Question

Each of the following transmutations produces a radionuclide used in positron emission tomography (PET). (a) In equations (i) and (ii), identify the species signified as "X." (b) In equation (iii), one of the species is indicated as "d." What do you think it represents? (i) \({ }^{14} \mathrm{~N}(\mathrm{p}, \alpha) \mathrm{X}\) (ii) \({ }^{18} \mathrm{O}(\mathrm{p}, \mathrm{X})^{18} \mathrm{~F}\) (iii) \({ }^{14} \mathrm{~N}(\mathrm{~d}, \mathrm{n}){ }^{15} \mathrm{O}\)

Short answer

In the given nuclear reactions, the species X in equation (i) is Carbon-11 (\({ }^{11}\mathrm{C}\)), while the species X in equation (ii) is a neutron (\({ }^{1}\mathrm{n}\)). In equation (iii), the species "d" represents a deuteron (\({ }^{2}\mathrm{H}\)), a stable isotope of hydrogen. These results were obtained using the conservation of mass number (A) and atomic number (Z) in nuclear reactions.

Step-by-step solution

Identify Variables

In each reaction, we are given the reactants and the products, and we need to identify the unknown species. We are also given the mass number (A) and atomic number (Z) of the elements involved.

Solve equation (i)

For equation (i): \({ }^{14} \mathrm{N}(\mathrm{p}, \alpha) \mathrm{X}\) We know that \(A_{N} = 14\) and \(Z_{N} = 7\) for nitrogen, \(A_{p} = 1\) and \(Z_{p} = 1\) for proton, and \(A_{\alpha} = 4\) and \(Z_{\alpha} = 2\) for alpha particle. Using conservation of mass number (A) and atomic number (Z), we get \(A_{N} + A_{p} = A_{\alpha} + A_{X}\) and \(Z_{N} + Z_{p} = Z_{\alpha} + Z_{X}\). Solve for \(A_{X}\) and \(Z_{X}\): \(A_{X} = 14 + 1 - 4 = 11\) \(Z_{X} = 7 + 1 - 2 = 6\) Thus, the species X in equation (i) is Carbon-11 (\({ }^{11}\mathrm{C}\)).

Solve equation (ii)

For equation (ii): \({ }^{18} \mathrm{O}(\mathrm{p}, \mathrm{X})^{18}\mathrm{F}\) We know that \(A_{O} = 18\) and \(Z_{O} = 8\) for oxygen, \(A_{p} = 1\) and \(Z_{p} = 1\) for proton, and \(A_{F} = 18\) and \(Z_{F} = 9\) for fluorine. Using conservation of mass number (A) and atomic number (Z), we get \(A_{O} + A_{p} = A_{F} + A_{X}\) and \(Z_{O} + Z_{p} = Z_{F} + Z_{X}\). Solve for \(A_{X}\) and \(Z_{X}\): \(A_{X} = 18 + 1 - 18 = 1\) \(Z_{X} = 8 + 1 - 9 = 0\) Thus, the species X in equation (ii) is a neutron (\({ }^{1}\mathrm{n}\)).

Identify "d" in equation (iii)

For equation (iii): \({ }^{14} \mathrm{N}(\mathrm{d}, \mathrm{n}){ }^{15}\mathrm{O}\) We are given all the species in this reaction except for "d." Using the same conservation rules, we are looking for an atomic nucleus (not an atomic element) with an atomic number and mass number adding up to the difference in the reaction. Using conservation of mass number (A) and atomic number (Z), we can deduce that "d" represents an atomic nucleus with: \(A_{d} = A_{O} + A_{n} - A_{N} = 15 + 1 - 14 = 2\) \(Z_{d} = Z_{O} - Z_{N} = 8 - 7 = 1\) Thus, "d" represents an atomic nucleus with mass number 2 and atomic number 1. This is known as a deuteron (\({ }^{2}\mathrm{H}\)) which is a stable isotope of hydrogen.

Key concepts

Nuclear Transmutation

Nuclear transmutation is a fascinating process where the nucleus of an atom changes and becomes a different element. This transformation is not just any change—it happens at the core of an atom. This is like re-designing the atom from its heart. For example, during a nuclear reaction, one element can lose particles like protons or neutrons, resulting in a new element. This process is fundamental in fields like medicine and energy production. In positron emission tomography (PET), specific radionuclides are formed through such transmutations.

• Nuclear reactions involve high-energy bombardment, usually by particles like protons or neutrons.
• Elements can be transformed into other elements with different properties and atomic structures.

Understanding this concept involves imagining how atomic nuclei undergo changes, resulting in new and exciting compounds with diverse applications.

Radionuclides

Radionuclides are atoms with unstable nuclei, which means they have too much energy and seek stability by emitting radiation. In the context of PET scans, these radionuclides are crucial. They are used as tracers that emit positrons while decaying, enabling doctors to see inside the body in remarkable ways. These atoms have varying lifespans, from a few seconds to millions of years, depending on their type and composition.

Key considerations include:

• Radionuclides help in diagnosing and monitoring diseases by providing images of tissues through emitted radiation.
• The selection of an appropriate radionuclide is critical due to its decay rate and the necessary half-life for the medical examination.
• In PET, commonly used radionuclides like Fluorine-18 decay by emitting positrons.

This profound use makes radionuclides invaluable in both medical diagnostics and treatment processes, transforming them from mere isotopes to lifesaving tools.

Conservation of Mass and Atomic Number

When chemical reactions occur at the atomic level, they follow a very specific rule: the conservation of mass and atomic number. This principle ensures that in nuclear reactions, like those observed in PET transmutations, the sum of mass numbers and atomic numbers remains constant. Imagine this as a balanced puzzle where, despite changes, the total count of pieces (protons and neutrons) stays the same.

This conservation is crucial because:

• It aids in predicting the products of nuclear reactions accurately.
• Ensures stability and consistency, preserving the simple yet powerful foundation of nuclear chemistry.
• Facilitates accurate identification of unknown species in elegant reactions, as demonstrated in several PET radionuclide processes.

The understanding of this conservation provides insight into how these atomic equations can be balanced just like regular arithmetic, allowing the precise crafting of essential medical isotopes used for patient diagnostics.