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 1 4 _ o =9 10⁹ in SI units)

Step-by-step solution

At closest approach, KE converts to PE: KE = 1 4 ₀ (2e)(Ze) r_ min r_ min = 9 10^9 2 79 (1.6 10⁻¹⁹)^2 7.7 10^6 1.6 10⁻¹⁹ = 9 2 79 1.6 10⁻¹⁹ 7.7 10³ = 2275.2 7.7 10⁻¹⁶ = 2.95 10⁻¹⁴ m