An elementary particle of mass m and charge +e is projected with velocity v at a much more massive particle of charge Ze, where Z > 0. What is the closest possible approach of the incident particle

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$\large \frac{{Ze}}{{4\pi {\varepsilon _0}m{v^2}}}$

$\large \frac{{Ze}}{{8\pi {\varepsilon _0}m{v^2}}}$

$\large \frac{{Z{e^2}}}{{2\pi {\varepsilon _0}m{v^2}}}$

$\large \frac{{Z{e^2}}}{{8\pi {\varepsilon _0}m{v^2}}}$

answer is A.

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Detailed Solution

Suppose distance of closest approach is r, and according to energy conservation applied for elementary charge.
Energy at the time of projection = Energy at the distance of closest approach
$\large \Rightarrow \frac{1}{2}m{v^2} = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{{(Ze).e}}{r} \Rightarrow r = \frac{{Z{e^2}}}{{2\pi {\varepsilon _0}m{v^2}}}$

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