Asked by anonymous

A moving electron passes near the nucleus of a gold atom, which contains 79 protons and 118 neutrons. At a particular moment the electron is a distance of 7.5 × 10−9 m from the gold nucleus.

What is the magnitude of the force exerted by the gold nucleus on the electron?

Answers

Answered by drwls
I seem to recall answering this already.

F = k*Q1*Q2/R^2 = 79k*e^2/R^2

(Coulomb's law; k = Coulomb's constant)

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Answered by apm
I'm not getting the answer using this formula. Can you explain it a little more. Thanks
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