advanced30 min studyLesson 571 in Physics

Nuclear sizes and charge distributions

Infer a charge radius from a normalized form factor while separating sharp-sphere, charge, and matter radii.

Before you begin

On this page

The question for today

What you’ll understand

Normalize a spherical charge density and calculate its mean-square radius.
Relate the low-momentum form-factor slope to charge radius.
Identify shape and scattering approximations needed for a radius extraction.

A nucleus has no perfectly sharp surface. Its reported radius depends on the distribution being probed and the definition used to summarize that distribution. Electrons predominantly probe electric charge, whereas strongly interacting projectiles and weak probes carry different sensitivities. A useful analysis therefore begins with an observable and a normalization convention, rather than treating every measured length as the boundary of one hard sphere. Simple density models provide tractable interpretations whose assumptions can be tested.

Define the distribution before the radius

Let a spherical charge density integrate to the total nuclear charge $Ze$ . Its mean-square radius weights each volume element by distance squared, then divides by total charge. A small amount of charge far from the center can therefore affect the radius appreciably. The root-mean-square radius is the square root of this moment. It is not generally the half-density radius of a diffuse profile, nor the radius of a sharp uniform sphere chosen to approximate it.

For a uniform sphere the radial volume element introduces a factor of radius squared into both integrals. Evaluating them gives mean-square radius three fifths of the sharp radius squared. The familiar scaling of sharp radius with the cube root of nucleon number expresses approximately constant bulk density, not an exact law for every isotope. Surface diffuseness, deformation, light-nucleus clustering, and weakly bound tails can modify the relationship, so a fitted radius parameter must retain its model definition.

\int\rho_{\rm ch}(\mathbf r)\,d^3r=Ze,\qquad \langle r^2\rangle_{\rm ch}=\frac{1}{Ze}\int r^2\rho_{\rm ch}(\mathbf r)\,d^3r,\qquad \langle r^2\rangle_{\rm uniform}=\frac35R^2

Charge density has units of charge per volume; R and the rms radius have length units. The last identity assumes a uniform spherical density with a sharp boundary.↔ Scroll wide equations horizontally.

Use elastic scattering as a Fourier probe

In a static nonrelativistic density interpretation, the normalized elastic form factor is the Fourier transform of the charge density. Define q as momentum transfer divided by the reduced Planck constant, so it has inverse-length units. For spherical density, angular integration produces the spherical sinc factor. At zero q the form factor equals one, fixing the normalization independently of the details of the density.

Expanding the sinc factor at small q shows that the first correction is quadratic and proportional to the mean-square radius. In a plane-wave Born treatment of elastic scattering from a suitable spin-zero target, the point-charge cross section is multiplied by the squared form factor, with recoil and other kinematic factors treated consistently. Coulomb distortion, radiative effects, magnetic contributions, and relativistic interpretation complicate real extractions. A small measured cross-section deficit cannot simply be equated to the form-factor deficit without the square.

F(q)=\frac{4\pi}{Ze}\int_0^\infty r^2\rho_{\rm ch}(r)\frac{\sin(qr)}{qr}\,dr=1-\frac{q^2\langle r^2\rangle_{\rm ch}}6+O(q^4)

F is dimensionless and q may be expressed in inverse femtometers when radii are in femtometers. The expansion assumes finite moments and momentum transfer small on the density length scale.↔ Scroll wide equations horizontally.

Distinguish what each probe measures

A measured charge density includes the finite electromagnetic structure of protons and neutrons, as well as corrections from nuclear motion and currents. A point-proton distribution removes those internal-nucleon contributions within a stated convention. A matter radius weights nucleons rather than electric charge and may be more sensitive to neutrons. These quantities are related through modeling, but they are not interchangeable just because all are quoted in femtometers.

Finite experimental momentum coverage also limits spatial reconstruction. Many density shapes can reproduce a restricted low-q interval, particularly if only the radius-sensitive slope is resolved. Diffraction minima add information about the larger-scale profile, but locating a minimum is not a complete image of a nucleus. A defensible fit specifies its functional form, statistical uncertainty, correction scheme, and shape dependence. Comparing several admissible forms is more informative than reporting many digits from one assumed shape.

F_{\rm sphere}(q)=3\frac{\sin(qR)-qR\cos(qR)}{(qR)^3},\qquad F_{\rm sphere}(0)=1

This exact transform is for the uniform sharp sphere only. The apparent zero-over-zero at q=0 is removed by taking its continuous limit.↔ Scroll wide equations horizontally.

Work it through

A small-q radius estimate

A toy uniform charge sphere has sharp radius $R=3\,\mathrm{fm}$ . Find its rms radius, estimate F at $q=0.10\,\mathrm{fm}^{-1}$ to quadratic order, and estimate the cross-section ratio $|F|^2$ .

The mean-square radius is $(3/5)9=5.4\,\mathrm{fm}^2$ , so the rms radius is $\sqrt{5.4}\approx2.324\,\mathrm{fm}$ . This is smaller than the specified sharp boundary.
The quadratic form-factor correction is $(0.10)^2(5.4)/6=0.009$ . Thus $F\approx0.991$ ; the dimensionless parameter $qR=0.30$ indicates a low-q expansion rather than an exact evaluation.
Squaring the estimated real amplitude gives $|F|^2\approx0.982081$ . Its leading deficit is about twice the amplitude deficit. This ratio refers to the stipulated Born comparison, excluding distortion and additional responses.

The result

The toy rms radius is 2.324 fm; the quadratic amplitude estimate is 0.991 and its squared ratio is approximately 0.9821.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

Inspect a quadratic small-transfer correction

Use dimensionless $u=q\sqrt{\langle r^2\rangle/6}$ so the leading correction is $F-1=-u^2$ . Reset and set coefficient $a=-1$ , position $x=0.2$ ; predict y equal to minus 0.04. For a fresh comparison Reset and set $a=-1,x=0.4$ ; predict minus 0.16. The lab plots one quadratic, not a charge density, form factor, or scattering solver.

y = ax²TangentSigned integral

y at chosen x: 0.608
Tangent slope · 2ax: 1.35
Signed area · ax³/3: 0.182

Coefficient a0.75Position x0.9

t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forDoubling the small-transfer coordinate quadruples its leading quadratic correction; the approximation still requires small u.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

Integrating the uniform density with the three-dimensional volume element gives mean-square radius three fifths R squared. Taking its square root produces a length, unlike the third option.

Your turn

Try it yourself

Compare a uniform sphere and an infinitesimally thin spherical charge shell, each with outer radius $4\,\mathrm{fm}$ . Compute their rms radii and their low-q slope coefficients in $F=1-cq^2+\cdots$ .

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

All charge in the thin shell sits at one radius; the uniform sphere averages over its interior.

Compare with the solution

The uniform sphere has $\langle r^2\rangle=9.6\,\mathrm{fm}^2$ and rms radius $\sqrt{9.6}\approx3.098\,\mathrm{fm}$ . Its coefficient is $c=9.6/6=1.6\,\mathrm{fm}^2$ . The shell has mean-square radius $16\,\mathrm{fm}^2$ , rms radius $4\,\mathrm{fm}$ , and $c=16/6\approx2.667\,\mathrm{fm}^2$ . Equal outer radii do not imply equal elastic slopes, illustrating the dependence on the full charge distribution.

Keep the idea

Take it with you

A radius must identify its distribution and averaging convention.
The low-q form-factor slope measures a mean-square charge radius under stated corrections.
Charge, point-proton, matter, and sharp-surface radii are distinct quantities.

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