Question

The distance of closest approach of an $\alpha$-particle projected towards a nucleus with momentum p is r. What will be the distance of closest approach, when the momentum of projected $\alpha$-particle is 2p?

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

Answer 1

The correct option is D $\frac{r}{4}$

At the distance of closest approach:$KE=PE$

$\frac{p^2}{2m}=\frac{k2eq}{r}$

When the momentum chnages to 2p:

$\frac{(2p{)}^2}{2m}=k\frac{2eq}{r^{\prime }}$

$\frac{4p^2}{2m}=k\frac{2eq}{r^{\prime }}$

$\frac{2p^2}{m}=k\frac{2eq}{r^{\prime }}$

$r^{\prime }=k\frac{2eq}{p^2}\times \frac{m}{2}$

$=\frac{r}{2m}\times \frac{m}{2}$

$r^{\prime }=\frac{r}{4}$