= Uniform-sphere nuclear charge form factor
{title2=$F(q)=3[\sin(qR)-qR\cos(qR)]/(qR)^3$}
Consider an ideal <nuclear charge distribution> that is constant inside a ball of radius $R>0$ and zero outside:
$$
\rho_{\mathrm{ch}}(r)=\begin{cases}3Ze/(4\pi R^3),&0\le r\le R,\\0,&r>R.\end{cases}
$$
This is a deliberately sharp-edged model. Substitution into the spherical <nuclear charge form factor> gives, for $q\ne0$,
$$
F(q)=\frac3{R^3}\int_0^R r^2\frac{\sin(qr)}{qr}\,dr
=\frac3{qR^3}\left[-\frac{r\cos(qr)}q+\frac{\sin(qr)}{q^2}\right]_0^R
=3\frac{\sin(qR)-qR\cos(qR)}{(qR)^3}.
$$
At $q=0$ take the continuous limit $F(0)=1$. Direct radial integration gives $\langle r^2\rangle_{\mathrm{ch}}=3R^2/5$, and the <Taylor series> provides an independent check:
$$
F(q)=1-\frac{(qR)^2}{10}+\frac{(qR)^4}{280}+O((qR)^6).
$$
The first positive zero solves $\sin x-x\cos x=0$, equivalently $\tan x=x$, with $x=qR\simeq4.49340946$. 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> is
$$
\rho_G(r)=\frac{Ze}{(2\pi\sigma^2)^{3/2}}e^{-r^2/(2\sigma^2)},\qquad
\sigma^2=R^2/5,\qquad F_G(q)=e^{-q^2R^2/10}.
$$
The <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 $F\simeq0.653097$ and the <Gaussian function> gives $F_G\simeq0.670320$. Measuring only a <nuclear charge radius> cannot distinguish these two <nuclear charge distributions>.
\Image[/nuclear-charge-sphere-and-gaussian-form-factors.png]
{title=Equal-radius charge distributions can have very different diffraction patterns}
{description=Original calculated model comparison; no experimental points or fitted nuclear parameters. Left: the signed <nuclear charge form factor>. Right: its squared magnitude, the charge factor in the stated <Born approximation>. The models share $\langle r^2\rangle=3R^2/5$; the sharp sphere has zeros while the smooth <Gaussian function> does not.}
As a chosen scale illustration, take $R=3.00\,\mathrm{fm}$. The predicted first zero is $q=1.49780\,\mathrm{fm}^{-1}$, or $\hbar cq\simeq295.557\,\mathrm{MeV}$ using $\hbar c\simeq197.327\,\mathrm{MeV}\,\mathrm{fm}$. With a $200\,\mathrm{MeV}$ ultrarelativistic <Electron> beam and negligible <nuclear recoil>, it occurs at $\theta\simeq95.27^\circ$. These are consequences of chosen model parameters, not measured properties of carbon or a claim that all <atomic nuclei> are uniform balls.
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