Question

The distance of closest approach of an $ \alpha $ -particle fired towards a nucleus with momentum $p$, is $r$. If the momentum of the a-particle is $2p$, the corresponding distance of closest approach is

• A. $ \frac{r}{2} $
• B. 2r
• C. 4r
• D. $ \frac{r}{8} $
• E. $ \frac{r}{4} $

Answer 1

Answer: (E)

$ p\propto \frac{1}{\sqrt{r}} $
$ r\propto \frac{1}{{{p}^{2}}} $
Hence, $ \frac{{{r}_{1}}}{{{r}_{2}}}={{\left( \frac{{{p}_{2}}}{{{p}_{1}}} \right)}^{2}} $
$ ={{\left( \frac{p}{2p} \right)}^{2}}=\frac{1}{4} $ $ {{r}_{1}}=\frac{{{r}_{2}}}{4} $