Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Fluorine-18 is an isotope used in Positron Emission Tomography (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
The decay time is approximately 91 hours.

Step by step solution

01

Understand the problem

The problem asks how long it will take for a mass of Fluorine-18 to decay from \(1.50 \mu \text{g}\) to \(1.0 \text{ ng}\). The half-life of Fluorine-18 is given as 109.8 minutes. We need to convert both masses to the same units and then use the decay formula.
02

Convert units

Convert the initial mass from micrograms to nanograms: 1 microgram = 1000 nanograms, so \(1.50 \mu \text{g} = 1500 \text{ ng}\). The final mass is already given in nanograms as 1.0 ng.
03

Apply the decay formula

The decay of a radioactive substance is given by the formula \(N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}\), where \(N(t)\) is the remaining mass, \(N_0\) is the initial mass, \(t\) is the time elapsed, and \(T_{1/2}\) is the half-life. Here, \(N_0 = 1500 \text{ ng}\), \(N(t) = 1.0 \text{ ng}\), and \(T_{1/2} = 109.8 \text{ min}\).
04

Solve for time t

Set up the equation: \(1.0 = 1500 \times \left(\frac{1}{2}\right)^{\frac{t}{109.8}}\). Divide both sides by 1500 to get \(\frac{1}{1500} = \left(\frac{1}{2}\right)^{\frac{t}{109.8}}\). To isolate \(t\), take the natural logarithm of both sides: \(\ln\left(\frac{1}{1500}\right) = \frac{t}{109.8} \ln\left(\frac{1}{2}\right)\).
05

Calculate t in minutes

Calculate \(\ln\left(\frac{1}{1500}\right)\) and \(\ln\left(\frac{1}{2}\right)\), then solve for \(t\): \(t = \frac{\ln\left(\frac{1}{1500}\right)}{\ln\left(\frac{1}{2}\right)} \times 109.8\).
06

Convert t to hours

After calculating \(t\) in minutes, convert to hours: \(\text{time in hours} = \frac{t}{60}\). Calculate this to determine which answer fits best.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Half-life Calculation
Radioactive decay happens because unstable isotopes emit particles, turning into different elements or isotopes. An important concept related to radioactive decay is the half-life, which is the time it takes for half of a sample to decay. This means if you begin with a certain amount of a radioactive substance, half of it will have decayed after one half-life period. The half-life is crucial in calculations because it helps predict how long a sample takes to reach a particular state of decay.

For instance, the decay of a substance can be described mathematically as follows: \[N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}\]where:
  • \( N(t) \): Remaining mass at time \( t \)
  • \( N_0 \): Initial mass
  • \( T_{1/2} \): Half-life of the substance
Using this formula, students can predict the time it takes for a particular amount of substance to decay by rearranging the formula and solving for \( t \). This involves logarithmic calculations, where the natural logarithm is often used to isolate the variable \( t \), allowing us to determine the time at which a particular mass remains.
Fluorine-18
Fluorine-18 is a radioisotope significant in the field of medicine. It is a positron emitter, which makes it highly suitable for imaging techniques like Positron Emission Tomography (PET).

Fluorine-18 has a relatively short half-life of approximately 109.8 minutes. This is ideal for medical imaging because the isotope decays quickly enough to minimize long-term radiation exposure to patients but allows sufficient time to conduct a PET scan. During its decay, Fluorine-18 releases positrons. When these positrons encounter electrons, they annihilate, producing gamma rays that the PET scanner detects to create detailed images of the body's internal structures and functions.

In practice, this means that handling and storing Fluorine-18 requires stringent safety protocols due to its radioactive nature. Nevertheless, its short lifespan demands efficient logistics to ensure that it is utilized in medical scans soon after production.
Positron Emission Tomography (PET)
Positron Emission Tomography (PET) is a powerful imaging technique in medical diagnostics.

PET scans take advantage of the properties of substances like Fluorine-18, which emit positrons during decay. These positrons travel short distances before encountering electrons, resulting in the emission of gamma rays. PET scanners detect these gamma rays to produce comprehensive images of metabolic processes.

PET is predominantly used in oncology, cardiology, and neurology. The scans allow healthcare professionals to detect abnormalities and monitor disease progression by revealing changes in cellular activity. This makes PET a valuable tool for early detection of cancers, brain disorders, and heart diseases, offering insights that other imaging techniques might not reveal.

The precision of PET is remarkable as it doesn't just show structural changes but can illustrate functional activities within tissues. This aspect is pivotal in advancing personalized medicine and treatment plans tailored specifically to a patient's unique physiological conditions.
Unit Conversion in Chemistry
Unit conversion is a fundamental skill in chemistry, essential for solving problems accurately. It allows chemists and students to express quantities in different units that align with the context of the problem.

In the context of radioactive decay, converting units like micrograms to nanograms can be crucial. For instance:
  • 1 microgram (\(\mu\text{g}\)) = 1000 nanograms (ng)
  • This fact helps standardize units for calculations involving changes in mass
Using correct units is vital. Whether you're calculating the remaining quantity of a substance or solving for time, consistent unit usage avoids errors that could result in incorrect scientific interpretations.

Learning to adeptly navigate between units often involves dimensional analysis, allowing for smooth transitions between complex conversions in chemistry problems. Over time, this skill enhances accuracy in laboratory settings and when working with theoretical problems in scientific research.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In a first order reaction the concentration of reactant decreases from \(800 \mathrm{~mol} / \mathrm{dm}^{3}\) to \(50 \mathrm{~mol} / \mathrm{dm}^{3}\) in \(2 \times\) \(10^{4} \mathrm{sec}\). The rate constant of reaction in \(\mathrm{sec}^{-1}\) is a. \(2 \times 10^{4}\) b. \(3.45 \times 10^{-5}\) c. \(1.386 \times 10^{-4}\) d. \(2 \times 10^{-4}\)

When the concentration of \(\mathrm{A}\) is doubled, the rate for the reaction: \(2 \mathrm{~A}+\mathrm{B} \rightarrow 2 \mathrm{C}\) quadruples. When the concentration of \(\mathrm{B}\) is doubled the rate remains the same. Which mechanism below is consistent with the experimental observations? a. Step I: \(2 \mathrm{~A} \rightleftharpoons \mathrm{D}\) (fast equilibrium) Step II: \(\mathrm{B}+\mathrm{D} \rightarrow \mathrm{E}\) (slow) Step III: \(\mathrm{E} \rightarrow 2 \mathrm{C}\) (fast) b. Step I: \(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{D}\) (fast equilibrium) Step II: \(\mathrm{A}+\mathrm{D} \rightarrow 2 \mathrm{C}\) (slow) c. Step \(\mathrm{I}: \mathrm{A}+\mathrm{B} \rightarrow \mathrm{D}\) (slow) Step II: \(\mathrm{A}+\mathrm{D} \rightleftharpoons 2 \mathrm{C}\) (fast equilibrium) d. Step I: \(2 \mathrm{~A} \rightarrow \mathrm{D}\) (slow) Step II: \(\mathrm{B}+\mathrm{D} \rightarrow \mathrm{E}\) (fast) Step III: \(\mathrm{E} \rightarrow 2 \mathrm{C}\) (fast)

Hydrogen iodide decomposes at \(800 \mathrm{~K}\) via a second order process to produce hydrogen and iodine according to the following chemical equation. \(2 \mathrm{HI}(\mathrm{g}) \rightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g})\) At \(800 \mathrm{~K}\) it takes 142 seconds for the initial concentration of \(\mathrm{HI}\) to decrease from \(6.75 \times 10^{-2} \mathrm{M}\) to \(3.50 \times 10^{-2} \mathrm{M}\). What is the rate constant for the reaction at this temperature? a. \(6.69 \times 10^{-3} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) b. \(7.96 \times 10^{-2} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) c. \(19.6 \times 10^{-3} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) d. \(9.69 \times 10^{-2} \mathrm{M}^{-1} \mathrm{~s}^{-1}\)

Observe the reaction given below \(\mathrm{A}+2 \mathrm{~B} \rightarrow 3 \mathrm{C}\) \(\begin{array}{lll}\mathrm{g} & \mathrm{g} & \mathrm{g}\end{array}\) If the rate of this reaction \(-\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}\) is \(2 \times 10^{-3} \mathrm{~mol} \mathrm{lit}^{-1} \mathrm{~min}^{-1}\) than the value of \(\frac{\mathrm{dB}}{\mathrm{dt}}\) and \(\frac{\mathrm{dC}}{\mathrm{dt}}\) will be respectively a. \(1 \times 10^{-3}, 2 / 3 \times 10^{-3}\) b. \(4 \times 10^{-3}, 6 \times 10^{-3}\) c. \(6 \times 10^{-3}, 4 \times 10^{-3}\) d. \(2 / 3 \times 10^{-3}, 1 \times 10^{-3}\)

Which of the following graphs for a first order reaction ( \(\mathrm{A} \rightarrow\) Products) would be straight line? a. Rate vs time b. Rate vs \([\mathrm{A}]\) c. Rate vs \(\log [\mathrm{A}]\) d. \(\log [\mathrm{A}]\) vs time

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free