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Nuclear, particle & frontier orientation · Interactive lesson

Nuclear size and the charge form factor

Electrons are the cleanest ruler we have for a nucleus: no strong force, a vertex QED gives exactly, and a resolution you set with the beam energy. This is how a diffraction pattern in a spectrometer becomes a radius, a surface thickness, and a density that refuses to move.

UPIV-13.01 · Lesson 1 of 9 in this unit. Position is not a completion score.
Starting model: Nuclear size and the charge form factorFermi profile ρ(r) = ρ₀/[1 + exp((r − c)/a)], the measured charge shape scaled to A nucleons, r running 0 to 10 fm along the base. The circle marks the half-density point r = c, the arrow spans the 90-to-10% surface t = 2.42 fm, the upper dashed line is 0.16 fm⁻³ saturation, and the dashed step is the uniform sphere R = 7.11 fm. Interior 0.162 nucleons fm⁻³.t = 2.42 fmρ(r) = ρ₀ / [1 + exp((r − c)/a)]A = 208 · c = 6.60 fm · a = 0.55 fmdashed step: R = 1.2 A¹⁄³mean ρ = 0.138 fm⁻³ for all Anucleon density / fm⁻³saturation ≈ 0.16 fm⁻³r = 0r = 10 fm
Start with a picture. Change the model in the lab below.

Nuclear, particle & frontier orientation. These numbered units provide nuclear, particle and frontier orientation, not a substitute for a complete specialist course. Check your university’s required depth and prerequisites.

1

Build the model

Divide out the point-charge answer; what is left is the shape.

The key idea

The nuclear charge form factor F(q) is the measured elastic electron-scattering cross-section divided by the point-charge Mott cross-section, and in the Born approximation it is the Fourier transform of the nuclear charge distribution, normalised so that F(0) = 1.

Example. For ²⁰⁸Pb at 250 MeV the first minimum of |F|² sits near θ = 29°, where q/ħ = 2E sin(θ/2)/ħc = 0.634 fm⁻¹; setting qR/ħ = 4.493 gives R = 7.08 fm, within 0.4% of 1.2 × 208¹⁄³ = 7.11 fm.

The complete physical model and assumptions

A nucleus cannot be looked at; it can only be bounced off, and the pattern read. Electrons are the right projectile because they feel only the electromagnetic current, so anything unexpected in the data belongs to the target and not to the probe. Drive them hard enough that the resolution ħ/q falls below a fermi and the scattering stops looking point-like: the cross-section departs from the Mott prediction, and the ratio of the two defines the form factor F(q). In first Born approximation F is the Fourier transform of the charge density — one dimensionless function of momentum transfer, normalised to F(0) = 1, carrying shape and no Z at all. Its slope at small q returns ⟨r²⟩ with no model assumed; its minima return a radius; a wide enough sweep in q inverts to ρ(r) itself. What comes back is a two-parameter Fermi profile whose half-density radius grows as A¹⁄³ while its interior sits near 0.16 nucleons per fm³ for everything from carbon to uranium. That saturation is direct evidence that the nuclear force is short-ranged and self-limiting, and the liquid-drop model is built on it. The cost is written into the method: Born maps charge, so this locates protons rather than nucleons, and at Z = 82 the expansion parameter Zα = 0.60 is not small, so the single Fourier transform must give way to a full phase-shift analysis.

1Why an electron, and how hard you have to hit

An electron has no strong interaction and no measurable size of its own, so it couples to the nucleus through a single exchanged photon whose electron end is fixed by theory rather than fitted: nothing about the probe has to be subtracted before the target can be read. What the beam energy buys is resolution: the probe sees structure of order ħ/q, and since q/ħ = 2E sin(θ/2)/ħc for E ≫ mₑc², with ħc = 197.3 MeV fm, a momentum transfer of 0.634 fm⁻¹ resolves only 1.6 fm. Getting below a fermi means hundreds of MeV — Hofstadter's Stanford measurements of the 1950s ran from about 100 to 550 MeV, and it is no accident that nuclear structure appeared in scattering data only once accelerators reached that range. One further number must be checked before any of this is used: the Born expansion parameter Zα. For ¹²C it is 6/137 = 0.044 and single-photon exchange is an excellent approximation. For ²⁰⁸Pb it is 82/137 = 0.60, and it is not.

2Divide out Mott, and keep what refuses to cancel

The point-charge prediction is Rutherford with two relativistic repairs. Electron spin supplies a factor cos²(θ/2), because helicity conservation forbids exact backscattering of an ultrarelativistic electron, and target recoil supplies 1/[1 + (2E/Mc²) sin²(θ/2)]. Both are smooth and monotonic in θ: the Mott cross-section has no minima, no shoulders, nothing an experimenter could mistake for structure. Nor is the recoil factor doing heavy lifting here — at 250 MeV and 29° it amounts to 1.6 × 10⁻⁴ for ²⁰⁸Pb and 2.8 × 10⁻³ for ¹²C. So divide the measured elastic cross-section by the Mott value point by point in angle. The ratio is |F(q)|²: it starts at 1 in the forward direction and, for ²⁰⁸Pb at 250 MeV, falls by more than two orders of magnitude on the way into a first minimum near 29°. That fall is the entire measurement.

3Read the transform three different ways

For a spherically symmetric density the angular integral in F(q) collapses the exponential to sin(qr/ħ)/(qr/ħ), and three readings follow. Expand for small argument: F ≈ 1 − q²⟨r²⟩/6ħ², so the initial slope of F against q² gives the mean-square charge radius with no profile assumed — the only genuinely model-independent number here. Next, the positions of the minima fix a radius, once a shape is assumed. Third, data spanning a wide enough range of q can be inverted directly to ρ(r), which is what produced the charge-density curves in every textbook figure. A consistency test comes free with the first Born approximation: F depends on q alone, so 250 MeV data and 500 MeV data must land on one curve when plotted against q rather than against θ. Where they separate, Born is failing and the extracted radius is not trustworthy.

4Spacing gives the radius, envelope gives the surface

For a uniform sphere F(x) = 3(sin x − x cos x)/x³ with x = qR/ħ, vanishing at x = 4.493, 7.725 and 10.90, the roots of tan x = x — the three-dimensional counterpart of a circular aperture's dark rings, and for the same diffraction reason. One minimum angle therefore fixes R, and the spacing of the later minima tests whether a sphere was a fair description. It is not quite. Real data show the maxima between minima dying away much faster than a sphere predicts, because it is a sharp edge that keeps a diffraction pattern alive. Softening that edge over a diffuseness a multiplies the envelope of F by about (πqa/ħ)/sinh(πqa/ħ), which dies away as exp(−πqa/ħ) at large q; at a = 0.55 fm and q/ħ = 2 fm⁻¹ that is 0.22, so the maxima of the measured |F|² sit some twenty times below a sharp sphere's. Spacing measures the radius; the rate of decay measures the surface. Two independent numbers out of one pattern is exactly why a two-parameter profile is enough.

5Saturation: 0.138 and 0.160 are both A-independent

Fits return the two-parameter Fermi (Woods-Saxon) form ρ(r) = ρ₀/[1 + exp((r − c)/a)] with a ≈ 0.55 fm for essentially everything, so the 90%-to-10% surface thickness t = 4a ln3 = 2.4 fm hardly moves from ¹⁶O to ²³⁸U while c grows as A¹⁄³. Two densities fall out of that, and they are not the same. Normalising, A = (4π/3)ρ₀c³[1 + π²(a/c)²], which for ²⁰⁸Pb with c = 6.62 fm and a = 0.55 fm gives ρ₀ = 0.160 nucleons fm⁻³ — the interior value, with light nuclei within about 10% of it. Separately, the equivalent uniform sphere R = √(5/3)√⟨r²⟩ follows R ≈ 1.2A¹⁄³ fm, a fit good to a few percent above A ≈ 90 that drifts to 1.39A¹⁄³ at carbon; its mean density is A/[(4π/3)(1.2)³A] = 0.138 nucleons fm⁻³, with A cancelling identically. Both constants say one thing — nucleons pack at a fixed density because the force saturates — and they differ by 15% only because the equivalent radius is pushed outward to cover a diffuse tail.

6What the Born approximation cannot hand you

Three limits are built into the method. First, F is the transform of the charge density, so electron scattering locates protons; the neutron distribution needs another probe, and the parity-violating PREX measurement puts the ²⁰⁸Pb neutron rms radius about 0.28 fm outside the proton one. Second, what is measured is ρch, which already has the proton's own 0.84 fm charge radius folded into it; that has to be unfolded before the result is a point-nucleon distribution. Third, one-photon exchange assumes Zα ≪ 1, and at Z = 82 it is 0.60: the incoming and outgoing waves are distorted by the nuclear Coulomb field, and the measured minima are shifted inward by several percent and partly filled in. A defensible heavy-nucleus analysis is an iterated phase-shift calculation against a trial ρ(r), not a single Fourier transform. Quoting a lead radius from a plane-wave fit is a systematic error, not a rounding one.

2

Key equations

Read the units, assumptions and limits before substituting.

Momentum transfer sets the resolution

q/ħ = 2E sin(θ/2) / ħc, with ħc = 197.3 MeV fm

The probe resolves detail of order ħ/q, so 0.634 fm⁻¹ sees 1.6 fm — a fermi needs hundreds of MeV.

E in MeV for E ≫ mₑc², recoil neglected; q/ħ in fm⁻¹. 250 MeV at 29° gives 0.634 fm⁻¹.

Mott cross-section for a point nucleus

(dσ/dΩ)M = (Zαħc)² cos²(θ/2) / [4E² sin⁴(θ/2) (1 + (2E/Mc²) sin²(θ/2))]

This is the denominator you divide the data by. It is smooth in θ, so every feature that survives belongs to the nucleus.

αħc = 1.440 MeV fm; cos²(θ/2) is electron spin, the bracket is target recoil, Mc² the nuclear rest energy.

Form factor from the measured ratio

|F(q)|² = (dσ/dΩ)ₑₓₚ / (dσ/dΩ)M, F(q) = (1/Ze) ∫ ρch(r) exp(iq⋅r/ħ) d³r

One curve replaces every beam energy: 250 and 500 MeV data must fall on the same F plotted against q.

ρch in e fm⁻³; F is dimensionless with F(0) = 1, so Z cancels and F carries shape only.

Spherical density and the small-q slope

F(q) = (4πħ/Zeq) ∫ ρch(r) sin(qr/ħ) r dr ≈ 1 − q²⟨r²⟩/6ħ²

The one model-independent number in the subject: ¹²C returns √⟨r²⟩ = 2.47 fm from that slope alone.

The expansion holds while qR/ħ ≲ 1; the slope of F against q² as q → 0 assumes no profile at all.

Uniform sphere and its diffraction zeros

F(x) = 3(sin x − x cos x)/x³, x = qR/ħ, zeros at x = 4.493, 7.725, 10.90

One minimum angle fixes R with no curve fitting; the spacing of later minima then tests the shape you assumed.

R is the equivalent uniform radius, R = √(5/3) √⟨r²⟩. Measured minima are filled in, never true zeros.

Fermi profile, radius, and saturation

ρ(r) = ρ₀ / [1 + exp((r − c)/a)], R = 1.2A¹⁄³ fm, t = 4a ln3

Only c moves with A. A fixed interior density is saturation, and it is the assumption the liquid-drop model rests on.

c ≈ 1.1A¹⁄³ fm, a ≈ 0.55 fm, so t ≈ 2.4 fm; ρ₀ ≈ 0.16 nucleons fm⁻³ and A cancels.
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

208
12 to 238; steps of 2.
6.60 fm
2 to 8 fm; steps of 0.05.
0.55 fm
0.3 to 0.9 fm; steps of 0.05.

Try this. Hold a = 0.55 fm, set A = 208, c = 6.60 fm: the plateau lands on 0.162 fm⁻³. Pull c to 5.50 fm and it leaps to 0.272, denser than any nucleus. Now try A = 16, c = 2.60 fm, a = 0.50 fm: c is 60% smaller yet the plateau holds at 0.159. Saturation — c tracks A¹⁄³, the interior does not.

Live results

INTERIOR DENSITY
0.162 fm⁻³
c / A¹⁄³
1.114 fm
SURFACE t = 4a ln3
2.42 fm
UNIFORM R = 1.2A¹⁄³
7.11 fm

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Physics IV modelFermi profile ρ(r) = ρ₀/[1 + exp((r − c)/a)], the measured charge shape scaled to A nucleons, r running 0 to 10 fm along the base. The circle marks the half-density point r = c, the arrow spans the 90-to-10% surface t = 2.42 fm, the upper dashed line is 0.16 fm⁻³ saturation, and the dashed step is the uniform sphere R = 7.11 fm. Interior 0.162 nucleons fm⁻³.t = 2.42 fmρ(r) = ρ₀ / [1 + exp((r − c)/a)]A = 208 · c = 6.60 fm · a = 0.55 fmdashed step: R = 1.2 A¹⁄³mean ρ = 0.138 fm⁻³ for all Anucleon density / fm⁻³saturation ≈ 0.16 fm⁻³r = 0r = 10 fm
Fermi profile ρ(r) = ρ₀/[1 + exp((r − c)/a)], the measured charge shape scaled to A nucleons, r running 0 to 10 fm along the base. The circle marks the half-density point r = c, the arrow spans the 90-to-10% surface t = 2.42 fm, the upper dashed line is 0.16 fm⁻³ saturation, and the dashed step is the uniform sphere R = 7.11 fm. Interior 0.162 nucleons fm⁻³.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Mass number A: 208; Half-density radius c: 6.60 fm; Surface diffuseness a: 0.55 fm

Baseline 208; choose 12 to 238, in steps of 2.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

The commonest error is to read R = 1.2A¹⁄³ fm as an edge. It is not: it is the radius of the uniform sphere with the same mean-square radius as a distribution whose charge density slides from 90% to 10% of its central value over t ≈ 2.4 fm — a surface almost as thick as the radius of ¹⁶O. Two things follow that students routinely miss. The mean density of that equivalent sphere, 0.138 nucleons fm⁻³, is not the interior density, which is nearer 0.160; the equivalent radius is inflated outward to cover the tail. And the surface is what damps the diffraction envelope, so two nuclei with the same c but different a give minima in nearly the same places with very different maxima between them. Spacing measures c, decay measures a. Quote a radius without saying which one you mean and the two get silently swapped.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

Use R = 1.2A¹⁄³ fm to find the radius of ²⁰⁸Pb and of ¹⁶O, then the mean nucleon density of each, and convert one of them to kg m⁻³.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 4 steps revealed.

MediumExample 2

A 250 MeV electron beam scatters elastically from ¹²C. At θ = 10.0° the measured cross-section is 0.905 of the Mott value. Find the rms charge radius from that single point, and say which way the truncated expansion pushes the answer.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

HardExample 3

Elastic scattering of 250 MeV electrons from ²⁰⁸Pb puts the first minimum of |F|² at θ = 29.0°. Treat the nucleus as a uniform sphere to extract R, check it against 1.2A¹⁄³, then compare that sphere's mean density with the interior density of the fitted Fermi profile c = 6.62 fm, a = 0.55 fm.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

6

Quick check

Select an answer and explain it before checking.

At one fixed beam energy, elastic electron scattering puts the first diffraction minimum at a larger angle for ¹⁶O than for ²⁰⁸Pb. What does that tell you?
7

Review & continue

The nuclear charge form factor F(q) is the measured elastic electron-scattering cross-section divided by the point-charge Mott cross-section, and in the Born approximation it is the Fourier transform of the nuclear charge distribution, normalised so that F(0) = 1.

  • Explain the relationship without looking at the equation.
  • Check a sign, a unit and a model assumption in your working.
  • Return to the lab or worked example if a step is still unclear.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.