• anonymous
In using classical mechanics to determine the Ruthoford model, why is coulombs law applied where the electron charge in the numerator is squared? If Z= atomic numberand e= electron charge, then why is the (q1q2) part of his equation (Ze^2) instead of (Ze)?
  • Stacey Warren - Expert
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  • katieb
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  • UnkleRhaukus
does the incident electron only get close enough to One of the targets electrons for the repulsive coulombs force to be regarded?
  • anonymous
You're forgetting that Z is just a number, and the charge on the nucleus is Ze (Z times e, the charge on a single proton), not just Z. Hence the numerator of Coulomb's Law, which is q1 q2, is equal to (Ze)(-e) = - Ze^2, where Ze is the charge on the nucleus and -e is the charge on the electron.

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