In Rutherford scattering experiment, if a proton is taken instead of an alpha particle, then for same distance of closest approach, how much K.E. in comparison to K.E. of `alpha` particle will be required?

Explanation

As distance of closest approach for `alpha`-particle =`d_0`
`d_0=1/(4piepsilon_0)((Ze)(2e))/K`
`K=1/(4piepsilon_0)((Ze)(2e))/d_0`
`K_alpha =1/(4piepsilon_0)((Ze)(2e))/d_0`…(i)
`K_P=1/(4piepsilon_0)((Ze)(e))/d_0` ….(ii)
From (i) & (ii) we can say `K_P=K_alpha/2`
Proton will have half kinetic energy than `alpha`-particle.