Read the following passage carefully and answer the questions The approximate size of the nucleus can be calculated by using energy conservation theorem in Rutherford's `alpha-`scattering experiment. If an `alpha-`perticle is projected from infinity with speed v, towards the nucleus having z protons then the `apha-`perticle which is reflected back or which is deflected by `180^(@)` must have approach closest to the nucleus. It can be approximated that `alpha-`particles collides with the nucleus and gets back. Now if we apply the energy conservation at initial and collision point then: `("Total Energy")_("initial") = ("Total Enregy")_("final")` `(KE)_(i) +(PE)_(i)=(KE)_(f)+(PE)_(f)` `(PE)_(i) =0, "since"PE` of two charge system separated by infinite distance is zero, finally the particle stops and then starts coming back. `1/2m_(alpha)v_(alpha)^(2) + 0=0 + (kq_(1)q_(2))/(R)` `rArr1/2m_(alpha)v_(alpha)^(2)=k(2exxze)/(R)rArrR=(4kze^(2))/(m_(alpha)v_(alpha)^(2)` Thus the radius of nucleus can be calculated using above equation. The nucleus is so small a particle that we can't define a sharp boundary for it . Experiments show that the average radius R fo a nucleus may be written as `R=R_(0)(A)^(1//3)` where `R_(0) = 1.2xx10^(-15)m` A= atomic mass number R=radius of nucleus If `alpha-`particle with speed `v_(0)` is projected from infinity and it approaches upto `r_(0)` distance from the nuclei. Then the speed of `alpha-`particle which approaches `2r