Final answer:
The minimum initial speed of an alpha particle to approach a gold nucleus to a distance of 300 fm is calculated by equating its kinetic energy with the electric potential energy at the closest approach. Specific values for the charges and the kinetic energy are required to solve for initial speed.
Step-by-step explanation:
The question asks about the minimum initial speed an alpha particle must have to approach a gold nucleus to a distance of 300 femtometers (fm) without being deflected. This is a problem in classical nuclear physics that involves electrostatic forces between charged particles—the alpha particle and the gold nucleus. The electrical potential energy at the closest distance can be equated to the kinetic energy of the alpha particle given its initial speed. The formula to use is:
K.E. = (1/2)mv² = k(Q1Q2)/r,
where K.E. is the kinetic energy, m is the mass of the alpha particle, v is the initial speed, k is Coulomb's constant, Q1 and Q2 are the charges of the alpha particle and gold nucleus, respectively, and r is the closest approach distance. If the energy of the doubly charged alpha nucleus was 5.00 MeV, we can solve for the required speed. However, the exact numerical answer is not provided in the question or reference information. We would need specific values, especially the charge of the alpha particle, the charge of the gold nucleus, and the distance (which is provided, 300 fm) to compute the initial speed.