# Simple Scattering

1. Dec 27, 2003

### yxgao

As every physics major knows, elastic scattering from a hard sphere of radius a can be modeled by the following potential:
U(r) = 0 for r>a,
and infinity for r<a.

Consider the effective potential l^2/(2mr^2) + U(r). And consider scattering with impact parameter b less than a. Where is the turning point of the motion?
Then, how would you use this line of reasoning to show that the differential cross section is Sigma(theta) = a^2/4?

I know how to find the differential cross section geometrically, by finding b(theta), but I don't know how to do this with effective potentials.

This isn't homework. It's a problem from pa.uky.edu/~cvj/PHY404G/ 1998 final exam.

Thanks!!

Last edited: Dec 27, 2003