In the head-on collision of two alpha particles ₁ and ₂ with the gold nucleus, the closest approaches are 31.4 fermi and 94.2 fermi respectively. Then the ratio of the energy possessed by the alpha particles ₂ / ₁ 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 formula r₀ = 1 4 ₀ 2Ze^2 K . From this expression, it is evident that r₀ 1 K , which implies K 1 r₀ . Given the closest approaches for the two alpha particles are r₁ = 31.4 fermi and r₂ = 94.2 fermi, the ratio of their kinetic energies K₂ / K₁ is given by: K₂ K₁ = r₁ r₂ Substituting the given values: K₂ K₁ = 31.4 94.2 = 1 3 Answer: 1: 3