The closest approach of an alpha particle when it make a head on collision with a gold nucleus is 10 10⁻¹⁴ ~m , then the kinetic energy of the alpha particle is :

Step-by-step solution

The distance of closest approach r₀ for an alpha particle of kinetic energy K colliding head-on with a nucleus of atomic number Z is given by the conservation of energy: K = 1 4 ₀ (Ze)(2e) r₀ For a gold nucleus, the atomic number Z = 79 . The value of the Coulomb constant is 1 4 ₀ = 9 10^9 N m ^2 C ⁻² . The charge of an electron is e = 1.6 10⁻¹⁹ C . Substituting the given values r₀ = 10 10⁻¹⁴ m = 10⁻¹³ m : K = (9 10^9) (79 2) (1.6 10⁻¹⁹)^2 10⁻¹³ K = 9 10^9 158 2.56 10⁻³⁸ 10⁻¹³ K = 9 158 2.56 10^ 9 - 38 + 13 K = 364