Calculating Closest Approach of Colliding Protons

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Homework Help Overview

The discussion revolves around calculating the closest approach of colliding protons, specifically focusing on a scenario where two protons collide head-on with an energy of 2 keV each. The original poster expresses difficulty in arriving at the correct answer and raises concerns about their calculations and unit conversions.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to relate potential energy to the closest approach using the formula V = q1q2 / r and expresses confusion regarding the use of electronvolts and joules in their calculations. Other participants question the consistency of units being used and clarify the conversion from eV to joules.

Discussion Status

Participants are actively engaging in clarifying unit conversions and the appropriate application of formulas. Some guidance has been provided regarding the use of potential energy in electronvolts, but there is no explicit consensus on the correct approach yet.

Contextual Notes

There is mention of the need for consistent units, with discussions around using cgs units versus joules and electronvolts. The original poster is uncertain about their understanding of energy conversions and the implications for their calculations.

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

Thanks for your help,

Greg
 
If you use eV for the energy, then you should onlly use one factor of e in the potential energy. That is U=ke/r is the PE for two protons in the unit eV.
 

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