Question

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

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

Answer 1

Answer: (C)

According to law of conservation of energy, kinetic energy of $\alpha -$ particle = potential energy of $\alpha -$ particle at distance of closest approach i.e., $\frac{1}{2}m{v}^2=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r}$ $\therefore$ $5MeV=\frac{9\times {10}^9\times (2e)\times (92e)}{r}$ $\left(\because \frac{1}{2}m{v}^2=5MeV\right)$ $\Rightarrow$ $r=\frac{9\times {10}^9\times 2\times 92\times {(1.6\times {10}^{-19})}^2}{5\times {10}^6\times 1.6\times {10}^{-19}}$ $\therefore$ $r=5.3\times {10}^{-14}m\approx {10}^{-12}cm$