Question

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

Answer 1

At the point of the closest approach, the kinetic energy of the alpha particle must be equal to the potential energy of interaction of the alpha particle with the copper nucleus.

According to the problem:-

Consider the equation

$\frac{1}{2}m{u}^2=\frac{1}{4\pi {\in }_{\circ }}\times \frac{Z{e}^2}{r}$

Implies that

$u^2=\frac{Z{e}^2}{4\pi \in \circ .m.r}$

To find $u$,

$u^2=\frac{29\times {\left(1.6\times {10}^{-19}\right)}^2}{3.14\times 8.85\times {10}^{-12}\times \left(4\times 1.672\times {10}^{-27}\right)\times {10}^{-13}}$

$u=6.3\times {10}^{6}m/\sec$