A narrow beam of alpha particles with kinetic energy `T= 600 ke V` falls normally on a golden foil incorporating `n = 1.1.10^(19)` nuclei`//cm^(2)`. Find the fraction of alpha particles scattered thtough the angles `theta lt theta_(0)= 20^(@)`

Explanation

Because of the `"cosec"^(4)(theta)/(2)` dependence of the scattering, the number of particles (or fraction) scattered through `theta lt theta_(0)` cannot be calculated directy. But we can write theis fracrtion as
`P(theta_(0))= 1-Q(theta_(0))`
where `Q (theta_(0))` is the fraction of particles scattered through `theta ge theta_(0)`. This fraction haas been calculated before and is (see the results of 6.12(b))
`Q(theta_(0))=pin((Ze^(2))/((4pi epsilon_(0))T))^(2)"cot"^(2)(theta_(0))/(2)`
where `n` here is number of nuclei`//cm^(2)`. Using the data we get
`Q= 0.4`
`P(theta_(0))=0.6`