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
Radius of a particular nucleus is calculated by the projection of `alpha-`particle from infinity at a particular speed. Let this radius be the true radius. If the radius calculation for the same nucleus is made by , `alpha-`particle with half of the earlier speed then the percentage error involved in the radius calculation is :