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Technetium-99 has been used as a radiographic agent in bone scans \(\left({ }_{4,}^{99} \mathrm{Tc}\right.\) is absorbed by bones). If \({ }_{43}^{99} \mathrm{Tc}\) has a half-life of \(6.0\) hours, what fraction of an administered dose of \(100 . \mu \mathrm{g}_{43}^{99} \mathrm{Tc}\) remains in a patient's body after \(2.0\) days?

Short Answer

Expert verified
After \(2.0\) days, approximately \(0.390\mu g\) of \({}_{43}^{99}\mathrm{Tc}\) remains in the patient's body, which is a fraction of \(\frac{1}{256}\) of the initial administered dose of \(100\mu g\).

Step by step solution

01

Convert time from days to hours

First, we need to convert the given time from days to hours, as the half-life is provided in hours. Given time, \(t = 2.0\) days. There are \(24\) hours in a day, so we have: \[t = 2.0 \times 24 = 48\; \mathrm{hours}\]
02

Determine the number of half-lives

Now, we need to find out how many half-lives have passed in \(48\) hours. To do this, we will divide the total time passed (48 hours) by the half-life of Technetium-99 (6 hours). \[\text{Number of Half-lives} = \frac{\text{Total Time Passed}}{\text{Half-life Period}} = \frac{48}{6} = 8\]
03

Calculate the remaining fraction

We know that after 1 half-life, half of the initial substance remains. Therefore, after 2 half-lives, half of the remaining half will remain, and so on. To find the remaining fraction of the initial dose after 8 half-lives, we will simply calculate the following: \[ \text{Remaining Fraction} = \left(\frac{1}{2}\right)^\textit{n}\] where \(\textit{n}\) is the number of half-lives. Substituting the value of \(\textit{n} = 8\): \[ \text{Remaining Fraction} = \left(\frac{1}{2}\right)^8 = \frac{1}{256}\]
04

Find the remaining dose of Technetium-99

Now that we have the remaining fraction, we can calculate the remaining dose of Technetium-99 after 2 days. To do this, we simply multiply the initial dose (100µg) by the remaining fraction: \[ \text{Remaining Dose} = \text{Initial Dose} \times \text{Remaining Fraction} = 100\mu g \times \frac{1}{256}\] \[ \text{Remaining Dose} = \frac{100\mu g}{256} \approx 0.390\mu g\] So, after 2 days, approximately \(0.390\) micrograms of \({}_{43}^{99}\mathrm{Tc}\) remains in the patient's body.

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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 crucial in the study of radioactive substances. It refers to the time it takes for half of a given amount of a radioactive material to decay. This decay process is exponential, meaning it happens at a consistent rate relative to the remaining amount of substance. For example, if a substance has a half-life of 6 hours, after 6 hours, only half of the original quantity will remain. After another 6 hours, a quarter will remain, and so on.
Understanding half-life helps in various fields such as medicine, where it is used to determine how long a radioactive substance will remain active in the body, which is vital for patient safety in treatments and diagnostics.
Technetium-99
Technetium-99 ( _{43}^{99}Tc) is a widely used radioisotope in the field of nuclear medicine, particularly in diagnostic imaging. It emits gamma rays, which can be detected by imaging devices to create a visual representation of structures and functions inside the body.
It has a relatively short half-life of 6 hours, which is advantageous because it limits the patient's exposure to radiation. Technetium-99 is also notable for being easily incorporated into other molecules, making it versatile for targeting different bodily organs, such as bones, lungs, and the heart, among others.
Radioisotopes
Radioisotopes are variants of an element that have unstable nuclei, emitting radiation as they decay to achieve stability. They have numerous applications due to these properties, particularly in medical diagnostics and treatment.
Their ability to decay and emit radiation allows them to be used as tracers in imaging techniques, where they can highlight abnormalities in the body. They are also employed in radiotherapy to target and destroy cancer cells. Understanding each radioisotope's characteristics, such as half-life and type of radiation emitted, is crucial for their effective and safe application.
Bone Scans
Bone scans are a diagnostic tool that uses radioactive substances to assess bone health. A common isotope used in these scans is Technetium-99 because of its properties that make it ideal for highlighting bone abnormalities. When injected into the body, Technetium-99 tends to accumulate in areas of high bone activity.
This accumulation can be detected with a special camera, providing images that can reveal changes in the bones, such as fractures, infections, or cancerous growths. Bone scans are particularly useful because they can detect problems earlier than X-rays, allowing for prompt diagnosis and treatment.

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Most popular questions from this chapter

The most stable nucleus in terms of binding energy per nucleon is \({ }^{56} \mathrm{Fe}\). If the atomic mass of \({ }^{56} \mathrm{Fe}\) is \(55.9349 \mathrm{amu}\), calculate the binding energy per nucleon for \({ }^{56} \mathrm{Fe}\).

A positron and an electron can annihilate each other on colliding, producing energy as photons: $${ }_{-1}^{0} \mathrm{e}+{ }_{+1}^{0} \mathrm{e} \longrightarrow 2{ }_{0}^{0} \gamma$$ Assuming that both \(\gamma\) rays have the same energy, calculate the wavelength of the electromagnetic radiation produced.

The radioactive isotope \({ }^{247} \mathrm{Bk}\) decays by a series of \(\alpha\) -particle and \(\beta\) -particle productions, taking \({ }^{247} \mathrm{Bk}\) through many transformations to end up as \({ }^{207} \mathrm{~Pb}\). In the complete decay series, how many \(\alpha\) particles and \(\beta\) particles are produced?

Predict whether each of the following nuclides is stable or unstable (radioactive). If the nuclide is unstable, predict the type of radioactivity you would expect it to exhibit. a. \({ }_{19} \mathrm{~K}\) b. \({ }_{26}^{56} \mathrm{Fe}\) c. \({ }_{11}^{20} \mathrm{Na}\) d. \({ }_{81}^{194} \mathrm{Tl}\)

Write an equation describing the radioactive decay of each of the following nuclides. (The particle produced is shown in parentheses, except for electron capture, where an electron is a reactant.) a. \({ }^{3} \mathrm{H}(\beta)\) b. \({ }_{3}^{8} \mathrm{Li}(\beta\) followed by \(\alpha\) ) c. \({ }_{4}^{7} \mathrm{Be}\) (electron capture) d. \({ }_{5}^{8} \mathrm{~B}\) (positron)

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