Chemistry - Atomic Structure Question with Solution | TestHub

The alpha-particle's initial kinetic energy is converted into electrostatic potential energy at the distance of closest approach.

The charge of an alpha-particle is $q_1=+2e$.

The charge of a copper nucleus (atomic number $Z=29$) is $q_2=+Ze=+29e$.

The distance of closest approach is $r=10^{-12}$ m.

The mass of an alpha-particle is $m=4\times 1.672\times 10^{-27}$ kg.

Using conservation of energy:

$$\frac{1}{2}m{v}^2=\frac{k{q}_1{q}_2}{r}$$

$$v=\sqrt{\frac{2k{q}_1{q}_2}{mr}}$$

Substitute values ($k=9\times 10^{9}N{m}^2/C^2$, $e=1.6\times 10^{-19}$ C):

$$v=\sqrt{\frac{2\times 9\times 10^9\times (2\times 1.6\times 10^{-19})\times (29\times 1.6\times 10^{-19})}{4\times 1.672\times 10^{-27}\times 10^{-12}}}$$

$$v\approx 2.0\times 10^{6}m/s$$

Comparing with $A\times 10^{6}m/s$, the value of $A$ is 2.