- (0,3.9) node[above, black!70] {density};
  % profile: flat then logistic falloff
  \draw[acc, very thick]
    (0,3.2) -- (2.7,3.2)
    .. controls (3.3,3.2) and (3.5,0.2) .. (4.6,0.2)
    -- (6.8,0.06);
  % rho0 level
  \draw[black, dashed] (0,3.2) -- (2.7,3.2);
  \node[anchor=east, black!70] at (-0.05,3.2) {peak};
  % half density
  \draw[black, dashed] (0,1.7) -- (3.63,1.7);
  \node[anchor=east, black, font=\scriptsize] at (-0.05,1.7) {half};
  \draw[black, dashed] (3.63,0) -- (3.63,1.7);
  \node[anchor=north, black!70] at (3.63,-0.05) {$c$};
  % 0.9 and 0.1 levels
  \draw[black, dashed] (0,2.9) -- (3.15,2.9);
  \draw[black, dashed] (0,0.32) -- (4.35,0.32);
  \node[anchor=east, black, font=\scriptsize] at (-0.05,2.9) {0.9};
  \node[anchor=east, black, font=\scriptsize] at (-0.05,0.32) {0.1};
  % skin bracket
  \draw[black] (3.15,1.05) -- (4.35,1.05);
  \draw[black] (3.15,0.95) -- (3.15,1.15);
  \draw[black] (4.35,0.95) -- (4.35,1.15);
  \node[black, anchor=south, font=\scriptsize] at (3.75,1.1) {$t$};
\end{tikzpicture}
$$

## 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 $E_C = \tfrac35 (Z^2 e^2/4\pi\epsilon_0 R)$, so the Coulomb-energy
  difference between mirror partners is
  $$
  \Delta E_C = \frac{3}{5}\frac{e^2}{4\pi\epsilon_0 R}\bigl[Z^2 - (Z-1)^2\bigr]
  = \frac{3}{5}\frac{e^2}{4\pi\epsilon_0 R}(2Z - 1).
  $$
  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 $\langle r^2\rangle$, 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, $\delta\nu_{\text{FS}} \propto |\psi(0)|^2\,\delta\langle r^2\rangle$,
  on top of a **mass shift** from the finite nuclear mass. Separating the two yields
  $\delta\langle r^2\rangle$ along an isotopic chain, tracking how the radius grows as
  neutrons are added.

$$
% caption: Root-mean-square charge radii from electron scattering, muonic X-rays, and
% mirror-nucleus Coulomb energies all fall on a common line proportional to A^(1/3),
% confirming constant nuclear density.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (7.4,0) node[right, black!70] {cube root of $A$};
  \draw[->, black] (0,0) -- (0,4.4) node[above, black!70] {radius (fm)};
  % axis ticks
  \foreach \x/\lab in {1.5/2,3.0/4,4.5/6,6.0/8} \node[anchor=north, black, font=\scriptsize] at (\x,-0.05) {\lab};
  \foreach \y/\lab in {1/2,2/4,3/6,4/8} {\draw[black] (-0.07,\y) -- (0.07,\y); \node[anchor=east, black, font=\scriptsize] at (-0.05,\y) {\lab};}
  % best-fit line R = 0.94 A^{1/3}
  \draw[acc, very thick] (0.3,0.28) -- (6.6,3.35);
  \node[acc, anchor=west, font=\scriptsize] at (4.2,2.85) {trend line};
  % data points from different methods (line style / shade distinguishes them)
  \fill[black] (1.5,0.75) circle (1.7pt);
  \fill[black] (2.6,1.28) circle (1.7pt);
  \fill[black] (3.7,1.82) circle (1.7pt);
  \fill[black] (5.0,2.47) circle (1.7pt);
  \draw[black, thick] (1.9,0.98) circle (1.9pt);
  \draw[black, thick] (3.1,1.55) circle (1.9pt);
  \draw[black, thick] (4.4,2.18) circle (1.9pt);
  \draw[black, thick] (5.8,2.9) circle (1.9pt);
  % legend
  \fill[black] (0.6,4.0) circle (1.7pt);
  \node[black!70, anchor=west, font=\scriptsize] at (0.8,4.0) {electron scattering};
  \draw[black, thick] (0.6,3.55) circle (1.9pt);
  \node[black!70, anchor=west, font=\scriptsize] at (0.8,3.55) {muonic X-ray / mirror};
\end{tikzpicture}
$$

## 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

$$
R(\theta) = R_0\Bigl[1 + \beta_2\,Y_{20}(\theta)\Bigr],
$$

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](/nuclear-physics/nuclear-properties/nuclear-moments-multipoles) 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.

$$
% caption: A spherical charge cloud (left) deforms into a prolate spheroid for positive
% quadrupole deformation (center) and an oblate spheroid for negative (right); the
% surface radius varies with polar angle as R0 times one plus a quadrupole term.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  % sphere
  \draw[very thick] (0,0) circle (0.85);
  \draw[black, dashed] (0,-1.05) -- (0,1.05);
  \node[anchor=north, black!70] at (0,-1.25) {spherical};
  % prolate
  \begin{scope}[xshift=3.4cm]
    \draw[very thick] (0,0) ellipse (0.6 and 1.05);
    \draw[black, dashed] (0,-1.25) -- (0,1.25);
    \node[anchor=north, black!70] at (0,-1.45) {prolate};
    \node[black, anchor=west, font=\scriptsize] at (0.7,0.6) {elongated};
  \end{scope}
  % oblate
  \begin{scope}[xshift=6.8cm]
    \draw[very thick] (0,0) ellipse (1.05 and 0.6);
    \draw[black, dashed] (0,-0.8) -- (0,0.8);
    \node[anchor=north, black!70] at (0,-1.45) {oblate};
    \node[black, anchor=west, font=\scriptsize] at (1.15,0.35) {squashed};
  \end{scope}
\end{tikzpicture}
$$

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.

[^krane-radius]: **Krane**, _Introductory Nuclear Physics_, §3.1 (The Nuclear Radius) and §3.4 (The Distribution of Nuclear Charge): the Born-approximation form factor, the Woods-Saxon parametrization ($c \approx 1.07\,A^{1/3}\,\mathrm{fm}$, $a \approx 0.54\,\mathrm{fm}$, $t \approx 2.3\,\mathrm{fm}$), and the electron-scattering, muonic-atom, and mirror-nucleus radius determinations. Evaluated charge radii are tabulated by the NNDC, [https://www.nndc.bnl.gov/](https://www.nndc.bnl.gov/). The form-factor treatment follows **Povh, Rith, Scholz, Zetsche**, _Particles and Nuclei_, Ch. 5.