/*! 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 33 An antimatter reactor. When a pa... [FREE SOLUTION] | 91Ó°ÊÓ

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An antimatter reactor. When a particle meets its antipar- ticle (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 \(\left(7.86 / \mathrm{cm}^{3}\right)\) 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
The yearly mass consumption is 1.11 × 10³ kg, and the cube pile height is approximately 0.53 m.

Step by step solution

01

Understanding the problem

We are given the total energy consumption of the United States in a year, which is approximately \(1.0 \times 10^{20} \mathrm{J}\). The problem asks us to calculate the mass consumed if all this energy was produced by an antimatter reactor, applying Einstein's mass-energy equivalence principle.
02

Applying Einstein's Mass-Energy Equivalence

Einstein's formula for mass-energy equivalence is \(E = mc^2\). We need to rearrange this to solve for mass \(m\) as \(m = \frac{E}{c^2}\), where \(c\) is the speed of light \(3.0 \times 10^8 \text{m/s}\).
03

Calculating the Mass (Part a)

Substitute the given energy into the formula: \(m = \frac{1.0 \times 10^{20} \text{ J}}{(3.0 \times 10^8 \text{ m/s})^2}\). Calculate this to find the mass. The result is \(m = \frac{1.0 \times 10^{20}}{9.0 \times 10^{16}} = 1.11 \times 10^3 \text{ kg}\).
04

Understanding the density problem

For part (b), the problem asks us to find how big the stack of antimatter would be if it were as dense as iron (Fe) at \(7.86 \text{ g/cm}^3\). Convert this density to \(7.86 \times 10^3 \text{ kg/m}^3\) for consistency with the mass units.
05

Calculating the volume

Use the density formula \( \text{density} = \frac{\text{mass}}{\text{volume}} \) and rearrange to find the volume: \( \text{volume} = \frac{\text{mass}}{\text{density}} \). Substitute in the mass (1.11 \times 10^3 kg) and density (7860 kg/m^3): \( volume = \frac{1.11 \times 10^3}{7.86 \times 10^3} \approx 0.141 \text{ m}^3 \).
06

Calculate the side of the cube (width)

To find the height of the cubical pile, calculate the cube root of the volume since the volume of a cube is \( ext{volume} = ext{side}^3 \). Therefore, \( ext{side} = \sqrt[3]{0.141} \approx 0.53 \text{ m} \).

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Key Concepts

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

Mass-Energy Equivalence
Mass-energy equivalence is a fascinating concept that connects mass and energy as two sides of the same coin. This was first introduced by Albert Einstein in his theory of relativity. It suggests that mass can be converted into energy and vice versa. This is particularly important in processes like nuclear reactions and, as mentioned, hypothetically in antimatter reactions too.
When matter and antimatter meet, they annihilate each other, converting their entire mass into energy. This concept is pivotal when imagining futuristic technologies like antimatter reactors as mentioned in the exercise. Knowing this principle allows us to calculate how much mass is needed to produce a specific amount of energy, which we do using Einstein's famous formula.
Einstein's Formula
Einstein's mass-energy equivalence is elegantly captured by the simple and powerful formula: \[ E = mc^2 \]Where:
  • \(E\) is the energy, measured in joules (J).
  • \(m\) is the mass, measured in kilograms (kg).
  • \(c\) is the speed of light in a vacuum, approximately \(3.0 \times 10^8 \text{ m/s}\).
This equation tells us that a small amount of mass can be transformed into a massive amount of energy due to the speed of light being such a large number. Rearranging this formula helps calculate the mass from known energy values.In the exercise, by rearranging to \[ m = \frac{E}{c^2} \] we can determine how much mass needs to be annihilated to yield the energy demands of the United States if powered solely by antimatter.
Density of Materials
Density plays a crucial role in determining how much space a specific mass will occupy. The density of a material is defined as its mass per unit volume, typically expressed in kilograms per cubic meter (kg/m³) or grams per cubic centimeter (g/cm³). In the exercise, the density of iron (Fe) is provided as \(7.86\, \text{g/cm}^3\), which is converted to \(7860\, \text{kg/m}^3\) for calculation purposes.
This density determines how tightly packed the material's molecules are in a given object. So for antimatter fuel with the density of Fe, knowing the density allows us to calculate the volume necessary to store the given mass of antimatter. Understanding this concept illuminates how real-life physics applies to theoretical scenarios like antimatter storage.
Volume Calculation
Calculating volume is essential when dealing with physical objects. Volume is the measure of how much space an object or substance occupies and is often calculated in cubic meters (m³) or centimeters cubed (cm³).
For a substance with a known mass and density, the formula to calculate volume is:\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]In the given exercise, after finding the mass of antimatter needed (\(1.11 \times 10^3\, \text{kg}\)), the volume occupied by this mass with the density of iron is determined. This results in a volume of approximately \(0.141\, \text{m}^3\).
To find the height of a cubical stack of this antimatter, we use the formula for the volume of a cube, \(\text{Volume} = \text{Side}^3\), to determine the length of one side of the cube. For this exercise, the side length of the cube, or the height, comes out to approximately \(0.53\, \text{m}\). This allows one to visualize the physical height of such an antimatter stack.

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Most popular questions from this chapter

Space travel? Travel to the stars requires hundreds or thousands of years, even at the speed of light. Some people have suggested that we can get around this difficulty by accelerating the rocket (and its astronauts) to very high speeds so that they will age less due to time dilation. The fly in this ointment is that it takes a great deal of energy to do this. Suppose you want to go to the immense red giant Betelgeuse, which is about 500 light years away. (A light year is the distance that light travels in one year.) You plan to travel at constance that in a 1000 kg rocket ship (a little over a ton), which, in reality, is far too small for this purpose. In each case that follows, calculate the time for the trip, as measured by people on earth and by astronauts in the rocket ship, the energy needed in joules, and the energy needed as a percent of U.S. yearly use (which is \(1.0 \times 10^{20} \mathrm{J} ) .\) For comparison, arrange your results in a table showing \(v_{\text { Rocket }}, t_{\text { Earth }}, t_{\text { Rocket }}, E(\) in \(\mathrm{J}),\) and \(E\) (as \(\%\) of U.S. use \() .\) The rocket ship's speed is (a) \(0.50 c,\) (b) 0.99\(c\) , and (c) 0.9999\(c .\) On the basis of your results, does it seem likely that any government will invest in such high- speed space travel any time soon?

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