${\mathit{A}}^{\mathbf{7}}\mathit{L}\mathit{i}$nucleus with a kinetic energy of 3.00 MeV is sent toward a ${\mathit{a}}^{\mathbf{232}}\mathit{T}\mathit{h}$nucleus. What is the least center-to-center separation between the two nuclei, assuming that the${\mathbf{(}\mathbf{more}\mathbf{}\mathbf{massive}\mathbf{)}}^{\mathbf{232}}\mathit{T}\mathit{h}$nucleus does not move?

The kinetic energy ${\mathrm{of}}^7\mathrm{Li}$nucleus is , $\mathrm{K}.\mathrm{E}.=3.00\mathrm{MeV}\left(\frac{{10}^6\mathrm{eV}}{1\mathrm{MeV}}\right)=3.00\times {10}^6\mathrm{eV}$

Assuming that${\mathrm{the}}^{232}\mathrm{Th}$ nucleus does not move.

From the given kinetic energy of the lithium nucleus, we can see that the classical formula is that the speed of the nucleus is less than the speed of light. Thus, using the concept of energy conservation, we can see that the kinetic energy of the lithium nucleus is transferred to the final potential energy of the system as the thorium nucleus does not move due to the bombardment of the lithium nucleus into the thorium nucleus. We can get the most minor center-to-center separation between the nuclei using the basic potential formula for two given two charged bodies.

Formula:

The electric potential energy between two charged bodies, $\mathrm{V}=\frac{{\mathrm{kq}}_1{\mathrm{q}}_2}{\mathrm{r}}$ (1)

Where is the separation between their centers or nuclei.

Kinetic energy (we use the classical formula since v is much less than c is converted into potential energy. (K = V)

Now, we can get the charge of a particle from the concept that

$\mathrm{q}=\mathrm{Ze}$,where Z is the atomic number

Thus, the charge of lithium,${\mathrm{q}}_1=3\mathrm{e}$

Again for thorium,${\mathrm{q}}_2=90\mathrm{e}$

Now, using the given data in equation (1), we can get the least center-to-center separation between the two nuclei as follows:

$$\begin{gathered} \mathrm{r}=\frac{{\mathrm{kq}}_1{\mathrm{q}}_2}{\mathrm{K}} \\ =\frac{\left(9\times {10}^9\mathrm{V}.\mathrm{m}/\mathrm{C}\right)\left(3\times 1.6\times {10}^{19}\mathrm{C}\right)\left(90\mathrm{e}\right)}{\left(3.00\times {10}^6\mathrm{eV}\right)} \\ =1.3\times {10}^{-13}\mathrm{m}\left(\frac{1\mathrm{fm}}{{10}^{-15}\mathrm{m}}\right) \\ =130\mathrm{fm} \end{gathered}$$

Hence, the value of the least separation is 130 fm .