= Rutherford differential cross-section
{c}
{title2=$\frac{d\sigma}{d\Omega}=\frac{\kappa^2}{16E^2}\csc^4\frac\theta2$}
For an incident beam uniform across its relevant <impact parameters>, an incident <annulus> has <area> $d\sigma=2\pi b|db|$. Rotational symmetry maps it to outgoing <solid angle> $d\Omega=2\pi\sin\theta\,d\theta$. The <central-potential classical differential cross-section> and the <Rutherford impact-parameter relation> therefore give
$$
\left|\frac{db}{d\theta}\right|=\frac a2\csc^2\frac\theta2,\qquad
\frac{d\sigma}{d\Omega}
=\frac{b}{\sin\theta}\left|\frac{db}{d\theta}\right|
=\frac{a^2}{4}\csc^4\frac\theta2
=\boxed{\frac{\kappa^2}{16E^2}\csc^4\frac\theta2}.
$$
This is a <differential scattering cross-section>, with units of <area> per <steradian>, conventionally $\mathrm{m^2\,sr^{-1}}$, not a detector count and not an <area> per degree. It applies for $0<\theta<\pi$, with a finite limiting value at $\pi$. Doubling incident <kinetic energy> divides it by four at fixed <angle>; doubling either <electric charge> number multiplies it by four.
A useful independent check integrates over all azimuths between two positive <angles>. For $0<\theta_1<\theta_2\leq\pi$,
$$
\sigma_{[\theta_1,\theta_2]}
=\int_{\theta_1}^{\theta_2}\frac{d\sigma}{d\Omega}\,2\pi\sin\theta\,d\theta
=\pi a^2\left(\cot^2\frac{\theta_1}{2}-\cot^2\frac{\theta_2}{2}\right).
$$
This equals the incident <annulus> <area> $\pi(b_1^2-b_2^2)$ directly. A detector covering only a small patch at one <angle> does not collect this full azimuthal band.
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