What is the distance of closest approach when a 5 MeV proton approaches a gold nucleus ?

Explanation

At the distance `r_(0)` of closest approach
K.E of a Proton = P.E. Of proton and the gold nucleus
` K = 1/2 mv^(2) = 1/(4 pi epsilon _(0)) (Ze.e)/(r_(0))`
`r_(0) = 1/(4 pi epsilon_(0)) (Ze^(2))/(K)`
But K = 5 MeV =` 5 xx 1.6 xx 10^(-13)`J.
For gold, Z = 79
`r_(0) = (9 xx 10^(9) xx 79 xx(1.6 xx 10^(-19))^(2))/(5 xx 1.6 xx 10^(-13))`
` = 2.28 xx 10^(-14)`m
`r_(0) = 2.3 xx 10^(-14)m`