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Chapter 21: Problem 27

Why do radioactive decay series obey first-order kinetics?

Short Answer

Expert verified
Radioactive decay series follow first-order kinetics because the rate of decay is directly proportional to the number of parent nuclides present at any given time, which fits the definition of first-order kinetics. This means that each decay event is independent of all others and does not depend on the history of the sample, but only on the number of parent nuclides present.

Step by step solution

01

Understanding Radioactive Decay

Radioactive decay is a random process at the level of single atoms, in that, according to quantum theory, it is impossible to predict when a particular atom will decay. However, the chance that a given atom will decay is constant over time. For a large number of atoms, the overall decay can be seen to follow a predictable pattern. This pattern forms the basis for the principles of radioactivity and notably the first-order kinetics.
02

Understanding First-Order Kinetics

First-order kinetics is a concept in physical chemistry where the rate of reaction is directly proportional to the concentration of one of the products. It's called 'first order' because only the first power of the concentration affects the reaction rate. The mathematical representation of first-order kinetics is \(-\frac{d[A]}{dt} = k[A]\), where [A] represents the concentration of the material, k is the rate constant, and t represents time.
03

Applying First-Order Kinetics to Radioactive Decay

In a radioactive decay series, each parent nuclide spontaneously decays into a daughter nuclide. The decay rate (which is the change in the number of parent nuclides over time) is directly proportional to the number of parent nuclides present. This fits the definition of first-order kinetics, as the rate of decay depends on the first power of the number of parent nuclides present and remains constant over time. As such, we can describe radioactive decay using the mathematical model of first-order kinetics.

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Most popular questions from this chapter

(a) Calculate the energy released when a \({ }^{238} \mathrm{U}\) isotope decays to \({ }^{234} \mathrm{Th} .\) The atomic masses are given by: \(^{238} \mathrm{U}: 238.0508 \mathrm{amu} ;{ }^{234} \mathrm{Th}: 234.0436 \mathrm{amu} ;{ }^{4} \mathrm{He}\) 4.0026 amu. (b) The energy released in (a) is transformed into the kinetic energy of the recoiling \({ }^{234} \mathrm{Th}\) nucleus and the \(\alpha\) particle. Which of the two will move away faster? Explain.

Define nuclear fusion, thermonuclear reaction, and plasma.

Nuclear waste disposal is one of the major concerns of the nuclear industry. In choosing a safe and stable environment to store nuclear wastes, consideration must be given to the heat released during nuclear decay. As an example, consider the \(\beta\) decay of \({ }^{90} \mathrm{Sr}\) \((89.907738 \mathrm{amu})\) $$ { }_{38}^{90} \mathrm{Sr} \longrightarrow{ }_{39}^{90} \mathrm{Y}+{ }_{-1}^{0} \beta \quad t_{\frac{1}{2}}=28.1 \mathrm{yr} $$ The \({ }^{90} \mathrm{Y}(89.907152 \mathrm{amu})\) further decays as follows: $$ { }_{39}^{90} \mathrm{Y} \longrightarrow{ }_{40}^{90} \mathrm{Zr}+{ }_{-1}^{0} \beta \quad t_{\frac{1}{2}}=64 \mathrm{~h} $$ Zirconium-90 (89.904703 amu) is a stable isotope. (a) Use the mass defect to calculate the energy released (in joules) in each of the preceding two decays. (The mass of the electron is \(5.4857 \times\) \(10^{-4}\) amu. ( b) Starting with 1 mole of \({ }^{90}\) Sr, calculate the number of moles of \(9^{9}\) Sr that will decay in a year. (c) Calculate the amount of heat released (in kilojoules) corresponding to the number of moles of \({ }^{90} \mathrm{Sr}\) decayed to \({ }^{90} \mathrm{Zr}\) in \((\mathrm{b})\)

From the definition of curie, calculate Avogadro's number, given that the molar mass of \({ }^{226} \mathrm{Ra}\) is \(226.03 \mathrm{~g} / \mathrm{mol}\) and that it decays with a half-life of \(1.6 \times 10^{3} \mathrm{yr}\)

Why is strontium-90 a particularly dangerous isotope for humans?
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