The effective cross section of a uranium nucleus corresponding to the scattering of monoenergetic alpha particles with an the angular interval from `90^(@)` to `180^(@)` is equal to `Delta sigma= 0.50kb`. Find ,
(a) the energy of alpha particles,
(b) the differntial cross section of scattering `d singma//dOmega(kb//sr)` coresponding to the angle `theta=60^(@)`

Explanation

(a) From the previous formula
`Delta sigma=((Ze^(2))/((4piepsilon_(0))2T))^(2) pi "cot"^(2)(theta_(0))/(2)`
or `T=(Ze^(2))/(4pi epsilon_(0))"cot" (theta_(0))/(2)sqrt((pi)/(Deltasigma))` Substituting the values with `Z= 79` we get `(theta_(0)= 90^(@))`
`T= 0.903MeV`
The differential scattering cross section is
`(d singma)/(d Omega)= C "cosec"^(4)(theta)/(2)`
where `Delta sigma(theta gt theta_(0))= 4pi C "cot"^(2)(theta_(0))/(2)` Thus from the given data
`C=(500)/(4pi)b= 39.79b//sr`
So `(dsigma)/(d Omega)(theta=60^(@))= 39.79xx16b//sr= 0.637kb//sr`