7.9 MeV -particle scatters from a target material of atomic number 79. From the given data the estimated diameter of nuclei of the target material is (approximately) _ _ _ _ m. [ 1 4 _ o =9 10⁹ Nm ² / C ² . and electron charge .=1.6 10⁻¹⁹ C ]

Step-by-step solution

For Rutherford scattering, at the closest approach all kinetic energy converts to electrostatic potential energy. For alpha particle-nucleus collision: K.E. = 1 4 ₀ Z₁ Z₂ e^2 r_ min Where: Z₁ = 2 (alpha), Z₂ = 79 (gold), K.E. = 7.9 MeV r_ min = 9 10^9 2 79 (1.6 10⁻¹⁹)^2 7.9 10^6 1.6 10⁻¹⁹ r_ min = 2.88 10⁻¹⁴ m The estimated nuclear diameter is approximately: d = 2 r_ min = 5.76 10⁻¹⁴ m