Atoms and Nuclei Part 3

Concept:At the distance of closest approach, the entire kinetic energy of the alpha particle converts into the electrostatic potential energy of the alpha particle–nucleus system.Formula:$$K=\frac{1}{4\pi\epsilon_0}\frac{Z_1 Z_2 e^2}{r_0}\;\Rightarrow\; r_0=\frac{1}{4\pi\epsilon_0}\frac{Z_1 Z_2 e^2}{K}$$Given:$K=7.7$ MeV $=7.7\times 10^{6}\times 1.6\times 10^{-19}=1.232\times 10^{-12}$ J.Charge of alpha particle: $q_1=Z_1 e=+2e$.Charge of gold nucleus: $q_2=Z_2 e=+79e$.Thus $Z_1 Z_2=2\times 79=158$.Also $\frac{1}{4\pi\epsilon_0}=9\times 10^{9}$ N m²/C².Solution:$$r_0=\frac{9\times 10^{9}\times 158\times (1.6\times 10^{-19})^2}{1.232\times 10^{-12}}$$$$e^2=(1.6\times 10^{-19})^2=2.56\times 10^{-38}$$Numerator $=9\times 10^{9}\times 158\times 2.56\times 10^{-38}=3640.32\times 10^{-29}=3.64032\times 10^{-26}$.$$r_0=\frac{3.64032\times 10^{-26}}{1.232\times 10^{-12}}=2.955\times 10^{-14}\ \text{m}$$$$r_0\approx 2.95\times 10^{-14}\ \text{m}$$Answer:$2.95\times 10^{-14}$ m, which matches option (C).Hence the marked answer option (C) is correct.