University Physics V · Nuclear Physics · 14.2
Nuclear Radius from the Charge Form Factor
Electron scattering is the cleanest ruler physics has for the nucleus: a structureless probe, an interaction we can write down exactly, first-order perturbation theory good enough to use. This lesson is about what that ruler hands back — a squared modulus over a finite range of q — and how much of the density survives the trip home.
Build the model
Connect the measurement to the mechanism.
Accelerate electrons until their reduced wavelength is smaller than a nucleus and the first Born approximation applies: the amplitude is the Fourier transform of the charge distribution, so the measured cross-section is the point-charge Mott cross-section times |F(q)|², with F the normalised transform sampled at momentum transfer q = 2k sin(θ/2). Sweep the angle, sweep q, and you are reading off the transform of ρch. That is the promise.
The bill arrives in three parts. A detector counts rates, so what is recorded is |F|² and never F — for a real spherical density F is real and flips sign at every diffraction zero, and those signs are not measured. The beam energy caps q at qₘₐₓ = 2E/ħc, so the transform is truncated and structure finer than about π/qₘₐₓ is simply absent.
And the Born approximation itself fails for heavy nuclei, where the electron is pulled tens of MeV deeper before it scatters. What comes out is therefore never the density but a fitted shape — a two-parameter Fermi profile, or a Fourier–Bessel series carrying a stated incompleteness band — from which a half-density radius near 1.1 A¹⁄³ fm, a surface thickness near 2.4 fm and a saturated interior near 0.16 nucleons per fm³ follow. Only one length escapes the convention: the rms radius, which the slope of F at q → 0 pins down whatever shape ρ has.
- Simple definition
- The charge form factor F(q) is the normalised Fourier transform of a nucleus's charge density, defined so that the elastic electron cross-section is the point-charge Mott cross-section multiplied by |F(q)|², with F(0) = 1.
- Example
- For ¹²C at 200 MeV and θ = 20°, q = 2E sin(θ/2)/ħc = 0.352 fm⁻¹, and a measured ratio to Mott of 0.778 gives F = 0.882 — a 12% dip below unity that the low-q expansion turns into a radius of about 2.4 fm.
Divide the data by Mott and every nuclear property left in the plot sits inside |F|².
Mott carries the point charge Ze, the spin factor cos²(θ/2) and recoil, in mb/sr
200 MeV at 20° is q = 0.352 fm⁻¹; the resolvable detail is δr ≈ π/qₘₐₓ, 1.55 fm at 200 MeV.
ħc = 197.33 MeV fm, so E in MeV gives q in fm⁻¹, and qₘₐₓ = 2E/ħc at backscattering
One sine transform in one variable — invertible in principle, if you had F over all q with its signs.
ρch in e fm⁻³ normalised to ∫ρch d³r = Ze, so F(0) = 1, with q in fm⁻¹
The one model-independent length in the topic: no shape is assumed to get it.
⟨r²⟩ in fm², and the series is usable only while q²⟨r²⟩ stays well below 6
The minima locate R from angles alone — no absolute normalisation of the beam is needed.
⟨r²⟩ = 3R²/5, and the convention R = 1.2 A¹⁄³ fm is tuned to reproduce measured ⟨r²⟩
Two numbers fit almost every measured nucleus — and hide the model dependence inside a choice of shape.
c ≈ 1.1 A¹⁄³ fm, a ≈ 0.55 fm, t ≈ 2.4 fm, ρ₀ ≈ 0.16 nucleons fm⁻³
Divide by Mott before you interpret anything
The electron is the right probe because it has no strong interaction and no structure of its own, so the whole vertex is electromagnetic and one order of perturbation theory is enough. In the first Born approximation the amplitude is the Fourier transform of the potential, so for a static charge distribution the cross-section factorises: dσ/dΩ = (dσ/dΩ)Mott |F(q)|². The Mott factor is the point-charge answer — Rutherford, with the relativistic spin factor cos²(θ/2) and a recoil denominator — and it holds no nuclear structure whatever. Everything the nucleus contributes lives in |F(q)|². So the first operation on any data set is a division: plot measured cross-section over Mott and you have |F|² on a scale where F(0) = 1 by construction. Nothing has yet been assumed about ρ beyond a static, spherical source.
The variable is q, and the beam energy caps it
Angle is what the apparatus turns; q is what the physics depends on. For an ultrarelativistic electron k = E/ħc, and elastic scattering through θ transfers q = 2k sin(θ/2), so with ħc = 197.33 MeV fm a 200 MeV beam at 20° gives q = 0.352 fm⁻¹. Two consequences follow. The reach is largest at backscattering, qₘₐₓ = 2E/ħc: 2.03 fm⁻¹ at 200 MeV, 5.07 fm⁻¹ at 500 MeV. And a truncated transform cannot resolve structure finer than roughly δr ≈ π/qₘₐₓ — 1.55 fm at 200 MeV, 0.62 fm at 500 MeV. That is why radius work is done at several hundred MeV and above: run it lower and you are measuring a blurred nucleus, and no fitting procedure invents detail the kinematics never delivered.
Two reads of a radius, answering different questions
Expand the transform at small q: F(q) = 1 − q²⟨r²⟩/6 + q⁴⟨r⁴⟩/120 − …, a series in q², so ⟨r²⟩ = −6 dF/d(q²) at q = 0 is model-independent — the slope is the radius whatever shape ρ has. That is the number quoted as the rms charge radius: 2.470 fm for ¹²C, 5.501 fm for ²⁰⁸Pb, and it carries no convention at all. The second read is the diffraction pattern. A uniform sphere has F = 3[sin(qR) − qR cos(qR)]/(qR)³, vanishing at qR = 4.4934, so the first minimum gives R = 4.4934/q₁ from angles alone. For ²⁰⁸Pb at 250 MeV that minimum sits at θ = 30.5°, q₁ = 0.667 fm⁻¹, hence R = 6.74 fm and √(3/5)R = 5.22 fm — 5% short of 5.50. The reads disagree because the second assumed a shape and the first did not.
The phase problem, and what a fit is really doing
A detector counts rates, so the measurement is |F(q)|², while for a real spherical density F is real and changes sign at every zero. Those signs are supplied by the model, not by the experiment. Add truncation at qₘₐₓ and the inversion is ill-posed twice over: densities agreeing below qₘₐₓ and differing above it fit the data identically, and so do densities differing by a sign flip on one lobe. Practice takes one of two routes. Fit a shape with few parameters — the two-parameter Fermi profile — and quote c and a with their covariance, accepting the shape as an assumption. Or expand ρ in a Fourier–Bessel series inside a fixed cutoff radius, fit the coefficients qₘₐₓ supports, and bound the unmeasured tail, quoting the resulting incompleteness error as a band around ρ(r). Both are honest. Neither is an inversion.
Saturation, and why two radii give two densities
Fit the Fermi profile across the chart and c comes out near 1.1 A¹⁄³ fm with a near 0.55 fm, so the 90-to-10 surface thickness t = 4a ln3 ≈ 2.4 fm hardly changes from carbon to lead: nuclear matter saturates, and the interior density is close to A-independent. But the number you quote for that interior depends on which radius convention you used, and the two in common use are different lengths. For ²⁰⁸Pb the Fermi fit c = 6.60 fm, a = 0.55 fm gives ρ₀ = 3A/(4πc³[1 + (πa/c)²]) = 0.162 nucleons fm⁻³. The sharp-sphere convention R = 1.2 A¹⁄³ fm = 7.11 fm reproduces the same rms radius (√0.6 × 7.11 = 5.51 fm) but spreads 208 nucleons through a bigger volume: 0.138 fm⁻³, 15% lower. Each is right inside its own definition; taking c from one and a density from the other is not.
Where plane waves fail: Coulomb distortion
The Born approximation treats both electron waves as plane waves, which needs Zα ≪ 1. For lead Zα = 82/137 = 0.60, and the failure is quantitative, not cosmetic. An electron at the centre of a uniformly charged sphere sits 3Zαħc/2R ≈ 24.9 MeV deeper than at infinity, so its local wavenumber inside exceeds the asymptotic k — a 10% effect at 250 MeV, 5% at 500 MeV. The diffraction minima therefore appear at smaller angles than plane-wave Born predicts, and such a fit reads the radius about 10% high: 0.7 fm on 7.1 fm, against data good to thousandths of a femtometre. Distortion also partly fills the minima, so they stop being true zeros. The cure is to stop approximating — solve the Dirac equation for the electron in the combined Coulomb-plus-nuclear potential, do a phase-shift analysis, and adjust ρ until the computed cross-section matches.
Change one variable at a time
Make the relationship visible.
Drag c and every minimum slides together — their spacing measures the radius and nothing else. Then drag a: the minima stay put while the envelope tilts, which is how surface thickness is read. Pull qₘₐₓ left and count the lobes you lose.
FIRST ZERO q10.681 fm⁻¹
RMS RADIUS (model)5.34 fm
SURFACE 90-10 t2.31 fm
DETAIL LIMIT π/qmax1.05 fm
Live interpretationFIRST ZERO q1: 0.681 fm⁻¹. RMS RADIUS (model): 5.34 fm. SURFACE 90-10 t: 2.31 fm. DETAIL LIMIT π/qmax: 1.05 fm
Catch the common trap
Explain before calculating.
A measurement on ²⁰⁸Pb reaches qₘₐₓ = 2.0 fm⁻¹ and resolves three diffraction minima. The beam energy is doubled, qₘₐₓ reaches 4.0 fm⁻¹, and three more minima appear. What does the extra data most directly buy?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyElastic scattering of 200 MeV electrons from ¹²C at θ = 20° gives a cross-section 0.778 times the Mott value. Take ħc = 197.33 MeV fm. Find the momentum transfer, the form factor, and the rms charge radius from the leading term of the low-q expansion — then say what that truncation costs.
- Kinematics first: k = E/ħc = 200/197.33 = 1.0135 fm⁻¹, so q = 2k sin(θ/2) = 2 × 1.0135 × sin 10° = 2 × 1.0135 × 0.17365 = 0.3520 fm⁻¹, and q² = 0.1239 fm⁻².
- The detector measured a modulus squared. On the first lobe, before any zero, F is positive, so F = √0.778 = 0.8820.
- Keep only the q² term of F = 1 − q²⟨r²⟩/6 + q⁴⟨r⁴⟩/120 − …: ⟨r²⟩ = 6(1 − F)/q² = 6 × 0.1180/0.1239 = 5.71 fm², so √⟨r²⟩ = 2.39 fm.
- The accepted value is 2.470 fm. The dropped q⁴ term is q⁴⟨r⁴⟩/120 = 0.01535 × 62/120 = 0.0079 with ⟨r⁴⟩ ≈ 62 fm⁴ — only 0.9% of F, but 6.7% of (1 − F), and (1 − F) is the entire signal.
- Restore it: ⟨r²⟩ = 6(1 − F + 0.0079)/q² = 6.10 fm², √⟨r²⟩ = 2.47 fm. So the fix is smaller q, not a longer series — fit F against q² over the lowest angles and take the slope at q → 0, where ⟨r²⟩ = −6 dF/d(q²) is exact.
Answerq = 0.352 fm⁻¹ and F = 0.882. The one-point leading-order read gives √⟨r²⟩ = 2.39 fm, 3.2% below the accepted 2.470 fm, because the q⁴ term is 6.7% of the signal (1 − F) at this q.
MediumElastic scattering of 250 MeV electrons from ²⁰⁸Pb puts the first minimum of |F(q)|² at θ = 30.5°. Read a sharp-sphere radius from it and convert to an rms radius. Compare with the accepted 5.501 fm and with the convention R = 1.2 A¹⁄³ fm, and decide whether one sharp sphere can satisfy both.
- k = 250/197.33 = 1.2669 fm⁻¹ and sin(15.25°) = 0.26303, so the minimum sits at q₁ = 2 × 1.2669 × 0.26303 = 0.6665 fm⁻¹.
- The uniform sphere's F = 3[sin(qR) − qR cos(qR)]/(qR)³ first vanishes where tan(qR) = qR, at qR = 4.4934. So R = 4.4934/0.6665 = 6.742 fm.
- For that sphere √⟨r²⟩ = √(3/5) R = 0.7746 × 6.742 = 5.222 fm, which is 5.1% below the accepted 5.501 fm.
- The convention R = 1.2 × 208¹⁄³ = 1.2 × 5.925 = 7.110 fm does reproduce the rms radius: 0.7746 × 7.110 = 5.507 fm. But a sharp sphere that large would put its first minimum at q = 4.4934/7.110 = 0.632 fm⁻¹, that is θ = 28.9°, a degree and a half from where it is.
- No single sharp sphere fits both facts, because the real surface is diffuse. A two-parameter Fermi profile with c = 6.60 fm and a = 0.55 fm delivers both at once: first zero at 0.667 fm⁻¹ and √⟨r²⟩ = 5.51 fm.
Answerq₁ = 0.667 fm⁻¹, R = 6.74 fm, √⟨r²⟩ = 5.22 fm — 5% under the measured 5.50 fm. Fitting the minimum and fitting the rms radius give sharp-sphere radii 5% apart, so the sharp sphere is a convention, not a shape.
HardFor ²⁰⁸Pb a two-parameter Fermi fit gives c = 6.60 fm and a = 0.55 fm. (a) Find the central density ρ₀ in nucleons fm⁻³. (b) Find the density implied instead by R = 1.2 A¹⁄³ fm and account for the difference. (c) Estimate how badly a plane-wave analysis of a 250 MeV measurement would misread the radius, and name the fix.
- (a) Normalising the Fermi profile gives A = (4π/3)ρ₀c³[1 + (πa/c)²] to better than a part in 10⁴ here, so ρ₀ = 3A/(4πc³[1 + (πa/c)²]). With c³ = 287.5 fm³ and (πa/c)² = (1.7279/6.60)² = 0.0685: ρ₀ = 208/(1204.2 × 1.0685) = 208/1286.7 = 0.162 nucleons fm⁻³.
- (b) The sharp sphere has R = 1.2 × 5.925 = 7.110 fm and volume (4π/3)R³ = 1505.6 fm³, so ρ = 208/1505.6 = 0.138 nucleons fm⁻³ — 15% lower. Both radii reproduce √⟨r²⟩ = 5.51 fm, but R is pushed outwards to mimic the diffuse tail, whereas c marks where the density has already fallen to ρ₀/2 and the surface then decays over t = 4a ln3 = 2.42 fm.
- (c) Coulomb distortion: Zα = 82/137.04 = 0.598, and an electron at the centre of the charge sphere is bound 3Zαħc/2R = 3 × 0.598 × 197.33/(2 × 7.110) = 24.9 MeV deeper than at infinity.
- At E = 250 MeV that is a 10.0% boost in the local wavenumber, so the minima move to smaller angles and a plane-wave fit returns R about 10% high — roughly 0.7 fm on 7.1 fm, against data good to thousandths of a femtometre. Doubling the beam to 500 MeV only halves it to 5%.
- The fix is not a correction factor but a different calculation: solve the Dirac equation for the electron in the combined Coulomb-plus-nuclear potential, build the cross-section from a phase-shift sum, and vary ρ until it matches the data.
Answerρ₀ = 0.162 nucleons fm⁻³ from the Fermi fit against 0.138 fm⁻³ from the sharp sphere — a 15% gap between two conventions that agree on ⟨r²⟩. Coulomb distortion at 250 MeV misreads the radius by about 10%, so lead is never analysed with plane waves.