91Ó°ÊÓ

Based on the following atomic mass values - \({ }^{1} \mathrm{H}, 1.00782\) amu; \({ }^{2} \mathrm{H}, 2.01410 \mathrm{amu} ;{ }^{3} \mathrm{H}, 3.01605 \mathrm{amu} ;{ }^{3} \mathrm{He}, 3.01603 \mathrm{amu} ;\) \({ }^{4} \mathrm{He}, 4.00260 \mathrm{amu}-\) 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: (a) \({ }_{1}{ }_{1} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}+{ }_{0}^{1} \mathrm{n}\) (b) \({ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{2}^{3} \mathrm{He}+{ }_{0}^{1} \mathrm{n}\) (c) \({ }_{1}^{2} \mathrm{H}+{ }_{2}^{3} \mathrm{He} \longrightarrow{ }_{2}^{4} \mathrm{He}+{ }_{1}^{1} \mathrm{H}\) \(21.53\) Which of the following nuclei is likely to have the largest mass defect per nucleon: (a) \({ }^{59} \mathrm{Co}\), (b) \({ }^{11} \mathrm{~B}\), (c) \({ }^{118} \mathrm{Sn}\), (d) \({ }^{243} \mathrm{Cm}\) ? Explain your answer.

Step by step solution

01

Understand mass defect and fusion reactions

Mass defect refers to the difference between the mass of a nucleus and the sum of the masses of its individual nucleons. In nuclear reactions, mass is converted into energy as described by Einstein's famous equation, \(E = mc^2\), where \(E\) is energy, \(m\) is mass difference, and \(c\) is the speed of light. In fusion reactions, energy is released as the lighter nuclei combine to form a heavier nucleus.

02

Calculate energy released per mole for reaction (a)

For reaction (a), we will convert the given atomic masses into mass defect and then use the \(E = mc^2\) equation to calculate the energy released per mole. Reaction (a): \({ }_{1}{ }_{1} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}+{ }_{0}^{1} \mathrm{n}\) Mass defect (a) = \([1.00782 + 3.01605] - [4.00260 + 1.00867]\) amu = ‑0.0004 amu (mass of a neutron is 1.00867 amu) Energy released per mole (a) = \((-0.0004 \text{ amu})(1.6605\times10^{-27} \text{ kg/amu})(6.022\times10^{23} \text{mol}^{-1})\) \((3\times10^8 \text{ m/s})^2\) = \(1.33\times10^{12}\) J/mol

03

Calculate energy released per mole for reaction (b)

Similarly, for reaction (b): Reaction (b): \({ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{2}^{3} \mathrm{He}+{ }_{0}^{1} \mathrm{n}\) Mass defect (b) = \([2.01410 + 2.01410] - [3.01603 + 1.00867]\) amu = \(-0.0035\) amu Energy released per mole (b) = \((-0.0035 \text{ amu})(1.6605\times10^{-27} \text{ kg/amu})(6.022\times10^{23} \text{mol}^{-1})\) \((3\times10^8 \text{ m/s})^2\) = \(1.02\times10^{13}\) J/mol

04

Calculate energy released per mole for reaction (c)

Finally, for reaction (c): Reaction (c): \({ }_{1}^{2} \mathrm{H}+{ }_{2}^{3} \mathrm{He} \longrightarrow{ }_{2}^{4} \mathrm{He}+{ }_{1}^{1} \mathrm{H}\) Mass defect (c) = \([2.01410 + 3.01603] - [4.00260 + 1.00782]\) amu = \(0.01971\) amu Energy released per mole (c) = \(0.01971 \text{ amu}(1.6605\times10^{-27} \text{ kg/amu})(6.022\times10^{23} \text{mol}^{-1})\) \((3\times10^8 \text{ m/s})^2\) = \(1.90\times10^{13}\) J/mol

05

Compare the calculated energies and answer the question

Now that we have calculated the energy released per mole in each of the nuclear reactions, we can analyze which nuclei is likely to have the largest mass defect per nucleon by comparing the energy values: Energy released per mole: (a) \(1.33\times10^{12}\) J/mol (b) \(1.02\times10^{13}\) J/mol (c) \(1.90\times10^{13}\) J/mol From the calculations above, we can see that reaction (c) releases the most energy per mole. Therefore, the nuclei \({ }^{118} \mathrm{Sn}\) is likely to have the largest mass defect per nucleon.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass Defect

The concept of mass defect centers around the observation that the total mass of a nucleus is not just the simple sum of its constituent nucleons. In fact, the nucleus tends to have less mass than expected. This "missing" mass, known as the mass defect, is important because it is converted into energy during nuclear reactions. Mass defect provides an insight into how much energy is made available during nuclear fusion or fission, where nuclear particles come together or break apart. This is foundational to understanding nuclear energy as it reveals the hidden potential energy trapped within atomic structures, aiding in harnessing energy from nuclear processes.

Energy Release

Energy release in nuclear fusion is a fascinating concept where considerable amounts of energy are unleashed when light nuclei merge to form a heavier nucleus. This release is due to the conversion of mass into energy, as explained by Albert Einstein's equation, \(E = mc^2\). Here, \(E\) is the energy released, \(m\) is the mass defect, and \(c\) is the speed of light. The energy comes from the binding energy that holds the nucleus together. As nuclei bond and some of their mass turns into energy, this phenomenon naturally powers stars, such as the sun, and has implications for producing cleaner and more sustainable energy sources on Earth.

Atomic Mass

Atomic mass is a critical factor in understanding nuclear reactions. It is the mass of an atom expressed in atomic mass units (amu), serving as a measure of the total number of protons and neutrons in an atom's nucleus. Since atomic masses offer a relative scale for elements, they are key to determining mass defects in nuclear reactions, such as those in nuclear fusion, where mass and energy balances are crucial. By precisely knowing atomic masses, scientists can perform accurate calculations to predict the energy outcomes of nuclear processes and the stability of different atomic nuclei.

Fusion Reaction Calculations

Calculations for fusion reactions involve understanding how merging nuclei result in energy release. Consider a typical reaction like \({ }_{1}^{2}\text{H} + { }_{2}^{3}\text{He} \to { }_{2}^{4}\text{He} + { }_{1}^{1}\text{H}\). Begin by determining the mass defect: the difference between the initial sum of masses and the resultant fusion product masses. Using the mass defect, the energy release \(E\) can be calculated using Einstein’s equation. Converting the mass defect from amu to kilograms using the conversion factor \(1.6605 \times 10^{-27} \text{ kg/amu}\), and then applying \(E = mc^2\), where \(c = 3 \times 10^8 \text{ m/s}\), yields the energy released per mole. These calculations reveal the massive amounts of energy that can be released from very small amounts of fusion fuel, making nuclear fusion a potential powerhouse for future energy needs.