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## Homework Statement

An alpha particle is accelerated from rest through a potential dierence of 20 kV. It travels directly towards a stationary Beryllium nucleus (4 protons, 5 neutrons). Calculate the distance of closest approach.

The problem definitely wants a solution that takes into account the recoil of the stationary nucleus on approach.

## Homework Equations

V

_{CM}=m1v1+m2v2/m1+m2

U=Q1Q2/4πε

_{0}d

T=mv

^{2}/2

## The Attempt at a Solution

Let the oncoming alpha particle have speed v. V

_{CM}=4v/13. In the CM frame, we have a particle of mass 4m, charge 2e, speed 9v/13 heading towards on of mass 9m, charge 4e, speed -4v/13.

Initially, T=18mv

^{2}/3. This all becomes PE which will be 2e

^{2}/πε

_{0}d for closest approach d. Equating given d=e

^{2}/3πε

_{0}mv

^{2}. T=qV for a p.d V so v

^{2}=2qV/m. Then d=e/3πε

_{0}V. Plugging in the numbers given d=96fm.

I don't have a solution so would appreciate somebody having a check. I'm unsure about my approach because it assumes that the initial scenario (i.e as soon as the alpha particle has finished being accelerated through the p.d) has the PE as zero, which isn't realistic and the question doesn't indicate this is valid. Thanks.