Question

A 4.78-MeV alpha particle from a \(^{226}\)Ra decay makes a head-on collision with a uranium nucleus. A uranium nucleus has 92 protons. (a) What is the distance of closest approach of the alpha particle to the center of the nucleus? Assume that the uranium nucleus remains at rest and that the distance of closest approach is much greater than the radius of the uranium nucleus. (b) What is the force on the alpha particle at the instant when it is at the distance of closest approach?

Short answer

The distance of closest approach is approximately 3.22 x 10^-14 meters. The force at that point is about 6.53 x 10^-10 Newtons.

Step-by-step solution

Understanding the Problem

We need to find the distance of closest approach for an alpha particle colliding head-on with a uranium nucleus. We will use the conservation of energy and electrostatic force concepts to solve the problem. Given the initial kinetic energy of the alpha particle and the charges, we can calculate the distance where the kinetic energy is completely converted into potential energy.

Conservation of Energy

The initial kinetic energy of the alpha particle is completely converted into electrostatic potential energy at the distance of closest approach. The initial kinetic energy (KE) is 4.78 MeV. We need to express this in joules:\[KE = 4.78 \text{ MeV} = 4.78 \times 1.602 \times 10^{-13} \text{ J}\]At the distance of closest approach, this kinetic energy is equal to the electrostatic potential energy (U) between the alpha particle and the uranium nucleus:\[U = \frac{k \cdot (2e) \cdot (92e)}{r}\]where \(k\) is Coulomb's constant, \(e\) is the charge of an electron, and \(r\) is the distance of closest approach.

Calculate the Distance of Closest Approach

Set the kinetic energy equal to the potential energy and solve for \(r\):\[KE = \frac{k \cdot (2e) \cdot (92e)}{r}\]Substitute the values:\[4.78 \times 1.602 \times 10^{-13} = \frac{8.99 \times 10^{9} \cdot 2 \times 1.602 \times 10^{-19} \cdot 92 \times 1.602 \times 10^{-19}}{r}\]Solve for \(r\):\[r \approx \frac{8.99 \times 10^{9} \cdot 2 \cdot 92 \cdot (1.602 \times 10^{-19})^2}{4.78 \times 1.602 \times 10^{-13}} \approx 3.22 \times 10^{-14} \text{ m}\]

Calculate the Force on the Alpha Particle

The force on the alpha particle at the distance of closest approach can be calculated using Coulomb's Law:\[F = \frac{k \cdot (2e) \cdot (92e)}{r^2}\]Substituting the previously found value for \(r\), and the other known values:\[F = \frac{8.99 \times 10^{9} \cdot 2 \times 1.602 \times 10^{-19} \cdot 92 \times 1.602 \times 10^{-19}}{(3.22 \times 10^{-14})^2}\]After calculation:\[F \approx 6.53 \times 10^{-10} \text{ N}\]

Conclusion

The distance of closest approach of the alpha particle to the center of the uranium nucleus is approximately \(3.22 \times 10^{-14} \text{ m}\). At this distance, the force on the alpha particle is approximately \(6.53 \times 10^{-10} \text{ N}\).

Key concepts

Conservation of Energy

Energy is never created or destroyed. It can transform from one form to another. This fundamental principle is known as the conservation of energy. Imagine an alpha particle flying towards a heavy uranium nucleus. It starts with a certain kinetic energy, due to its motion. As it gets closer to the nucleus, it slows down because it's being affected by the electrostatic force. This happens because like charges repel each other! At the closest point of approach, the alpha particle's kinetic energy is completely converted into electrostatic potential energy. To find this closest distance, we set the original kinetic energy equal to the electrostatic potential energy formula. In this scenario, the kinetic energy is given in MeV, which converts to joules, and the potential energy depends on the charges and their separation distance. Solving the equation gives us the exact distance where the kinetic energy has changed entirely into potential energy.

Alpha Particle Collision

An alpha particle collision with a nucleus is like a tiny game of cosmic pool. Alpha particles are essentially helium nuclei, which means they have two protons and two neutrons. In the given problem, the alpha particle approaches a uranium nucleus head-on. When two charged entities collide head-on, they interact via electrostatic forces. Alpha particles, in high-energy environments, can make such direct impact events. As the collision occurs, the repulsion between the positive charges of the alpha and uranium nuclei slows down the alpha particle until it momentarily stops. At this instance, it's at the closest point to the nucleus before it rebounds back. The collision does not involve the two nuclei merging. Instead, it's an interaction dictated by their electromagnetic repulsion due to the positive charges of the protons in both nuclei.

Electrostatic Potential Energy

Electrostatic potential energy arises from interactions between charged particles. When an alpha particle approaches a uranium nucleus, they both experience a force due to their charges, governed by Coulomb's Law. Coulomb's Law describes how two charged objects either attract or repel each other with a force proportional to the product of their charges and inversely proportional to the square of the distance between them. The formula for electrostatic potential energy, which depends on their charges and separation distance, tells us how energy varies as the distance changes. As our alpha particle approaches the uranium nucleus, the potential energy increases due to the repulsive force. This energy change is crucial because, as per the conservation of energy, it directly affects the kinetic energy of the alpha particle. The energy conversion between kinetic and potential energy allows us to calculate meaningful distances like the closest approach point.”}]} ponents. This section introduces these concepts in a clear and concise way, helping you deeply understand how they interrelate. Now, let's look at each topic in detail. rápidamente. Aquí tienes una introducción a mantener la energía potencial electrostática y el potencial constante para las partículas alfa. Como un resumen del potencial constante a lo largo de este trayecto en la partícula. Esto es debido a cómo cambia la energía mecánica total.