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Jadehaan
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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([tex]\theta[/tex]/2)
The Attempt at a Solution
I can not figure out where to start.
Thanks for any help,
Jim