/*! 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 53 Nuclear fusion reactions require... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 23: Problem 53

Nuclear fusion reactions require that positively charged nuclei be brought into close proximity, against the electrostatic repulsion. As a simple example, suppose a proton is fired at a second, stationary proton from a large distance away. What kinetic energy must be given to the moving proton to get it to come within \(1.00 \cdot 10^{-15} \mathrm{~m}\) of the target? Assume that there is a head-on collision and that the target is fixed in place.

Short Answer

Expert verified
Answer: The initial kinetic energy required for the proton is \(2.30 \cdot 10^{-13} \mathrm{~J}\).

Step by step solution

01

Identify the knowns and what we need to find

We know the distance that the protons need to come within, as well as the electric charge on a proton. We need to find the initial kinetic energy required for the moving proton.
02

Calculate the electric potential energy

When the protons are at the closest separation of \(1.00 \cdot 10^{-15} \mathrm{~m}\), the electric potential energy can be calculated using Coulomb's Law: \(U = \frac{k \cdot q_1 \cdot q_2}{r}\) where \(U\) is the potential energy, \(k = 8.99 \cdot 10^9 \mathrm{~N m^2/C^2}\) is the electrostatic constant, \(q_1 = q_2 = 1.60 \cdot 10^{-19} \mathrm{~C}\) are the charges of the protons, and \(r = 1.00 \cdot 10^{-15} \mathrm{~m}\) is the distance between the protons. Calculate the energy: \(U = \frac{(8.99 \cdot 10^9) \cdot (1.60 \cdot 10^{-19})^2}{1.00 \cdot 10^{-15}}\)
03

Solve for the electric potential energy

Performing the calculation, we get: \(U \approx 2.30 \cdot 10^{-13} \mathrm{~J}\) So, the electric potential energy when the two protons are \(1.00 \cdot 10^{-15} \mathrm{~m}\) apart is \(2.30 \cdot 10^{-13} \mathrm{~J}\).
04

Use energy conservation to find the initial kinetic energy

Since energy is conserved, the initial kinetic energy of the moving proton, \(K\), must be equal to the sum of the final kinetic energy and the final electric potential energy. However, the final kinetic energy of the moving proton is assumed to be zero, so the initial kinetic energy is equal to the electric potential energy: \(K = U = 2.30 \cdot 10^{-13} \mathrm{~J}\)
05

Write the final answer

The initial kinetic energy required for the moving proton to come within \(1.00 \cdot 10^{-15} \mathrm{~m}\) of the stationary proton is \(2.30 \cdot 10^{-13} \mathrm{~J}\).

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!

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

Consider an electron in the ground state of the hydrogen atom, separated from the proton by a distance of \(0.0529 \mathrm{nm}\) a) Viewing the electron as a satellite orbiting the proton in the electrostatic potential, calculate the speed of the electron in its orbit. b) Calculate an effective escape speed for the electron. c) Calculate the energy of an electron having this speed, and from it determine the energy that must be given to the electron to ionize the hydrogen atom.

A negatively charged particle revolves in a clockwise direction around a positively charged sphere. The work done on the negatively charged particle by the electric field of the sphere is a) positive. b) negative. c) zero.

A proton is placed midway between points \(A\) and \(B\). The potential at point \(A\) is \(-20 \mathrm{~V}\), and the potential at point \(B\) \(+20 \mathrm{~V}\). The potential at the midpoint is \(0 \mathrm{~V}\). The proton will a) remain at rest. b) move toward point \(B\) with constant velocity. c) accelerate toward point \(A\). d) accelerate toward point \(B\). e) move toward point \(A\) with constant velocity.

The electric field, \(\vec{E}(\vec{r}),\) and the electric potential \(V(\vec{r}),\) are calculated from the charge distribution, \(\rho(\vec{r}),\) by integrating Coulomb's Law and then the electric field. In the other direction, the field and the charge distribution are determined from the potential by suitably differentiating. Suppose the electric potential in a large region of space is given by \(V(r)=V_{0} \exp \left(-r^{2} / a^{2}\right),\) where \(V_{0}\) and \(a\) are constants and \(r=\sqrt{x^{2}+y^{2}+z^{2}}\) is the distance from the origin. a) Find the electric field \(\vec{E}(\vec{r})\) in this region. b) Determine the charge density \(\rho(\vec{r})\) in this region, which gives rise to the potential and field. c) Find the total charge in this region. d) Roughly sketch the charge distribution that could give rise to such an electric field.

Use \(V=\frac{k q}{r}, E_{x}=-\frac{\partial V}{\partial x}, E_{y}=-\frac{\partial V}{\partial y},\) and \(E_{z}=-\frac{\partial V}{\partial z}\) to derive the expression for the electric field of a point charge, \(q\)
See all solutions

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

Read Explanation

Classical Mechanics

Read Explanation

Fundamentals of Physics

Read Explanation

Electricity

Read Explanation

Circular Motion and Gravitation

Read Explanation

Waves Physics

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.