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Chapter 14: Problem 105

Americium-241 is used in smoke detectors. It has a first order rate constant for radioactive decay of \(k=1.6 \times 10^{-3} \mathrm{yr}^{-1}\). By contrast, iodine- \(125,\) which is used to test for thyroid functioning, has a rate constant for radioactive decay of \(k=0.011\) day \(^{-1}\). (a) What are the halflives of these two isotopes? (b) Which one decays at a faster rate? (c) How much of a 1.00 -mg sample of each isotope remains after 3 half-lives? (d) How much of a 1.00 -mg sample of each isotope remains after 4 days?

Short Answer

Expert verified
(a) Half-lives: Am-241: \(t_{1/2} \approx 433\ \mathrm{yr}\), I-125: \(t_{1/2} \approx 63\ \mathrm{days}\). (b) I-125 decays at a faster rate (shorter half-life). (c) Remaining amounts after 3 half-lives: Am-241: \(0.125\ \mathrm{mg}\), I-125: \(0.125\ \mathrm{mg}\). (d) Remaining amounts after 4 days: Am-241: \(0.991\ \mathrm{mg}\), I-125: \(0.656\ \mathrm{mg}\).

Step by step solution

01

Find the half-lives of Am-241 and I-125

To find the half-life of a radioactive isotope, we can use the relationship between the first-order rate constant (k) and the half-life (t½) : \[ t_{1/2} = \frac{0.693}{k} \] For Americium-241 (Am-241): \( k = 1.6 \times 10^{-3}\ \mathrm{yr}^{-1} \) For Iodine-125 (I-125): \( k = 0.011\ \mathrm{day}^{-1} \) Now we can calculate the half-lives: Half-life of Am-241: \( t_{1/2} = \frac{0.693}{1.6 \times 10^{-3}\ \mathrm{yr}^{-1}} \) Half-life of I-125: \( t_{1/2} = \frac{0.693}{0.011\ \mathrm{day}^{-1}} \)
02

Compare the decay rates of Am-241 and I-125

We can compare the decay rates by comparing their half-lives. A shorter half-life means a faster decay rate.
03

Calculate the remaining amounts of isotopes after 3 half-lives

Using the half-life and the initial mass (1 mg), we can calculate the remaining amounts of each isotope after 3 half-lives. For a first-order decay, the remaining amount after n half-lives can be calculated using the equation: \( remaining\ amount = initial\ amount \times (\frac{1}{2})^n \) where, remaining amount = remaining amount of the isotope after n half-lives initial amount = initial amount of the isotope (1 mg) n = the number of half-lives
04

Calculate the remaining amounts of isotopes after 4 days

For this step, we need to use the decay constant (k) and the initial mass (1 mg) to calculate the remaining amounts of each isotope after 4 days. The remaining amount of an isotope after a certain time can be calculated using the equation: \( remaining\ amount = initial\ amount \times e^{-kt} \) where, remaining amount = remaining amount of the isotope after time t initial amount = initial amount of the isotope (1 mg) k = first-order rate constant t = time of decay (4 days) Here, we need to convert the time in days for Am-241. 1 year = 365 days, so 4 days = \(\frac{4}{365}\) years Now, we can calculate the remaining amounts of Am-241 and I-125 after 4 days.

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Most popular questions from this chapter

Consider the following reaction: $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) The rate law for this reaction is first order in \(\mathrm{H}_{2}\) and second order in NO. Write the rate law. (b) If the rate constant for this reaction at \(1000 \mathrm{~K}\) is \(6.0 \times 10^{4} \mathrm{M}^{-2} \mathrm{~s}^{-1}\), what is the reaction rate when \([\mathrm{NO}]=0.035 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=0.015 \mathrm{M} ?(\mathrm{c}) \mathrm{What}\) is the reaction rate at \(1000 \mathrm{~K}\) when the concentration of \(\mathrm{NO}\) is increased to \(0.10 \mathrm{M},\) while the concentration of \(\mathrm{H}_{2}\) is \(0.010 \mathrm{M}\) ?

The \(\mathrm{NO}_{x}\) waste stream from automobile exhaust includes species such as \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\). Catalysts that convert these species to \(\mathrm{N}_{2}\) are desirable to reduce air pollution. (a) Draw the Lewis dot and VSEPR structures of \(\mathrm{NO}, \mathrm{NO}_{2},\) and \(\mathrm{N}_{2} .(\mathbf{b})\) Using a resource such as Table 8.4 , look up the energies of the bonds in these molecules. In what region of the electromagnetic spectrum are these energies? (c) Design a spectroscopic experiment to monitor the conversion of \(\mathrm{NO}_{x}\) into \(\mathrm{N}_{2}\), describing what wavelengths of light need to be monitored as a function of time.

Dinitrogen pentoxide \(\left(\mathrm{N}_{2} \mathrm{O}_{5}\right)\) decomposes in chloroform as a solvent to yield \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\). The decomposition is first order with a rate constant at \(45^{\circ} \mathrm{C}\) of \(1.0 \times 10^{-5} \mathrm{~s}^{-1}\). Calculate the partial pressure of \(\mathrm{O}_{2}\) produced from \(1.00 \mathrm{~L}\) of \(0.600 \mathrm{M} \mathrm{N}_{2} \mathrm{O}_{5}\) solution at \(45^{\circ} \mathrm{C}\) over a period of \(20.0 \mathrm{hr}\) if the gas is collected in a \(10.0-\mathrm{L}\) container. (Assume that the products do not dissolve in chloroform.)

As described in Exercise \(14.43,\) the decomposition of sulfuryl chloride \(\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right)\) is a first-order process. The rate constant for the decomposition at \(660 \mathrm{~K}\) is \(4.5 \times 10^{-2} \mathrm{~s}^{-1}\). (a) If we begin with an initial \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) pressure of 450 torr, what is the pressure of this substance after \(60 \mathrm{~s} ?\) (b) At what time will the pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decline to one-tenth its initial value?

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