when the \alpha particle is placed at a distance of 10-13m to the Cu atom, there will be repulsion between the two nuclei
potential energy due to repulsion =2ze2/4∩kr (k = epsilon)
this shud be equal to K.e of \alpha particle approaching the nucleus =1/2mv2
mv2/2 = 2ze2/4∩kr
v2=4ze2/4∩kmr
solvinf we get
v=6.32x106