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  4. If an α - particle of mass m, charge q and velocity v is incident on a nucleus of charge Q and mass m, then the distance of closest approach is

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Q. If an $\alpha$ - particle of mass $m$, charge $q$ and velocity $v$ is incident on a nucleus of charge $Q$ and mass $m$, then the distance of closest approach is

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Solution:

For the distance of closest approach kinetic energy will totally be converted to potential energy.
Hence, $\frac{1}{2} mv ^{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{ Qq }{ r _{0}}$
$ \Rightarrow r _{0}=\frac{ Qq }{2 \pi \varepsilon_{0} mv ^{2}}$