Atoms and Nuclei Part 3

Concept:Energy conservation: The initial kinetic energy of the alpha particle is completely converted into electrostatic potential energy at the distance of closest approach.Explanation:Kinetic energy $K_0 = 7.9 \ \text{MeV} = 7.9 \times 10^6 \ \text{eV}$.Convert to joules: $K_0 = (7.9 \times 10^6) \times (1.6 \times 10^{-19}) = 1.264 \times 10^{-12} \ \text{J}$.Charge of alpha particle $q_1 = +2e$.Charge of target nucleus $q_2 = +79e$.Electrostatic potential energy $U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}$.At closest approach $r_0$: $K_0 = \frac{1}{4\pi\epsilon_0} \frac{(2e)(79e)}{r_0}$.Solve for $r_0$: $r_0 = \frac{1}{4\pi\epsilon_0} \frac{158 e^2}{K_0}$.Substitute values: $r_0 = \frac{(9 \times 10^9) \times 158 \times (1.6 \times 10^{-19})^2}{1.264 \times 10^{-12}}$.Simplify: $r_0 = 2.88 \times 10^{-14} \ \text{m}$.This $r_0$ is an estimate of the nuclear radius.Diameter of nucleus = $2 \times r_0 = 2 \times 2.88 \times 10^{-14} = 5.76 \times 10^{-14} \ \text{m}$.Answer:B. $5.76 \times 10^{-14} \ \text{m}$