Problem 33 - College Physics

Question

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.)

Step-by-Step Solution

Answer

The yearly mass consumption is 1.11 × 10³ kg, and the cube pile height is approximately 0.53 m.

Step 1: 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.

Step 2: 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 \(3.0 \times 10^8 \text{m/s}\).

Step 3: 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 \(m = \frac{1.0 \times 10^{20}}{9.0 \times 10^{16}} = 1.11 \times 10^3 \text{ kg}\).

Step 4: 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.

Step 5: 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): \( volume = \frac{1.11 \times 10^3}{7.86 \times 10^3} \approx 0.141 \text{ m}^3 \).

Step 6: 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, \( ext{side} = \sqrt[3]{0.141} \approx 0.53 \text{ m} \).

Key Concepts

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.