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Chapter 11: Problem 46

Iodine-131 is used to treat tumors in the thyroid. Its first-order half-life is \(8.1\) days. If a patient is given a sample containing \(5.00 \mathrm{mg}\) of \(\mathrm{I}-131\), how long will it take for \(25 \%\) of the isotope to remain in her system?

Short Answer

Expert verified
Answer: It takes approximately 10.63 days for only 25% of the initial amount of Iodine-131 isotope to remain.

Step by step solution

01

Find the decay constant (k)

We can find the decay constant using the half-life for Iodine-131. The relation between decay constant (k) and half-life (t_1/2) is as follows: \( t_{1/2} = \frac{ln(2)}{k} \) We are given half-life, so let's rearrange the formula to get k: \( k = \frac{ln(2)}{t_{1/2}} \) Now, plug in the half-life value for Iodine-131: \( k = \frac{ln(2)}{8.1\, \text{days}} \approx 0.0855\; \text{day}^{-1} \)
02

Determine the remaining percentage of Iodine-131 after the decay

We are given that we want 25% of the initial amount to remain. First, let's express this remaining percentage as a fraction: \( \frac{A(t)}{A_0} = \frac{25}{100} = 0.25 \) Where A(t) is the amount of Iodine-131 after time t, and A0 is the initial amount of Iodine-131 (5mg).
03

Use the integrated rate law to find the time (t)

The first-order integrated rate law equation is: \( ln\left(\frac{A(t)}{A_0}\right) = -kt \) Now, plug in the values for \(\frac{A(t)}{A_0}\) and \(k\): \( ln(0.25) = -0.0855\; t \) Calculate the value of t: \( t = \frac{ln(0.25)}{-0.0855} \approx 10.63\; \text{days} \)
04

Final Answer

The time it takes for 25% of Iodine-131 to remain in the patient's system is approximately 10.63 days.

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Most popular questions from this chapter

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