In a geiger - marsden experiment. Find the distance of closest approach to the nucleus of a 7.7 me v `alpha`- particle before it comes momentarily to rest and reverses its direction. (z for gold nucleus = 79) .

Explanation

They key idea here is that the total mechanical energy of the system consisting of an `alpha-` particle and a gold nucleus is conserved.
The intial energy `E_(i)` is just the kinetic energy K of the incoming `alpha-` particle. The final energy `E_(f)` is just the electric potential energy U of the system T.
Let d be the centre - to - centre distance between the `alpha-` particle and the gold nucleus when `alpha-` particle is at its stopping point. Then we can write the conservation of energy `E_(i)=E_(f)` as
`K_(alpha)=(1)/(4piepsilon_(0))((2e)(Ze))/(r_(0))=(2Ze^(2))/(4piepsilon_(0)r_(0))`
Distance of closest approach
`r_(0)=(2Ze^(2))/(4piepsilon_(0)K)`
here `K_(alpha)=7.7MeV =1.2xx10^(-12)J`
`(1)/(4piepsilon_(0))=9xx10^(9)" MKS unit"`
`"Z = 79 for Gold"`
`"So, "r_(0)=(2xx9xx10^(9)xx79xx(1.6xx10^(-19))^(2))/(1.2xx10^(-12))`
`r_(0)=3xx10^(-14)m="30 fm"`.
Radius of gold nucleus in actual is 6 fm where distance of closest approach is 30 fm. This discrepancy is due to the fact that distance of closest approach `alpha` is larger than sum of radii of the gold nucleus and the `alpha-` particle.