In alpha particle scattering experiment, if \(v\) is the initial velocity of the particle, then the distance of closest approach is \(d\). If the velocity is doubled, then the distance of closest approach changes to:

If \(r_0\) be the distance of closest approach, then
\((\mathrm{KE})_\alpha=\frac{(Z e)(2 e)}{4 \pi \varepsilon_0 \cdot\left(r_0\right)_\alpha} \Rightarrow \frac{1}{2} m v_\alpha^2=\frac{(Z e)(2 e)}{4 \pi \varepsilon_0 \cdot\left(r_0\right)_\alpha}\)
\(\Rightarrow \quad v_\alpha^2 \propto \frac{1}{\left(r_0\right)_\alpha} \Rightarrow\left(r_0\right)_\alpha \propto \frac{1}{v_\alpha^2}\)
\(\Rightarrow \quad \frac{\left(r_0\right)_{\alpha_i}}{\left(r_0\right)_{\alpha_f}}=\frac{v_{\alpha_f}^2}{v_{\alpha i}^2}=\frac{\left(2 v_{\alpha_i}\right)^2}{v_{\alpha_i}^2}=4\)
\(\therefore \quad\left(r_0\right)_{\alpha_f}=\frac{\left(r_0\right) \alpha_i}{4}=\frac{d}{4} \quad\left[\because\left(r_0\right)_{\alpha_i}=d\right]\)