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Chapter 18: Problem 33

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

Short Answer

Expert verified
After 2 days (48 hours), approximately \(3.18\%\) of the administered dose of \(100 \: \mu g\) of Technetium-99 remains in the patient's body.

Step by step solution

01

Convert the given time to hours

Since we are given the half-life in hours, we need to convert the given time of 2.0 days to hours. There are 24 hours in a day, so 2.0 days are equal to \( 2 \times 24 \) hours. 2 days = \(2 \times 24 \) hours = 48 hours.
02

Calculate the decay constant

The decay constant, denoted as λ, can be found using the formula: λ = \( \frac{log(2)}{t_{1/2}} \), where \(t_{1/2}\) is the half-life of the radioactive isotope. Using the half-life of 6.0 hours, we get: λ = \( \frac{log(2)}{6} \) ≈ 0.1155 h⁻¹.
03

Apply the decay formula

Now, we will use the decay formula to find the remaining amount of Technetium-99 after 48 hours: \(N_t = N_0 \times e^{-\lambda t}\), where: - \(N_t\) is the remaining amount of Technetium-99 after time \(t\), - \(N_0\) is the initial amount of Technetium-99, - \(\lambda\) is the decay constant, and - \(t\) is the time in hours. We use \(N_0 = 100\) µg, \(\lambda = 0.1155\) h⁻¹, and \(t = 48\) h: \(N_t = 100\times e^{-0.1155 \times 48}\) ≈ 3.18 µg.
04

Calculate the fraction of the remaining dose

Next, we calculate the fraction of the remaining dose by dividing the remaining amount of Technetium-99 by the initial administered dose: Fraction = \( \frac{N_t}{N_0} \) = \( \frac{3.18}{100} \) = 0.0318. To express this as a percentage, multiply by 100: Percentage = 0.0318 × 100 ≈ 3.18%. Therefore, after 2 days (48 hours), about 3.18% of the administered dose of Technetium-99 remains in the patient's body.

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Most popular questions from this chapter

Consider the following information: i. The layer of dead skin on our bodies is sufficient to protect us from most \(\alpha\) -particle radiation. ii. Plutonium is an \(\alpha\) -particle producer. iii. The chemistry of \(\mathrm{Pu}^{4+}\) is similar to that of \(\mathrm{Fe}^{3+}\). iv. Pu oxidizes readily to \(\mathrm{Pu}^{4+}\) Why is plutonium one of the most toxic substances known?

When using a Geiger-Müller counter to measure radioactivity, it is necessary to maintain the same geometrical orientation between the sample and the Geiger-Müller tube to compare different measurements. Why?

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The curie (Ci) is a commonly used unit for measuring nuclear radioactivity: 1 curie of radiation is equal to \(3.7 \times 10^{10}\) decay events per second (the number of decay events from 1 g radium in 1 s). A 1.7 -mL sample of water containing tritium was injected into a 150 -lb person. The total activity of radiation injected was \(86.5 \mathrm{mCi}\). After some time to allow the tritium activity to equally distribute throughout the body, a sample of blood plasma containing \(2.0 \mathrm{mL}\) water at an activity of \(3.6 \mu \mathrm{Ci}\) was removed. From these data, calculate the mass percent of water in this 150 -lb person.
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