Max Scattering Angle from Thompson Plum Pudding Model

[James Brady]

1. Problem Statement:

Consider the scattering of an alpha particle from the positively charged part of the Thomson plum-pudding model. Let the kinetic energy of the α particle be K (nonrelativistic) and let the atomic radius be R.

• (a) Assuming that the maximum transverse Coulomb force acts on the α particle for a time Δt = 2R/v (where v is the initial speed of the α particle), show that the largest scattering angle we can expect from a single atom is
  
  $$\theta = \frac{2z_2e^2}{4\pi\epsilon_0KR}$$

(b) Evaluate θ for an 8.0-MeV α particle scattering from a gold atom of radius 0.135 nm.

Homework Equations

$$ΔP = \int F_{ΔP}dt$$[/B]

The Attempt at a Solution

By writing out the units for the above equation, I get that the units all cancel each other out, so it it dimensionless like an angle would be... But I'm really not seeing how to make the connection between what looks like some sort of Coulomb's Law application and angle is made.

Up to this point, when I've solved for angles, there has always been some sort of arctan or arcsine function. I understand that sometimes, when θ is very small, sin(θ) ≈ tan(θ) ≈ θ. But we are talking about large angles in this example. So I am pretty much lost. Any help to get me started on this would be appreciated.

 

[TSny]

Small angle approximations should be valid here. In the plum pudding model, an alpha particle with kinetic energy of the order of an Mev would not be expected to deflect much.

It appears that you are asked to construct a simple, approximate (hand-waving) derivation of θ. Assume the transverse force on the alpha particle is constant for the time Δt given and that the magnitude of the force is equal to the maximum force that the particle could experience from the sphere of positive charge.

 

[James Brady]

Alright man, I figured maybe they were expecting small-angle approximation, because the only other technique I know is a polynomial expansion which didn't seem valid in this case. Thanks for the help.