Question

In the Rutherford scattering experiment 400 MeV alpha particles scatter off gold nuclei (containing 79 protons and 118 neutrons). Assume a particular alpha particle moves directly toward the gold nucleus and scatters backward at 118 degrees, and that the gold nucleus remains fixed throughout the entire process. Determine (a) the distance of closest approach of the alpha particle to the gold nucleus and (b) the maximum force exerted on the alpha particle

Answer 1

Given,

Kinetic energy of alpha particle K =400Mev$=400\times 1.6\times 10^{-19}\times 10^6 Joule$

Atomic number of gold nuclie Z=79

(a) Distance of closest approach r = $\dfrac{1}{4\pi \epsilon_{o}}\dfrac{2ze^2}{K}$

=$9\times 10^9\times \dfrac{2\times 79\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{400\times 1.6\times 10^{-19}\times 10^6}$

=$5.668\times 10^{-16}m$

(b) Maximum force exerted on alpha particle F$_{max}$ = $\dfrac{1}{4\pi \epsilon_{o}}\dfrac{2ze^2}{r^2}$

=$9\times 10^9\times \dfrac{2\times 79\times 2.56\times 10^{-38}}{(5.688\times 10^{-16})^2}$

=$\dfrac{18\times 79\times 2.56\times 10^{-29}}{(10^{-32}\times (5.688)^2)}$

=$1.1125\times 10^5N$