An `alpha`-particle having kinetic energy of 7.5 MeV is scattered by gold (Z=79) mucleus through `180^(@)`. Calculate the distance of closest approach.

Explanation

Kinetic energy of `alpha-`particle `=7.5 Mev =7.5 xx10^(6) xx1.6xx10^(-19) J =1.20xx10^(-12)`
Charge on the electron, `e=1.6xx10^(-19) C`, Permittivity constant, `K=9.0xx10^(9) Nm^(2)C^(-2)`
`r_(0)=(4KZe^(2))/(mv^(2)) =(9xx10^(9) Nm^(2) C^(-2) xx 2xx79xx(1.6xx10^(-19) C)^(2))/(1.20xx10^(-12) J) =3.033xx10^(-14) m`