Problem 37 - University Physics with Modern Physics

Question

When a particle meets its antiparticle, they annihilate each other and their mass is converted to light energy. The United States uses approximately \(1.0 \times 10^{19} \mathrm{J}\) of energy per year (a) If all this energy came from a futuristic ant-matter reactor, how much mass of matter and antimatter fuel would be consumed yearly? (b) If this fuel had the density of iron \(\left(7.86 \mathrm{g} / \mathrm{cm}^{3}\right)\) 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.)

Step-by-Step Solution

Answer

111 kg of matter and antimatter; cube side length approximately 24.1 cm.

Step 1: Determine Energy-Mass Equivalence

Using Einstein's mass-energy equivalence principle, we know \( E = mc^2 \), where \( E \) is energy, \( m \) is mass, and \( c \) is the speed of light (\( 3 \times 10^8 \text{ m/s} \)). Rearrange to find mass: \( m = \frac{E}{c^2} \).

Step 2: Calculate Total Mass Annihilated

The total energy needed per year is \( E = 1.0 \times 10^{19} \text{ J} \). Plug this into the equation to find mass: \[ m = \frac{1.0 \times 10^{19} \text{ J}}{(3 \times 10^8 \text{ m/s})^2} = \frac{1.0 \times 10^{19}}{9 \times 10^{16}} = 1.11 \times 10^2 \text{ kg} \]. This mass represents the combined total of matter and antimatter, so each contributes \( 55.5 \text{ kg} \).

Step 3: Determine the Volume of the Mass

To find the volume of this mass when stacked into a cubical shape, we use the density formula \( \rho = \frac{m}{V} \), where \( \rho = 7.86 \text{ g/cm}^3 = 7860 \text{ kg/m}^3 \). Solving for volume \( V \), we get: \[ V = \frac{111 \text{ kg}}{7860 \text{ kg/m}^3} = 0.0141 \text{ m}^3 \].

Step 4: Calculate the Height of the Cube

The volume of a cube is given by \( V = s^3 \), where \( s \) is the side length of the cube. Thus, \[ s = \sqrt[3]{0.0141} = 0.241 \text{ m} \]. Converting to centimeters, \( s \approx 24.1 \text{ cm} \).

Step 5: Conclusion: Interpret the Results

If all the energy needed by the United States in a year were derived from an antimatter reactor, a total of \( 111 \) kg of matter and antimatter would be annihilated. This would fit into a cube approximately \( 24.1 \) cm on each side.

Key Concepts

Mass-Energy Equivalence

Einstein's mass-energy equivalence principle is a cornerstone of modern physics. It provides the relationship between mass and energy, famously expressed by the formula \(E = mc^2\). Here, \(E\) represents energy, \(m\) stands for mass, and \(c\) is the constant speed of light in a vacuum, approximately \(3 \times 10^8 \text{ m/s}\). This equation tells us that mass and energy are interchangeable; they can be converted into one another. In the context of antimatter annihilation, when matter and antimatter come into contact, they destroy each other, converting their entire mass into energy. By rearranging the equation to \(m = \frac{E}{c^2}\), we find the amount of mass that needs to be annihilated to generate a specific amount of energy. For instance, to fulfill the entire yearly energy consumption of the United States, which is about \(1.0 \times 10^{19} \text{ J}\), approximately \(111 \text{ kg}\) of combined matter and antimatter must be annihilated annually. This concept not only shows the potential power source that antimatter represents but also demonstrates the vast energy locked within even small amounts of mass.

Energy Consumption

The problem of energy consumption is significant, especially when considering the vast amount of energy used annually by a country like the United States. In this exercise, the total energy usage is approximated as \(1.0 \times 10^{19} \text{ J}\). Such figures help us appreciate the growing demand for energy and the need for revolutionary energy sources, such as those potentially offered by antimatter reactors. Currently, energy is primarily harnessed through methods like burning fossil fuels, nuclear power, and renewable sources. Each method has its limitations and environmental consequences.
Antimatter, being the most efficient way of energy conversion, is enticing because of its potential to produce enormous energy from a tiny mass. However, the practical application of this remains uncertain due to technical challenges and costs involved in producing and storing antimatter safely. Despite these hurdles, examining such futuristic possibilities helps us push the boundaries of physics and energy resource management.

Density and Volume Calculation

When we discuss density and volume calculation in this context, we focus on understanding how a mass, like the \(111 \text{ kg}\) of combined matter and antimatter, occupies space. Density is defined as mass per unit volume, \(\rho = \frac{m}{V}\). Knowing the density of the material, which in this case is the same as iron with a density of \(7.86 \text{ g/cm}^3\), we can calculate the volume required for this mass. For this exercise, using the formula \(V = \frac{m}{\rho}\), we substitute in the mass of \(111 \text{ kg}\) and the density of \(7860 \text{ kg/m}^3\). The resultant volume is \(0.0141 \text{ m}^3\). This volume, when formed into a cube, has each side measuring approximately \(0.241 \text{ m}\), which converts to \(24.1 \text{ cm}\). Such calculations illustrate how even a substance with a high density, like iron, can occupy a surprisingly small space when the mass is kept low, as in this scenario. Understanding these principles of density and volume is vital in fields such as material science and engineering, where efficient space usage is crucial.