Question

An $\alpha$- particle of energy $5MeV$ is scattered through ${180}^o$ by a fixed uranium nucleus. The distance of the closest approach is of the order of:

• A. $1\mathring{A}$
• B. ${10}^{-10}cm$
• C. ${10}^{-12}cm$
• D. ${10}^{-15}cm$

Answer 1

Answer: (C)

$\begin{array}{c}\text{Given-}\ast \text{Energy of}\alpha \text{-particle}=5MeV\\ \text{By conservation of energy, we can say that}\\ \text{for the distance of closest approach all the}\\ \text{kinetic energy is converted into potential energy.}\\ \Rightarrow {KE}_{\text{initial}}=P{E}_{\text{final}}\end{array}$

$\begin{array}{cc}\Rightarrow \frac{1}{2}m{v}^2=\frac{k{q}_1{q}_2}{r}& ,q_1=\text{charge of uranium}\\ & \text{nucleus}\\ & q_2=\text{charge of}\alpha -\text{particle}\end{array}$

$\begin{array}{c}\Rightarrow 5\times {10}^6\times 1.6\times {10}^{-19}J=\frac{9\times {10}^9\times 2\times 1.6\times {10}^{-19}\times 92\times 1.6\times {10}^{-19}}{1}\\ \Rightarrow r_0=5.3\times {10}^{-14}m\\ r_0=5.3\times {10}^{-12}cm\\ \text{Hence, radius of closest approach is of the}\\ \text{order of}{10}^{-12}cm\text{.}\\ \text{Hence, option (c) is correct.}\end{array}$