91Ó°ÊÓ

When${\mathit{Q}}_{\mathbf{1}}$ is the charge of one point and${\mathit{Q}}_{\mathbf{2}}$ is the charge of the second point and r is the distance between two points, then the electric potential energy and electric force for a system of two point charges are given by:

role="math" localid="1664011253268" $$\begin{gathered} \mathit{U}\mathbf{=}\frac{{\mathbf{Q}}_{\mathbf{1}}{\mathbf{Q}}_{\mathbf{2}}}{\mathbf{4}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}\mathbf{r}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(1\right) \\ \mathit{F}\mathbf{=}\frac{{\mathbf{Q}}_{\mathbf{1}}{\mathbf{Q}}_{\mathbf{2}}}{\mathbf{4}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}\mathbf{r}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(2\right) \end{gathered}$$

Let the initial potential is zero.

By the principle of conservation of energy, the kinetic energy of alpha particle is converted into potential energy.

$$\begin{gathered} ∴U=K \\ =4.78×{10}^{6}eV \\ =4.78×{10}^6×\left(1.602×{10}^{-19}\right) \\ =7.6584×{10}^{-13}J \end{gathered}$$

The positive charge on the alpha particle is:

$$\begin{gathered} Q_1=2e \\ =2×1.602×{10}^{-19} \\ =3.204×{10}^{-19}C \end{gathered}$$

The positive charge on the nucleus is:

$$\begin{gathered} Q_1=92e \\ =92×1.602×{10}^{-19} \\ =1.474×{10}^{-17}C \end{gathered}$$

$$\begin{gathered} =\frac{\left(3.204×{10}^{-19}\right)\left(1.474×{10}^{-17}\right)}{4\mathrm{π}\left(8.854×{10}^{-12}\right)×\left(7.6584×{10}^{-13}\right)} \\ =5.53×{10}^{-14}m \end{gathered}$$

Hence, the distance of closest approach of the alpha particle to the center of the nucleus is$5.543×{10}^{-14}m$ .

$$\begin{gathered} F=\frac{Q_1{Q}_2}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2} \\ =\frac{\left(3.204×{10}^{-19}\right)\left(1.474×{10}^{-17}\right)}{4\mathrm{Ï€}\left(8.854×{10}^{-12}\right)×{\left(5.543×{10}^{-14}\right)}^2} \\ =13.82N \end{gathered}$$

Hence, the force on the alpha particle at the instant when it is at the distance of the closest approach is 13.82N .