Atoms and Nuclei: JEE Main 2024 Physics question

In an alpha particle scattering experiment distance of closest approach for the $\alpha$ particle is $4.5 \times 10^{-14} \mathrm{~m}$. If target nucleus has atomic number 80 , then maximum velocity of $\alpha$-particle is __________ $\times 10^5 \mathrm{~m} / \mathrm{s}$ approximately. ($\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{SI}$ unit, mass of $\alpha$ particle $=6.72 \times 10^{-27} \mathrm{~kg}$)