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NEET UG PHYSICS MCQ: ATOMS

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An elementary particle of mass mm and charge +e+ e is projected with velocity vv at a much more massive particle of charge Ze,Ze, where Z>0.Z > 0. What is the closest possible approach of the incident particle
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Correct option: A
Ze22πε0mv2\frac{{Z{e^2}}}{{2\pi {\varepsilon _0}m{v^2}}}
Option A (Ze22πε0mv2)\frac{{Z{e^2}}}{{2\pi {\varepsilon _0}m{v^2}}}) is correct because it represents the distance of closest approach derived from equating the initial kinetic energy to the electrostatic potential energy at the closest point.

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