QuestionJune 17, 2025

13) Consider a sample of iodinc -131 a. How many half-lives would it take for the sample to decay until less than 1% of the original isotope remained?

13) Consider a sample of iodinc -131 a. How many half-lives would it take for the sample to decay until less than 1% of the original isotope remained?
13) Consider a sample of iodinc -131 a. How many half-lives would it take for the sample to decay until less than 1% of the original isotope remained?

Solution
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Answer

Approximately 7 half-lives are needed. Explanation 1. Define the decay formula Use the formula for exponential decay: N = N_0 \left(\frac{1}{2}\right)^{t/T}, where N is the remaining quantity, N_0 is the initial quantity, t is time, and T is the half-life. 2. Set up inequality for less than 1% We need N \frac{-2}{\log_{10}(0.5)}. 5. Calculate numerical value Compute \log_{10}(0.5) \approx -0.3010. Therefore, t/T > \frac{-2}{-0.3010} \approx 6.64.

Explanation

1. Define the decay formula<br /> Use the formula for exponential decay: $N = N_0 \left(\frac{1}{2}\right)^{t/T}$, where $N$ is the remaining quantity, $N_0$ is the initial quantity, $t$ is time, and $T$ is the half-life.<br /><br />2. Set up inequality for less than 1%<br /> We need $N < 0.01 \times N_0$. Substitute into the decay formula: $N_0 \left(\frac{1}{2}\right)^{t/T} < 0.01 \times N_0$.<br /><br />3. Simplify the inequality<br /> Divide both sides by $N_0$: $\left(\frac{1}{2}\right)^{t/T} < 0.01$.<br /><br />4. Solve for number of half-lives<br /> Take logarithm base 10 on both sides: $\log_{10}\left(\left(\frac{1}{2}\right)^{t/T}\right) < \log_{10}(0.01)$.<br /> Simplify using log properties: $\frac{t}{T} \cdot \log_{10}(0.5) < -2$.<br /> Solve for $t/T$: $t/T > \frac{-2}{\log_{10}(0.5)}$.<br /><br />5. Calculate numerical value<br /> Compute $\log_{10}(0.5) \approx -0.3010$. Therefore, $t/T > \frac{-2}{-0.3010} \approx 6.64$.
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