/*! 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 42 An electron and a positron have ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An electron and a positron have masses of \(9.11 \times 10^{-31} \mathrm{kg}\). They collide and both vanish, with only electromagnetic radiation appearing after the collision. If each particle is moving at a speed of \(0.20 \mathrm{c}\) relative to the laboratory before the collision, determine the energy of the electromagnetic radiation.

Short Answer

Expert verified
The energy of the electromagnetic radiation is approximately \(1.6712 \times 10^{-13} \text{ J}\).

Step by step solution

01

Rest Energy Calculation

First, we calculate the rest energy of one electron and one positron. The rest energy can be found using the equation \( E = mc^2 \) where \( m \) is the mass of the particle and \( c \) is the speed of light \( (3\times10^8 \, \text{m/s}) \). Thus, the rest energy of one electron or positron is \( E_0 = (9.11 \times 10^{-31} \, \text{kg})(3 \times 10^8 \, \text{m/s})^2 = 8.19 \times 10^{-14} \, \text{J} \). Since we have both an electron and a positron, the total rest energy is twice this value: \( 2 \times 8.19 \times 10^{-14} \, \text{J} = 1.638 \times 10^{-13} \, \text{J} \).
02

Kinetic Energy Calculation

Next, we find the kinetic energy of each particle before collision. The formula for relativistic kinetic energy is \( KE = (\gamma - 1)mc^2 \), where \( \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} \) is the Lorentz factor. Here, \( v = 0.20c \), hence, \( \gamma = \frac{1}{\sqrt{1 - (0.20)^2}} = \frac{1}{\sqrt{0.96}} \approx 1.0204 \). The kinetic energy for one particle is therefore \( KE = (1.0204 - 1)(9.11 \times 10^{-31})(3 \times 10^8)^2 \approx 1.66 \times 10^{-15} \, \text{J} \). For both particles, the total kinetic energy is \( 2 \times 1.66 \times 10^{-15} = 3.32 \times 10^{-15} \, \text{J} \).
03

Total Pre-Collision Energy

We calculate the total energy present before the collision, combining rest energy and kinetic energy. We have already calculated the total rest energy to be \( 1.638 \times 10^{-13} \, \text{J} \) and the total kinetic energy to be \( 3.32 \times 10^{-15} \, \text{J} \). Thus, the total pre-collision energy is \( 1.638 \times 10^{-13} \, \text{J} + 3.32 \times 10^{-15} \, \text{J} \approx 1.6712 \times 10^{-13} \, \text{J} \).
04

Post-Collision Energy

Since energy is conserved during the collision and both the electron and positron are converted entirely into electromagnetic radiation, the total energy calculated prior to collision will be converted into the electromagnetic radiation energy. Thus, the energy of the electromagnetic radiation produced is also \( 1.6712 \times 10^{-13} \, \text{J} \).

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 91Ó°ÊÓ!

Key Concepts

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

Relativistic Energy
When dealing with particles moving at speeds close to the speed of light, it is essential to consider relativistic effects. This is because classical physics does not apply to particles moving at such high velocities. Relativistic energy includes not just the energy from motion but also the intrinsic energy contained in the mass of the particles. The total relativistic energy of a particle is given by the equation:\[ E = \gamma mc^2 \]where \( E \) is the total energy, \( m \) is the mass, \( c \) is the speed of light, and \( \gamma \) is the Lorentz factor. This equation shows how energy increases significantly as an object's speed approaches the speed of light. Relativistic energy is particularly important when calculating energies involved in high-speed particle interactions, such as electron-positron annihilation.
Rest Energy
Every particle possesses rest energy, which is the energy inherent to the mass of the particle when it's not in motion. This energy is given by Einstein’s famous equation:\[ E_0 = mc^2 \]Here, \( E_0 \) represents the rest energy, \( m \) is the rest mass of the particle, and \( c \) is the speed of light. Rest energy is the most basic form of energy a particle can have, and for the electron and the positron, this is a substantial portion of their total energy. Understanding rest energy is crucial in particle physics because it sets a baseline for conversion during particle reactions, such as in annihilation, where rest mass contributes to radiation energy.
Kinetic Energy
Kinetic energy in the realm of relativity is a bit different from the classical perspective because it includes the Lorentz factor. At high velocities, kinetic energy is no longer simply \(1/2 mv^2\). Instead, we use:\[ KE = (\gamma - 1)mc^2 \]This formula shows how, as particles move faster and closer to the speed of light, their kinetic energy increases disproportionately compared to what Newtonian physics would predict. For an electron and positron moving at \(0.20c\), this results in additional energy being available for conversion during their annihilation, which adds to the overall energy seen in the resulting electromagnetic radiation.
Lorentz Factor
The Lorentz factor, denoted as \( \gamma \), is a crucial component when discussing relativistic dynamics. It accounts for the effects of time dilation and length contraction at relativistic speeds, given by:\[ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \]Where \( v \) is the velocity of the particle and \( c \) is the speed of light. As the speed of a particle approaches the speed of light, \( \gamma \) increases rapidly. This factor modifies both kinetic energy and momentum in relativistic equations. For example, an electron moving at \(0.20c\) has a \( \gamma \) of approximately 1.0204, introducing corrections to the energy calculations that are necessary to accurately describe the behavior of particles at relativistic speeds.
Energy Conservation
In physics, one of the most fundamental principles is the conservation of energy, which states that energy in an isolated system remains constant. During electron-positron annihilation, this principle is perfectly illustrated. Initially, the system's energy consists of both the rest energy and the kinetic energy of the particles. When they collide and annihilate, all of this energy is transformed into electromagnetic radiation.The equation that represents this conservation in an annihilation scenario is:\[ E_{\text{initial}} = E_{\text{final}} \]Here, \( E_{\text{initial}} \) is the total energy before collision, including both rest and kinetic energy, and \( E_{\text{final}} \) is the energy of the resulting radiation. For the electron and positron in the exercise, their combined energy of \(1.6712 \times 10^{-13} \text{ J} \) validates the principle of conservation, as the same amount is converted into radiant energy post-annihilation.

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

ssm A particle known as a pion lives for a short time before breaking apart into other particles. Suppose that a pion is moving at a speed of \(0.990 c\), and an observer who is stationary in a laboratory measures the pion's lifetime to be \(3.5 \times 10^{-8}\) s. (a) What is the lifetime according to a hypothetical person who is riding along with the pion? (b) According to this hypothetical person, how far does the laboratory move before the pion breaks apart?

There are many astonishing consequences of special relativity, two of which are time dilation and length contraction. Problem 50 reviews these important concepts in the context of a golf game in a hypothetical world where the speed of light is only a little faster than that of a golf cart. Other important consequences of special relativity are the equivalence of mass and energy, and the dependence of kinetic energy on the total energy and on the rest energy. Problem 51 serves as a review of the roles played by mass and energy in special relativity. Imagine playing golf in a world where the speed of light is only \(c=3.40 \mathrm{m} / \mathrm{s} .\) Golfer \(\mathrm{A}\) drives a ball down a flat horizontal fairway for a distance that he measures as \(75.0 \mathrm{m}\). Golfer \(\mathrm{B}\), riding in a cart, happens to pass by just as the ball is hit (see the figure). Golfer A stands at the tee and watches while golfer \(\mathrm{B}\) moves down the fairway toward the ball at a constant speed of \(2.80 \mathrm{m} / \mathrm{s} .\) Concepts: (i) Who measures the proper length of the drive, and who measures the contracted length? (ii) Who measures the proper time interval, and who measures the dilated time interval? Calculations: (a) How far is the ball hit according to golfer \(\mathrm{B} ?\) (b) According to each golfer, how much time does it take golfer \(B\) to reach the ball?

A spaceship travels at a constant speed from earth to a planet orbiting another star. When the spacecraft arrives, 12 years have elapsed on earth, and 9.2 years have elapsed on board the ship. How far away (in meters) is the planet, according to observers on earth?

A radar antenna is rotating and makes one revolution every 25 s, as measured on earth. However, instruments on a spaceship moving with respect to the earth at a speed \(v\) measure that the antenna makes one revolution every 42 s. What is the ratio o\(/ \mathrm{c}\) of the speed \(v\) to the speed \(c\) of light in a vacuum?

Three particles are listed in the table. The mass and speed of each particle are given as multiples of the variables \(m\) and \(v,\) which have the values \(m=1.20 \times 10^{-8} \mathrm{kg}\) and \(v=0.200 \mathrm{c} .\) The speed of light in a vacuum is \(c=3.00 \times 10^{8} \mathrm{m} / \mathrm{s} .\) Determine the momentum for each particle according to special relativity. $$ \begin{array}{ccc} \text { Particle } & \text { Mass } & \text { Speed } \\ \hline \mathbf{a} & m & v \\ \hline \mathbf{b} & \frac{1}{2} m & 2 v \\ \hline \mathbf{c} & \frac{1}{4} m & 4 v \\ \hline \end{array} $$

See all solutions

Recommended explanations on Physics Textbooks

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.