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Chapter 40: Problem 69

\(^{214} \mathrm{Pb}\) has a half-life of \(26.8 \mathrm{~min}\). How many minutes must elapse for \(90.0 \%\) of a given sample of \({ }^{214} \mathrm{~Pb}\) atoms to decay?

Short Answer

Expert verified
Answer: It takes approximately 89.16 minutes for 90% of a given sample of 214Pb atoms to decay.

Step by step solution

01

Understand the given information and the half-life formula

We are given the half-life of \(^{214}\mathrm{Pb}\) atoms as 26.8 minutes, and we need to find the time it takes for 90% of the sample to decay. We can use the half-life formula: \(N_t = N_0 \times \left( \frac{1}{2} \right)^\frac{t}{t_{1/2}}\) where: - \(N_t\) is the remaining amount of substance at time \(t\) - \(N_0\) is the initial amount of substance - \(t\) is the time elapsed - \(t_{1/2}\) is the half-life of the substance Since 90% of the substance has decayed, it means that 10% is remaining. So, we can write \(N_t\) as \(0.1N_0\). Now, we need to find the time \(t\).
02

Setup the equation and isolate time

We can rewrite the half-life formula with the given values for the initial and remaining amounts of \(^{214}\mathrm{Pb}\) atoms: \(0.1N_0 = N_0 \times \left( \frac{1}{2} \right)^\frac{t}{26.8}\) Since we are looking for the time, we can eliminate the initial amount \(N_0\) from both sides: \(0.1 = \left( \frac{1}{2} \right)^\frac{t}{26.8}\) Now, we need to solve for \(t\).
03

Solve for time

To solve for \(t\), we can take the natural logarithm of both sides of the equation: \(\ln(0.1) = \ln \left(\frac{1}{2}\right)^{\frac{t}{26.8}}\) Using the properties of logarithms, we can bring the exponent to the front: \(\ln(0.1) = \frac{t}{26.8} \times \ln \left( \frac{1}{2} \right)\) Now, we can solve for \(t\) by isolating it: \(t = \frac{26.8 \times \ln(0.1)}{\ln\left(\frac{1}{2}\right)}\) Calculating the value for \(t\): \(t \approx 89.16 \mathrm{~minutes}\)
04

Conclusion

It takes approximately 89.16 minutes for 90% of a given sample of \(^{214}\mathrm{Pb}\) atoms to decay.

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Most popular questions from this chapter

The most common isotope of uranium, \({ }_{92}^{238} \mathrm{U},\) produces radon \({ }_{86}^{222} \mathrm{Rn}\) through the following sequence of decays: $$\begin{array}{c}{ }^{238} \mathrm{U} \rightarrow{ }^{234} \mathrm{Th}+\alpha,{ }^{234} \mathrm{Th} \rightarrow{ }^{234} \mathrm{~Pa}+\beta^{-}+\bar{\nu}_{e}, \\\\{ }_{91}^{234} \mathrm{~Pa} \rightarrow{ }_{92}^{234} \mathrm{U}+\beta+\bar{\nu}_{e},{ }^{234} \mathrm{U} \rightarrow{ }^{230} \mathrm{Th}+\alpha ,\\\\{ }_{91}^{230} \mathrm{Th} \rightarrow{ }_{90}^{226} \mathrm{Ra}+\alpha,{ }_{88}^{226} \mathrm{Ra} \rightarrow{ }_{86}^{222} \mathrm{Rn}+\alpha,\end{array}$$. A sample of \({ }_{92}^{238} \mathrm{U}\) will build up equilibrium concentrations of its daughter nuclei down to \({ }_{88}^{226} \mathrm{Ra} ;\) the concentrations of each are such that each daughter is produced as fast as it decays. The \({ }_{88}^{226} \mathrm{Ra}\) decays to \({ }_{86}^{222} \mathrm{Rn},\) which escapes as a gas. (The \(\alpha\) particles also escape, as helium; this is a source of much of the helium found on Earth.) In high concentrations, the radon is a health hazard in buildings built on soil or foundations containing uranium ores, as it can be inhaled. a) Look up the necessary data, and calculate the rate at which \(1.00 \mathrm{~kg}\) of an equilibrium mixture of \({ }_{92}^{238} \mathrm{U}\) and its first five daughters produces \({ }_{86}^{222} \mathrm{Rn}\) (mass per unit time). b) What activity (in curies per unit time) of radon does this represent?

What is more dangerous, a radioactive material with a short half-life or a long one?

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A drug containing \({ }_{43}^{99} \mathrm{Tc}\left(t_{1 / 2}=6.05 \mathrm{~h}\right)\) with an activity of \(1.50 \mu \mathrm{Ci}\) is to be injected into a patient at \(9.30 \mathrm{a} . \mathrm{m} .\) You are to prepare the sample \(2.50 \mathrm{~h}\) before the injection (at 7: 00 a.m.). What activity should the drug have at the preparation time (7:00 a.m.)?

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