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Chapter 19: Problem 38

Iodine- 131 is used in the diagnosis and treatment of thyroid disease and has a half-life of 8.0 days. If a patient with thyroid disease consumes a sample of \(\mathrm{Na}^{131} \mathrm{I}\) containing $10 . \mu \mathrm{g}^{131 \mathrm{I}}\( how long will it take for the amount of \)^{131} \mathrm{I}$ to decrease to 1\(/ 100\) of the original amount?

Short Answer

Expert verified
It will take approximately 53.26 days for the amount of Iodine-131 to decrease to 1/100th of the original amount.

Step by step solution

01

Identify the given values from the problem

We are given the following data: - \(N_0\) (initial amount consumed) = \(10\mu g\) - \(T_{1/2}\) (half-life of Iodine 131) = 8.0 days - We need to find the time \(t\) for the amount \(N(t)\) to decrease to 1/100th of the original amount, which is \(N(t) = \frac{N_0}{100}\).
02

Substitute the values into the exponential decay formula

We know that \(N(t) = N_0 \cdot (1/2)^{\frac{t}{T_{1/2}}}\), and we want to find the time \(t\) when \(N(t) = \frac{N_0}{100}\). Plugging in the values, we get: \[\frac{N_0}{100} = N_0 \cdot (1/2)^{\frac{t}{8}}\]
03

Simplify and solve for t

To solve for \(t\), we'll proceed as follows: 1. Divide both sides by \(N_0\): \[\frac{1}{100} = (1/2)^{\frac{t}{8}}\] 2. Take the natural logarithm of both sides to get the exponent down: \[\ln \left(\frac{1}{100}\right) = \ln \left( (1/2)^{\frac{t}{8}}\right)\] 3. Use the logarithm property to bring down the exponent: \[\ln \left(\frac{1}{100}\right) = \frac{t}{8} \cdot \ln(1/2)\] 4. To find \(t\), divide both sides by \(\ln(1/2)\): \[t = \frac{8\cdot \ln \left(\frac{1}{100}\right)}{\ln(1/2)}\] 5. Finally, calculate the value of \(t\): \[t \approx \frac{8\cdot (-4.605)}{-0.693} \approx 53.26\]
04

Interpret the result

The time it takes for the amount of Iodine-131 to decrease to 1/100th of the original amount is approximately 53.26 days.

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