Question

When an alpha particle of mass M moving with velocity V bombards a heavy nucleus of charge Ze its distance of closest approach from the nucleus depends on M As:

Answer 1

Step 1: Given Information:

• The velocity of an alpha particle is V.
• The mass of an alpha particle is M.
• Therefore, its initial Kinetic energy will be $\frac{1}{2}{\mathrm{MV}}^2$.
• Then potential energy of the alpha particle and the heavy nucleus will be approximately zero.

Step 2: Calculation for potential energy:

• The potential energy at the closest distance of approach will be:
• $\mathrm{PE}=\frac{{\mathrm{kq}}_1{\mathrm{q}}_2}{\mathrm{r}}=\frac{\mathrm{k}\left(2\mathrm{e}\right)\left(\mathrm{Ze}\right)}{{\mathrm{r}}_0}$, where $2\mathrm{e}$ is the charge on the alpha particle and $\mathrm{Ze}$ is the charge on heavy nuclei and ${\mathrm{r}}_0$ is the distance of closest approach.

Step 3: Application of conversation of mass:

• On applying the law of conversation of mass.
• The kinetic energy is equal to the potential energy.

$$\begin{gathered} \mathrm{KE}=\mathrm{PE} \\ \Rightarrow \frac{1}{2}{\mathrm{MV}}^2=\mathrm{k}\frac{\left(2\mathrm{e}\right)\left(\mathrm{Ze}\right)}{{\mathrm{r}}_0} \\ \Rightarrow {\mathrm{r}}_0=\frac{4{\mathrm{kZe}}^2}{{\mathrm{MV}}^2} \\ \therefore {\mathrm{r}}_0\propto \frac{1}{\mathrm{M}} \end{gathered}$$

Thus, the distance of the closest approach in α−particle bombarding is inversely proportional to its mass.