Minimum separation, impact parameter, and scattering angle

[Jadehaan]

Homework Statement

Alpha particles of kinetic energy 9.6MeV are incident on a silver foil of thickness 7.0 micro meters. For a certain value of the impact parameter, the alpha particles lose exactly half their incident kinetic energy when they reach their minimum separation from the nucleus. Find the minimum separation, the impact parameter, and the scattering angle.

K=9.6 MeV, t=7.0 x 10^(-6) meters, density; n=5.68 x 10^28 atoms/m^3
z=47

Homework Equations

b=(zZ/2K)(1.44MeV nm)cot(\theta/2)

The Attempt at a Solution

I can not figure out where to start.



Thanks for any help,
Jim

 

[gabbagabbahey]

Use conservation of energy. If the alpha particle has initial kinetic energy of K, what is its kinetic energy at minimum separation from a silver nucleus? What is its potential energy? What does that tell you?