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