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Which member of the following pairs of nuclides would you expect to be most abundant in natural sources: (a) \(_{20}^{40} \mathrm{Ca}\) or \(_{20}^{42} \mathrm{Ca} ;\) (b) \(_{15}^{31} \mathrm{P}\) or \(_{15}^{32} \mathrm{P} ;\) (c) \(_{30}^{63} \mathrm{Zn}\) or 64 \(\mathrm{Zn} ?\) Explain your reasoning?

Short Answer

Expert verified
The most abundant isotopes are generally those that are closest to having a balanced number of protons and neutrons. So, \(_{20}^{40} \mathrm{Ca}\), _{15}^{31} \mathrm{P}, and _{30}^{64}\mathrm{Zn} are the most abundant.

Step by step solution

01

Problem (a)

Consider \(_{20}^{40} \mathrm{Ca}\) and \(_{20}^{42} \mathrm{Ca}\). Both have 20 protons, but differ when it comes to neutrons. The first has 20 neutrons (40-20), while the second has 22 neutrons (42-20). The isotopes with approximately equal number of protons and neutrons are the most abundant. Thus, the answer is \(_{20}^{40} \mathrm{Ca}\), because it has equal number of protons and neutrons
02

Problem (b)

Consider \(_{15}^{31} \mathrm{P}\) and \(_{15}^{32} \mathrm{P}\). The first isotope has 16 neutrons (31-15), and the second has 17 neutrons (32-15). In this case, \(_{15}^{31} \mathrm{P}\) is more abundant because it has a more evenly balanced number of protons to neutrons.
03

Problem (c)

Consider \(_{30}^{63}\mathrm{Zn}\) and \(_{30}^{64}\mathrm{Zn}\) . The first has 33 neutrons (63-30), and the second has 34 neutrons (64-30). The answer is \(_{30}^{64}\mathrm{Zn}\) since it is closest to having a balanced number of protons and neutrons.

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Key Concepts

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

Isotopes
Isotopes are variations of the same chemical element that have the same number of protons but different numbers of neutrons. This variation in neutron number gives each isotope a different atomic mass. For example, carbon-12 (_{6}^{12}_{}C) and carbon-14 (_{6}^{14}_{}C) are both isotopes of carbon, sharing the same atomic number (6 protons) but having different neutron counts of 6 and 8, respectively. Isotopes exhibit nearly identical chemical behavior but can have vastly different physical properties, such as radioactivity or stability. In natural sources, the abundance of isotopes depends on factors like nuclear stability and the processes that gave birth to the elements, such as nucleosynthesis in stars.

Neutron Number
The neutron number, often denoted as N, represents the count of neutrons in an atomic nucleus. It is a key factor in determining the isotope of an element. Since neutrons and protons are similar in mass, and the atomic mass is the sum of these particles, variations in neutron number create isotopes with different atomic masses. Neutron number is calculated by subtracting the number of protons (atomic number) from the atomic mass number. The balance between neutron and proton numbers is crucial for nuclear stability; having too many or too few neutrons in relation to protons can make an isotope radioactive or unstable. Generally, a balanced ratio leads to greater natural abundance, such as seen with _{20}^{40}Ca, which has an equal number of protons and neutrons.

Proton Number
The proton number, also known as the atomic number, is the count of protons in the nucleus of an atom, and it determines the chemical element. Each element on the periodic table possesses a unique proton number which dictates its placement on the table and its chemical properties. For instance, carbon always has 6 protons, and any atom with 6 protons is carbon. The proton number is crucial in identifying an element, and it remains constant across all isotopes of that element. In isotopic abundance, the proton number is consistent while the neutron number may vary, differentiating isotopes of the same element.

Nuclear Stability
Nuclear stability refers to how well an atom's nucleus is held together and is determined by the forces among the protons and neutrons. Stable nuclei have a balance of protons to neutrons that allows the atom to exist without spontaneously altering its structure through nuclear reactions such as radioactive decay. For lighter elements, a stable isotope often has a balanced or nearly balanced number of protons and neutrons, while heavier elements require a greater number of neutrons to offset the electrical repulsion between protons. As an isotope becomes more imbalanced, it becomes less stable and more prone to radioactive decay, which reduces its natural abundance. Stable isotopes, such as _{20}^{40}Ca and _{15}^{31}P, are more likely to be found in nature due to their enduring nuclear configurations.

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Most popular questions from this chapter

A lunar rock was analyzed for argon by mass spectrometry and for potassium by atomic absorption. The results of these analyses showed that the sample contained \(3.02 \times 10^{-5} \mathrm{mL} \mathrm{g}^{-1}\) of argon and \(0.083 \%\) of potassium. The half-life of potassium- 40 is \(1.248 \times\) \(10^{9} \mathrm{y} \cdot\) Calculate the age of the lunar rock.

For medical uses, radon-222 formed in the radioactive decay of radium-226 is allowed to collect over the radium metal. Then, the gas is withdrawn and sealed into a glass vial. Following this, the radium is allowed to disintegrate for another period, when a new sample of radon- 222 can be withdrawn. The procedure can be continued indefinitely. The process is somewhat complicated by the fact that radon-222 itself undergoes radioactive decay to polonium- 218 , and so on. The half-lives of radium-226 and radon-222 are \(1.60 \times 10^{3}\) years and 3.82 days, respectively.(a) Beginning with pure radium- \(226,\) the number of radon-222 atoms present starts at zero, increases for a time, and then falls off again. Explain this behavior. That is, because the half-life of radon-222 is so much shorter than that of radium- \(226,\) why doesn't the radon-222 simply decay as fast as it is produced, without ever building up to a maximum concentration?(b) Write an expression for the rate of change \((d \mathrm{D} / d t)\) in the number of atoms (D) of the radon- 222 daughter in terms of the number of radium- 226 atoms present initially ( \(\mathrm{P}_{0}\) ) and the decay constants of the parent \(\left(\lambda_{\mathrm{p}}\right)\) and daughter \(\left(\lambda_{\mathrm{d}}\right)\) (c) Integration of the expression obtained in part (b) yields the following expression for the number of atoms of the radon-222 daughter (D) present at a time \(t\).$$\mathrm{D}=\frac{\mathrm{P}_{0} \lambda_{\mathrm{p}}\left(\mathrm{e}^{-\lambda_{\mathrm{p}} \times t}-\mathrm{e}^{-\lambda_{\mathrm{d}} \times t}\right)}{\lambda_{\mathrm{d}}-\lambda_{\mathrm{p}}}$$,Starting with \(1.00 \mathrm{g}\) of pure radium- \(226,\) approximately how long will it take for the amount of radon222 to reach its maximum value: one day, one week, one year, one century, or one millennium?

A nuclide has a decay rate of \(2.00 \times 10^{10} \mathrm{s}^{-1} .\) After 25.0 days, its decay rate is \(6.25 \times 10^{8} \mathrm{s}^{-1}\). What is the nuclide's half-life? (a) 25.0 d; (b) 12.5 d; (c) 50.0 d; (d) \(5.00 \mathrm{d} ;\) (e) none of these.

Explain why the rem is more satisfactory than the rad as a unit for measuring radiation dosage.

Radioactive decay and mass spectrometry are often used to date rocks after they have cooled from a magma. \(^{87} \mathrm{Rb}\) has a half-life of \(4.8 \times 10^{10}\) years and follows the radioactive decay $$^{87} \mathrm{Rb} \longrightarrow^{87} \mathrm{Sr}+\beta^{-}$$ A rock was dated by assaying the product of this decay. The mass spectrum of a homogenized sample of rock showed the \(^{87} \mathrm{Sr} /^{86} \mathrm{Sr}\) ratio to be \(2.25 .\) Assume that the original \(^{87} \mathrm{Sr} /^{86} \mathrm{Sr}\) ratio was 0.700 when the rock cooled. Chemical analysis of the rock gave \(15.5 \mathrm{ppm}\) Sr and 265.4 ppm \(\mathrm{Rb},\) using the average atomic masses from a periodic table. The other isotope ratios were \(^{86} \mathrm{Sr} /^{88} \mathrm{Sr}=\) 0.119 and \(^{84} \mathrm{Sr} /^{88} \mathrm{Sr}=0.007 .\) The isotopic ratio for \(^{87} \mathrm{Rb} /^{85} \mathrm{Rb}\) is 0.330. The isotopic masses are as follows:Calculate the following: (a) the average atomic mass of Sr in the rock (b) the original concentration of \(\mathrm{Rb}\) in the rock in \(\mathrm{ppm}\) (c) the percentage of rubidium- 87 decayed in the rock (d) the time since the rock cooled.

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