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Each of the following transmutations produces a radionuclide used in positron emission tomography (PET). (a) Inequations (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? \begin{equation}\begin{array}{l}{(\mathrm{i})^{14} \mathrm{N}(\mathrm{p}, \alpha) \mathrm{X}} \\ {(\mathrm{ii})^{18} \mathrm{O}(\mathrm{p}, \mathrm{X})^{18} \mathrm{F}} \\ {\text { (iii) }^{14} \mathrm{N}(\mathrm{d}, \mathrm{n})^{15} \mathrm{O}}\end{array}\end{equation}

Short Answer

Expert verified
The species "X" in the given reactions can be identified by conserving atomic and mass numbers. In reaction (i), the species X is \(_{6}^{11}\mathrm{C}\). In reaction (ii), the species X is a proton (\(\mathrm{p}\)). In reaction (iii), "d" represents a deuteron (\(_{1}^{2}\mathrm{D}\)), consisting of one proton and one neutron.

Step by step solution

01

Identify "X" in Reaction (i)

To identify "X" in reaction (i), follow the conservation of mass numbers (A) and atomic numbers (Z). The mass numbers should be the same on both sides of the reaction equation, and similarly, the atomic numbers should balance as well. Reaction (i) is given by: \[\,^{14}\mathrm{N}(\mathrm{p},\alpha) \mathrm{X}\] The atomic numbers (Z) and mass numbers (A) of the particles in the reaction are as follows: Nitrogen-14 has Z=7 and A=14, a proton has Z=1 and A=1, and an alpha particle has Z=2 and A=4. Now balance the equation: - On the left side, the total atomic number is 7 (N) + 1 (p) = 8, and the mass number is 14 (N) + 1 (p) = 15. - On the right side, the atomic number should be 8 - 2 (alpha) = 6, and the mass number should be 15 - 4 (alpha) = 11. Thus, "X" should have Z = 6 and A = 11, making species X: \[\,^{11}\mathrm{C}\] Since carbon has an atomic number of 6.
02

Identify "X" in Reaction (ii)

To identify "X" in reaction (ii), again follow the conservation of mass numbers (A) and atomic numbers (Z). Reaction (ii) is given by: \[\,^{18}\mathrm{O}(\mathrm{p}, \mathrm{X})^{18}\mathrm{F}\] The atomic numbers and mass numbers of the particles in the reaction are given as follows: Oxygen-18 has Z=8 and A=18, a proton has Z=1 and A=1, and Fluorine-18 has Z=9 and A=18. Now balance the equation: - On the left side, the total atomic number is 8 (O) + 1 (p) = 9, and the mass number is 18 (O) + 1 (p) = 19. - On the right side, the atomic number should be 9 (F), and the mass number should be 19 - 18 (F) = 1. Thus, "X" should have Z = 1 and A = 1, making species X: \[\mathrm{p}\] Since a proton has an atomic number of 1 and a mass number of 1.
03

Identify "d" in Reaction (iii)

Now, we need to identify what "d" represents in reaction (iii). Reaction (iii) is given by: \[\,^{14}\mathrm{N}(\mathrm{d}, \mathrm{n})^{15}\mathrm{O}\] In the context of nuclear reactions, "d" stands for "deuteron" or "deuterium nucleus," which consists of one proton and one neutron. It has an atomic number of Z = 1 since it has one proton and a mass number of A = 2 due to the proton and neutron. Hence, "d" represents a deuteron and is denoted by \(_{1}^{2}\mathrm{D}\).

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Key Concepts

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

Nuclear Transmutation
Nuclear transmutation is a process where the nucleus of an atom changes its identity by altering the number of protons it contains. This change can be precipitated by bombarding nuclei with particles such as protons, neutrons, or alpha particles. In a typical transmutation, a target nucleus absorbs an incoming particle, and the resulting compound nucleus emits other particles, resulting in a completely different element or isotope.

For instance, in positron emission tomography (PET) imaging, radioactive isotopes, which emit positrons, are produced through nuclear transmutation. These isotopes are created when a stable atom captures a positively charged particle like a proton. The compound nucleus may then release an alpha particle (a helium nucleus comprising two protons and two neutrons), leaving behind a different atom that is radioactively unstable and will decay by emitting a positron during its transition to stability.

Such transformations are crucial in medicine as they provide radionuclides used in PET scans to diagnose and monitor diseases. Understanding nuclear transmutation allows for the safe and effective creation of these vital medical isotopes.
Conservation of Mass Number
The conservation of mass number is a fundamental principle in nuclear chemistry. The mass number, represented by the letter 'A', is the total number of protons and neutrons in an atom's nucleus. During nuclear reactions, like those used in creating radionuclides for PET scans, the total mass number must remain constant before and after the reaction takes place.

To visualize this, think of the nuclear reaction as a precise balancing scale. When a nucleus is bombarded with a particle, the sum of the mass numbers on the reactant side must equal the sum on the product side. If we take the reaction where Nitrogen-14 (icefrac{^{14}N}) absorbs a proton (icefrac{^1H}) and emits an alpha particle (icefrac{^4He}), the remaining product must have a mass number of 11 (icefrac{^{14}N} + icefrac{^1H} - icefrac{^4He} = icefrac{^{11}C}). This ensures that despite the transmutation of elements, the total mass number remains conserved.
Conservation of Atomic Number
Equally important to the principle of conservation of the mass number is the conservation of the atomic number. Associated with the letter 'Z', the atomic number is the count of protons in the nucleus and defines the identity of the element. During nuclear reactions, such as the ones used to generate imaging agents for PET, the total atomic number must also remain unchanged.

A clear understanding of this principle helps in identifying the new elements formed during nuclear transmutations. In the reaction where a proton merges with Nitrogen-14, the addition of the proton increases the atomic number by one. However, when an alpha particle is released, the atomic number is reduced by two, resulting in the production of a nucleus with an atomic number of six, which corresponds to carbon (icefrac{^6C}). This illustrates the process of changing one element into another, while respecting the conservation law which insists that the arithmetic sum of atomic numbers should be coherent before and after the transformation.

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Most popular questions from this chapter

One of the nuclides in each of the following pairs is radioactive. Predict which is radioactive and which is stable: \((\mathbf{a})_{19}^{39} \mathrm{K}\) and \(_{19}^{40} \mathrm{K},\) \((\mathbf{b})^{209} \mathrm{Bi}\) and \(^{208} \mathrm{Bi}\) \((\mathbf{c})\) nickel-58 and nickel-65.

In 2010, a team of scientists from Russia and the United States reported creation of the first atom of element 117, which is named tennessine, and whose symbol is Ts. The synthesis involved the collision of a target of \(_{97}^{249} \mathrm{Bk}\) with accelerated ions of an isotope which we will denote Q. The product atom, which we will call Z, immediately releases neutrons and forms \(_{97}^{249} \mathrm{Bk} :\) $$_{97}^{249} \mathrm{Bk}+\mathrm{Q} \longrightarrow \mathrm{Z} \longrightarrow_{117 \mathrm{Ts}}^{294 \mathrm{Ts}}+3_{0}^{1} \mathrm{n}$$ (a) What are the identities of isotopes Q and Z? (b) Isotope Q is unusual in that it is very long-lived (its half-life is on the order of 1019 yr) in spite of having an unfavorable neutron-to-proton ratio (Figure 21.1). Can you propose a reason for its unusual stability? (c) Collision of ions of isotope Q with a target was also used to produce the first atoms of livermorium, Lv. The initial product of this collision was \(_{116}^{296} \mathrm{Zn}\). What was the target isotope with which Q collided in this experiment?

Radium-226, which undergoes alpha decay, has a half-life of 1600 yr. (a) How many alpha particles are emitted in 5.0 min by a 10.0 -mg sample of \(^{226} \mathrm{Ra}\) ? (b) What is the activity of the sample in mCi?

Indicate the number of protons and neutrons in the following nuclei: \((\mathbf{a}) _{24}^{56} \mathrm{Cr},(\mathbf{b})^{193} \mathrm{Tl},(\mathbf{c})\) argon-\(38.\)

Complete and balance the following nuclear equations by supplying the missing particle: \begin{equation}\begin{array}{l}{\text { (a) }_{98}^{252} \mathrm{Cf}+_{5}^{10} \mathrm{B} \longrightarrow 3_{0}^{1} \mathrm{n}+?} \\\ {\text { (b) }_{1}^{2} \mathrm{H}+_{2}^{3} \mathrm{He} \longrightarrow_{2}^{4} \mathrm{He}+?}\\\ {\text { (c) }_{1}^{1} \mathrm{H}+_{5}^{11} \mathrm{B} \longrightarrow 3?} \\ {\text { (d) }_{53}^{122} \mathrm{I}\longrightarrow_{54}^{122} \mathrm{Xe}+?}\\\ {\text { (e) }_{26}^{59} \mathrm{Fe}\longrightarrow_{-1}^{0} \mathrm{e}+?}\end{array}\end{equation}

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