Question

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

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

Answer 1

Answer: (D)

$\frac{r}{4}$

At the distance of closest approach, r
$K=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{2eZ{e}^2}{K}$
$\because K=\frac{P^2}{2m}$
$\therefore r=\frac{4}{4\pi {\epsilon }_0}\cdot \frac{2Z{e}^2}{\frac{p^2}{pm}}$
$\frac{1}{4\pi {\epsilon }_0}\cdot \frac{4mZ{e}^2}{p^2}$
Thus, $r\propto \frac{1}{p^2}$
$\therefore \frac{r_1}{r_2}=\frac{p_2^2}{p_1^2}=\frac{(2p{)}^2}{p^2}$
$\frac{r_1}{r_2}=\frac{4}{1}\Rightarrow \therefore r_2=\frac{r_1}{4}=\frac{r}{4}$