Calculate the distance of closest approach for a head-on collision between an 5.30 MeV alpha particle and a copper nucleus.

Kinetic energy of an alpha particle, K.E. = 5.30 Me V

The collision happens between an alpha particle and a copper nucleus.

The distance of the closest approach refers to the minimum distance of the charged particle from the nucleus at which initial kinetic energy is the same as the potential energy of the nucleus. This is very similar to the Ruther ford scattering experiment calculations. Thus, considering the concept, we calculate the distance using the potential energy formula.

Formula:

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

Where, is the separation between their centers or nuclei.

As per the concept, the kinetic energy of the alpha particle is same as the potential energy of the system having the both particles (K = U) .

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

q = Ze, where Z is the atomic number

Thus, charge of alpha particle,$q_1=4e$

Again for copper nucleus,$q_2=29e$

Thus, using the given data in equation (1), we can get the distance of closest approach for a head-on collision between the particles as follows:

$$\begin{gathered} \mathrm{r}=\frac{{\mathrm{kq}}_1{\mathrm{q}}_2}{\mathrm{K}} \\ =\frac{\left(9\times {10}^9\mathrm{V}.\mathrm{m}/\mathrm{c}\right)\left(4\times 1.6\times {10}^{-19}\mathrm{C}\right)\left(29\mathrm{e}\right)}{\left(5.30\times {10}^6\mathrm{eV}\right)} \\ =1.58\times {10}^{-14} \\ =15.8\mathrm{fm} \end{gathered}$$

Hence, the value of the distance of closest approach is 15.8 fm.