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When a particle meets its antiparticle (more about this in Chapter 30 ), they annihilate each other and their mass is converted to light energy. The United States uses approximately \(1.0 \times 10^{20} \mathrm{~J}\) of energy per year. (a) If all this energy came from a futuristic antimatter reactor, how much mass would be consumed yearly? (b) If this antimatter fuel had the density of Fe \(\left(7.86 \mathrm{~g} / \mathrm{cm}^{3}\right)\) and were stacked in bricks to form a cubical pile, how high would it be? (Before you get your hopes up, antimatter reactors are a long way in the future-if they ever will be feasible.)

Short Answer

Expert verified
a) 1.11 × 10³ kg of mass is consumed yearly. b) The pile would be approximately 0.53 meters high.

Step by step solution

01

Understand the Energy-Mass Conversion

When a particle and its antiparticle annihilate, their mass is converted entirely into energy, described by Einstein's mass-energy equivalence formula: \( E = mc^2 \). Here, \( E \) is the energy, \( m \) is the mass, and \( c \) is the speed of light in a vacuum \((3.00 \times 10^8 \, \text{m/s})\). We will use this formula to calculate the mass consumed.
02

Calculate the Mass Consumed

Using the formula from the previous step, set \( E = 1.0 \times 10^{20} \, \text{J} \). Rearrange the equation \( E = mc^2 \) to solve for \( m \): \[ m = \frac{E}{c^2} = \frac{1.0 \times 10^{20}}{(3.00 \times 10^8)^2} \approx 1.11 \times 10^3 \, \text{kg} \]. This is the mass of matter-antimatter consumed yearly.
03

Understand Volume Calculation Using Density

Density is defined as mass per unit volume \( (\rho = \frac{m}{V}) \). We can rearrange this equation to find the volume \( V = \frac{m}{\rho} \). We are given the density \( \rho = 7.86 \text{g/cm}^3 \), which is equivalent to \( 7860 \text{kg/m}^3 \) in SI units.
04

Calculate the Volume of Antimatter Consumed

Substitute the known mass \( m = 1.11 \times 10^3 \, \text{kg} \) and \( \rho = 7860 \text{kg/m}^3 \) into the equation for volume: \[ V = \frac{1.11 \times 10^3}{7860} \approx 0.141 \text{ m}^3 \]. This is the volume of the antimatter consumed annually.
05

Calculate the Height of the Cubical Pile

If we assume this volume forms a cube, the side length \( s \) of the cube can be found by \( s^3 = V \). Solving for \( s \), we find: \[ s = \sqrt[3]{0.141} \approx 0.53 \, \text{m} \]. This is the height of the cube.

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Key Concepts

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

Antimatter
Antimatter is the mirror image of ordinary matter. If matter consists of particles, like protons and electrons, then antimatter consists of antiparticles, such as antiprotons and positrons. These antiparticles look the same as their matter counterparts but have opposite charges. When matter and antimatter particles come into contact, they annihilate each other.
This means that they vanish and the mass that made them up is converted directly into energy. This process follows Einstein's mass-energy equivalence formula, which we'll touch on later.
  • Annihilation releases large amounts of energy, much more than typical chemical reactions.
  • The idea of using antimatter as an energy source is popular in science fiction, but in reality, it is currently impractical due to the difficulties in creating and storing antimatter.
  • Researchers study antimatter to learn about the universe's origins, as the Big Bang should have produced equal amounts of matter and antimatter.
Einstein's Equation
Einstein's mass-energy equivalence principle is one of the most famous equations in physics: \( E = mc^2 \).
This formula shows the relationship between mass (\( m \)) and energy (\( E \)), with \( c \) representing the speed of light in a vacuum, approximately \( 3.00 \times 10^8 \) meters per second. Here’s a breakdown of what this means:
  • Mass and energy are interchangeable; meaning mass can be converted into energy and vice versa.
  • The speed of light squared (\( c^2 \)) is a huge number, implying that even a small amount of mass can be converted into a large amount of energy.
  • This concept is critical in understanding nuclear reactions, such as those in stars or nuclear power plants.
  • The equation highlights why antimatter could theoretically be a powerful energy source, as whole mass transforms into energy.
In practical terms, using \( E = mc^2 \), we can calculate the mass of antimatter needed to provide a year’s worth of energy in our example exercise. By setting \( E \) (energy needed) and solving for \( m \), we find out how much antimatter would be required to meet the energy demands.
Density Calculation
Density is a measure of how much mass is contained in a given volume, mathematically expressed as \( \rho = \frac{m}{V} \). Here, \( \rho \) is the density, \( m \) is the mass, and \( V \) is the volume. This concept is crucial when thinking about how materials are organized in space.

To calculate how much space a given mass of antimatter would take up, given its density, you rearrange the formula to find volume: \( V = \frac{m}{\rho} \). Consider these key points:
  • The density of a material tells us how tightly its mass is packed, impacting its overall volume.
  • In practical scenarios, like the exercise, density helps us visualize how a particular mass fits into three-dimensional space.
  • For instance, with antimatter density similar to iron, we can find that the antimatter would occupy a very small volume because of the large energy it represents.
In the exercise, after calculating the volume using the known density and mass, the next step is geometry. Assuming the antimatter forms a cube, the length of each side is the cube root of the volume. This enables us to understand how the abstract concept of mass-energy equivalence translates into a physical form.

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Most popular questions from this chapter

(a) Through what potential difference does an electron have to be accelerated, starting from rest, to achieve a speed of \(0.980 c ?\) (b) What is the kinetic energy of the electron at this speed? Express your answer in joules and in electronvolts.

A particle of mass \(m\) is moving at a speed \(v .\) (a) At what fraction of the speed of light is the momentum of the particle twice the nonrelativistic Newtonian momentum \(p=m v ?\) (b) Can the magnitude of the relativistic momentum ever be less than that of the Newtonian momentum?

A muon is created \(55.0 \mathrm{~km}\) above the surface of the earth (as measured in the earth's frame). The average lifetime of a muon, measured in its own rest frame, is \(2.20 \mu \mathrm{s}\), and the muon we are considering has this lifetime. In the frame of the muon, the earth is moving toward the muon with a speed of \(0.9860 \mathrm{c}\). (a) In the muon's frame, what is its initial height above the surface of the earth? (b) In the muon's frame, how much closer does the earth get during the lifetime of the muon? What fraction is this of the muon's original height, as measured in the muon's frame? (c) In the earth's frame, what is the lifetime of the muon? In the earth's frame, how far does the muon travel during its lifetime? What fraction is this of the muon's original height in the earth's frame?

A proton (rest mass \(1.67 \times 10^{-27} \mathrm{~kg}\) ) has total energy that is 4.00 times its rest energy. What are (a) the kinetic energy of the proton; (b) the magnitude of the momentum of the proton; and (c) the speed of the proton?

An enemy spaceship is moving toward your starfighter with a speed of \(0.400 c,\) as measured in your reference frame. The enemy ship fires a missile toward you at a speed of \(0.700 c\) relative to the enemy ship. (See Figure \(27.24 .)\) (a) What is the speed of the missile relative to you? Express your answer in terms of the speed of light. (b) If you measure the enemy ship to be \(8.00 \times 10^{6} \mathrm{~km}\) away from you when the missile is fired, how much time, measured in your frame, will it take the missile to reach you?

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