Answer: (4)
All of the kinetic energy of incident particles is converted to the potential energy of the system at the distance of closest approach.
PE = KE
\(\frac{p^{2}}{2m} = \frac{1}{4\pi \epsilon _{0}}\times \frac{q_{1}q_{2}}{r}\)Initially
\(\frac{p^{2}}{2m}= \frac{2eq}{4\pi \epsilon _{0}r}\)………………………..(1)Now
\(\frac{2p^{2}}{2m}= \frac{2eq}{4\pi \epsilon _{0}r^{1}}\)………………….(2)Equation (1) divided by (2)
1/4 = r1/r
r1 = r/4
