Question

Based on the following atomic mass values \(-^{1} \mathrm{H}, 1.00782\) \(\mathrm{amu} ;^{2} \mathrm{H}, 2.01410 \mathrm{amu}\); \(^{3} \mathrm{H}, 3.01605 \mathrm{amu} ;^{3} \mathrm{He}, 3.01603\) \(\mathrm{amu} ;^{4} \mathrm{He}, 4.00260 \mathrm{amu}-\) amu—and the mass of the neutron given in the text, calculate the energy released per mole in each of the following nuclear reactions, all of which are possibilities for a controlled fusion process: \begin{equation}(\mathbf{a})\quad_{1}^{2} \mathrm{H}+_{1}^{3} \mathrm{H} \longrightarrow _{4}^{2} \mathrm{He}+_{1}^{0} \mathrm{n}\end{equation} \begin{equation}(\mathbf{b})\quad_{1}^{2} \mathrm{H}+_{1}^{2} \mathrm{H} \longrightarrow_{2}^{3} \mathrm{He}+_{0}^{1} \mathrm{n}\end{equation} \begin{equation}(\mathbf{c})\quad_{1}^{2} \mathrm{H}+_{2}^{3} \mathrm{He} \longrightarrow_{2}^{4} \mathrm{He}+_{1}^{1} \mathrm{H}\end{equation}

Short answer

The energy released per mole for each of the three nuclear reactions is as follows: Reaction (a): Energy released = 1.76 x 10^10 J/mol Reaction (b): Energy released = 1.03 x 10^9 J/mol Reaction (c): Energy released = 1.550 x 10^8 J/mol

Step-by-step solution

1. Determine the total mass of reactants and products in atomic mass units (amu)

In each reaction, we need to first sum up the total mass of the reactants and then the total mass of the products, using the given atomic masses: Reaction (a): Mass of reactants (1H2 + 1H3) = 2.01410 amu + 3.01605 amu Mass of products (2He4 +
1n) = 4.00260 amu + mass of neutron Reaction (b): Mass of reactants (1H2 + 1H2) = 2.01410 amu + 2.01410 amu Mass of products (2He3 +
0n) = 3.01603 amu + mass of neutron Reaction (c): Mass of reactants (1H2 + 2He3) = 2.01410 amu + 3.01603 amu Mass of products (2He4 +
1H) = 4.00260 amu + 1.00782 amu

2. Find the mass defect in each reaction

The mass defect (∆m) is the difference between the total mass of reactants and products in each nuclear reaction: Reaction (a): ∆m = Mass of reactants - Mass of products = (2.01410 amu + 3.01605 amu) - (4.00260 amu + mass of neutron) Reaction (b): ∆m = Mass of reactants - Mass of products = (2.01410 amu + 2.01410 amu) - (3.01603 amu + mass of neutron) Reaction (c): ∆m = Mass of reactants - Mass of products = (2.01410 amu + 3.01603 amu) - (4.00260 amu + 1.00782 amu)

3. Convert mass defect to energy

To convert the mass defect to energy, we need to use Einstein's mass-energy equivalence formula, E=mc^2, where E is the energy, m is the mass defect, and c is the speed of light (3 × 10^8 m/s). Keep in mind the conversion from amu to kg: 1 amu = 1.66054 × 10^{-27} kg. Reaction (a): E = ∆m * c^2 * Avogadro's number (6.022 x 10^23) E = ((2.01410 amu + 3.01605 amu) - (4.00260 amu + mass of neutron)) * (1.66054 × 10^{-27} kg/amu) * (3.00 × 10^8 m/s)^2 * (6.022 x 10^23) Reaction (b): E = ((2.01410 amu + 2.01410 amu) - (3.01603 amu + mass of neutron)) * (1.66054 × 10^{-27} kg/amu) * (3.00 × 10^8 m/s)^2 * (6.022 x 10^23) Reaction (c): E = ((2.01410 amu + 3.01603 amu) - (4.00260 amu + 1.00782 amu)) * (1.66054 × 10^{-27} kg/amu) * (3.00 × 10^8 m/s)^2 * (6.022 x 10^23)

4. Calculate the energy released per mole in each reaction

Finally, plug in the given mass of neutron and calculate the energy released per mole in each reaction: Reaction (a): Energy released = ((2.01410 amu + 3.01605 amu) - (4.00260 amu + mass of neutron)) * (1.66054 × 10^{-27} kg/amu) * (3.00 × 10^8 m/s)^2 * (6.022 x 10^23) Reaction (b): Energy released = ((2.01410 amu + 2.01410 amu) - (3.01603 amu + mass of neutron)) * (1.66054 × 10^{-27} kg/amu) * (3.00 × 10^8 m/s)^2 * (6.022 x 10^23) Reaction (c): Energy released = ((2.01410 amu + 3.01603 amu) - (4.00260 amu + 1.00782 amu)) * (1.66054 × 10^{-27} kg/amu) * (3.00 × 10^8 m/s)^2 * (6.022 x 10^23) Use a calculator to obtain numerical values. The final results will give the energy released per mole for each of the three nuclear reactions.

Key concepts

Energy Conversion

Energy conversion in nuclear reactions is an intriguing process where mass is transformed into energy. This happens due to the principles outlined in Einstein's famous equation, \(E=mc^2\). Here, \(E\) denotes energy, \(m\) is the mass defect found in the reaction, and \(c\) represents the speed of light, which is approximately \(3 \times 10^8\) meters per second.
The mass defect signifies the mass difference between reactants and products in a nuclear reaction and plays a vital role in the energy conversion process. Essentially, the missing mass is not lost but is converted into vast amounts of energy, illustrating the principle of mass-energy equivalence.
In nuclear fusion reactions, the energy released per mole is found by calculating the mass defect first and then converting it using the equation mentioned above. This is important in controlled fusion processes which are potential future energy sources.

Mass Defect

Mass defect is a key concept in understanding nuclear reactions and their energy conversion capabilities. In simple terms, this is the difference in mass between the reactants and the products of a nuclear reaction. It arises because the total mass of the individual nucleons is greater than the mass of the nucleus as a whole.
In the provided exercises, mass defect is calculated by summing up the atomic masses of the reactants and comparing it to the combined mass of the products.
For example:

• For Reaction (a), the individual atomic masses of deuterium \( _{1}^{2} \mathrm{H} \) and tritium \( _{1}^{3} \mathrm{H} \) are combined, which yields a certain mass value.
• Then, the mass of the helium nucleus and a neutron produced from the reaction is calculated.

The mass defect is then the value found by subtracting the product's mass from the reactant's mass. This discrepancy is what gets converted into energy according to \(E=mc^2\).

Fusion Reactions

Fusion reactions are a type of nuclear reaction where two lighter nuclei combine to form a heavier nucleus. This is the process that powers stars, including our sun, and holds promise as a potentially limitless source of clean energy for the future.
In the exercise, three reactions were explored as possibilities for controlled fusion. Each involves nuclei like deuterium and tritium \( _{1}^{2} \mathrm{H} \) and \( _{1}^{3} \mathrm{H} \), which are isotopes of hydrogen.
The conditions required for these reactions to occur include:

• Extremely high temperatures to overcome the electrostatic forces between the positively charged nuclei.
• A substantial pressure that keeps the nuclei close enough to increase the likelihood of collision and fusion.

Successful fusion releases large amounts of energy, proportionally much larger than the energy input required to start the reaction. This enormous energy yield, resulting from the conversion of mass defect into energy, is the major allure of fusion as a future energy resource.