Chapter 16: Problem 40

How many half-lives does it take for a \(10-\mathrm{g}\) sample of \({ }_{53}^{123} \mathrm{I}\) to drop to \(0.039 \mathrm{~g} ?\) What length of time is this? [The half-life of \({ }_{53}^{123} \mathrm{I}\) is \(13.1 \mathrm{~h}\). \(]\)

Short Answer

Expert verified

It takes approximately 7.4 half-lives for a 10g sample of iodine-123 to drop to 0.039g. This process takes about 96.94 hours.

Step by step solution

Identify the given values

We are given the following values: - Initial amount, \(A_0 = 10 g\) - Final amount, \(A = 0.039 g\) - Half-life, \(t_{1/2} = 13.1 h\)

Set up the radioactive decay formula

Using the formula for radioactive decay, we can set up the following equation with the given values: \(0.039 = 10 (1/2)^n\) Now, we need to solve for \(n\).

Solve for the number of half-lives, n

To solve for \(n\), we will first divide both sides by 10: \(\frac{0.039}{10} = (1/2)^n\) Now, we can take the logarithm of both sides, base (1/2): \(n = \log_{1/2} \frac{0.039}{10}\) Using a calculator, we find that: \(n \approx 7.4\) This means that it takes approximately 7.4 half-lives for the 10g sample of iodine-123 to drop to 0.039g.

Calculate the time for the decay

Now that we know the number of half-lives, we can find the total time \(t\) using the following formula: \(t = n \times t_{1/2}\) Plugging in the values, we get: \(t = 7.4 \times 13.1 \, h\) \(t \approx 96.94 \, h\) Therefore, it takes approximately 96.94 hours for a 10g sample of iodine-123 to drop to 0.039g.

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How many half-lives does it take for a \(10-\mathrm{g}\) sample of \({ }_{53}^{123} \mathrm{I}\) to drop to \(0.039 \mathrm{~g} ?\) What length of time is this? [The half-life of \({ }_{53}^{123} \mathrm{I}\) is \(13.1 \mathrm{~h}\). \(]\)

Short Answer

Step by step solution

Identify the given values

Set up the radioactive decay formula

Solve for the number of half-lives, n

Calculate the time for the decay

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