A narrow beam of alpha particles with kinetic energy `1.0 MeV` falls normally on a platinum foil `1.0 mu` thick. The scattered particles are observed at an angle of `60^(@)` to the incident beam direction by means of a counter with a circular inlet area `1.0 cm^(2)` located at the distacne `10 cm`from the scattering section of the foil. What fraction of scattered alpha particles reaches the counter inlet?

Explanation

From the formula (6.1b) of the booke
`(dN)/(N)=n((Ze^(e ))/((4 pi epsilon_(0))2T))^(2) (dOmega)/("sin"^(4)(theta)/(2)`
we have put `q_(1)= 2e,q_(2)=Ƶe` here. Also `n=` no. of `Pt` nuclei in the foil per unit area
`=underset("the foil")underset ("mass of")underset(darr)(Atrho)underset("unit mass")underset("nuclei per")underset ("no.of")underset(darr)((N_(A))/(A_(pt))).(1)/(A)=(N_(A)rhot)/(A_(pt))`
Using the value `A_(pt)= 195,rho= 21.5xx10^(3)kg//m^(3)`
`N_(A)= 6.023xx10^(26)//"kilo mole"`
we get `n= 6.641xx10^(22)perm^(2)`
Also `dOmega=(dS_(n))/(r^(2))= 10^(-2)sr`
Substituting we get `(dN)/(N)= 3.36xx10^(-5)`