= Nuclear elastic electron scattering cross-section
{title2=$d\sigma/d\Omega\simeq(d\sigma/d\Omega)_{\mathrm{point}}|F(q)|^2$}
For an ultrarelativistic <Electron> scattered elastically by a static, spherical, spin-zero <atomic nucleus>, a plane-wave <Born approximation> retains a single electromagnetic interaction. The <Fourier transform> of its <Coulomb potential> is proportional to $ZeF(q)/q^2$. Relative to the same point charge, the scattering amplitude is therefore multiplied by $F(q)$ and the <differential scattering cross-section> by $|F(q)|^2$:
$$
\frac{d\sigma}{d\Omega}\simeq
\left(\frac{Z\alpha\hbar c}{2E}\right)^2
\frac{\cos^2(\theta/2)}{\sin^4(\theta/2)}|F(q)|^2,
\qquad q\simeq\frac{2E}{\hbar c}\sin\frac\theta2.
$$
Here $E\gg m_ec^2$ is the incident total <Electron> <energy>, $\theta$ the <scattering angle>, $d\Omega$ a <solid angle> element, and $\alpha$ the <fine-structure constant>. The cosine factor already accounts for the projectile's <spin>; a spinless projectile would give a different point-charge factor. The formula neglects <nuclear recoil>, electron rest-mass corrections, <radiative corrections> and <Coulomb distortion in electron scattering>. The latter can be substantial at large $Z\alpha$: electron waves then require treatment in the target's <Coulomb potential>, rather than undistorted plane waves.
Nonzero target <spin> can introduce magnetic and additional electric contributions, so division by this point-charge expression does not generally isolate one charge form factor for a <proton> or <deuteron>. Even for spin zero, finite <energy> resolution, recoil and <Coulomb distortion in electron scattering> can change a measured minimum. A zero of the uniform-sphere model is a clear illustration of amplitude cancellation, not a universal prediction of a zero experimental counting rate.
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