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(C) The distance of closest approach $(r_0)$ is the distance at which the initial kinetic energy of the $\alpha$-particle is completely converted into

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$1/m$

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(C) The distance of closest approach $(r_0)$ is the distance at which the initial kinetic energy of the $\alpha$-particle is completely converted into electrostatic potential energy.Equating kinetic energy to potential energy:$\frac{1}{2}mv^2 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{(Ze)(2e)}{r_0}$Solving for $r_0$:$r_0 = \frac{2Ze^2}{4\pi\varepsilon_0 \cdot \frac{1}{2}mv^2} = \frac{Ze^2}{\pi\varepsilon_0 mv^2}$Since $Z$,$e$,$\varepsilon_0$,and $v$ are constants for a given experiment,we have:$r_0 \propto \frac{1}{m}$

When an $\alpha$-particle of mass $m$ moving with velocity $v$ bombards a heavy nucleus of charge $Ze$,its distance of closest approach from the nucleus depends on $m$ as

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