Question

An alpha particle with kinetic energy 15.0MeV makes a collision with lead nucleus, but it is not "aimed" at the center of the lead nucleus, and has an initial-nonzero angular momentum (with respect to the stationary lead nucleus) of magnitude L=p⁰

​b, where p⁰ is the magnitude of the initial momentum of the alpha particle and b=1.20×10⁻¹²m. (Assume that the lead nucleus remains stationary and that it may be treated as a point charge. The atomic number of lead is 82 . The alpha particie is a helium nucleus, with atomic number 2) What is the distance of closest approach?

Answer 1

The distance of closest approach between an alpha particle with specified kinetic energy and angular momentum and a lead nucleus is determined by equating the initial kinetic energy with the electric potential energy at the closest approach and then solving for the distance.

Step-by-step explanation:

To calculate the distance of closest approach to a lead nucleus by an alpha particle with a given initial angular momentum, we can use the conservation of energy and the concept of an alpha particle's interaction with a nucleus via the Coulomb force. First, we consider the kinetic energy of the alpha particle as it approaches the lead nucleus. When it is at the closest approach, all of its kinetic energy is converted to electric potential energy due to the repulsive Coulomb force between the positively charged alpha particle and the lead nucleus.

The conservation of energy states that the initial kinetic energy (KE) of the alpha particle is equal to the electric potential energy (PE) at the closest distance of approach (r): KEinitial = PE = (k_e)(Z_1Z_2e^2)/(r), where k_e is the Coulomb's constant, Z_1 and Z_2 are the charges of the alpha particle and the lead nucleus respectively, and e is the elementary charge. We then solve for r to find the distance of closest approach.