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Chapter 18: Problem 68

An alpha particle with a kinetic energy of \(10.0 \mathrm{MeV}\) makes a head-on collision with a gold nucleus at rest. What is the distance of closest approach of the two particles? (Assume that the gold nucleus remains stationary and that it may be treated as a point charge. The atomic number of gold is \(79,\) and an alpha particle is a helium nucleus consisting of two protons and two neutrons.)

Short Answer

Expert verified
The distance of closest approach is approximately \(3.14 \times 10^{-14}\, \text{m}\).

Step by step solution

01

Understand the Problem

The problem involves finding the distance of closest approach in a collision between an alpha particle and a gold nucleus. The alpha particle's initial kinetic energy is given as 10.0 MeV. The gold nucleus has a charge equivalent to 79 protons, while the alpha particle has the charge of 2 protons.
02

Use the Conservation of Energy

At the point of closest approach, all the kinetic energy of the alpha particle is converted into electrostatic potential energy. The conservation of energy can be expressed as:\[ K = \frac{k \cdot Q_1 \cdot Q_2}{r} \]where \(K = 10.0\, \text{MeV}\) is the kinetic energy, \(k = 8.99 \times 10^9\, \text{N m}^2/\text{C}^2\) is the Coulomb's constant, \(Q_1 = 2e\) is the charge of the alpha particle, \(Q_2 = 79e\) is the charge of the gold nucleus, and \(r\) is the distance of closest approach.
03

Convert Energy Units

Convert the kinetic energy from MeV to joules for consistency with the units of the Coulomb's constant. Use the conversion:\[ 1\, \text{MeV} = 1.60219 \times 10^{-13} \text{J} \]Thus, \( 10.0\, \text{MeV} = 1.60219 \times 10^{-12} \text{J} \).
04

Solve for Distance r

Rearrange the conservation of energy equation to solve for \(r\):\[ r = \frac{k \cdot Q_1 \cdot Q_2}{K} \]Substitute the values: \(k = 8.99 \times 10^9\, \text{N m}^2/\text{C}^2\), \(Q_1 = 2e = 2 \times 1.60219 \times 10^{-19}\, \text{C}\), \(Q_2 = 79e = 79 \times 1.60219 \times 10^{-19}\, \text{C}\), and \(K = 1.60219 \times 10^{-12}\, \text{J}\).
05

Simplify and Calculate

Plug the numbers into the equation:\[ r = \frac{(8.99 \times 10^9) \cdot (2 \times 1.60219 \times 10^{-19}) \cdot (79 \times 1.60219 \times 10^{-19})}{1.60219 \times 10^{-12}} \]Calculate to find \(r\). After computation, \(r \approx 3.14 \times 10^{-14}\, \text{m}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Alpha Particles
Alpha particles are small but quite significant in the world of nuclear physics. They are essentially helium nuclei, composed of two protons and two neutrons. These particles are emitted during radioactive decay processes like that of uranium or radium. Due to their positive charge, alpha particles interact strongly with other charged particles.
Their size and charge give them important properties:
  • Relatively massive compared to other subatomic particles, making them less likely to penetrate far into materials.
  • Very effective at ionizing atoms because they carry a double positive charge.
  • Their interactions with matter and other particles lead to interesting phenomena, such as in Rutherford's gold foil experiment.
Understanding alpha particles is crucial when studying nuclear reactions and phenomena.
Gold Nucleus
The gold nucleus presents an interesting partner in nuclear collision experiments. Gold, with its atomic number of 79, has a very large nucleus compared to many other elements. This means it has 79 protons and is typically surrounded by many neutrons, contributing to its overall stability and mass.
Gold nuclei are often used in experimental settings due to:
  • Their large charge, which is useful for experiments involving electrostatic forces.
  • High atomic weight, offering a stable target for collision experiments.
  • Rarity in naturally occurring radioactive forms, making them stable and safe to study under standard conditions.
In the realm of nuclear physics, gold nuclei help us understand fundamental processes and forces at play during atomic interactions.
Distance of Closest Approach
The distance of closest approach is key to understanding how charged particles, such as alpha particles, interact with other particles, like gold nuclei. At this point, the kinetic energy of the moving particle is completely converted into potential energy. This occurs because the electrostatic repulsive force becomes overwhelming as the particles approach each other closely.
Calculating the distance involves:
  • Understanding the conversion of kinetic energy to potential energy.
  • Recognizing that this distance is determined by the balance of kinetic energy and the repulsive electrostatic force.
  • Using mathematical equations, such as those derived from energy conservation principles, to find the exact distance.
This concept not only reveals insights into the interactions of subatomic particles but also showcases the nature of fundamental forces in action.
Conservation of Energy
The conservation of energy principle is a cornerstone in physics and crucial when analyzing particle collisions. In the case of nuclear collisions, this principle tells us that the total energy of the system remains the same, though it may transform from kinetic to potential energy.
For example, when considering a collision like that between an alpha particle and a gold nucleus:
  • The alpha particle's initial kinetic energy is completely converted into electrostatic potential energy at the closest approach point.
  • Mathematically, this means the initial kinetic energy equals the potential energy stored due to electrostatic forces at closest approach.
  • Using the equation for conservation of energy helps in calculating crucial values like the distance of closest approach.
This concept provides a vital framework for solving complex problems and understanding particle behavior in nuclear physics.
Coulomb's Law
Coulomb's law is an essential tool in understanding the forces between charged particles, such as alpha particles and gold nuclei during collisions. It describes how the electrostatic force between two charged objects works, depending on their charges and the distance between them:
  • The force is directly proportional to the product of the magnitudes of the two charges.
  • The force is inversely proportional to the square of the distance separating the charges.
  • This law allows us to calculate the potential energy stored due to electrostatic interactions, key to understanding energy transformation in nuclear collisions.
Using Coulomb's law in problems involving nuclear physics allows students to grasp how forces and energy behave at the microscopic level, providing a clearer picture of atomic interactions.

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Most popular questions from this chapter

In the text, it was shown that the energy stored in a capacitor \(C\) charged to a potential \(V\) is \(U=\frac{1}{2} Q V .\) Show that this energy can also be expressed as (a) \(U=Q^{2} / 2 C\) and (b) \(U=\frac{1}{2} C V^{2}\).

Cell membranes. Cell membranes (the walled enclosure around a cell) are typically about \(7.5 \mathrm{nm}\) thick. They are partially permeable to allow charged material to pass in and out, as needed. Equal but opposite charge densities build up on the inside and outside faces of such a membrane, and these charges prevent additional charges from passing through the cell wall. We can model a cell membrane as a parallel-plate capacitor, with the membrane itself containing proteins embedded in an organic material to give the membrane a dielectric constant of about 10\. (See Figure \(18.47 .)\) (a) What is the capacitance per square centimeter of such a cell wall? (b) In its normal resting state, a cell has a potential difference of \(85 \mathrm{mV}\) across its membrane. What is the electric field inside this membrane?

You make a capacitor by cutting the 15.0 -cm-diameter bottoms out of two aluminum pie plates, separating them by \(3.50 \mathrm{~mm}\), and connecting them across a \(6.00 \mathrm{~V}\) battery. (a) What's the capacitance of your capacitor? (b) If you disconnect the battery and separate the plates to a distance of \(3.50 \mathrm{~cm}\) without discharging them, what will be the potential difference between them?

You are working on an electronics project that requires a variety of capacitors, but you have only five capacitors available, each with a capacitance of \(100 \mathrm{nF}\). (a) What is the maximum capacitance that you can produce with these five capacitors? (b) What is the minimum capacitance, other than zero, that you can produce?

A capacitor consists of two parallel plates, each with an area of \(16.0 \mathrm{~cm}^{2},\) separated by a distance of \(0.200 \mathrm{~cm} .\) The material that fills the volume between the plates has a dielectric constant of \(5.00 .\) The plates of the capacitor are connected to a \(300 \mathrm{~V}\) battery. (a) What is the capacitance of the capacitor? (b) What is the charge on either plate? (c) How much energy is stored in the charged capacitor?
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