Physics - Modern Physics Question with Solution | TestHub

At distance of closest approach relative velocity of two particles is$v$. Here target is considered as stationary, so$\alpha$-particle comes to rest instantaneously at distance of closest approach. Let required distance is$r,$then from work energy-theorem.

$0-\frac{{mv}^2}{2}=-\frac{1}{{4\pi \epsilon }_0}\frac{Z_e\times Z_e}{r}$

$\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{r}\propto \frac{1}{m}$

$\propto \frac{1}{v^2}$

$\propto {Ze}^2$