Atoms And Nuclei

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 \(=6.72 \times 10^{-27} \mathrm{~kg}\))