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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Half-Life Calculation
The half-life of a substance is a key concept in understanding radioactive decay. It represents the time it takes for half of the radioactive isotopes in a sample to decay, reducing their quantity by half. In practical terms, if you start with 10 micrograms of a substance, a single half-life is the time it will take for only 5 micrograms to remain.
For Iodine-131, the half-life is given as 8.0 days. This means every 8 days, the amount of Iodine-131 will reduce by half. To find the decay constant (a crucial part in predicting decay over time), we use the known relation between half-life and decay constant: \[ T_{1/2} = \frac{ln(2)}{\lambda} \]where \( \lambda \) is the decay constant and \( T_{1/2} \) is the half-life. Plug in the half-life to find \( \lambda \).
Decay Constant
The decay constant, represented as \( \lambda \), is pivotal when discussing how quickly a radioactive isotope decays over time. It's derived from the relationship with half-life discussed earlier: \[ \lambda = \frac{ln(2)}{T_{1/2}} \]For Iodine-131 with a half-life of 8.0 days, calculate the decay constant:- Substitute 8.0 days into the formula: \( \lambda = \frac{ln(2)}{8.0} \).The decay constant offers insight into the rate of decay, allowing us to not only compute the remaining quantity at any given time but also reliably predict future quantities less than the initial amount due to decay.
The larger the decay constant, the quicker the decay process, indicating a faster decrease in the radioactive isotopes within a given timeframe.
Radioactive Isotopes
Radioactive isotopes, or radioisotopes, are variants of chemical elements that possess unstable nuclei and emit radiation during decay, transforming into isotopes of different elements. Such radioactive behavior is a natural phenomenon in numerous elements, including Iodine-131, commonly used in medicine. Key aspects of radioactive isotopes include: - **Instability**: Their nuclei tend to be unstable, which results in decay over time and the emission of particles. - **Applications**: Utilized extensively in fields like medicine (for diagnostic imaging or treatment), archaeology (carbon dating), and energy (nuclear power). - **Decay Types**: Can undergo alpha, beta, and gamma decay, each involving the release of specific types of particles or energy. Understanding these isotopes is essential for applications requiring precise management of the decay process, like in medical treatments where accurate dosage over time is crucial.
Exponential Decay Equation
The exponential decay equation is fundamental in predicting how much of a radioactive isotope remains after a certain time period. The equation:\[ N(t) = N_0 e^{-\lambda t} \]Where:- \( N(t) \) is the amount remaining at time \( t \).- \( N_0 \) is the original amount.- \( \lambda \) is the decay constant.- \( t \) is time.This formula captures the essence of how substances undergo exponential decay, showing a decreasing curve over time. For practical application, it’s used to determine how long it will take for a certain quantity of a substance to decay to a specific level. In our example, with initial Iodine-131 being 10.0 micrograms reducing to 0.1 micrograms (\(1/100\) of the original), plugging in these numbers and solving allows us to predict that roughly 26.57 days is required to reach this reduced level.
This predictable pattern of decay makes exponential functions powerful tools in radioactivity studies, allowing scientists and professionals to plan and track material usage or exposure over time.

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Most popular questions from this chapter

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

The bromine- 82 nucleus has a half-life of \(1.0 \times 10^{3}\) min. If you wanted \(1.0 \mathrm{~g}{ }^{82} \mathrm{Br}\) and the delivery time was \(3.0\) days, what mass of NaBr should you order (assuming all of the \(\mathrm{Br}\) in the \(\mathrm{NaBr}\) was \({ }^{82} \mathrm{Br}\) )?

Zirconium is one of the few metals that retains its structural integrity upon exposure to radiation. The fuel rods in most nuclear reactors therefore are often made of zirconium. Answer the following questions about the redox properties of zirconium based on the half-reaction \(\mathrm{ZrO}_{2} \cdot \mathrm{H}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}+4 \mathrm{e}^{-} \longrightarrow \mathrm{Zr}+4 \mathrm{OH}^{-} \quad \mathscr{E}^{\circ}=-2.36 \mathrm{~V}\) a. Is zirconium metal capable of reducing water to form hydrogen gas at standard conditions? b. Write a balanced equation for the reduction of water by zirconium. c. Calculate \(\mathscr{E}^{\circ}, \Delta G^{\circ}\), and \(K\) for the reduction of water by zirconium metal. d. The reduction of water by zirconium occurred during the accidents at Three Mile Island in \(1979 .\) The hydrogen produced was successfully vented and no chemical explosion occurred. If \(1.00 \times 10^{3} \mathrm{~kg}\) Zr reacts, what mass of \(\mathrm{H}_{2}\) is produced? What volume of \(\mathrm{H}_{2}\) at \(1.0 \mathrm{~atm}\) and \(1000 .{ }^{\circ} \mathrm{C}\) is produced? e. At Chernobyl in 1986, hydrogen was produced by the reaction of superheated steam with the graphite reactor core: $$ \mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_{2}(g) $$

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 \({ }^{232}\) Pu nuclei and 1 mole of \({ }^{231}\) Pa nuclei are each formed from their respective number of protons and neutrons.

Natural uranium is mostly nonfissionable \({ }^{238} \mathrm{U} ;\) it contains only about \(0.7 \%\) of fissionable \({ }^{235} \mathrm{U}\). For uranium to be useful as a nuclear fuel, the relative amount of \({ }^{235} \mathrm{U}\) 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 5, Section 5.7). Explain how Graham's law of effusion allows natural uranium to be enriched by the gaseous diffusion process.

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