Read the following passage carefully and answer the questions The approximate size of the nucleus
Channel: Doubtnut
Date: 2020-04-19

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 the diameter of two different nuclei are in the ratio 1:2 then their mass number are in the ratio: