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^(@)`

The effective cross section of a uranium nucleus corresponding to the

1.	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^(@)`
Answer» (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`

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