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Chapter 19: Problem 82

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

Short Answer

Expert verified
a. The rate constant (k) for the radioactive nuclide is approximately \(6.4 * 10^{-5}\) \(\mathrm{s}^{-1}\). b. The decay rate for 1.000 mole of this nuclide is approximately 3.85 x 10^19 decays/s.

Step by step solution

01

Convert half-life to seconds

Since the half-life is given in hours, let's first convert it to seconds. 1 hour = 3600 seconds Half-life = 3.00 hours = 3.00 * 3600 seconds
02

Calculate the rate constant (k)

The half-life is related to the rate constant for radioactive decay via the formula: Half-life (t) = \(\frac{0.693}{k}\) Where k is the rate constant of the decay. Rearrange this equation to find the rate constant: k = \(\frac{0.693}{t}\) Substitute the half-life value into the formula: k = \(\frac{0.693}{(3.00 * 3600) \, \text{s}}\) Now, calculate the value of k: k ≈ \(6.4 * 10^{-5}\) \(\mathrm{s}^{-1}\)
03

Calculate the decay rate

To calculate the decay rate in decays/s for 1.000 mole of this nuclide, we can use the following formula: Decay rate (R) = k * N Where R is the decay rate, k is the rate constant, and N is the number of radioactive atoms in the sample. First, convert 1.000 mole of the nuclide to the number of atoms using Avogadro's number (6.022 x 10^23 atoms/mol): N = 1.000 mol * (6.022 x 10^23 atoms/mol) = 6.022 x 10^23 atoms Now, substitute the values of k and N into the decay rate formula: R = \(6.4 * 10^{-5}\) \(\mathrm{s}^{-1}\) * 6.022 x 10^23 atoms Finally, calculate the decay rate: R ≈ 3.85 x 10^19 decays/s The decay rate in decays/s for 1.000 mole of this nuclide is approximately 3.85 x 10^19 decays/s.

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Most popular questions from this chapter

Many transuranium elements, such as plutonium-232 , have very short half- lives. (For \(^{232} \mathrm{Pu}\) , the half-life is 36 minutes.) However, some, like protactinium- 231 (half-life \(=3.34 \times 10^{4}\) years), have relatively long half-lives. Use the masses given in the following table to calculate the change in energy when 1 mole of \(^{232} \mathrm{Pu}\) nuclei and 1 mole of \(^{231} \mathrm{Pa}\) nuclei are each formed from their respective number of protons and neutrons. (Since the masses of \(^{232} \mathrm{Pu}\) and \(^{231} \mathrm{Pa}\) are atomic masses, they each include the mass of the electrons present. The mass of the nucleus will be the atomic mass minus the mass of the electrons.)

Radioactive copper-64 decays with a half-life of 12.8 days. a. What is the value of \(k\) in \(\mathrm{s}^{-1} ?\) b. A sample contains 28.0 \(\mathrm{mg}^{64} \mathrm{Cu}\) . How many decay events will be produced in the first second? Assume the atomic mass of $^{64} \mathrm{Cu}\( is 64.0 \)\mathrm{u} .$ c. A chemist obtains a fresh sample of \(^{64} \mathrm{Cu}\) and measures its radioactivity. She then determines that to do an experiment, the radioactivity cannot fall below 25\(\%\) of the initial measured value. How long does she have to do the experiment?

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 NO \(_{2} ?\) Assume that \(\mathrm{N}\) is the central atom in \(\mathrm{NO}_{3},\) assume only \(\mathrm{N}^{16} \mathrm{O}^{16} \mathrm{O}_{2}\) forms, and assume stoichiometric amounts of reactants are combined.

Radioactive cobalt-60 is used to study defects in vitamin \(\mathrm{B}_{12}\) absorption because cobalt is the metallic atom at the center of the vitamin \(\mathrm{B}_{12}\) molecule. The nuclear synthesis of this cobalt isotope involves a three-step process. The overall reaction is iron-58 reacting with two neutrons to produce cobalt-60 along with the emission of another particle. What particle is emitted in this nuclear synthesis? What is the binding energy in J per nucleon for the cobalt-60 nucleus (atomic masses: \(^{60} \mathrm{Co}=\) $59.9338 \mathrm{u} ;^{-1} \mathrm{H}=1.00782$ u)? What is the de Broglie wavelength of the emitted particle if it has a velocity equal to \(0.90 c,\) where \(c\) is the speed of light?

Each of the following isotopes has been used medically for the purpose indicated. Suggest reasons why the particular element might have been chosen for this purpose a. cobalt-57, for study of the body's use of vitamin \(\mathrm{B}_{12}\) b. calcium- 47 , for study of bone metabolism c. iron-59, for study of red blood cell function
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