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Chapter 7: Problem 61

Fluorine-1 18 is an isotope used in Positron Emission Tomography ( \(\mathrm{PET}\) ) to scan the brain. If a researcher has \(1.50 \mu \mathrm{g}\) of \({ }^{18} \mathrm{~F}\), how long before it decays to \(1.0\) ng? The half- life of \({ }^{18} \mathrm{~F}\) is \(109.8\) minutes. a. \(2.9 \times 10^{-2}\) hours b. 91 hours c. 39 hours d. 19 hours

Short Answer

Expert verified
It will take approximately 19 hours (option d) for the sample to decay to 1.0 ng.

Step by step solution

01

Understand the Problem

We need to find the time it takes for the sample of \(^{18} \mathrm{F}\) to decay from 1.50 \(\mu\mathrm{g}\) to 1.0 ng. Given the half-life of \(^{18} \mathrm{F}\) is 109.8 minutes, we'll use the formula for exponential decay based on half-lives.
02

Convert Units

First, convert micrograms to nanograms: 1 microgram (\(\mu\mathrm{g}\)) equals 1000 nanograms (ng). Therefore, 1.50 \(\mu\mathrm{g}\) is equal to 1500 ng.
03

Set Up the Exponential Decay Formula

The decay formula is \[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \], where \(N(t)\) is the final amount, \(N_0\) is the initial amount, \(T_{1/2}\) is the half-life, and \(t\) is time.
04

Substitute Known Values

Substitute into the decay formula: \( N_0 = 1500 \) ng, \( N(t) = 1 \) ng, and \( T_{1/2} = 109.8 \) minutes: \[ 1 = 1500 \left( \frac{1}{2} \right)^{\frac{t}{109.8}} \].
05

Isolate the Time Variable

First, divide both sides by 1500:\[ \left( \frac{1}{2} \right)^{\frac{t}{109.8}} = \frac{1}{1500} \].Now take the natural logarithm of both sides:\[ \ln\left( \left( \frac{1}{2} \right)^{\frac{t}{109.8}} \right) = \ln\left( \frac{1}{1500} \right) \].This simplifies to:\[ \frac{-t}{109.8} \ln(2) = \ln\left( \frac{1}{1500} \right) \].
06

Solve for Time \(t\)

Rearrange to solve for \(t\):\[ t = -109.8 \frac{\ln\left( \frac{1}{1500} \right)}{\ln(2)} \].Calculate:\( \ln\left( \frac{1}{1500} \right) \approx -7.3132 \) and \( \ln(2) \approx 0.6931 \).Substitute the values:\[ t = 109.8 \times \frac{7.3132}{0.6931} \approx 1157.5 \text{ minutes} \].
07

Convert Time to Hours

Convert minutes into hours by dividing by 60:\[ \frac{1157.5}{60} \approx 19.29 \text{ hours} \].

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Key Concepts

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

Half-life
The concept of half-life is pivotal when studying radioactive decay. Half-life is the time required for half of the radioactive atoms in a sample to decay. It鈥檚 a measure of how quickly a substance loses its radioactivity. Each radioactive isotope has a unique half-life. For instance, the half-life of Fluorine-18 (used in PET scans) is 109.8 minutes. This means every 109.8 minutes, the amount of Fluorine-18 will reduce to half.

Understanding half-life allows us to calculate how long it will take for a substance to decay to a desired amount. This is essential in various applications, especially in medical imaging. For the exercise above, knowing the half-life helps determine how quickly a specific amount of Fluorine-18 decays. By using exponential decay formulas, we can calculate the exact time needed for an isotope to reach a certain level.
Radioactive Isotopes
Radioactive isotopes are variants of elements that have an unstable nucleus. This instability causes the atom to emit radiation as it decays to achieve stability. These isotopes are commonly used in medicine, research, and industrial applications.

In medicine, isotopes play a crucial role in diagnosing and treating diseases. For example, Fluorine-18, a radioactive isotope with a relatively short half-life, is used in PET scans. These isotopes are labeled and introduced into the body, where they emit positrons. The emissions are then captured to create detailed images, offering insights into bodily functions.

It鈥檚 important to handle radioactive isotopes with care due to their potential health hazards. Understanding their half-lives and decay processes is key to using them safely and effectively.
Positron Emission Tomography (PET)
Positron Emission Tomography (PET) is a diagnostic technique that helps visualize functional processes in the body. A radioactive substance, like Fluorine-18, is administrated to the patient. This substance emits positrons, which collide with electrons, producing gamma rays detected by the PET scanner.

PET scans are unique because they provide insights into the body鈥檚 metabolic activity. Doctors use it to examine blood flow, oxygen use, and how glucose is metabolized in tissues. This is particularly useful for detecting cancers, evaluating brain disorders, and examining heart problems.
PET imaging is non-invasive and provides critical information that can influence treatment plans. The use of isotopes with shorter half-lives, like Fluorine-18, ensures that patients are not exposed to high levels of radiation. This balance between cutting-edge technology and safety makes PET a valuable tool in modern medicine.

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Most popular questions from this chapter

Which of the following statements are correct about half life period? (1) time required for \(99.9 \%\) completion of a reaction is 100 times the half life period. (2) time required for \(75 \%\) completion of a \(1^{\text {st }}\) order reaction is double the half life of the reaction. (3) average life \(=1.44\) times the half life for 1 s order reaction (4) it is proportional to initial concentration for zeroth ordera. 1,2 and 3 b. 2,3 and 4 c. 2 and 3 d. 3 and 4

For a first order reaction, which is/are correct here? a. The time taken for the completion of \(75 \%\) reaction is twice the \(t_{1 / 2}\) of the reaction b. The degree of dissociation is equal to \(1-\mathrm{e}^{-k t}\). c. A plot of reciprocal concentration of the reactant versus time gives a straight line d. The pre-exponential factor in the Arrhenius equation has the dimension of time, \(T^{-1}\).

Consider a bimolecular reaction in the gas phase. Which one of the following changes in conditions will not cause an increase in the rate of the reaction? a. Increase the volume at constant temperature. b. Increase the temperature at constant volume c. Add a catalyst d. All of the above will increase the rate of reaction

For the reaction \(\mathrm{W}+\mathrm{X} \rightarrow \mathrm{Y}+\mathrm{Z}\), the rate \((\mathrm{dx} / \mathrm{dt})\) when plotted against time ' \(\mathrm{t}\) ' gives a straight line parallel to time axis. The order and rate for this reaction are a. \(\mathrm{O}, \mathrm{K}\) b. \(1, \mathrm{~K}+1\) b. II, \(\mathrm{K}+1\) d. \(\mathrm{K}, \mathrm{K}+1\)

Which of the following statement about the Arrhenius equation is/are correct? a. On raising temperature, rate constant of the reaction of greater activation energy increases less rapidly than that of the reaction of smaller activation energy. b. The term \(\mathrm{e}^{-\mathrm{Ea} \mathrm{RT}}\) represents the fraction of the molecules having energy in excess of threshold value. c. The pre-exponential factor becomes equal to the rate constant of the reaction at extremely high temperature. d. When the activation energy of the reaction is zero, the rate becomes indenendent of temnerature
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