Question #270685

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


1
Expert's answer
2021-11-24T11:59:33-0500

By conservation of energy:


E=kq1q2r, r=kq1q2E=114 fm.E=k\frac{q_1q_2}{r},\\\space\\ r=\frac{kq_1q_2}{E}=114\text{ fm}.


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