Maximum energy transefer in COmpton effect - rutherford

Click For Summary
SUMMARY

The discussion focuses on calculating the impact parameter of a 4.57 MeV alpha particle detected at a 52.4-degree angle in a Rutherford scattering experiment. The user employs the formula b = (rmin/2)cot(theta/2), where rmin is calculated using the equation rmin = Z1Z2e^2/4πEok. The user arrives at a value of 3.46 × 10^-15 meters for the impact parameter, using constants such as the charge of the electron and the Coulomb's constant. The calculations are confirmed to be on the right track, indicating a solid understanding of the principles involved.

PREREQUISITES
  • Understanding of Rutherford scattering principles
  • Familiarity with the concept of impact parameter
  • Knowledge of Coulomb's law and electrostatic interactions
  • Basic proficiency in energy conversion between eV and Joules
NEXT STEPS
  • Study the derivation of the Rutherford scattering formula
  • Learn about the significance of the impact parameter in scattering experiments
  • Explore the relationship between kinetic energy and scattering angles
  • Investigate advanced topics in nuclear physics related to alpha particle interactions
USEFUL FOR

Students and researchers in nuclear physics, particularly those studying particle interactions and scattering experiments. This discussion is beneficial for anyone looking to deepen their understanding of the Rutherford experiment and its implications in modern physics.

PhatPartie
Messages
7
Reaction score
0
In the Rutherford scattering experiment a very thin gold foil target is bombarded with a beam of a particles of known kinetic energy. A detector which can be moved on a circle around the target counts the scattered particles. What was the impact parameter of a 4.57 MeV a particle if it was detected at 52.4 degree angle?

Im using b= (rmin/2)cot(theta/2) where rmin = Z1Z2e^2/4piEok

so i have ((2)(79)(1.6*10^-19 c)^2 / (4.57 *10^6 eV)(1.6*10^-9 J/eV) ) * 8.99*10^9 Jm/c ...and got 3.46 *10^-15 is this right so far? ...i got 8.99*10^9 by (1/4piEo) ...thanks
 
Physics news on Phys.org
DO NOT DOUBLE POST!:mad:This is a HW problem anyway.

Daniel.
 
my bad figured more exposure i would get quicker response
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
9K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 21 ·
Replies
21
Views
5K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
3K