Question

What is the distance of closest approach to the nucleus of an $\alpha$-particle which undergoes scattering by ${180}^o$ in Geiger-Marsden experiment?

Answer 1

$r_0=4.13fm$.
For closest approach.
$\frac{1}{2}m{v}^2=K\frac{Ze\times e}{r_0}$
For Rutherford experiment,
$\frac{1}{2}m{v}^2=5.5MeV=5.5\times {10}^6\times 1.6\times {10}^{-19}J=8.8\times {10}^{-13}J$
$8.8\times {10}^{-13}=\frac{9\times {10}^9\times 2\times 79\times {(1.6\times {10}^{-19})}^2}{r_0}$
$r_0=4.136\times {10}^{-15}m$
$r_0=4.13fm$.