Chemistry - Atomic Structure Question with Solution | TestHub

An α-particle is accelerated from rest through a potential difference of $2.0\times 10^{6}V$ and then approaches a copper nucleus $(Z=29)$ head on. Assuming the nucleus stays at rest and the distance of closest approach is much larger than its radius, the distance of closest approach is $(e^2/4\pi {\epsilon }_0=1.44$ MeV fm)