Scattering from a Hard Sphere

Assume a low energy beam is incident upon a small, hard sphere of radius \bgroup\color{black}$r_0$\egroup. We will assume that \bgroup\color{black}$\hbar kr_0<\hbar$\egroup so that only the \bgroup\color{black}$\ell=0$\egroup partial wave is significantly affected by the sphere. As with the particle in a box, the boundary condition on a hard surface is that the wavefunction is zero. Outside the sphere, the potential is zero and the wavefunction solution will have reached its form for large \bgroup\color{black}$r$\egroup. So we set

\left(e^{-i(kr_0-\ell\pi/2)}-e^{2i\delta_\ell(k)}e^{i(kr_0-\el...\vert^2 \\
{d\sigma\over d\Omega}={\sin^2(kr_0)\over k^2} \\

For very low energy, \bgroup\color{black}$kr_0«1$\egroup and

\begin{displaymath}\bgroup\color{black} {d\sigma\over d\Omega}\approx {(kr_0)^2\over k^2}=r_0^2 \egroup\end{displaymath}

The total cross section is then \bgroup\color{black}$\sigma=4\pi r_0^2$\egroup which is 4 times the area of the hard sphere.

Jim Branson 2013-04-22