# Energy conservation in an alpha-scattering experiment

Gold Member

## Homework Statement

In scattering experiment, find distance of closest approach if a 6 MeV alpha particle is used

2. The attempt at a solution
initially KE of alpha particle is 6 x 10^6 x e joules and 0 PE, finally its PE is kq1q2/d, k=9 x 10^9, q1=4e, q2=Ze=79e (assuming gold), d is distance of closest approach, e is charge on electron. Plugging in the values gives an incorrect answer. The correct answer (far from it) is 2 x 10^(-14) m. Is there a mistake I'm making?

Mentor
Which answer did you get and can you show the steps?
At closest approach both the alpha particle and the gold nucleus will move a bit (conservation of momentum!) but this is a small effect.

Cutter Ketch
Your approach is correct and I get your “correct” answer. I see only 1 definite mistake, the charge of an alpha particle isn’t 4. However I suspect you big problem is units. You have k in SI units and PE in MeV and two charges in electron charges. I’m sure you converted some of them, but did you get all of them? Is your answer off by 10^19? If so, you missed one.

Gold Member
Why am I not getting any alerts to replies to all my posts?
Anyway, I'll be back in a while and show my work

Gold Member
Your approach is correct and I get your “correct” answer. I see only 1 definite mistake, the charge of an alpha particle isn’t 4. However I suspect you big problem is units. You have k in SI units and PE in MeV and two charges in electron charges. I’m sure you converted some of them, but did you get all of them? Is your answer off by 10^19? If so, you missed one.
Right, its 2e. Thank you for pointing it out, I'll try that again

Gold Member
Which answer did you get and can you show the steps?
At closest approach both the alpha particle and the gold nucleus will move a bit (conservation of momentum!) but this is a small effect.
After correcting the charge, I'll redo my calculations and post here within the next 24 hrs