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Chapter 19: Problem 57

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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