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?

Options

A. `1.84xx10^(-12)J`
B. `184xx10^(-12)J`
C. `1.44xx10^(-12)J`
D. `14.4xx10^(-12)J`

Answer

a