/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 Calculate the height of the Coul... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 43: Problem 31

Calculate the height of the Coulomb barrier for the head-on collision of two deuterons, with effective radius \(2.1 \mathrm{fm}\).

Short Answer

Expert verified
The height of the Coulomb barrier for the collision of two deuterons is approximately 0.343 MeV.

Step by step solution

01

Understanding the Coulomb Barrier

The Coulomb barrier is the energy barrier due to electrostatic force that two positively charged nuclei must overcome to get close enough to undergo a nuclear reaction. In this exercise, we need to calculate this energy for two deuterons.
02

Formula for Coulomb Potential Energy

To calculate the Coulomb barrier, we use the formula for the Coulomb potential energy: \[ V = \frac{k \cdot e^2}{r} \]where:- \( V \) is the potential energy,- \( k \) is Coulomb's constant \( 8.9875 \times 10^9 \, \text{N m}^2/\text{C}^2 \),- \( e \) is the elementary charge \( 1.602 \times 10^{-19} \, \text{C} \),- \( r \) is the distance between the charges, which is twice the effective radius given \( r = 2.1 \, \text{fm} \times 2 \).
03

Convert Effective Radius to Meters

The given effective radius is \(2.1 \, \text{fm}\). We need to convert it into meters: \[ 1 \, \text{fm} = 1 \times 10^{-15} \, \text{m} \]Thus, the distance \( r \) is:\[ r = 2 \times 2.1 \, \text{fm} = 4.2 \, \text{fm} = 4.2 \times 10^{-15} \, \text{m} \]
04

Calculate Coulomb Potential Energy

Substitute the values into the Coulomb potential energy formula:\[ V = \frac{8.9875 \times 10^9 \, \text{N m}^2/\text{C}^2 \times (1.602 \times 10^{-19} \, \text{C})^2}{4.2 \times 10^{-15} \, \text{m}}\]Simplifying the above expression gives:\[ V \approx \frac{8.9875 \times 10^9 \times 2.5664 \times 10^{-38}}{4.2 \times 10^{-15}} \approx 5.5 \times 10^{-14} \, \text{J} \]
05

Convert Joules to MeV

To express the energy in MeV, we convert Joules to electronvolts and then MeV:\[ 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \]Thus, for the potential energy in eV:\[ V = \frac{5.5 \times 10^{-14} \, \text{J}}{1.602 \times 10^{-19} \, \text{J/eV}} \approx 3.43 \times 10^5 \, \text{eV} \approx 343 \text{ MeV} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Coulomb Potential Energy
Coulomb potential energy refers to the energy between two charged particles due to their electrostatic interaction. This energy is crucial when studying nuclear reactions, as it determines the barrier that must be overcome for reactions to occur.

The formula for Coulomb potential energy is given by:
  • \[ V = \frac{k \cdot e^2}{r} \]
  • where \( V \) is the potential energy,
  • \( k \) is Coulomb's constant (\( 8.9875 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
  • \( e \) is the elementary charge (\( 1.602 \times 10^{-19} \, \text{C} \)),
  • \( r \) is the distance between the charges.
The Coulomb potential energy plays a significant role in calculating the height of the Coulomb barrier, which defines the energy required to bring two charged particles, like deuterons, close enough to interact.
Nuclear Reaction
A nuclear reaction involves the transformation of atomic nuclei, resulting in the release or absorption of energy. This type of reaction is different from chemical reactions, which involve only the electrons orbiting a nucleus.

In nuclear reactions, like the fusion of deuterons, the energy barrier to be overcome is largely due to the Coulomb potential. The positive charges in the nuclei repel each other, creating a barrier that must be surpassed for them to collide and initiate a reaction. Overcoming this barrier can lead to the release of a significant amount of energy, often much larger than that of chemical reactions.
Deuterons Collision
Deuterons are nuclei of deuterium, an isotope of hydrogen, consisting of one proton and one neutron. In a deuteron collision leading to nuclear fusion, the deuterons must come very close to each other, defying their natural electrostatic repulsion.

During a head-on collision, calculations focus on determining the height of the Coulomb barrier that these deuterons need to overcome. This is effected by the attributes of the deuterons themselves, such as their effective radius and the distance at which the repulsion is significant. Calculating the potential energy helps to understand how feasible it is for deuterons to fuse given specific conditions.
Coulomb's Constant
Coulomb's constant \( k \) is a fundamental value in electrostatics that quantifies the strength of the electrostatic force between two point charges. The value of Coulomb's constant is approximately \( 8.9875 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

This constant is often used in calculations related to electric forces and energy between charged bodies, especially in nuclear physics. It helps in determining the force or potential energy as seen in the Coulomb barrier equation. Understanding its role is crucial for calculating energy barriers in nuclear reactions, allowing predictions of reaction conditions.
Elementary Charge
The elementary charge \( e \) is the fundamental unit of electric charge found in every proton and is approximately \( 1.602 \times 10^{-19} \, \text{C} \). It represents the smallest charge any free particle can have and is an important quantity in physics, especially when dealing with atomic and subatomic particles.

When two particles, each with an elementary charge, come close enough, the potential energy due to the Coulombic interaction is calculated using \( e^2 \) in the Coulomb potential energy formula. This highlights the role of the elementary charge in assessing interactions and forces at the microscopic level, especially significant in nuclear physics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Verify that, as stated in Module 43 -1, neutrons in equilibrium with matter at room temperature, \(300 \mathrm{~K}\), have an average kinetic energy of about \(0.04 \mathrm{eV}\).

The neutron generation time \(t_{\text {gen }}\) in a reactor is the average time needed for a fast neutron emitted in one fission event to be slowed to thermal energies by the moderator and then initiate another fission event. Suppose the power output of a reactor at time \(t=0\) is \(P_{0} .\) Show that the power output a time \(t\) later is \(P(t),\) where \(P(t)=P_{0} k^{\text {ligen}}\) and \(k\) is the multiplication factor. For constant power output, \(k=1\)

(a) A neutron of mass \(m_{\mathrm{n}}\) and kinetic energy \(K\) makes a head-on elastic collision with a stationary atom of mass \(m\). Show that the fractional kinetic energy loss of the neutron is given by $$ \frac{\Delta K}{K}=\frac{4 m_{\mathrm{n}} m}{\left(m+m_{\mathrm{n}}\right)^{2}} $$ Find \(\Delta K / K\) for each of the following acting as the stationary atom: (b) hydrogen, (c) deuterium, (d) carbon, and (e) lead. (f) If \(K=1.00 \mathrm{MeV}\) initially, how many such head-on collisions would it take to reduce the neutron's kinetic energy to a thermal value \((0.025 \mathrm{eV})\) if the stationary atoms it collides with are deuterium, a commonly used moderator? (In actual moderators, most collisions are not head-on.)

What is the \(Q\) of the following fusion process? \({ }^{2} \mathrm{H}_{1}+{ }^{1} \mathrm{H}_{1} \rightarrow{ }^{3} \mathrm{He}_{2}+\) photon Here are some atomic masses. \({ }^{2} \mathrm{H}_{1} \quad 2.014102 \mathrm{u} \quad{ }^{1} \mathrm{H}_{1} \quad 1.007825 \mathrm{u}\) \({ }^{3} \mathrm{He}_{2} \quad 3.016029 \mathrm{u}\)

A star converts all its hydrogen to helium, achieving a \(100 \%\) helium composition. Next it converts the helium to carbon via the triple-alpha process, $$ { }^{4} \mathrm{He}+{ }^{4} \mathrm{He}+{ }^{4} \mathrm{He} \rightarrow{ }^{12} \mathrm{C}+7.27 \mathrm{MeV} $$ The mass of the star is \(4.6 \times 10^{32} \mathrm{~kg}\), and it generates energy at the rate of \(5.3 \times 10^{30} \mathrm{~W}\). How long will it take to convert all the helium to carbon at this rate?
See all solutions

Recommended explanations on Physics Textbooks

Rotational Dynamics

Read Explanation

Translational Dynamics

Read Explanation

Force

Read Explanation

Atoms and Radioactivity

Read Explanation

Solid State Physics

Read Explanation

Linear Momentum

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.