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

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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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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

?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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