Problem 32 - College Physics

Question

\(\bullet\) In a hypothetical nuclear-fusion reactor, two deuterium nuclei combine or "fuse" to form one helium nucleus. The mass of a deuterium nucleus, expressed in atomic mass units (u), is 2.0136 u; that of a helium nucleus is 4.0015 u. \(\left(1 \mathrm{u}=1.661 \times 10^{-27} \mathrm{kg} .\right)\) (a) How much energy is released when 1.0 \(\mathrm{kg}\) of deuterium undergoes fusion? (b) The annual consumption of electrical energy in the United States is on the order of \(1.0 \times 10^{19} \mathrm{J} .\) How much deuterium must react to pro- duce this much energy?

Step-by-Step Solution

Answer

(a) 1 kg deuterium releases \(5.75 \times 10^{14}\) J; (b) 1.74 \(\times 10^4\) kg deuterium is needed for \(1.0 \times 10^{19}\) J.

Step 1: Calculate Mass Defect per Fusion Reaction

First, determine the mass difference between the initial and final nuclear states: two deuterium nuclei and one helium nucleus. The initial mass is \(2 \times 2.0136 \text{ u} = 4.0272 \text{ u}\). The final mass is \(4.0015 \text{ u}\). The mass defect is \(4.0272 \text{ u} - 4.0015 \text{ u} = 0.0257 \text{ u}\).

Step 2: Convert Mass Defect to Kilograms

Convert the mass defect from atomic mass units to kilograms. Using \(1 \text{ u} = 1.661 \times 10^{-27} \text{ kg}\), the mass defect in kilograms is \(0.0257 \text{ u} \times 1.661 \times 10^{-27} \text{ kg/u} = 4.271 \times 10^{-29} \text{ kg}\).

Step 3: Calculate Energy Released per Fusion

Use Einstein's mass-energy equivalence \(E=mc^2\) to find the energy released per fusion reaction. The energy is \(4.271 \times 10^{-29} \text{ kg} \times (3 \times 10^8 \text{ m/s})^2 = 3.844 \times 10^{-12} \text{ J}\).

Step 4: Calculate Number of Deuterium Nuclei in 1 kg

Find the number of deuterium nuclei in 1 kg. Knowing the mass of one deuterium nucleus is \(2.0136 \text{ u}\), convert it to kilograms: \(2.0136 \text{ u} \times 1.661 \times 10^{-27} \text{ kg/u} = 3.344 \times 10^{-27} \text{ kg}\). The number of deuterium nuclei in 1 kg is \(\frac{1 \text{ kg}}{3.344 \times 10^{-27} \text{ kg/nucleus}} = 2.992 \times 10^{26} \text{ nuclei}\).

Step 5: Calculate Total Energy Released for 1 kg of Deuterium

Calculate the total energy released when 1 kg of deuterium undergoes fusion. Given each reaction uses two deuterium nuclei, the number of reactions is \( \frac{2.992 \times 10^{26}}{2} = 1.496 \times 10^{26}\). Thus, the total energy is \(1.496 \times 10^{26} \times 3.844 \times 10^{-12} \text{ J} = 5.75 \times 10^{14} \text{ J}\).

Step 6: Calculate Deuterium Mass for Given Energy

Determine how much deuterium is needed to release \(1.0 \times 10^{19} \text{ J}\) of energy. Using the energy produced from 1 kg of deuterium, the required deuterium mass is \(\frac{1.0 \times 10^{19} \text{ J}}{5.75 \times 10^{14} \text{ J}} = 1.74 \times 10^4 \text{ kg}\).

Key Concepts

Mass-Energy Equivalence

Albert Einstein’s famous equation, \(E=mc^2\), describes mass-energy equivalence. This powerful concept shows that mass can be converted into energy. In the equation, \(E\) stands for energy, \(m\) is the mass, and \(c\) represents the speed of light, which is approximately \(3 \, \times \, 10^8 \, \text{m/s}\). This equation plays a crucial role in nuclear reactions, like the fusion of deuterium nuclei, where mass is transformed into energy. When two atomic nuclei merge, they form a new nucleus with slightly less mass than the sum of the original nuclei. The "missing" mass is converted into energy, which can be released as light or heat. Understanding this principle helps us grasp how energy can be harnessed in fusion power plants.

Deuterium Nuclei

Deuterium is a stable isotope of hydrogen, often called "heavy hydrogen." It carries a proton and a neutron in its nucleus, making it heavier than regular hydrogen, which has just one proton. The atomic mass of a deuterium nucleus is about \(2.0136 \, \text{u}\) (atomic mass units). Deuterium is naturally found in water molecules, about 1 in every 6,420 hydrogen atoms. This may sound rare, but when extracted, it forms valuable fuel for nuclear fusion. In fusion reactions, two deuterium nuclei can combine to form a helium nucleus. This process releases a significant amount of energy, making deuterium a promising candidate for clean, sustainable energy.

Energy Calculation

The release of energy during nuclear fusion can be calculated using the mass-energy equivalence principle. For fusion reactions involving deuterium, it’s crucial to compute the energy released by each reaction.

• Calculate the number of total fusion reactions: If 1 kg of deuterium is used, the number of deuterium nuclei involved can be determined and then used to find how many reactions occur.
• Compute the energy per reaction: Using the mass defect and Einstein's formula, determine the energy generated in a single fusion event.
• Total energy release: Multiply the energy per reaction by the total number of reactions, providing the total energy output.

This detailed approach highlights how nuclear fusion can potentially supply vast amounts of energy, by transforming small mass amounts into significant energy.

Mass Defect

The mass defect is a key concept in nuclear physics, referring to the difference between the mass of individual nucleons (like protons and neutrons) and the total mass of a nucleus. When two deuterium nuclei fuse to form one helium nucleus, the mass of the helium nucleus is less than the initial combined mass of the deuterium nuclei. This "missing" mass is known as the mass defect, measured as \(0.0257 \text{ u}\) in this specific fusion reaction. To convert this into kilograms for energy calculations, use the conversion factor \(1 \text{ u} = 1.661 \times 10^{-27} \text{ kg}\). Understanding the mass defect is essential to grasp why energy is released in fusion. It explains how the reduction in mass results in the production of energy, further demonstrating the incredible potential of nuclear fusion as an energy source.