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}

Step by step solution

01

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

02

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)

03

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)

04

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.

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