Uniform-sphere nuclear charge form factor ($F(q)=3[\sin (qR)-qR\cos (qR)]/(qR{)}^3$)

Consider an ideal nuclear charge distribution that is constant inside a ball of radius $R>0$ and zero outside: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.