Find the effective corss section of a uranium nucleus corresponding to the scattering ofalpha particles with kinetic energy `T= 1.5 MeV` through the angles exceeding `theta_(0)=60^(@)`.

Explanation

From the Rutherford scattering formula
`(d sigma)/(dOmega)=((Ze^(2))/((4piepsilon_(0))2T))^(2)(1)/("sin"^(4)(theta)/(2))`
or `d sigma=((Ze^(2))/((4pi epsilon_(0))2T))^(2) (2picos theta//2 d theta)/("sin"^(4)(theta)/(2))`
`=((Ze^(2))/((4pi epsilon_(0))T))^(2)pi( cos theta//2dtheta)/(sin^(3)theta//2)` ltbr Then intergrating from `theta = theta_(0) to theta= pi` we get the required cross section
`Delta sigma=((Ze^(2))/((4pi epsilon_(0))T))^(2)pi int_(theta_(0))^(x)(cos theta//2d theta)/("sin"^(3)(theta)/(2))`
`=((Ze^(2))/((4pi epsilon_(0))T))^(2) "cot"^(2)(theta_(0))/(2)`.
For `U` nucleus `Z= 92` and we get on putting the values
`Delta sigma= 737b= 0.737kb`.
`(1b= 1bam= 10^(-28)m^(2))`.