What is the distance of closest approach to the nucleus of an `alpha`partile which undergoes scattering by `180^(@)` is geiger-marsden experiment?

Answer

`r_(0)=4.13`fm

Explanation

For closest approch,
`(1)/(2)mv^(2)=K(Zexxe)/(r_(0))`
For Rutherford experiment ltBrgt `(1)/(2)mv^(2)=5.5MeV=5.5xx10^(6)xx1.6xx10^(-19)J=8.8xx10^(-13)J`
`8.8xx10(-13)=(9xx10^(9)xx2xx79xx(1.6xx10^(-19))^(2))/(r_(0))`
`r_(0)=4.136xx10^(-15)m`
`r_(0)=4.13` fm