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Chapter 19: Problem 32

Iodine-131 is used in the diagnosis and treatment of thyroid disease and has a half-life of \(8.0\) days. If a patient with thyroid disease consumes a sample of \(\mathrm{Na}^{131} \mathrm{I}\) containing \(10 . \mu \mathrm{g}{ }^{131} \mathrm{I}\), how long will it take for the amount of \({ }^{131} \mathrm{I}\) to decrease to \(1 / 100\) of the original amount?

Short Answer

Expert verified
It will take approximately \(26.57\) days for the amount of Iodine-131 to decrease to 1/100 of its original amount.

Step by step solution

01

Find the decay constant \(\lambda\)

We can find the decay constant \(\lambda\) using the following formula: $$T_{1/2} = \frac{ln(2)}{\lambda}$$ Where: - \(T_{1/2}\) is the half-life of the radioactive substance - \(\lambda\) is the decay constant We are given \(T_{1/2} = 8.0\) days. Plug this value into the formula to find \(\lambda\): $$\lambda = \frac{ln(2)}{T_{1/2}} = \frac{ln(2)}{8.0}$$
02

Set up the decay equation

Now that we have the decay constant \(\lambda\), we can set up the decay equation: \(N(t) = N_0 e^{-\lambda t}\) We are given the initial amount of Iodine-131 consumed by the patient: \(N_0 = 10 \mu g\). We are also given that the amount of Iodine-131 decreases to 1/100 of its initial amount. Therefore, \(N(t) = \frac{N_0}{100}\). Substitute these values into the decay equation: $$\frac{N_0}{100} = N_0 e^{-\lambda t}$$
03

Solve for the time \(t\)

Now, we need to solve this equation for \(t\). First, divide both sides by \(N_0\): $$\frac{1}{100} = e^{-\lambda t}$$ Next, take the natural logarithm of both sides: $$ln\left(\frac{1}{100}\right) = -\lambda t$$ Finally, divide by \(-\lambda\): $$t = \frac{ln\left(\frac{1}{100}\right)}{-\lambda}$$ Substitute the value of \(\lambda\) that we found earlier: $$t = \frac{ln\left(\frac{1}{100}\right)}{-\frac{ln(2)}{8.0}}$$
04

Calculate the time \(t\)

Now, calculate the time \(t\) using the expression we derived: $$t = \frac{ln\left(\frac{1}{100}\right)}{-\frac{ln(2)}{8.0}} \approx 26.57 \; \text{days}$$ So, it will take approximately 26.57 days for the amount of Iodine-131 to decrease to 1/100 of its original amount.

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Most popular questions from this chapter

There is a trend in the United States toward using coal-fired power plants to generate electricity rather than building new nuclear fission power plants. Is the use of coal-fired power plants without risk? Make a list of the risks to society from the use of each type of power plant.

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.

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.

Rubidium- 87 decays by \(\beta\) -particle production to strontium- 87 with a half-life of \(4.7 \times 10^{10}\) years. What is the age of a rock sample that contains \(109.7 \mu \mathrm{g}\) of \({ }^{87} \mathrm{Rb}\) and \(3.1 \mu \mathrm{g}\) of \({ }^{87} \mathrm{Sr}\) ? Assume that no \({ }^{87} \mathrm{Sr}\) was present when the rock was formed. The atomic masses for \({ }^{87} \mathrm{Rb}\) and \({ }^{87} \mathrm{Sr}\) are \(86.90919 \mathrm{u}\) and \(86.90888\) u, respectively.

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 \(B_{12}\) b. calcium- 47 , for study of bone metabolism c. iron-59, for study of red blood cell function
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