Impact Parameter of Alpha Particle Rutherford

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Homework Help Overview

The discussion revolves around the calculation of the impact parameter of an alpha particle in a Rutherford scattering experiment, where a beam of alpha particles bombards a thin gold foil target. The problem involves determining the impact parameter based on the kinetic energy of the alpha particles and the angle at which they are detected.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formula for calculating the impact parameter and the necessary values for the variables involved. There is an exploration of the numerical values used in the calculations and their implications on the final result.

Discussion Status

Some participants have provided feedback on the calculations, suggesting that the order of magnitude for certain parameters is appropriate. However, there is an indication that the final answer may not be correct, prompting further examination of the numerical values used in the calculations.

Contextual Notes

Participants are working under the constraints of using specific energy units and constants, and there is a focus on ensuring that the correct numerical values are applied in the calculations.

PhatPartie
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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
 
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The order of magnitude for r_{min} is good,indeed there is fermi/femtometer.Compute "b" now...

Daniel.
 
i then get an answe of 3.52 *10^-15 m ...but this is not the correct answer ... i don't know what i did wrong
 
Devious...Make sure you insert correct numerical values.The energy (which is supposed to be in the denominator) is 4.57MeV=4.57*1.6*10^{6}*10^{-19}J...

Daniel.
 

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