Calculate distance of closet approach by an `alpha `-particle of `KE=2.5 MeV` being scattered by gold nucleus `(Z=79)`.

Explanation

`Z_(1) = 2 , Z_(2) , = Z = 79`
Impact parameter, ` b = (Ze^(2) cot (theta//2))/(4 pi epsilon_(0) ((1)/(2) mv^(2)))`
Here` theta = 10^(@) , E = (1)/(2) mv^(2) = 5.0 xx 1.6 xx 10^(-13) `J
`(1)/(4 pi epsilon_(0)) = 9 xx 10^(-9) Nm^(2) C^(-2)`
`therefore b = (79 xx (1.6 xx 10^(-19))^(2) cot 5^(@) xx 9 xx 10^(@))/(5.0 xx 1.6 xx 10^(-13)) = 2.6 xx 10^(-13)` m