If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ____ m. (Atomic number of gold = 79 and frac14pi in_0 = 9 times 10^9$\frac{1}{4\pi \in_{0}} = 9 \times 10^{9}$ in SI units)

### Related Formula K_textinitial = U_textclosest approach$$K_{\text{initial}} = U_{\text{closest approach}}$$ K = frac14pi epsilon_0 frac(2e)(Ze)r_0$$K = \frac{1}{4\pi \epsilon_{0}} \frac{(2e)(Ze)}{r_0}$$ ### Core Logic By energy conservation, the entire kinetic energy of the alpha particle gets converted to electrostatic potential energy at the distance of closest approach (r_0$r_0$). K_i + U_i = K_f + U_f$K_i + U_i = K_f + U_f$ K_i + 0 = 0 + frac14piepsilon_0 frac(2e)(79e)r_0$$K_i + 0 = 0 + \frac{1}{4\pi\epsilon_0} \frac{(2e)(79e)}{r_0}$$ ### Step 1: Convert Energy and Solve Initial kinetic energy K_i = 7.7text MeV = 7.7 times 10^6 times 1.6 times 10^-19text J$K_i = 7.7\text{ MeV} = 7.7 \times 10^6 \times 1.6 \times 10^{-19}\text{ J}$. 7.7 times 10^6 times 1.6 times 10^-19 = frac9 times 10^9 times (2 times 1.6 times 10^-19) times (79 times 1.6 times 10^-19)r_0$$7.7 \times 10^6 \times 1.6 \times 10^{-19} = \frac{9 \times 10^9 \times (2 \times 1.6 \times 10^{-19}) \times (79 \times 1.6 \times 10^{-19})}{r_0}$$ r_0 = frac9 times 10^9 times 2 times 79 times (1.6 times 10^-19)^27.7 times 10^6 times 1.6 times 10^-19$$r_0 = \frac{9 \times 10^9 \times 2 \times 79 \times (1.6 \times 10^{-19})^2}{7.7 \times 10^6 \times 1.6 \times 10^{-19}}$$ r_0 = frac9 times 10^9 times 158 times 1.6 times 10^-197.7 times 10^6$$r_0 = \frac{9 \times 10^9 \times 158 \times 1.6 \times 10^{-19}}{7.7 \times 10^6}$$ r_0 = frac2275.2 times 10^-107.7 times 10^6 = 295.48 times 10^-16text m approx 2.95 times 10^-14text m$$r_0 = \frac{2275.2 \times 10^{-10}}{7.7 \times 10^6} = 295.48 \times 10^{-16}\text{ m} \approx 2.95 \times 10^{-14}\text{ m}$$ ### Pattern Recognition Distance of closest approach problem: Simply equate initial Kinetic Energy (in Joules) to Potential Energy k(Z_1e)(Z_2e)/r_0$k(Z_1e)(Z_2e)/r_0$. Alpha particle has charge 2e$2e$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Atoms