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Chapter 21: Problem 81

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

Expert verified
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

01

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.
02

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 (\({ }^{11}\mathrm{C}\)).
03

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 (\({ }^{1}\mathrm{n}\)).
04

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.

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Most popular questions from this chapter

A \(2.5-\mathrm{mL}\) sample of \(0.188 \mathrm{M}\) silver nitrate solution was mixed with \(2.5 \mathrm{~mL}\) of \(0.188 \mathrm{M}\) sodium chloride solution labeled with radioactive chlorine-36. The activity of the initial sodium chloride solution was \(2.46 \times 10^{6} \mathrm{~Bq} / \mathrm{mL}\). After the resultant precipitate was removed by filtration, the remaining filtrate was found to have an activity of 175 Bq/mL. (a) Write a balanced chemical equation for the reaction that occurred. (b) Calculate the \(K_{s p}\) for the precipitate under the conditions of the experiment.

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