A 4.78 - MeV alpha particle from a${}^{\mathbf{226}}\mathit{R}\mathit{a}$ decay makes a head-on collision with a uranium nucleus. A uranium nucleus has 92 protons. (a) What is the distance of closest approach of the alpha particle to the center of the nucleus? Assume that the uranium nucleus remains at rest and that the approach is much greater than the radius of the uranium nucleus. (b) What is the force on the alpha particle at the instant when it is at the distance of closest approach?

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" $$\begin{gathered} \mathit{U}\mathbf{=}\frac{{\mathbf{Q}}_{\mathbf{1}}{\mathbf{Q}}_{\mathbf{2}}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{r}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(1\right) \\ \mathit{F}\mathbf{=}\frac{{\mathbf{Q}}_{\mathbf{1}}{\mathbf{Q}}_{\mathbf{2}}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{r}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(2\right) \end{gathered}$$

Let the initial potential is zero.

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

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

The positive charge on the alpha particle is:

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

The positive charge on the nucleus is:

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

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

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

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

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