Answer to Question #270685 in Physics for Almas

Question #270685

Determine the distance of closest approach of 1.00-MeV protons incident on gold nuclei.

Expert's answer

By conservation of energy:

"E=k\\frac{q_1q_2}{r},\\\\\\space\\\\\nr=\\frac{kq_1q_2}{E}=114\\text{ fm}."

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