Atoms and Nuclei Part 3

Concept:The distance of closest approach is when the initial kinetic energy of the alpha particle is fully converted into electrostatic potential energy at the point of closest approach to the nucleus.Formula:$K = \frac{1}{4\pi\epsilon_0} \frac{Z_1 Z_2 e^2}{r_0}$Solution:Given $K = 7.7$ MeV = $7.7 \times 10^6$ eV.Convert to joules: $1$ eV = $1.6 \times 10^{-19}$ J.$K = 7.7 \times 10^6 \times 1.6 \times 10^{-19} = 1.232 \times 10^{-12}$ J.Charge of alpha particle $q_1 = +2e$.Charge of gold nucleus $q_2 = +79e$.Thus $Z_1 Z_2 = 2 \times 79 = 158$.Given $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9$ N m²/C².Substitute into formula:$r_0 = \frac{1}{4\pi\epsilon_0} \frac{Z_1 Z_2 e^2}{K} = \frac{9 \times 10^9 \times 158 \times (1.6 \times 10^{-19})^2}{1.232 \times 10^{-12}}$.$e^2 = (1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38}$.Numerator = $9 \times 10^9 \times 158 \times 2.56 \times 10^{-38} = 9 \times 158 \times 2.56 \times 10^{-29}$.$9 \times 158 = 1422$, then $1422 \times 2.56 = 3640.32$.Numerator ≈ $3.64032 \times 10^{-26}$? Wait, careful: $10^9 \times 10^{-38} = 10^{-29}$, so numerator = $9 \times 158 \times 2.56 \times 10^{-29} = 3640.32 \times 10^{-29} = 3.64032 \times 10^{-26}$? Actually $3640.32 \times 10^{-29} = 3.64032 \times 10^{-26}$. But dividing by $1.232 \times 10^{-12}$ gives $r_0 = (3.64032 \times 10^{-26}) / (1.232 \times 10^{-12}) = 2.955 \times 10^{-14}$ m.Thus $r_0 \approx 2.95 \times 10^{-14}$ m.Answer:$2.95 \times 10^{-14}$ m, which corresponds to option (C).