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

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SUMMARY

The discussion focuses on calculating the kinetic energy of alpha particles scattered by a 50μm thick aluminum plate. The scattering is analyzed using the differential cross-section formula, σdif, which incorporates the charge of the alpha particles and the aluminum target. Given a detection rate of 105 scattered particles between 40° and 41°, the relationship between the number of particles, density, and thickness is established to derive the kinetic energy. The density of aluminum is specified as 1850 kg/m³, which is crucial for the calculations.

PREREQUISITES
  • Understanding of Coulomb interactions in particle physics
  • Familiarity with differential cross-section calculations
  • Knowledge of scattering theory and its applications
  • Basic proficiency in algebra and integration techniques
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  • Study the derivation of the differential cross-section formula, σdif, in particle scattering
  • Learn about the principles of scattering theory and its relevance in nuclear physics
  • Explore the concept of particle density and its role in scattering experiments
  • Investigate the relationship between kinetic energy and scattering angles in particle physics
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This discussion is beneficial for physics students, researchers in nuclear physics, and anyone interested in understanding particle scattering phenomena and calculations related to kinetic energy in experimental setups.

jinn
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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

?
 

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