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 xx 10^(6)xx1.6 xx10^(-19)J = 1.20 xx 10^(-12)`
Charge on the electron, `e=1.6 xx 10^(-19)C`, Permittivity constant `K=9.0xx10^(9)Nm^(2)C^(-2)`
`r_(0)=(4KZ e^(2))/(mv^(2)) =(9xx10^(9) Nm^(2)C^(-2) xx 2xx79xx(1.6xx10^(-19)C)^(2))/(1.20xx10^(-12)J) =3.033 xx 10^(-14)m`