Determine the distance of closest approach when an alpha particle of kinetic energy 4.5 MeV strikes a nucleus of Z = 80, stops and reverses its direction.

Explanation

Let r be the centre distance between the alpha particles and the nucleus (Z = 80). When the alpha particle is at stopping point, then
`K=1/(4piepsi_(0))((Ze)(2e))/r`
(or) `r=1/(4piepsi_(0))*(2Ze^(2))/K`
= `(9xx10^(9)xx2xx80e^(2))/(4.5MeV)`
`=(9xx10^(9)xx2xx80xx(1.6xx10^(-19))^(2))/(4.5xx10^(6)xx1.6xx10^(-19)J)`
= `(9xx160xx1.6)/4.5xx10^(-16)=512xx10^(-16)m`
`= 5.12xx10^(-14)m`