With what velocity should an alpha particle travel towards the nucleus of a copper atom so as to arrive at a distance 10^-13 m from the nucleus of a copper atom?

As $\alpha$-particle can appraoch towards the Cu nucleus up to the distance r_$°$ which is equal to 10^-13m here.
To arrive at a distance r_$°$ from the nucleus, the kinetic energy of alpha particle should be equal to the potential energy of interaction of alpha particle with Cu nucleus.
or KE = PE
​1/2 m_$\alpha$v​_$\alpha$² = K q​$\alpha$q_Cu/r_$°$ ..................(1)
 m_$\alpha$ = ​mass of $\alpha$-particle = 4 x 1.67 x 10^-27 kg.
K = 9 x 10⁹ N$·$m² / C²
q​​_$\alpha$ = charge on alpha particle is +2 = 2 x (+1.6 x 10^-19C)​
q_Cu = 29 x (+1.6 x 10^-19C)​​
r⁰ = 10^–13 m​
Putting all the values in given equation (1) we get v​$\alpha$²

$$\begin{gathered} v_{\alpha }^2=\frac{2\times 9\times {10}^9 (2\times 1.6\times {10}^{-19})(29\times 1.6\times {10}^{-19})}{4\times 1.67\times {10}^{-27}\times {10}^{-13}} \\ =400.09\times {10}^{11} \\ =4000.9\times {10}^{10} \\ {v}_{\alpha }=\sqrt{4000.9\times {10}^{10}} \\ = 63.25\times {10}^5 m{s}^{-1} \end{gathered}$$