Physics - Modern Physics Question with Solution | TestHub

A rough estimate of size of the nucleus$r$is given by the distance of closest approach$d$of an

$\alpha -$particle incident head-on on a nucleus of charge$Z{e}^{}$;

$K=\frac{1}{2}m{v}^2=\frac{2Z{e}^2}{4\pi {\epsilon }_{0}d}\approx \frac{2Z{e}^2}{4\pi {\epsilon }_{0}r}$[where$K=6\text{ MeV}, Z=10$]



$r=\frac{2\times 100\times \left(1.6\times {10}^{-19}\right)\times 9\times {10}^9}{6\times {10}^6\times 1.6\times {10}^{-19}}$

$r=3\times 10^{-14} \mathrm{m}$