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Robert Hofstadter's Stanford installation selected Electrons up to $190\mathrm{MeV}$. Magnetic spectrometer scans counted shielded Lucite Cherenkov radiation pulses per monitor charge. Carbon-12 at $187\mathrm{MeV}$ and $80^{\circ }$ resolved $185.1\mathrm{MeV}$ elastic scattering and $180.7\mathrm{MeV}$ inelastic scattering ($4.43\mathrm{MeV}$ excitation). Calibration used forward-beam energy, hydrogen nuclear recoil and carbon-12 energy levels. Closed entrance slits checked background.^[ref]

The key distinction is between an Electron counter and an energy measurement. Here the counter registers particles admitted by the magnetic spectrometer; changing its magnetic field selects another outgoing momentum. A peak is therefore reconstructed from a scan, rather than read as one pulse whose height equals the scattered Electron energy. Dividing counts by monitored incident electric charge compensates for changes in beam intensity. A peak's height alone is insufficient: its area and the instrument's accepted momentum interval determine the selected yield.

An inelastic scattering peak corresponds to a different final nuclear quantum state, so including it in the elastic scattering yield would answer a different structural question. Conversely, finite energy resolution can hide sufficiently nearby excitations beneath an apparent elastic scattering peak. Background subtraction, target-loss corrections, angular acceptance and counter efficiency all enter before counts become a differential scattering cross-section. A dip in uncorrected counts is therefore not automatically a zero of a nuclear charge form factor.

In the static charge density model, define the normalized Fourier transformHere $\mathbf{q}=(\mathbf{p}-{\mathbf{p}}^{\prime })/\hslash$ is the spatial momentum transfer divided by the reduced Planck constant: $q=|\mathbf{q}|$ has units of inverse length. Physical momentum transfer is $\hslash \mathbf{q}$. We keep $\hslash$ and the speed of light $c$ explicit; compute $q=|\mathbf{p}-{\mathbf{p}}^{\prime }|c/(\hslash c)$, expressing the numerator in $\mathrm{MeV}$ and $\hslash c$ in $\mathrm{MeV}\mathrm{fm}$, to obtain $q$ in inverse femtometres. The spatial $q$ is distinct from the invariant four-momentum transfer used in relativistic scattering cross-sections.

For spherical symmetry, take the polar axis along $\mathbf{q}$ and use spherical coordinates. With $\mu$ the cosine of the polar angle,Every radial shell contributes an amplitude. At $q=0$ all shells add with the same sign; at larger $q$ their contributions can cancel. This cancellation is the origin of a diffraction minimum in the nuclear charge form factor, rather than an absence of electric charge in the nucleus.

The static Fourier transform interpretation is appropriate for the nuclear model and kinematic regime specified here. A fully relativistic composite system requires frame and current conventions; its electric form factor is not universally the transform of a frame-independent three-dimensional charge density.

The root mean square radius of a nuclear charge distribution, referred to its center. Define its charge-weighted second moment byFor a spherical charge density with finite fourth moment, insert the Taylor series $\sin u/u=1-u^2/6+O(u^4)$ into the nuclear charge form factor:Thus a radius is a low-momentum transfer slope, not necessarily the position of a sharp edge. In this real static model,The factor of two between the slopes of $F$ and $|F{|}^2$ matters when interpreting a differential scattering cross-section. A proton charge radius is the corresponding size concept for the proton; the spatial arrangement of many nucleons introduces an additional scale in a heavier atomic nucleus.

What is measured. A monoenergetic Electron beam is directed at a thin target. A movable magnetic spectrometer selects the scattering angle and outgoing momentum, separating elastic scattering from nuclear excitation and other backgrounds. Incident beam current and target thickness supply the normalization of the differential scattering cross-section. The Hofstadter nuclear electron scattering experiment is a classic implementation of this electromagnetic probe.

For a heavy, approximately spin-zero atomic nucleus, neglecting recoil and treating the electron in the Born approximation, the point-charge result is multiplied by the nuclear charge form factor:Here $q$ is the magnitude of the spatial momentum transfer in natural units. The cosine factor includes the Electron spin. Dividing the elastic data by the point-charge factor isolates $|F(q){|}^2$ within these approximations. Recoil, magnetic terms for targets with spin, radiative corrections and particularly Coulomb distortion in electron scattering must be included when significant.

From scattering to size and shape. For a spherical nuclear charge distribution,Thus the nuclear charge form factor is a normalized Fourier transform. Its low-$q$ expansion,measures the nuclear charge radius $r_{\mathrm{ch}}=\sqrt{⟨r^2{⟩}_{\mathrm{ch}}}$. Larger momentum transfers resolve the density profile through diffraction maxima and minima. For a uniform sphere of sharp radius $R$, the uniform-sphere nuclear charge form factor is $3[\sin (qR)-qR\cos (qR)]/(qR{)}^3$; its first zero is at $qR\simeq 4.493$, while $r_{\mathrm{ch}}=\sqrt{3/5}R$.

A practical extraction fits a family of densities to all measured cross sections, rather than treating a few measured values of $|F{|}^2$ as a complete density reconstruction. Finite angular coverage, resolution and the missing sign information make this an inverse problem.

Observed pattern. Apart from the lightest nuclei, the central density is approximately constant; the size grows approximately as $R\simeq 1.2A^{1/3}\mathrm{fm}$ and there is a diffuse surface. The coefficient describes an equivalent sharp nuclear radius, rather than the root-mean-square radius itself. Heavy nuclear charge distributions are often described by the two-parameter Fermi nuclear charge distribution, $\rho (r)={\rho }_0/[1+\exp ((r-c)/a)]$, with half-density radius $c$ scaling roughly as $A^{1/3}$ and diffuseness $a$ around $0.5\mathrm{fm}$. The surface thickness is then about $4a\ln 3\simeq 2.2\mathrm{fm}$. Shell effects, deformation and the lightest systems give departures from this smooth pattern. These qualitative features are established by the electron scattering density measurements.^[ref]

Complementary probes. Transition energies in a muonic atom are especially sensitive to nuclear size because the heavy negative muon orbits close to the charge. Electronic isotope shifts, after separating nuclear-mass and finite-size contributions, measure changes in $⟨r^2⟩$. Both require atomic-structure calculations. Electromagnetic excitation and transition strengths constrain deformation and charge multipoles. Hadronic scattering can also constrain nuclear dimensions, but carries uncertainties from the strong interaction and probes matter rather than purely electric charge.

Consider an ideal nuclear charge distribution that is constant inside a ball of radius $R>0$ and zero outside:This is a deliberately sharp-edged model. Substitution into the spherical nuclear charge form factor gives, for $q\ne 0$,At $q=0$ take the continuous limit $F(0)=1$. Direct radial integration gives $⟨r^2{⟩}_{\mathrm{ch}}=3R^2/5$, and the Taylor series provides an independent check:The first positive zero solves $\sin x-x\cos x=0$, equivalently $\tan x=x$, with $x=qR\simeq 4.49340946$. Consequently the first diffraction minimum moves toward smaller momentum transfer as $R$ increases.

For comparison, a smooth Gaussian function charge density with the same nuclear charge radius isThe Fourier transform of a Gaussian stays positive and has no zero. Both models have the same leading low-$q$ curvature, but they disagree at higher momentum transfer: at $qR=2$, the sphere gives $F\simeq 0.653097$ and the Gaussian function gives $F_G\simeq 0.670320$. Measuring only a nuclear charge radius cannot distinguish these two nuclear charge distributions.

As a chosen scale illustration, take $R=3.00\mathrm{fm}$. The predicted first zero is $q=1.49780{\mathrm{fm}}^{-1}$, or $\hslash cq\simeq 295.557\mathrm{MeV}$ using $\hslash c\simeq 197.327\mathrm{MeV}\mathrm{fm}$. With a $200\mathrm{MeV}$ ultrarelativistic Electron beam and negligible nuclear recoil, it occurs at $\theta \simeq 95.27^{\circ }$. These are consequences of chosen model parameters, not measured properties of carbon or a claim that all atomic nuclei are uniform balls.