Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 10

Suppose we have \(Z\) protons and have to distribute them into two nuclei, one with \(Z_{1}\) protons and the other with \(Z_{2}\) protons \(\left(Z=Z_{1}+Z_{2}\right)\) (a) What arrangements give the maximum and minimum Coulomb repulsion between the two nuclei? (b) What does this tell you about the types of fusion that are most likely to take place in stars?

Short Answer

Expert verified
Maximum repulsion occurs when \(Z_1 \approx Z_2\). Minimum repulsion occurs when one nucleus has all protons. Fusion is more likely with lighter elements.

Step by step solution

01

Understand the problem

We need to find the arrangements of distributing Z protons into two nuclei such that the Coulomb repulsion is maximized and minimized. We also need to determine the implications for fusion in stars.
02

Define Coulomb Repulsion

The Coulomb repulsion between two point charges is given by \[ F = k_e \frac{Z_1 Z_2}{r^2} \] where, \(F\) is the force, \(k_e\) is Coulomb's constant, \(Z_1\) and \(Z_2\) are the charges, and \(r\) is the distance between them. For simplicity, assume \(r\) to be constant.
03

Maximize Coulomb Repulsion

To maximize the repulsion, distribute the charges such that the product \(Z_1 Z_2\) is maximized. Since \(Z = Z_1 + Z_2\), maximum product is achieved when \(Z_1\) and \(Z_2\) are closest to equal. If \(Z\) is even, then \(Z_1 = Z_2 = Z/2\). If \(Z\) is odd, then \(Z_1 = \lfloor Z/2 \rfloor\) and \(Z_2 = \lceil Z/2 \rceil\).
04

Minimize Coulomb Repulsion

To minimize the repulsion, distribute the charges such that one nucleus has all the protons and the other has none. Thus, either \(Z_1 = Z\) and \(Z_2 = 0\) or vice versa.
05

Implications for Stellar Fusion

Fusion in stars is more likely when repulsion is minimized. Heavier nuclei (with higher proton counts) would need to overcome significant Coulomb repulsion to fuse. Hence, fusion processes involving lighter elements (like hydrogen) are more common in stellar environments.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Coulomb Repulsion
Coulomb repulsion refers to the electrostatic force that pushes two similarly charged particles away from each other. This is especially important in the context of nuclear fusion, where protons (positively charged particles) come close enough to fuse into heavier nuclei.

The formula for Coulomb repulsion is given by:






    The force ( F ) is directly proportional to the product of the charges ( Z 1 and 1 + 2 ) and inversely proportional to the square of the distance constant, k e , is Coulomb constant. In a stellar environment, minimizing Coulomb repulsion is critical for enabling nuclear fusion since a high repulsion force makes it harder for nuclei to come close enough to fuse.
Nuclear Charges Distribution
When discussing nuclear charges distribution, we are primarily concerned with how protons are spread out between different nuclei. This distribution directly affects Coulomb repulsion. For example:
  • Maximize Repulsion: If we want to maximize the repulsive force, protons should be distributed evenly between the two nuclei. For example, if Z = 6 , then Z 1 and Z 2 3 each.

If distribution is relatively equal, becomes
  • Minimize Repulsion: The repulsion is minimized when one nucleus contains all the protons. That is either Z 1 = Z and

    Coulomb repulsion is significantly with uneven proton when force in distribution.
Fusion in Stars
Fusion in stars is a process where lighter elements fuse to form heavier elements, releasing a vast amount of energy. This process is highly dependent on overcoming Coulomb repulsion between positively charged protons.

However, due to the significant repulsion between heavier nuclei, lighter elements like hydrogen are more likely to fuse. For example:
  • Hydrogen Fusion: This involves the fusion of hydrogen atoms to form helium. Because hydrogen has only one proton, the Coulomb repulsion is relatively low, making it easier to achieve the conditions necessary for fusion.
  • Helium Fusion: Fusion processes for heavier elements like helium require much higher temperatures and pressures due to increased Coulomb repulsion.


Therefore, fusion is more energy-efficient with lighter nuclei. This principle is what powers our Sun, where hydrogen fuses into helium, releasing energy in the form of light and heat.

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

Suppose the density of a star is given by. $$\rho(r)=\left\\{\begin{array}{ll} \rho_{0} & r

Find the gravitational potential energy of a sphere with a \(1 / r^{2}\) density distribution. Take the total mass of the sphere to be \(M,\) and let the density \(\rho(r)=\rho_{0} / r^{2}\) out to a radius, \(R\).
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Fluids

Read Explanation

Physics of Motion

Read Explanation

Oscillations

Read Explanation

Energy Physics

Read Explanation

Classical Mechanics

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.

Sign-up for free