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Chapter 18: Problem 74

A certain radioactive nuclide has a half-life of 3.00 hours. a. Calculate the rate constant in \(s^{-1}\) for this nuclide. b. Calculate the decay rate in decays/s for 1.000 mole of this nuclide.

Short Answer

Expert verified
The rate constant for the radioactive decay is \(6.42 × 10^{-5} s^{-1}\), and the decay rate for 1 mole of this radioactive nuclide is approximately \(3.864 × 10^{19} decays/s\).

Step by step solution

01

Calculate the rate constant

First, we need to calculate the rate constant for the radioactive decay. The half-life (t1/2) of a first-order reaction can be related to the rate constant (k) using the following equation: t1/2 = 0.693 / k Given the half-life is 3 hours, we can convert it into seconds by multiplying it by 3600 seconds/hour: t1/2 = 3.00 hours × 3600 s/h = 10,800 s Now, we can solve for the rate constant k: k = 0.693 / t1/2 k = 0.693 / 10,800 s k = 6.42 × 10^(-5) s^(-1) Therefore, the rate constant for the decay is 6.42 × 10^(-5) s^(-1).
02

Calculate the decay rate

Next, we need to calculate the decay rate for 1 mole of the nuclide. The decay rate (N) can be obtained using the following equation: N = k × N0 Where N0 is the initial amount of the radioactive substance. We know there is 1 mole of the nuclide, therefore we use Avogadro's number to calculate the number of radioactive nuclei: N0 = 1 mole × 6.022 × 10^(23) nuclei/mole ≈ 6.022 × 10^(23) nuclei Now, we can calculate the decay rate N: N = (6.42 × 10^(-5) s^(-1)) × (6.022 × 10^(23) nuclei) N = 3.864 × 10^(19) decays/s Therefore, the decay rate for 1 mole of this radioactive nuclide is approximately 3.864 × 10^(19) decays/s.

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Most popular questions from this chapter

Naturally occurring uranium is composed mostly of \(^{238} \mathrm{U}\) and \(^{235} \mathrm{U},\) with relative abundances of \(99.28 \%\) and \(0.72 \%,\) respectively. The half-life for \(^{238} \mathrm{U}\) is \(4.5 \times 10^{9}\) years, and the half-life for \(^{235} \mathrm{U}\) is \(7.1 \times 10^{8}\) years. Assuming that the earth was formed 4.5 billion years ago, calculate the relative abundances of the \(^{238} \mathrm{U}\) and \(^{235} \mathrm{U}\) isotopes when the earth was formed.

A chemist studied the reaction mechanism for the reaction $$2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)$$ by reacting \(\mathrm{N}^{16} \mathrm{O}\) with \(^{18} \mathrm{O}_{2}\). If the reaction mechanism is $$\begin{aligned} \mathrm{NO}+\mathrm{O}_{2} & \rightleftharpoons \mathrm{NO}_{3}(\text { fast equilibrium }) \\ \mathrm{NO}_{3}+\mathrm{NO} & \longrightarrow 2 \mathrm{NO}_{2}(\text { slow }) \end{aligned}$$ what distribution of \(^{18} \mathrm{O}\) would you expect in the \(\mathrm{NO}_{2} ? \)Assume that \(\mathrm{N}\) is the central atom in \(\mathrm{NO}_{3},\) assume only \(\mathrm{N}^{16} \mathrm{O}^{18} \mathrm{O}_{2}\) forms, and assume stoichiometric amounts of reactants are combined.

Natural uranium is mostly nonfissionable \(^{238} \mathrm{U} ;\) it contains only about \(0.7 \%\) of fissionable \(^{235}\) U. For uranium to be useful as a nuclear fuel, the relative amount of \(^{235}\) U must be increased to about \(3 \% .\) This is accomplished through a gas diffusion process. In the diffusion process, natural uranium reacts with fluorine to form a mixture of \(^{238} \mathrm{UF}_{6}(g)\) and \(^{235} \mathrm{UF}_{6}(g) .\) The fluoride mixture is then enriched through a multistage diffusion process to produce a \(3 \%^{235} \mathrm{U}\) nuclear fuel. The diffusion process utilizes Graham's law of effusion (see Chapter \(8,\) Section \(8-7\) ). Explain how Graham's law of effusion allows natural uranium to be enriched by the gaseous diffusion process.

Define "third-life" in a similar way to "half-life," and determine the "third- life" for a nuclide that has a half-life of 31.4 years.

Write an equation describing the radioactive decay of each of the following nuclides. (The particle produced is shown in parentheses, except for electron capture, where an electron is a reactant.) a. \(\frac{3}{1} \mathrm{H}(\beta)\) b. \(\frac{8}{3} L i(\beta \text { followed by } \alpha)\) c. \(^{7}_{4}\) Be (electron capture)d. \(^{8}_{5} B\) (positron)
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