Nuclear elastic electron scattering cross-section ($d\sigma /d\Omega \simeq (d\sigma /d\Omega {)}_{\mathrm{point}}|F(q){|}^2$)

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 $|F(q){|}^2$:Here $E\gg m_e{c}^2$ 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.