\(\frac{1}{2}m{v^2} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_1}{q_2}}}{{{r_0}}}\) \( = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{(Ze)(2e)}}{{{r_0}}}\)

\({r_0} = \frac{{4Z{e^2}}}{{4\pi {\varepsilon _0}m{v^2}}} = \) \(\frac{{Z{e^2}}}{{\pi {\varepsilon _0}{v^2}}}\left( {\frac{1}{m}} \right)\)

\(r_{0}\) is proportional to \(\left(\frac{1}{m}\right)\)