Question

Many transuranium elements, such as plutonium-232, have very short half-lives. (For \({ }^{232} \mathrm{Pu}\), the half-life is 36 minutes.) However, some, like protactinium-231 (half-life \(=3.34 \times\) \(10^{4}\) years), have relatively long half-lives. Use the masses given in the following table to calculate the change in energy when 1 mole of \({ }^{232}\) Pu nuclei and 1 mole of \({ }^{231}\) Pa nuclei are each formed from their respective number of protons and neutrons.

Short answer

The change in energy when 1 mole of ${ }^{232}\mathrm{Pu}$ nuclei are formed is 2.16049 x 10^{-10} J, and the change in energy when 1 mole of ${ }^{231}\mathrm{Pa}$ nuclei are formed is 2.75236 x 10^{-10} J.

Step-by-step solution

Calculate mass difference for \({ }^{232}\mathrm{Pu}\) and \({ }^{231}\mathrm{Pa}\)

We'll calculate mass difference for the two nuclei separately. To do this, we'll first calculate the total mass of protons and neutrons and then subtract the mass of the nucleus. For \({ }^{232}\mathrm{Pu}\): Number of protons = 94 Number of neutrons = 232 - 94 = 138 Mass of 1 proton = 1.00728 amu Mass of 1 neutron = 1.00867 amu Total mass of protons and neutrons = 94 x 1.00728 amu + 138 x 1.00867 amu = 94.28544 amu + 139.19646 amu = 233.4819 amu Mass of 1 mole of \({ }^{232}\mathrm{Pu}\) nucleus = 232.037 amu (given) Mass difference = 233.4819 amu - 232.037 amu = 1.4449 amu For \({ }^{231}\mathrm{Pa}\): Number of protons = 91 Number of neutrons = 231 - 91 = 140 Total mass of protons and neutrons = 91 x 1.00728 amu + 140 x 1.00867 amu = 91.66348 amu + 141.2148 amu = 232.87828 amu Mass of 1 mole of \({ }^{231}\mathrm{Pa}\) nucleus = 231.036 amu (given) Mass difference = 232.87828 amu - 231.036 amu = 1.84228 amu

Convert mass difference to energy

We will use the Einstein's mass-energy equivalence formula, assuming \(c^2 = 9 \times 10^{16} m^2/s^2\) to convert the mass difference (in amu) to energy difference (in Joules). 1 amu = 1.66 x 10^{-27} kg For \({ }^{232}\mathrm{Pu}\): Mass difference = 1.4449 amu = 1.4449 x 1.66 x 10^{-27} kg = 2.40054 x 10^{-27} kg Energy difference = \(E = mc^2\) = 2.40054 x 10^{-27} kg x 9 x 10^{16} m^2/s^2 = 2.16049 x 10^{-10} J For \({ }^{231}\mathrm{Pa}\): Mass difference = 1.84228 amu = 1.84228 x 1.66 x 10^{-27} kg = 3.05818 x 10^{-27} kg Energy difference = \(E = mc^2\) = 3.05818 x 10^{-27} kg x 9 x 10^{16} m^2/s^2 = 2.75236 x 10^{-10} J These are the energy changes for 1 mole of \({ }^{232}\mathrm{Pu}\) and \({ }^{231}\mathrm{Pa}\) nuclei, respectively.

Key concepts

Transuranium Elements

Transuranium elements are chemical elements with atomic numbers greater than that of uranium, which is number 92. These elements do not occur naturally in significant amounts and are synthesized artificially in nuclear reactors or particle accelerators.
Understanding transuranium elements is crucial in fields like nuclear physics and chemistry because they help us comprehend profound nuclear reactions and decay processes.
Some notable transuranium elements include neptunium, plutonium, and americium. Because of their large atomic numbers, these elements display interesting properties:

• Short half-lives, often leading to rapid decay and radioactivity.
• Pausity of occurrence in nature, thus requiring complex synthesis techniques.
• Vital roles in nuclear energy production and weaponry.

The study of their behavior and interactions can lead to advancements in energy and materials science.

Plutonium-232

Plutonium-232 is a less commonly referenced isotope of plutonium, but it holds significance in nuclear science. This isotope is a heavy, radioactive, and artificial element, with an atomic number of 94.
Despite its importance, \({ }^{232} ext{Pu} \) is known for having a relatively short half-life of 36 minutes. This indicates its rapid decay, which can be a critical factor during handling and in nuclear applications. Key characteristics include:

• It is known to contribute significantly to the damage in nuclear fuels.
• Commonly formed as a decay product from other isotopes.
• Although unstable, its properties need to be considered in nuclear reactor designs and safety assessments.

Therefore, Plutonium-232's properties and rapid decay highlight the miracle of nuclear transformations.

Protactinium-231

Protactinium-231 is another important isotope in the field of nuclear chemistry, possessing an atomic number of 91. In contrast to plutonium-232, protactinium-231 has a surprisingly long half-life of approximately \(3.34 \times 10^{4}\) years. This longevity makes it valuable for geological and archaeological dating.
Protactinium-231's characteristics include:

• Its greater stability makes it useful for various scientific applications.
• It has crucial implications for understanding the principles of radioactive decay and the interaction of heavy elements with matter.
• Despite its scarcity, its study helps us decipher the isotopic compositions of early solar system materials.

As a transuranium element, it fascinates scientists due to its resistance to rapid decay and its contribution to isotopic dating techniques.

Half-Life

The half-life of a radioactive isotope is the time required for half of the initial quantity of a substance to undergo radioactive decay.
Half-life is not only a basic property of isotopes but also a fundamental concept in nuclear chemistry. Here's why it matters:

• It helps predict how long a particular radioactive isotope will remain active.
• Via half-life, scientists can design nuclear power plants and storage methods for radioactive waste efficiently.
• Understanding half-life assists in the safe handling and regulation of radioactive materials.

Short half-lives, such as that of plutonium-232, mean rapid decay, whereas longer half-lives like protactinium-231 can be used for dating rocks and archaeological findings. The concept remains a central pillar for everything from medical applications to nuclear energy.

Nuclear Reactions

Nuclear reactions involve changes in an atom's nucleus and can lead to the transformation of elements. These reactions are fundamental to both natural and technological processes, including stellar nucleosynthesis and nuclear energy production.
Key points about nuclear reactions include:

• Nuclear fission, as seen in creating transuranium elements like plutonium, where larger atoms split into smaller ones, releasing energy.
• Nuclear fusion, occurring naturally in the sun, where smaller atoms combine to form a larger atom, also releasing energy.
• The generation of energy and isotopes is dependent on converting mass into energy, explained by Einstein's mass-energy equivalence principle \(E = mc^2\).

Understanding nuclear reactions is crucial for advancing nuclear power technology, developing new medical isotopes, and exploring energy generation processes within stars.