Question

The easiest fusion reaction to initiate is $${ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}+{ }_{0}^{1} \mathrm{n}$$ Calculate the energy released per \({ }_{2}^{4} \mathrm{He}\) nucleus produced and per mole of \({ }_{2}^{4}\) He produced. The atomic masses are \({ }_{1}^{2} \mathrm{H}, 2.01410 ;{ }_{1}^{3} \mathrm{H}\), \(3.01605\); and \({ }_{2}^{4}\) He, \(4.00260\). The masses of the electron and neutron are \(5.4858 \times 10^{-4}\) amu and \(1.00866\) amu, respectively.

Short answer

The mass defect of the reaction is calculated as: Mass defect = (2.01410 + 3.01605) - (4.00260 + 1.00866) = 0.01889 amu Converting the mass defect to kg: Mass defect (in kg) = 0.01889 amu x 1.66054 x 10^-27 kg/amu = 3.1378 x 10^-29 kg Calculating the energy released using E=mc^2: Energy released (in Joules) = (3.1378 x 10^-29 kg) x (2.998 x 10^8 m/s)^2 ≈ 2.818 x 10^-12 J Now converting the energy released to per mole basis: Energy released per mole of Helium-4 = (2.818 x 10^-12 J) x (6.022 x 10^23 mol^-1) ≈ 1.696 x 10^12 J/mol Therefore, the energy released per \(_{2}^{4}\) Helium nucleus produced is approximately \(2.818 x 10^{-12}\) Joules, and the energy released per mole of \(_{2}^{4}\) Helium nuclei produced is approximately \(1.696 x 10^{12}\) Joules per mole.

Step-by-step solution

Calculate the mass defect of the reaction

To calculate the mass defect, we need to find the difference between the sum of the masses of the reactants and the sum of the masses of the products. The given masses are: - Deuterium (\( _{1}^{2} \mathrm{H} \)): 2.01410 amu - Tritium (\( _{1}^{3} \mathrm{H} \)): 3.01605 amu - Helium-4 (\( _{2}^{4} \mathrm{He} \)): 4.00260 amu - Neutron (\( _{0}^{1} \mathrm{n} \)): 1.00866 amu Mass defect = (Mass of reactants) - (Mass of products) = (2.01410 + 3.01605) - (4.00260 + 1.00866)

Calculate the energy released using E=mc^2

Now that we have the mass defect, we can calculate the energy released using Einstein's famous equation, E=mc^2, where E represents energy, m represents mass, and c represents the speed of light (2.998 x 10^8 m/s). First, we need to convert the mass defect from atomic mass units (amu) to kilograms (kg) using the conversion factor 1 amu = 1.66054 x 10^-27 kg. Mass defect (in kg) = mass defect (in amu) x 1.66054 x 10^-27 kg/amu Energy released (in Joules) = Mass defect (in kg) x (speed of light)^2

Convert the energy released to per mole basis

To calculate the energy released per mole of Helium-4 nuclei produced, we need to multiply the energy released per nucleus by Avogadro's number (6.022 x 10^23 mol^-1). Energy released per mole of Helium-4 = Energy released per nucleus x Avogadro's number Now, let's put the values to calculate the desired results.

Key concepts

Mass Defect

The concept of mass defect is fundamental in nuclear fusion reactions. It refers to the difference in mass between the reactants and products of a nuclear reaction. In simple terms, during a fusion reaction, some of the mass of the reacting particles is converted into energy. This difference in mass, which appears to be 'lost', is what we call the mass defect.

In the provided fusion reaction:

• Deuterium (\( _{1}^{2} \mathrm{H} \)
• Tritium (\( _{1}^{3} \mathrm{H} \)
• Helium-4 (\( _{2}^{4} \mathrm{He} \))
• Neutron (\( _{0}^{1} \mathrm{n} \))

the mass of deuterium and tritium together (which are the reactants) is more than the combined mass of helium-4 and the neutron (the products).

To calculate the mass defect:

• Add the atomic masses of deuterium and tritium: \(2.01410 + 3.01605\, \text{amu}\)
• Add the atomic masses of helium-4 and neutron: \(4.00260 + 1.00866\, \text{amu}\)
• Subtract the total mass of products from the total mass of reactants to find the mass defect.

Energy Calculation

Once the mass defect is determined, the next step is to calculate the energy released. This is a crucial step because the energy released in fusion reactions, such as those happening in stars, forms the basis for their immense energy output.

The energy can be computed using the mass-energy equivalence principle. The conversion of mass defect to energy is done using Einstein's equation, which will be explained in the next section.

Before applying the equation, the mass defect must be converted to kilograms. This is necessary because the energy calculation in Einstein's equation requires mass in kilograms. To do this, use the conversion factor:

• 1 atomic mass unit (amu) = \(1.66054 \times 10^{-27} \text{ kg}\).

Multiply the mass defect in amu by this conversion factor to obtain the mass in kilograms. This value is then used in the energy calculation process.

Einstein's Equation

Einstein's equation, represented as \(E=mc^2\), is a pivotal concept in understanding how mass is related to energy. In the equation:- \(E\) stands for energy,- \(m\) is the mass in kilograms,- \(c\) is the speed of light, which is approximately \(2.998 \times 10^8 \text{ m/s}\).

With the mass defect converted to kilograms, you can substitute this value into Einstein's equation to find the energy released. The equation shows that even a small mass defect can yield a substantial amount of energy due to the large value of \(c^2\).

The calculation highlights how nuclear processes can be incredibly efficient in converting mass into energy, making nuclear fusion a potent source of power. Once the energy per reaction is calculated, you can further use Avogadro's number to find the energy released per mole, as described in the subsequent section.

Avogadro's Number

Avogadro's number is a fundamental constant in chemistry and physics that is used to scale microscopic measurements to macroscopic amounts, like moles. It is valued at approximately \(6.022 \times 10^{23} \text{ mol}^{-1}\).

In nuclear fusion calculations, after determining the energy released in a single fusion reaction, this number is essential for finding the energy released per mole of product, in this case, helium-4. By multiplying the energy per reaction (in Joules) by Avogadro's number, you can convert the energy released from the nuclear level to a more practical, macroscopic scale.

For example, if one helium-4 nucleus releases a certain amount of energy, multiplying this by Avogadro's number will give the total energy released when a mole of helium-4 nuclei is produced. This provides insights into the feasibility and potential of nuclear reactions for energy production.