What is the Kinetic Energy of Alpha Particles Scattered by an Aluminum Plate?

AI Thread Summary
The discussion revolves around calculating the kinetic energy of alpha particles scattered by a 50μm thick aluminum plate. Given the density of aluminum and the number of scattered particles detected at specific angles, the conversation highlights the use of scattering equations, including differential cross-section and solid angle calculations. The participants emphasize the importance of understanding the Coulomb interaction in this context. The problem requires integrating the differential cross-section over the specified angles to find the total scattering rate. Overall, the focus is on applying the relevant physics equations to derive the kinetic energy of the alpha particles.
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Homework Statement


Onto an Aluminium 50μm thick plate, we send a beam of alpha particles with unknown kinetic energy. the cross section of the beam is 2cm2, density 1013 / scm2 . Whats the kinetic energy of the particles if every second we sense 105 scattered particles between the angles 40° and 41°? Take into count only the Coulumb interaction and the density of Aluminium is 1850 kg/m3

Homework Equations

The Attempt at a Solution

 
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You have to show your attempt first! :)
 
first of all my understanding of scattering is shamefully low so please keep in mind...

σdif. = (z12*z22) / 16Tα2 * 1/ sin4(ϑ/2)
dΩ= 2πsinϑdϑ
σR=∫σdif dϑ σdif = dσ/dΩ and the integral borders are 40° and 41°

now those 105 = ndσr ? where d is the thickness and n = densiti times NA/M

?
 

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