An alpha particle of energy $\mathrm{K} \mathrm{MeV}$ is moving towards a nucleus of atomic number…

An alpha particle of energy $\mathrm{K} \mathrm{MeV}$ is moving towards a nucleus of atomic number $\mathrm{Z}$. The distance of closest approach of the alpha particle to the nucleus in metres is
  1. $7.2 \times 10^{-16} \frac{\mathrm{Z}}{\mathrm{K}}$
  2. $3.84 \times 10^{-16} \frac{\mathrm{Z}}{\mathrm{K}}$
  3. $14.4 \times 10^{-16} \frac{\mathrm{Z}}{\mathrm{K}}$
  4. $28.8 \times 10^{-16} \frac{\mathrm{Z}}{\mathrm{K}}$

Solution

At the distance of closest approach the whole kinetic energy get converted into potential energy. $\begin{aligned} & \mathrm{K} . \mathrm{E}=\frac{2 \mathrm{Ze}^2}{4 \pi \epsilon_0 \mathrm{r}} \\ & \Rightarrow \mathrm{r}=\frac{2 \mathrm{ze}^2}{4 \pi \epsilon_0 \mathrm{k}(\mathrm{mev})}=9 \times 10^9 \times \frac{2 \mathrm{eZ}}{\mathrm{k} \times 10^6} \\ & \mathrm{r}=28.8 \times 10^{-16} \frac{\mathrm{Z}}{\mathrm{k}} \end{aligned}$

Asked in: AP EAMCET 2023 (16 May Shift 2)

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