# Homework Help: Colliding Protons

1. Jan 30, 2006

### greggle

Hi Guys,

I was just wondering if you could help me, I seem to be having some problems getting to the correct answer (possibly because of my calculations). If two protons collide head on, both with energy 2keV... show that the closest approach is 7.2 x 10exp-11 cm. Now I know that when the protons are at closest approach then only the potential energy will be considered. Therefore V = q1q2 / r, which for a proton I know is e^2 / r, but for some reason I just cannot get to the right answer and I don't know if it is the way I am using electronvolts and joules. Please help.

Thanks
Greg

2. Jan 30, 2006

### Astronuc

Staff Emeritus
If one is using cgs units, then make sure all units are consistent, e.g. energy = ergs, displacement/distance = cm, charge = esu's.

3. Jan 30, 2006

### greggle

Hi, I'm working in joules and eV...I still can't get it to the right answer.... one question - to change from eV to joules it is simply multiplication of eV and 1.6 x 10exp-19? I haven't used eV for a very long time and can't quite remember how to do it.