The following text gives a qualitative idea about Rutherford's `alpha-` particle scattering
Channel: Doubtnut
Date: 2020-04-19

The following text gives a qualitative idea about Rutherford's `alpha-` particle scattering experiment.  Applying some trigonometry and calculus, Rutherford got the following equation:   `cot ((0)/(2)) = (Kxx(KE))/(Ze^(2))xxb`   `"where", K= "constant" =((1)/(9xx10))(c^(2))/(N-m^(2))`   KE = initial kinetic energy of `alpha-`particle,   `e=1.6xx10^(-19)C`   In an experment with gold nucleus (assume Z=80) with impact parameter equal to `10^(-4)m` the scattering angle was found to be `90^(@),` What should be the initial KE of `alpha-`particle?