An alpha particle with Kinetic energy `0.27 MeV` is deflected through an angle of `60^(@)` by a golden foil. Find the corresponding value of the aiming parameter.

Explanation

Equation(6.1a) of the book reads in `MKS` units
`tan theta//2 =((q_(1)q_(2))/(4pi epsilon_(0)))//2b T`
Thus `b=((q_(1)q_(2))/(4pi epsilon_(0))) (cot theta//2)/(2T)`
For `alpha` particle `q_(1)= 2e`, for gold `q_(2)=79e`
(In Gaussian units there is no factor `((1)/(4pi epsilon_(0)).)`
Substituting we get `b= 0.731p m`