A 10.2MeV Li nucleus is shot directly at the center of a Ds nucleus. At what center-to-center distance does the Li momentarily stop, assuming the does not move?

1. Kinetic energy ${\mathrm{of}}^7\mathrm{Li}$nucleus, K.E.= 10.2 MeV
2. A lithium nucleus is shot directly at the center of a Ds nucleus.

The electric potential energy between two charged bodies as follows:

$V=\frac{k{q}_1{q}_2}{r}$ ….. (i)

Here, r is the separation between their centers or nuclei.

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

Now, obtain the charge of a particle from the concept that q = Ze such that Z is the atomic number.

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

Again for Ds nucleus, ${\mathrm{q}}_{2}110\mathrm{e}$

Now, using the given data in equation (i), determine the center-to-center separation between the two nuclei as follows:

$r=\frac{k{q}_1{q}_2}{K}$

Substitute the values and solve as:

$$\begin{gathered} \mathrm{r}=\frac{\left(9\times {10}^9{\displaystyle \frac{\mathrm{V}.\mathrm{m}}{\mathrm{C}}}\right)\left(3\times 1.6\times {10}^{-19}\mathrm{C}\right)\left(110\mathrm{e}\right)}{\left(3\times {10}^6\mathrm{eV}\right)} \\ =4.65\times {10}^{-14}\mathrm{m} \\ =46.5\mathrm{fm} \end{gathered}$$

Hence, the value of the least separation is 46.5 fm.