Final answer:
The distance of closest approach in a nuclear physics experiment involving an alpha particle and a nucleus depends on their electric charge and kinetic energy. The distance can be calculated using the conservation of energy, comparing the initial kinetic energy to the potential energy at the point closest to the nucleus. An exact value requires the atomic number of the nucleus, which is not provided in the question.
Step-by-step explanation:
The question seeks to determine the distance of closest approach when an alpha particle with 4.5 MeV kinetic energy interacts with a nucleus via the Coulomb force. This scenario calls for the application of classical nuclear physics principles describing the behavior of charged particles under electrostatic forces. The determining factors for the closest approach distance include the kinetic energy of the incoming alpha particle and the electric charge of the target nucleus.
To solve for the closest distance of approach, one can use the formula for electric potential energy, which states that the kinetic energy of the particle will equate to its potential energy at the point of closest approach. The relevant formula is KE = (Z*Z'e2)/(4πε0r), where KE is the kinetic energy of the alpha particle, Z and Z' are the atomic numbers of the nucleus and alpha particle, respectively, e is the elementary charge, ε0 is the vacuum permittivity, and r is the distance of closest approach. However, an exact numeric value isn’t provided in the question, as it depends on the value of Z, representing the atomic number of the nucleus being struck by the alpha particle.