Problem 82 - University Physics with Modern Physics

Question

An alpha particle with kinetic energy 11.0 MeV makes a head-on collision with a lead nucleus at rest. What is the distance of closest approach of the two particles? (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 particle is a helium nucleus, with atomic number \(2 .\) )

Step-by-Step Solution

Answer

The distance of closest approach is approximately 35.5 femtometers.

Step 1: Understanding the Problem

The problem is asking for the distance of closest approach in a head-on collision between an alpha particle (helium nucleus) and a lead nucleus. The closest approach is determined by equating the initial kinetic energy of the alpha particle to the electrostatic potential energy at the point of closest approach.

Step 2: Express Kinetic Energy

The initial kinetic energy (KE) of the alpha particle is given as 11.0 MeV. We'll need to use this energy in the energy conservation equation to find the distance of closest approach.

Step 3: Formula for Electrostatic Potential Energy

The electrostatic potential energy (PE) between the alpha particle and the lead nucleus is given by \[ U = \frac{k \cdot Q_1 \cdot Q_2}{r} \]where \( k = 8.9875 \times 10^9 \text{ N m}^2/\text{C}^2 \) is Coulomb's constant, \( Q_1 = 2e \) is the charge of the alpha particle, \( Q_2 = 82e \) is the charge of the lead nucleus, and \( r \) is the distance of closest approach.

Step 4: Equate Kinetic and Potential Energy

At the closest approach, all the kinetic energy has been converted into electrostatic potential energy:\[ \text{KE} = \frac{k \cdot Q_1 \cdot Q_2}{r} \]Convert the kinetic energy into joules (1 MeV = 1.60218 x 10^-13 J) and solve for \( r \).

Step 5: Solve for r

Set up the equation:\[ (11.0 \times 1.60218 \times 10^{-13}) = \frac{(8.9875 \times 10^9) \cdot (2 \cdot 1.60218 \times 10^{-19}) \cdot (82 \cdot 1.60218 \times 10^{-19})}{r} \]Simplify and solve for \( r \) to find the distance of closest approach.

Step 6: Calculation

After solving, we get:\[ r = \frac{(8.9875 \times 10^9) \cdot (2 \cdot 1.60218 \times 10^{-19}) \cdot (82 \cdot 1.60218 \times 10^{-19})}{11.0 \times 1.60218 \times 10^{-13}} \]\[ r \approx 3.55 \times 10^{-14} \text{ meters} \]Therefore, the distance of closest approach is approximately 35.5 femtometers.

Key Concepts

Alpha Particle Collision

An alpha particle collision is a fascinating interaction between two atomic nuclei. It involves an alpha particle, which is essentially a helium nucleus made up of two protons and two neutrons. When this alpha particle collides with another nucleus, like that of a lead atom, interesting phenomena occur.
In our example, an alpha particle with a kinetic energy of 11.0 MeV approaches a lead nucleus. This is considered a head-on collision. The distance at which these particles are closest during the collision is of great interest, as it demonstrates the balance between their kinetic and potential energies.

• Understanding these collisions helps scientists study nuclear reactions and forces at play within atomic nuclei.
• Such collisions are instrumental in fields like nuclear physics, where they provide insights into both the structure and behavior of atomic particles.

A critical part of analyzing these collisions is determining the distance of closest approach, which is where an alpha particle's kinetic energy is entirely converted into electrostatic potential energy.

Coulomb's Law

Coulomb's Law is a fundamental principle that describes the electrostatic interaction between two charged particles. It is integral in calculating the potential energy created during an alpha particle collision.
Specifically, Coulomb's Law helps us calculate the electrostatic potential energy (U) between the colliding charges: the alpha particle and the lead nucleus. The equation is given by:\[ U = \frac{k \cdot Q_1 \cdot Q_2}{r} \]where:

• \( k \)is Coulomb's constant, approximately \( 8.9875 \times 10^9 \ \text{N m}^2/\text{C}^2 \)
• \( Q_1 \)is the charge of the alpha particle.
• \( Q_2 \)is the charge of the lead nucleus.
• \( r \)is the distance between the two charges at the closest approach.

Coulomb's Law reveals that the force between two charges is inversely proportional to the square of the distance between them. This is why the distance of closest approach is crucial for calculating potential energy in nuclear collisions.

Kinetic Energy Conversion

In physics, the concept of kinetic energy conversion is essential for understanding what happens during particle collisions. As an alpha particle approaches a lead nucleus, its kinetic energy is gradually converted into electrostatic potential energy.
At the point of closest approach, the alpha particle's kinetic energy is entirely converted into potential energy. This is described by the equation:\[ \text{KE} = \frac{k \cdot Q_1 \cdot Q_2}{r} \]

• The initial kinetic energy is supplied by the alpha particle's movement.
• When kinetic energy is zero at closest approach, all energy is stored as potential energy between the particle and the nucleus.
• This conversion is reversible: Moving away, the potential energy turns back into kinetic energy.

Understanding this energy transformation allows physicists to predict the dynamics of alpha particle collisions and similar interactions in nuclear processes.

Nuclear Physics

Nuclear physics explores the atomic nucleus and the elements that make it up, like protons and neutrons. It is a field rich with discoveries about forces and reactions within an atom.
Alpha particle collisions exemplify nuclear physics in action. They are not just about particles crashing into each other, but about revealing the forces that hold nuclei together.

• Nuclear reactions provide insights into elements' stability and the energy processes at the sun's core.
• The study of these collisions helps refine models of fundamental forces, like the strong nuclear force.
• Applications include nuclear energy, medical imaging techniques, and understanding universe origins.

Alpha particle collisions, therefore, contribute significantly to advancing nuclear physics, deepening our understanding of the atomic world.