In an alpha particle scattering experiment distance of closest...

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. $\left(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9\right.$ SI unit, mass of $\alpha$ particle $\left.=6.72 \times 10^{-27} \mathrm{~kg}\right)$