A Proton is fired from far away towards the nuclues of a mercury Atom

[Blitzp22]

Homework Statement

A proton is fired from far away towards the nucleus of a mercury atom. Mercury is element number 80, and the diameter of the nucleus is 14.0 fm. If the proton is fired at a speed of 32100000 m/s, what is its closest approach to the surface of the nucleus (in fm)? Assume that the nucleus remains at rest.

Homework Equations

U=kq1q2/r
kinematic equation
K = 1/2 m v^2

The Attempt at a Solution

So I got this far
1/2(1.67E-27)(3.21E7)^2=(9E9)(q1)(q2)/7fm

but I don't know what to do about the q1 and q2 to get my final answer. Hmmmm toughy

 

[Delphi51]

The q's are the charges on the nucleii. You can look them up on a periodic table. Remember, the charge shown on the table is in terms of electron or proton charges which you must convert into the standard unit of Coulombs. No doubt you have the charge of an electron in one of your lists.

Your calc is right, but difficult for me or your teacher to understand. Maybe you, too. You really should start with a general statement that shows your method of attack. Something like this:

Kinetic energy far out is entirely converted to potential energy at the closest approach
or KE far out = PE @ turnaround

 

[richnut27]

yeah but in response to Delphi51's comments, then what are searching for? you would've filled all the unknowns in the equation

 

[Delphi51]

Welcome to PF, richnut.
You would still be looking for the separation distance, which is the r in the original post. It just makes more sense to write:
KE far out = PE @ turnaround
1/2 m v^2 = kq1q2/r
The first step clarifies what the second means. Then you go on to solve for r.