In an alpha particle scattering experiment distance of closest approach for the \( \alpha \) particle is \( 4.5 \times 10^{-14} \, \text{m} \). If target nucleus has atomic number \( 80 \), then maximum velocity of \( \alpha \)-particle is ______ \( \times 10^{5} \, \text{m} / \text{s} \) approximately.
[\( \dfrac{1}{4\pi \varepsilon_{0}} = 9 \times 10^{9} \) SI unit, mass of \( \alpha \) particle \( = 6.72 \times 10^{-27} \, \text{kg} \)]
