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## Homework Statement

An alpha particle is a nucleus of helium. It has twice the charge and four times the mass of the proton. When they were very far away from each other, but headed toward directly each other, a proton and an alpha particle each had an initial speed of 3.6×10−3

*c*, where

*c*is the speed of light. What is their distance of closest approach? There are

*two*conserved quantities. Make use of both of them. (

*c*= 3.00 × 10^8 m/s,

*k*= 1/4

*πε*0 = 8.99 × 10^9 N · m2/C2,

*e*= 1.60 × 10^-19 C,

*m*proton = 1.67 x 10^-27

*kg*)

## Homework Equations

1/2 m v^2 + 1/2 m v^2 = q(kq / r)

## The Attempt at a Solution

I used the above equation to get 9.46 * 10^-14 as the radius, but that is not the correct answer. I made the assumption that the electric potential far away is 0 and that when the distance is closest they have no velocity.